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archives/1098994933_python.zip | matrix/tests/test_matrix_operation.py | """
Testing here assumes that numpy and linalg is ALWAYS correct!!!!
If running from PyCharm you can place the following line in "Additional Arguments" for the pytest run configuration
-vv -m mat_ops -p no:cacheprovider
"""
# standard libraries
import sys
import numpy as np
import pytest
import logging
# Custom/local libraries
from matrix import matrix_operation as matop
mat_a = [[12, 10], [3, 9]]
mat_b = [[3, 4], [7, 4]]
mat_c = [[3, 0, 2], [2, 0, -2], [0, 1, 1]]
mat_d = [[3, 0, -2], [2, 0, 2], [0, 1, 1]]
mat_e = [[3, 0, 2], [2, 0, -2], [0, 1, 1], [2, 0, -2]]
mat_f = [1]
mat_h = [2]
logger = logging.getLogger()
logger.level = logging.DEBUG
stream_handler = logging.StreamHandler(sys.stdout)
logger.addHandler(stream_handler)
@pytest.mark.mat_ops
@pytest.mark.parametrize(('mat1', 'mat2'), [(mat_a, mat_b), (mat_c, mat_d), (mat_d, mat_e),
(mat_f, mat_h)])
def test_addition(mat1, mat2):
if (np.array(mat1)).shape < (2, 2) or (np.array(mat2)).shape < (2, 2):
with pytest.raises(TypeError):
logger.info(f"\n\t{test_addition.__name__} returned integer")
matop.add(mat1, mat2)
elif (np.array(mat1)).shape == (np.array(mat2)).shape:
logger.info(f"\n\t{test_addition.__name__} with same matrix dims")
act = (np.array(mat1) + np.array(mat2)).tolist()
theo = matop.add(mat1, mat2)
assert theo == act
else:
with pytest.raises(ValueError):
logger.info(f"\n\t{test_addition.__name__} with different matrix dims")
matop.add(mat1, mat2)
@pytest.mark.mat_ops
@pytest.mark.parametrize(('mat1', 'mat2'), [(mat_a, mat_b), (mat_c, mat_d), (mat_d, mat_e),
(mat_f, mat_h)])
def test_subtraction(mat1, mat2):
if (np.array(mat1)).shape < (2, 2) or (np.array(mat2)).shape < (2, 2):
with pytest.raises(TypeError):
logger.info(f"\n\t{test_subtraction.__name__} returned integer")
matop.subtract(mat1, mat2)
elif (np.array(mat1)).shape == (np.array(mat2)).shape:
logger.info(f"\n\t{test_subtraction.__name__} with same matrix dims")
act = (np.array(mat1) - np.array(mat2)).tolist()
theo = matop.subtract(mat1, mat2)
assert theo == act
else:
with pytest.raises(ValueError):
logger.info(f"\n\t{test_subtraction.__name__} with different matrix dims")
assert matop.subtract(mat1, mat2)
@pytest.mark.mat_ops
@pytest.mark.parametrize(('mat1', 'mat2'), [(mat_a, mat_b), (mat_c, mat_d), (mat_d, mat_e),
(mat_f, mat_h)])
def test_multiplication(mat1, mat2):
if (np.array(mat1)).shape < (2, 2) or (np.array(mat2)).shape < (2, 2):
logger.info(f"\n\t{test_multiplication.__name__} returned integer")
with pytest.raises(TypeError):
matop.add(mat1, mat2)
elif (np.array(mat1)).shape == (np.array(mat2)).shape:
logger.info(f"\n\t{test_multiplication.__name__} meets dim requirements")
act = (np.matmul(mat1, mat2)).tolist()
theo = matop.multiply(mat1, mat2)
assert theo == act
else:
with pytest.raises(ValueError):
logger.info(f"\n\t{test_multiplication.__name__} does not meet dim requirements")
assert matop.subtract(mat1, mat2)
@pytest.mark.mat_ops
def test_scalar_multiply():
act = (3.5 * np.array(mat_a)).tolist()
theo = matop.scalar_multiply(mat_a, 3.5)
assert theo == act
@pytest.mark.mat_ops
def test_identity():
act = (np.identity(5)).tolist()
theo = matop.identity(5)
assert theo == act
@pytest.mark.mat_ops
@pytest.mark.parametrize('mat', [mat_a, mat_b, mat_c, mat_d, mat_e, mat_f])
def test_transpose(mat):
if (np.array(mat)).shape < (2, 2):
with pytest.raises(TypeError):
logger.info(f"\n\t{test_transpose.__name__} returned integer")
matop.transpose(mat)
else:
act = (np.transpose(mat)).tolist()
theo = matop.transpose(mat, return_map=False)
assert theo == act
| [] | [] | [] |
archives/1098994933_python.zip | networking_flow/ford_fulkerson.py | # Ford-Fulkerson Algorithm for Maximum Flow Problem
"""
Description:
(1) Start with initial flow as 0;
(2) Choose augmenting path from source to sink and add path to flow;
"""
def BFS(graph, s, t, parent):
# Return True if there is node that has not iterated.
visited = [False]*len(graph)
queue=[]
queue.append(s)
visited[s] = True
while queue:
u = queue.pop(0)
for ind in range(len(graph[u])):
if visited[ind] == False and graph[u][ind] > 0:
queue.append(ind)
visited[ind] = True
parent[ind] = u
return True if visited[t] else False
def FordFulkerson(graph, source, sink):
# This array is filled by BFS and to store path
parent = [-1]*(len(graph))
max_flow = 0
while BFS(graph, source, sink, parent) :
path_flow = float("Inf")
s = sink
while(s != source):
# Find the minimum value in select path
path_flow = min (path_flow, graph[parent[s]][s])
s = parent[s]
max_flow += path_flow
v = sink
while(v != source):
u = parent[v]
graph[u][v] -= path_flow
graph[v][u] += path_flow
v = parent[v]
return max_flow
graph = [[0, 16, 13, 0, 0, 0],
[0, 0, 10 ,12, 0, 0],
[0, 4, 0, 0, 14, 0],
[0, 0, 9, 0, 0, 20],
[0, 0, 0, 7, 0, 4],
[0, 0, 0, 0, 0, 0]]
source, sink = 0, 5
print(FordFulkerson(graph, source, sink)) | [] | [] | [] |
archives/1098994933_python.zip | networking_flow/minimum_cut.py | # Minimum cut on Ford_Fulkerson algorithm.
test_graph = [
[0, 16, 13, 0, 0, 0],
[0, 0, 10, 12, 0, 0],
[0, 4, 0, 0, 14, 0],
[0, 0, 9, 0, 0, 20],
[0, 0, 0, 7, 0, 4],
[0, 0, 0, 0, 0, 0],
]
def BFS(graph, s, t, parent):
# Return True if there is node that has not iterated.
visited = [False] * len(graph)
queue = [s]
visited[s] = True
while queue:
u = queue.pop(0)
for ind in range(len(graph[u])):
if visited[ind] == False and graph[u][ind] > 0:
queue.append(ind)
visited[ind] = True
parent[ind] = u
return True if visited[t] else False
def mincut(graph, source, sink):
"""This array is filled by BFS and to store path
>>> mincut(test_graph, source=0, sink=5)
[(1, 3), (4, 3), (4, 5)]
"""
parent = [-1] * (len(graph))
max_flow = 0
res = []
temp = [i[:] for i in graph] # Record orignial cut, copy.
while BFS(graph, source, sink, parent):
path_flow = float("Inf")
s = sink
while s != source:
# Find the minimum value in select path
path_flow = min(path_flow, graph[parent[s]][s])
s = parent[s]
max_flow += path_flow
v = sink
while v != source:
u = parent[v]
graph[u][v] -= path_flow
graph[v][u] += path_flow
v = parent[v]
for i in range(len(graph)):
for j in range(len(graph[0])):
if graph[i][j] == 0 and temp[i][j] > 0:
res.append((i, j))
return res
if __name__ == "__main__":
print(mincut(test_graph, source=0, sink=5))
| [] | [] | [] |
archives/1098994933_python.zip | neural_network/back_propagation_neural_network.py | #!/usr/bin/python
# encoding=utf8
'''
A Framework of Back Propagation Neural Network（BP） model
Easy to use:
* add many layers as you want ！！！
* clearly see how the loss decreasing
Easy to expand:
* more activation functions
* more loss functions
* more optimization method
Author: Stephen Lee
Github : https://github.com/RiptideBo
Date: 2017.11.23
'''
import numpy as np
import matplotlib.pyplot as plt
def sigmoid(x):
return 1 / (1 + np.exp(-1 * x))
class DenseLayer():
'''
Layers of BP neural network
'''
def __init__(self,units,activation=None,learning_rate=None,is_input_layer=False):
'''
common connected layer of bp network
:param units: numbers of neural units
:param activation: activation function
:param learning_rate: learning rate for paras
:param is_input_layer: whether it is input layer or not
'''
self.units = units
self.weight = None
self.bias = None
self.activation = activation
if learning_rate is None:
learning_rate = 0.3
self.learn_rate = learning_rate
self.is_input_layer = is_input_layer
def initializer(self,back_units):
self.weight = np.asmatrix(np.random.normal(0,0.5,(self.units,back_units)))
self.bias = np.asmatrix(np.random.normal(0,0.5,self.units)).T
if self.activation is None:
self.activation = sigmoid
def cal_gradient(self):
if self.activation == sigmoid:
gradient_mat = np.dot(self.output ,(1- self.output).T)
gradient_activation = np.diag(np.diag(gradient_mat))
else:
gradient_activation = 1
return gradient_activation
def forward_propagation(self,xdata):
self.xdata = xdata
if self.is_input_layer:
# input layer
self.wx_plus_b = xdata
self.output = xdata
return xdata
else:
self.wx_plus_b = np.dot(self.weight,self.xdata) - self.bias
self.output = self.activation(self.wx_plus_b)
return self.output
def back_propagation(self,gradient):
gradient_activation = self.cal_gradient() # i * i 维
gradient = np.asmatrix(np.dot(gradient.T,gradient_activation))
self._gradient_weight = np.asmatrix(self.xdata)
self._gradient_bias = -1
self._gradient_x = self.weight
self.gradient_weight = np.dot(gradient.T,self._gradient_weight.T)
self.gradient_bias = gradient * self._gradient_bias
self.gradient = np.dot(gradient,self._gradient_x).T
# ----------------------upgrade
# -----------the Negative gradient direction --------
self.weight = self.weight - self.learn_rate * self.gradient_weight
self.bias = self.bias - self.learn_rate * self.gradient_bias.T
return self.gradient
class BPNN():
'''
Back Propagation Neural Network model
'''
def __init__(self):
self.layers = []
self.train_mse = []
self.fig_loss = plt.figure()
self.ax_loss = self.fig_loss.add_subplot(1,1,1)
def add_layer(self,layer):
self.layers.append(layer)
def build(self):
for i,layer in enumerate(self.layers[:]):
if i < 1:
layer.is_input_layer = True
else:
layer.initializer(self.layers[i-1].units)
def summary(self):
for i,layer in enumerate(self.layers[:]):
print('------- layer %d -------'%i)
print('weight.shape ',np.shape(layer.weight))
print('bias.shape ',np.shape(layer.bias))
def train(self,xdata,ydata,train_round,accuracy):
self.train_round = train_round
self.accuracy = accuracy
self.ax_loss.hlines(self.accuracy, 0, self.train_round * 1.1)
x_shape = np.shape(xdata)
for round_i in range(train_round):
all_loss = 0
for row in range(x_shape[0]):
_xdata = np.asmatrix(xdata[row,:]).T
_ydata = np.asmatrix(ydata[row,:]).T
# forward propagation
for layer in self.layers:
_xdata = layer.forward_propagation(_xdata)
loss, gradient = self.cal_loss(_ydata, _xdata)
all_loss = all_loss + loss
# back propagation
# the input_layer does not upgrade
for layer in self.layers[:0:-1]:
gradient = layer.back_propagation(gradient)
mse = all_loss/x_shape[0]
self.train_mse.append(mse)
self.plot_loss()
if mse < self.accuracy:
print('----达到精度----')
return mse
def cal_loss(self,ydata,ydata_):
self.loss = np.sum(np.power((ydata - ydata_),2))
self.loss_gradient = 2 * (ydata_ - ydata)
# vector (shape is the same as _ydata.shape)
return self.loss,self.loss_gradient
def plot_loss(self):
if self.ax_loss.lines:
self.ax_loss.lines.remove(self.ax_loss.lines[0])
self.ax_loss.plot(self.train_mse, 'r-')
plt.ion()
plt.xlabel('step')
plt.ylabel('loss')
plt.show()
plt.pause(0.1)
def example():
x = np.random.randn(10,10)
y = np.asarray([[0.8,0.4],[0.4,0.3],[0.34,0.45],[0.67,0.32],
[0.88,0.67],[0.78,0.77],[0.55,0.66],[0.55,0.43],[0.54,0.1],
[0.1,0.5]])
model = BPNN()
model.add_layer(DenseLayer(10))
model.add_layer(DenseLayer(20))
model.add_layer(DenseLayer(30))
model.add_layer(DenseLayer(2))
model.build()
model.summary()
model.train(xdata=x,ydata=y,train_round=100,accuracy=0.01)
if __name__ == '__main__':
example()
| [] | [] | [] |
archives/1098994933_python.zip | neural_network/convolution_neural_network.py | #-*- coding: utf-8 -*-
'''
- - - - - -- - - - - - - - - - - - - - - - - - - - - - -
Name - - CNN - Convolution Neural Network For Photo Recognizing
Goal - - Recognize Handing Writting Word Photo
Detail：Total 5 layers neural network
* Convolution layer
* Pooling layer
* Input layer layer of BP
* Hiden layer of BP
* Output layer of BP
Author: Stephen Lee
Github: 245885195@qq.com
Date: 2017.9.20
- - - - - -- - - - - - - - - - - - - - - - - - - - - - -
'''
import pickle
import numpy as np
import matplotlib.pyplot as plt
class CNN():
def __init__(self, conv1_get, size_p1, bp_num1, bp_num2, bp_num3, rate_w=0.2, rate_t=0.2):
'''
:param conv1_get: [a,c,d]，size, number, step of convolution kernel
:param size_p1: pooling size
:param bp_num1: units number of flatten layer
:param bp_num2: units number of hidden layer
:param bp_num3: units number of output layer
:param rate_w: rate of weight learning
:param rate_t: rate of threshold learning
'''
self.num_bp1 = bp_num1
self.num_bp2 = bp_num2
self.num_bp3 = bp_num3
self.conv1 = conv1_get[:2]
self.step_conv1 = conv1_get[2]
self.size_pooling1 = size_p1
self.rate_weight = rate_w
self.rate_thre = rate_t
self.w_conv1 = [np.mat(-1*np.random.rand(self.conv1[0],self.conv1[0])+0.5) for i in range(self.conv1[1])]
self.wkj = np.mat(-1 * np.random.rand(self.num_bp3, self.num_bp2) + 0.5)
self.vji = np.mat(-1*np.random.rand(self.num_bp2, self.num_bp1)+0.5)
self.thre_conv1 = -2*np.random.rand(self.conv1[1])+1
self.thre_bp2 = -2*np.random.rand(self.num_bp2)+1
self.thre_bp3 = -2*np.random.rand(self.num_bp3)+1
def save_model(self, save_path):
#save model dict with pickle
model_dic = {'num_bp1':self.num_bp1,
'num_bp2':self.num_bp2,
'num_bp3':self.num_bp3,
'conv1':self.conv1,
'step_conv1':self.step_conv1,
'size_pooling1':self.size_pooling1,
'rate_weight':self.rate_weight,
'rate_thre':self.rate_thre,
'w_conv1':self.w_conv1,
'wkj':self.wkj,
'vji':self.vji,
'thre_conv1':self.thre_conv1,
'thre_bp2':self.thre_bp2,
'thre_bp3':self.thre_bp3}
with open(save_path, 'wb') as f:
pickle.dump(model_dic, f)
print('Model saved： %s'% save_path)
@classmethod
def ReadModel(cls, model_path):
#read saved model
with open(model_path, 'rb') as f:
model_dic = pickle.load(f)
conv_get= model_dic.get('conv1')
conv_get.append(model_dic.get('step_conv1'))
size_p1 = model_dic.get('size_pooling1')
bp1 = model_dic.get('num_bp1')
bp2 = model_dic.get('num_bp2')
bp3 = model_dic.get('num_bp3')
r_w = model_dic.get('rate_weight')
r_t = model_dic.get('rate_thre')
#create model instance
conv_ins = CNN(conv_get,size_p1,bp1,bp2,bp3,r_w,r_t)
#modify model parameter
conv_ins.w_conv1 = model_dic.get('w_conv1')
conv_ins.wkj = model_dic.get('wkj')
conv_ins.vji = model_dic.get('vji')
conv_ins.thre_conv1 = model_dic.get('thre_conv1')
conv_ins.thre_bp2 = model_dic.get('thre_bp2')
conv_ins.thre_bp3 = model_dic.get('thre_bp3')
return conv_ins
def sig(self, x):
return 1 / (1 + np.exp(-1*x))
def do_round(self, x):
return round(x, 3)
def convolute(self, data, convs, w_convs, thre_convs, conv_step):
#convolution process
size_conv = convs[0]
num_conv =convs[1]
size_data = np.shape(data)[0]
#get the data slice of original image data, data_focus
data_focus = []
for i_focus in range(0, size_data - size_conv + 1, conv_step):
for j_focus in range(0, size_data - size_conv + 1, conv_step):
focus = data[i_focus:i_focus + size_conv, j_focus:j_focus + size_conv]
data_focus.append(focus)
#caculate the feature map of every single kernel, and saved as list of matrix
data_featuremap = []
Size_FeatureMap = int((size_data - size_conv) / conv_step + 1)
for i_map in range(num_conv):
featuremap = []
for i_focus in range(len(data_focus)):
net_focus = np.sum(np.multiply(data_focus[i_focus], w_convs[i_map])) - thre_convs[i_map]
featuremap.append(self.sig(net_focus))
featuremap = np.asmatrix(featuremap).reshape(Size_FeatureMap, Size_FeatureMap)
data_featuremap.append(featuremap)
#expanding the data slice to One dimenssion
focus1_list = []
for each_focus in data_focus:
focus1_list.extend(self.Expand_Mat(each_focus))
focus_list = np.asarray(focus1_list)
return focus_list,data_featuremap
def pooling(self, featuremaps, size_pooling, type='average_pool'):
#pooling process
size_map = len(featuremaps[0])
size_pooled = int(size_map/size_pooling)
featuremap_pooled = []
for i_map in range(len(featuremaps)):
map = featuremaps[i_map]
map_pooled = []
for i_focus in range(0,size_map,size_pooling):
for j_focus in range(0, size_map, size_pooling):
focus = map[i_focus:i_focus + size_pooling, j_focus:j_focus + size_pooling]
if type == 'average_pool':
#average pooling
map_pooled.append(np.average(focus))
elif type == 'max_pooling':
#max pooling
map_pooled.append(np.max(focus))
map_pooled = np.asmatrix(map_pooled).reshape(size_pooled,size_pooled)
featuremap_pooled.append(map_pooled)
return featuremap_pooled
def _expand(self, datas):
#expanding three dimension data to one dimension list
data_expanded = []
for i in range(len(datas)):
shapes = np.shape(datas[i])
data_listed = datas[i].reshape(1,shapes[0]*shapes[1])
data_listed = data_listed.getA().tolist()[0]
data_expanded.extend(data_listed)
data_expanded = np.asarray(data_expanded)
return data_expanded
def _expand_mat(self, data_mat):
#expanding matrix to one dimension list
data_mat = np.asarray(data_mat)
shapes = np.shape(data_mat)
data_expanded = data_mat.reshape(1,shapes[0]*shapes[1])
return data_expanded
def _calculate_gradient_from_pool(self, out_map, pd_pool,num_map, size_map, size_pooling):
'''
calcluate the gradient from the data slice of pool layer
pd_pool: list of matrix
out_map: the shape of data slice(size_map*size_map)
return: pd_all: list of matrix, [num, size_map, size_map]
'''
pd_all = []
i_pool = 0
for i_map in range(num_map):
pd_conv1 = np.ones((size_map, size_map))
for i in range(0, size_map, size_pooling):
for j in range(0, size_map, size_pooling):
pd_conv1[i:i + size_pooling, j:j + size_pooling] = pd_pool[i_pool]
i_pool = i_pool + 1
pd_conv2 = np.multiply(pd_conv1,np.multiply(out_map[i_map],(1-out_map[i_map])))
pd_all.append(pd_conv2)
return pd_all
def train(self, patterns, datas_train, datas_teach, n_repeat, error_accuracy, draw_e = bool):
#model traning
print('----------------------Start Training-------------------------')
print((' - - Shape: Train_Data ',np.shape(datas_train)))
print((' - - Shape: Teach_Data ',np.shape(datas_teach)))
rp = 0
all_mse = []
mse = 10000
while rp < n_repeat and mse >= error_accuracy:
alle = 0
print('-------------Learning Time %d--------------'%rp)
for p in range(len(datas_train)):
#print('------------Learning Image: %d--------------'%p)
data_train = np.asmatrix(datas_train[p])
data_teach = np.asarray(datas_teach[p])
data_focus1,data_conved1 = self.convolute(data_train,self.conv1,self.w_conv1,
self.thre_conv1,conv_step=self.step_conv1)
data_pooled1 = self.pooling(data_conved1,self.size_pooling1)
shape_featuremap1 = np.shape(data_conved1)
'''
print(' -----original shape ', np.shape(data_train))
print(' ---- after convolution ',np.shape(data_conv1))
print(' -----after pooling ',np.shape(data_pooled1))
'''
data_bp_input = self._expand(data_pooled1)
bp_out1 = data_bp_input
bp_net_j = np.dot(bp_out1,self.vji.T) - self.thre_bp2
bp_out2 = self.sig(bp_net_j)
bp_net_k = np.dot(bp_out2 ,self.wkj.T) - self.thre_bp3
bp_out3 = self.sig(bp_net_k)
#--------------Model Leaning ------------------------
# calcluate error and gradient---------------
pd_k_all = np.multiply((data_teach - bp_out3), np.multiply(bp_out3, (1 - bp_out3)))
pd_j_all = np.multiply(np.dot(pd_k_all,self.wkj), np.multiply(bp_out2, (1 - bp_out2)))
pd_i_all = np.dot(pd_j_all,self.vji)
pd_conv1_pooled = pd_i_all / (self.size_pooling1*self.size_pooling1)
pd_conv1_pooled = pd_conv1_pooled.T.getA().tolist()
pd_conv1_all = self._calculate_gradient_from_pool(data_conved1,pd_conv1_pooled,shape_featuremap1[0],
shape_featuremap1[1],self.size_pooling1)
#weight and threshold learning process---------
#convolution layer
for k_conv in range(self.conv1[1]):
pd_conv_list = self._expand_mat(pd_conv1_all[k_conv])
delta_w = self.rate_weight * np.dot(pd_conv_list,data_focus1)
self.w_conv1[k_conv] = self.w_conv1[k_conv] + delta_w.reshape((self.conv1[0],self.conv1[0]))
self.thre_conv1[k_conv] = self.thre_conv1[k_conv] - np.sum(pd_conv1_all[k_conv]) * self.rate_thre
#all connected layer
self.wkj = self.wkj + pd_k_all.T * bp_out2 * self.rate_weight
self.vji = self.vji + pd_j_all.T * bp_out1 * self.rate_weight
self.thre_bp3 = self.thre_bp3 - pd_k_all * self.rate_thre
self.thre_bp2 = self.thre_bp2 - pd_j_all * self.rate_thre
# calculate the sum error of all single image
errors = np.sum(abs((data_teach - bp_out3)))
alle = alle + errors
#print(' ----Teach ',data_teach)
#print(' ----BP_output ',bp_out3)
rp = rp + 1
mse = alle/patterns
all_mse.append(mse)
def draw_error():
yplot = [error_accuracy for i in range(int(n_repeat * 1.2))]
plt.plot(all_mse, '+-')
plt.plot(yplot, 'r--')
plt.xlabel('Learning Times')
plt.ylabel('All_mse')
plt.grid(True, alpha=0.5)
plt.show()
print('------------------Training Complished---------------------')
print((' - - Training epoch: ', rp, ' - - Mse: %.6f' % mse))
if draw_e:
draw_error()
return mse
def predict(self, datas_test):
#model predict
produce_out = []
print('-------------------Start Testing-------------------------')
print((' - - Shape: Test_Data ',np.shape(datas_test)))
for p in range(len(datas_test)):
data_test = np.asmatrix(datas_test[p])
data_focus1, data_conved1 = self.convolute(data_test, self.conv1, self.w_conv1,
self.thre_conv1, conv_step=self.step_conv1)
data_pooled1 = self.pooling(data_conved1, self.size_pooling1)
data_bp_input = self._expand(data_pooled1)
bp_out1 = data_bp_input
bp_net_j = bp_out1 * self.vji.T - self.thre_bp2
bp_out2 = self.sig(bp_net_j)
bp_net_k = bp_out2 * self.wkj.T - self.thre_bp3
bp_out3 = self.sig(bp_net_k)
produce_out.extend(bp_out3.getA().tolist())
res = [list(map(self.do_round,each)) for each in produce_out]
return np.asarray(res)
def convolution(self, data):
#return the data of image after convoluting process so we can check it out
data_test = np.asmatrix(data)
data_focus1, data_conved1 = self.convolute(data_test, self.conv1, self.w_conv1,
self.thre_conv1, conv_step=self.step_conv1)
data_pooled1 = self.pooling(data_conved1, self.size_pooling1)
return data_conved1,data_pooled1
if __name__ == '__main__':
pass
'''
I will put the example on other file
'''
| [] | [] | [] |
archives/1098994933_python.zip | neural_network/perceptron.py | '''
Perceptron
w = w + N * (d(k) - y) * x(k)
Using perceptron network for oil analysis,
with Measuring of 3 parameters that represent chemical characteristics we can classify the oil, in p1 or p2
p1 = -1
p2 = 1
'''
import random
class Perceptron:
def __init__(self, sample, exit, learn_rate=0.01, epoch_number=1000, bias=-1):
self.sample = sample
self.exit = exit
self.learn_rate = learn_rate
self.epoch_number = epoch_number
self.bias = bias
self.number_sample = len(sample)
self.col_sample = len(sample[0])
self.weight = []
def training(self):
for sample in self.sample:
sample.insert(0, self.bias)
for i in range(self.col_sample):
self.weight.append(random.random())
self.weight.insert(0, self.bias)
epoch_count = 0
while True:
erro = False
for i in range(self.number_sample):
u = 0
for j in range(self.col_sample + 1):
u = u + self.weight[j] * self.sample[i][j]
y = self.sign(u)
if y != self.exit[i]:
for j in range(self.col_sample + 1):
self.weight[j] = self.weight[j] + self.learn_rate * (self.exit[i] - y) * self.sample[i][j]
erro = True
#print('Epoch: \n',epoch_count)
epoch_count = epoch_count + 1
# if you want controle the epoch or just by erro
if erro == False:
print(('\nEpoch:\n',epoch_count))
print('------------------------\n')
#if epoch_count > self.epoch_number or not erro:
break
def sort(self, sample):
sample.insert(0, self.bias)
u = 0
for i in range(self.col_sample + 1):
u = u + self.weight[i] * sample[i]
y = self.sign(u)
if y == -1:
print(('Sample: ', sample))
print('classification: P1')
else:
print(('Sample: ', sample))
print('classification: P2')
def sign(self, u):
return 1 if u >= 0 else -1
samples = [
[-0.6508, 0.1097, 4.0009],
[-1.4492, 0.8896, 4.4005],
[2.0850, 0.6876, 12.0710],
[0.2626, 1.1476, 7.7985],
[0.6418, 1.0234, 7.0427],
[0.2569, 0.6730, 8.3265],
[1.1155, 0.6043, 7.4446],
[0.0914, 0.3399, 7.0677],
[0.0121, 0.5256, 4.6316],
[-0.0429, 0.4660, 5.4323],
[0.4340, 0.6870, 8.2287],
[0.2735, 1.0287, 7.1934],
[0.4839, 0.4851, 7.4850],
[0.4089, -0.1267, 5.5019],
[1.4391, 0.1614, 8.5843],
[-0.9115, -0.1973, 2.1962],
[0.3654, 1.0475, 7.4858],
[0.2144, 0.7515, 7.1699],
[0.2013, 1.0014, 6.5489],
[0.6483, 0.2183, 5.8991],
[-0.1147, 0.2242, 7.2435],
[-0.7970, 0.8795, 3.8762],
[-1.0625, 0.6366, 2.4707],
[0.5307, 0.1285, 5.6883],
[-1.2200, 0.7777, 1.7252],
[0.3957, 0.1076, 5.6623],
[-0.1013, 0.5989, 7.1812],
[2.4482, 0.9455, 11.2095],
[2.0149, 0.6192, 10.9263],
[0.2012, 0.2611, 5.4631]
]
exit = [-1, -1, -1, 1, 1, -1, 1, -1, 1, 1, -1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, 1, 1, -1, -1, 1, -1, 1]
network = Perceptron(sample=samples, exit = exit, learn_rate=0.01, epoch_number=1000, bias=-1)
network.training()
if __name__ == '__main__':
while True:
sample = []
for i in range(3):
sample.insert(i, float(input('value: ')))
network.sort(sample)
| [] | [] | [] |
archives/1098994933_python.zip | other/anagrams.py | import collections, pprint, time, os
start_time = time.time()
print('creating word list...')
path = os.path.split(os.path.realpath(__file__))
with open(path[0] + '/words') as f:
word_list = sorted(list(set([word.strip().lower() for word in f])))
def signature(word):
return ''.join(sorted(word))
word_bysig = collections.defaultdict(list)
for word in word_list:
word_bysig[signature(word)].append(word)
def anagram(myword):
return word_bysig[signature(myword)]
print('finding anagrams...')
all_anagrams = {word: anagram(word)
for word in word_list if len(anagram(word)) > 1}
print('writing anagrams to file...')
with open('anagrams.txt', 'w') as file:
file.write('all_anagrams = ')
file.write(pprint.pformat(all_anagrams))
total_time = round(time.time() - start_time, 2)
print(('Done [', total_time, 'seconds ]'))
| [] | [] | [] |
archives/1098994933_python.zip | other/binary_exponentiation.py | """
* Binary Exponentiation for Powers
* This is a method to find a^b in a time complexity of O(log b)
* This is one of the most commonly used methods of finding powers.
* Also useful in cases where solution to (a^b)%c is required,
* where a,b,c can be numbers over the computers calculation limits.
* Done using iteration, can also be done using recursion
* @author chinmoy159
* @version 1.0 dated 10/08/2017
"""
def b_expo(a, b):
res = 1
while b > 0:
if b&1:
res *= a
a *= a
b >>= 1
return res
def b_expo_mod(a, b, c):
res = 1
while b > 0:
if b&1:
res = ((res%c) * (a%c)) % c
a *= a
b >>= 1
return res
"""
* Wondering how this method works !
* It's pretty simple.
* Let's say you need to calculate a ^ b
* RULE 1 : a ^ b = (a*a) ^ (b/2) ---- example : 4 ^ 4 = (4*4) ^ (4/2) = 16 ^ 2
* RULE 2 : IF b is ODD, then ---- a ^ b = a * (a ^ (b - 1)) :: where (b - 1) is even.
* Once b is even, repeat the process to get a ^ b
* Repeat the process till b = 1 OR b = 0, because a^1 = a AND a^0 = 1
*
* As far as the modulo is concerned,
* the fact : (a*b) % c = ((a%c) * (b%c)) % c
* Now apply RULE 1 OR 2 whichever is required.
"""
| [] | [] | [] |
archives/1098994933_python.zip | other/binary_exponentiation_2.py | """
* Binary Exponentiation with Multiplication
* This is a method to find a*b in a time complexity of O(log b)
* This is one of the most commonly used methods of finding result of multiplication.
* Also useful in cases where solution to (a*b)%c is required,
* where a,b,c can be numbers over the computers calculation limits.
* Done using iteration, can also be done using recursion
* @author chinmoy159
* @version 1.0 dated 10/08/2017
"""
def b_expo(a, b):
res = 0
while b > 0:
if b&1:
res += a
a += a
b >>= 1
return res
def b_expo_mod(a, b, c):
res = 0
while b > 0:
if b&1:
res = ((res%c) + (a%c)) % c
a += a
b >>= 1
return res
"""
* Wondering how this method works !
* It's pretty simple.
* Let's say you need to calculate a ^ b
* RULE 1 : a * b = (a+a) * (b/2) ---- example : 4 * 4 = (4+4) * (4/2) = 8 * 2
* RULE 2 : IF b is ODD, then ---- a * b = a + (a * (b - 1)) :: where (b - 1) is even.
* Once b is even, repeat the process to get a * b
* Repeat the process till b = 1 OR b = 0, because a*1 = a AND a*0 = 0
*
* As far as the modulo is concerned,
* the fact : (a+b) % c = ((a%c) + (b%c)) % c
* Now apply RULE 1 OR 2, whichever is required.
"""
| [] | [] | [] |
archives/1098994933_python.zip | other/detecting_english_programmatically.py | import os
UPPERLETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
LETTERS_AND_SPACE = UPPERLETTERS + UPPERLETTERS.lower() + ' \t\n'
def loadDictionary():
path = os.path.split(os.path.realpath(__file__))
englishWords = {}
with open(path[0] + '/dictionary.txt') as dictionaryFile:
for word in dictionaryFile.read().split('\n'):
englishWords[word] = None
return englishWords
ENGLISH_WORDS = loadDictionary()
def getEnglishCount(message):
message = message.upper()
message = removeNonLetters(message)
possibleWords = message.split()
if possibleWords == []:
return 0.0
matches = 0
for word in possibleWords:
if word in ENGLISH_WORDS:
matches += 1
return float(matches) / len(possibleWords)
def removeNonLetters(message):
lettersOnly = []
for symbol in message:
if symbol in LETTERS_AND_SPACE:
lettersOnly.append(symbol)
return ''.join(lettersOnly)
def isEnglish(message, wordPercentage = 20, letterPercentage = 85):
"""
>>> isEnglish('Hello World')
True
>>> isEnglish('llold HorWd')
False
"""
wordsMatch = getEnglishCount(message) * 100 >= wordPercentage
numLetters = len(removeNonLetters(message))
messageLettersPercentage = (float(numLetters) / len(message)) * 100
lettersMatch = messageLettersPercentage >= letterPercentage
return wordsMatch and lettersMatch
import doctest
doctest.testmod()
| [] | [] | [] |
archives/1098994933_python.zip | other/euclidean_gcd.py | # https://en.wikipedia.org/wiki/Euclidean_algorithm
def euclidean_gcd(a, b):
while b:
t = b
b = a % b
a = t
return a
def main():
print("GCD(3, 5) = " + str(euclidean_gcd(3, 5)))
print("GCD(5, 3) = " + str(euclidean_gcd(5, 3)))
print("GCD(1, 3) = " + str(euclidean_gcd(1, 3)))
print("GCD(3, 6) = " + str(euclidean_gcd(3, 6)))
print("GCD(6, 3) = " + str(euclidean_gcd(6, 3)))
if __name__ == '__main__':
main()
| [] | [] | [] |
archives/1098994933_python.zip | other/fischer_yates_shuffle.py | #!/usr/bin/python
# encoding=utf8
"""
The Fisher–Yates shuffle is an algorithm for generating a random permutation of a finite sequence.
For more details visit
wikipedia/Fischer-Yates-Shuffle.
"""
import random
def FYshuffle(LIST):
for i in range(len(LIST)):
a = random.randint(0, len(LIST)-1)
b = random.randint(0, len(LIST)-1)
LIST[a], LIST[b] = LIST[b], LIST[a]
return LIST
if __name__ == '__main__':
integers = [0,1,2,3,4,5,6,7]
strings = ['python', 'says', 'hello', '!']
print('Fisher-Yates Shuffle:')
print('List',integers, strings)
print('FY Shuffle',FYshuffle(integers), FYshuffle(strings))
| [] | [] | [] |
archives/1098994933_python.zip | other/frequency_finder.py | # Frequency Finder
# frequency taken from http://en.wikipedia.org/wiki/Letter_frequency
englishLetterFreq = {'E': 12.70, 'T': 9.06, 'A': 8.17, 'O': 7.51, 'I': 6.97,
'N': 6.75, 'S': 6.33, 'H': 6.09, 'R': 5.99, 'D': 4.25,
'L': 4.03, 'C': 2.78, 'U': 2.76, 'M': 2.41, 'W': 2.36,
'F': 2.23, 'G': 2.02, 'Y': 1.97, 'P': 1.93, 'B': 1.29,
'V': 0.98, 'K': 0.77, 'J': 0.15, 'X': 0.15, 'Q': 0.10,
'Z': 0.07}
ETAOIN = 'ETAOINSHRDLCUMWFGYPBVKJXQZ'
LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def getLetterCount(message):
letterCount = {'A': 0, 'B': 0, 'C': 0, 'D': 0, 'E': 0, 'F': 0, 'G': 0, 'H': 0,
'I': 0, 'J': 0, 'K': 0, 'L': 0, 'M': 0, 'N': 0, 'O': 0, 'P': 0,
'Q': 0, 'R': 0, 'S': 0, 'T': 0, 'U': 0, 'V': 0, 'W': 0, 'X': 0,
'Y': 0, 'Z': 0}
for letter in message.upper():
if letter in LETTERS:
letterCount[letter] += 1
return letterCount
def getItemAtIndexZero(x):
return x[0]
def getFrequencyOrder(message):
letterToFreq = getLetterCount(message)
freqToLetter = {}
for letter in LETTERS:
if letterToFreq[letter] not in freqToLetter:
freqToLetter[letterToFreq[letter]] = [letter]
else:
freqToLetter[letterToFreq[letter]].append(letter)
for freq in freqToLetter:
freqToLetter[freq].sort(key = ETAOIN.find, reverse = True)
freqToLetter[freq] = ''.join(freqToLetter[freq])
freqPairs = list(freqToLetter.items())
freqPairs.sort(key = getItemAtIndexZero, reverse = True)
freqOrder = []
for freqPair in freqPairs:
freqOrder.append(freqPair[1])
return ''.join(freqOrder)
def englishFreqMatchScore(message):
'''
>>> englishFreqMatchScore('Hello World')
1
'''
freqOrder = getFrequencyOrder(message)
matchScore = 0
for commonLetter in ETAOIN[:6]:
if commonLetter in freqOrder[:6]:
matchScore += 1
for uncommonLetter in ETAOIN[-6:]:
if uncommonLetter in freqOrder[-6:]:
matchScore += 1
return matchScore
if __name__ == '__main__':
import doctest
doctest.testmod()
| [] | [] | [] |
archives/1098994933_python.zip | other/game_of_life.py | '''Conway's Game Of Life, Author Anurag Kumar(mailto:anuragkumarak95@gmail.com)
Requirements:
- numpy
- random
- time
- matplotlib
Python:
- 3.5
Usage:
- $python3 game_o_life <canvas_size:int>
Game-Of-Life Rules:
1.
Any live cell with fewer than two live neighbours
dies, as if caused by under-population.
2.
Any live cell with two or three live neighbours lives
on to the next generation.
3.
Any live cell with more than three live neighbours
dies, as if by over-population.
4.
Any dead cell with exactly three live neighbours be-
comes a live cell, as if by reproduction.
'''
import numpy as np
import random, sys
from matplotlib import pyplot as plt
from matplotlib.colors import ListedColormap
usage_doc='Usage of script: script_nama <size_of_canvas:int>'
choice = [0]*100 + [1]*10
random.shuffle(choice)
def create_canvas(size):
canvas = [ [False for i in range(size)] for j in range(size)]
return canvas
def seed(canvas):
for i,row in enumerate(canvas):
for j,_ in enumerate(row):
canvas[i][j]=bool(random.getrandbits(1))
def run(canvas):
''' This function runs the rules of game through all points, and changes their status accordingly.(in the same canvas)
@Args:
--
canvas : canvas of population to run the rules on.
@returns:
--
None
'''
canvas = np.array(canvas)
next_gen_canvas = np.array(create_canvas(canvas.shape[0]))
for r, row in enumerate(canvas):
for c, pt in enumerate(row):
# print(r-1,r+2,c-1,c+2)
next_gen_canvas[r][c] = __judge_point(pt,canvas[r-1:r+2,c-1:c+2])
canvas = next_gen_canvas
del next_gen_canvas # cleaning memory as we move on.
return canvas.tolist()
def __judge_point(pt,neighbours):
dead = 0
alive = 0
# finding dead or alive neighbours count.
for i in neighbours:
for status in i:
if status: alive+=1
else: dead+=1
# handling duplicate entry for focus pt.
if pt : alive-=1
else : dead-=1
# running the rules of game here.
state = pt
if pt:
if alive<2:
state=False
elif alive==2 or alive==3:
state=True
elif alive>3:
state=False
else:
if alive==3:
state=True
return state
if __name__=='__main__':
if len(sys.argv) != 2: raise Exception(usage_doc)
canvas_size = int(sys.argv[1])
# main working structure of this module.
c=create_canvas(canvas_size)
seed(c)
fig, ax = plt.subplots()
fig.show()
cmap = ListedColormap(['w','k'])
try:
while True:
c = run(c)
ax.matshow(c,cmap=cmap)
fig.canvas.draw()
ax.cla()
except KeyboardInterrupt:
# do nothing.
pass
| [] | [] | [] |
archives/1098994933_python.zip | other/linear_congruential_generator.py | __author__ = "Tobias Carryer"
from time import time
class LinearCongruentialGenerator(object):
"""
A pseudorandom number generator.
"""
def __init__( self, multiplier, increment, modulo, seed=int(time()) ):
"""
These parameters are saved and used when nextNumber() is called.
modulo is the largest number that can be generated (exclusive). The most
efficent values are powers of 2. 2^32 is a common value.
"""
self.multiplier = multiplier
self.increment = increment
self.modulo = modulo
self.seed = seed
def next_number( self ):
"""
The smallest number that can be generated is zero.
The largest number that can be generated is modulo-1. modulo is set in the constructor.
"""
self.seed = (self.multiplier * self.seed + self.increment) % self.modulo
return self.seed
if __name__ == "__main__":
# Show the LCG in action.
lcg = LinearCongruentialGenerator(1664525, 1013904223, 2<<31)
while True :
print(lcg.next_number()) | [] | [] | [] |
archives/1098994933_python.zip | other/nested_brackets.py | '''
The nested brackets problem is a problem that determines if a sequence of
brackets are properly nested. A sequence of brackets s is considered properly nested
if any of the following conditions are true:
- s is empty
- s has the form (U) or [U] or {U} where U is a properly nested string
- s has the form VW where V and W are properly nested strings
For example, the string "()()[()]" is properly nested but "[(()]" is not.
The function called is_balanced takes as input a string S which is a sequence of brackets and
returns true if S is nested and false otherwise.
'''
def is_balanced(S):
stack = []
open_brackets = set({'(', '[', '{'})
closed_brackets = set({')', ']', '}'})
open_to_closed = dict({'{':'}', '[':']', '(':')'})
for i in range(len(S)):
if S[i] in open_brackets:
stack.append(S[i])
elif S[i] in closed_brackets:
if len(stack) == 0 or (len(stack) > 0 and open_to_closed[stack.pop()] != S[i]):
return False
return len(stack) == 0
def main():
S = input("Enter sequence of brackets: ")
if is_balanced(S):
print((S, "is balanced"))
else:
print((S, "is not balanced"))
if __name__ == "__main__":
main()
| [] | [] | [] |
archives/1098994933_python.zip | other/palindrome.py | # Program to find whether given string is palindrome or not
def is_palindrome(str):
start_i = 0
end_i = len(str) - 1
while start_i < end_i:
if str[start_i] == str[end_i]:
start_i += 1
end_i -= 1
else:
return False
return True
# Recursive method
def recursive_palindrome(str):
if len(str) <= 1:
return True
if str[0] == str[len(str) - 1]:
return recursive_palindrome(str[1:-1])
else:
return False
def main():
str = 'ama'
print(recursive_palindrome(str.lower()))
print(is_palindrome(str.lower()))
if __name__ == '__main__':
main()
| [] | [] | [] |
archives/1098994933_python.zip | other/password_generator.py | """Password generator allows you to generate a random password of length N."""
from random import choice
from string import ascii_letters, digits, punctuation
def password_generator(length=8):
"""
>>> len(password_generator())
8
>>> len(password_generator(length=16))
16
>>> len(password_generator(257))
257
>>> len(password_generator(length=0))
0
>>> len(password_generator(-1))
0
"""
chars = tuple(ascii_letters) + tuple(digits) + tuple(punctuation)
return ''.join(choice(chars) for x in range(length))
# ALTERNATIVE METHODS
# ctbi= characters that must be in password
# i= how many letters or characters the password length will be
def alternative_password_generator(ctbi, i):
# Password generator = full boot with random_number, random_letters, and
# random_character FUNCTIONS
pass # Put your code here...
def random_number(ctbi, i):
pass # Put your code here...
def random_letters(ctbi, i):
pass # Put your code here...
def random_characters(ctbi, i):
pass # Put your code here...
def main():
length = int(
input('Please indicate the max length of your password: ').strip())
print('Password generated:', password_generator(length))
print('[If you are thinking of using this passsword, You better save it.]')
if __name__ == '__main__':
main()
| [] | [] | [] |
archives/1098994933_python.zip | other/primelib.py | # -*- coding: utf-8 -*-
"""
Created on Thu Oct 5 16:44:23 2017
@author: Christian Bender
This python library contains some useful functions to deal with
prime numbers and whole numbers.
Overview:
isPrime(number)
sieveEr(N)
getPrimeNumbers(N)
primeFactorization(number)
greatestPrimeFactor(number)
smallestPrimeFactor(number)
getPrime(n)
getPrimesBetween(pNumber1, pNumber2)
----
isEven(number)
isOdd(number)
gcd(number1, number2) // greatest common divisor
kgV(number1, number2) // least common multiple
getDivisors(number) // all divisors of 'number' inclusive 1, number
isPerfectNumber(number)
NEW-FUNCTIONS
simplifyFraction(numerator, denominator)
factorial (n) // n!
fib (n) // calculate the n-th fibonacci term.
-----
goldbach(number) // Goldbach's assumption
"""
from math import sqrt
def isPrime(number):
"""
input: positive integer 'number'
returns true if 'number' is prime otherwise false.
"""
# precondition
assert isinstance(number,int) and (number >= 0) , \
"'number' must been an int and positive"
status = True
# 0 and 1 are none primes.
if number <= 1:
status = False
for divisor in range(2,int(round(sqrt(number)))+1):
# if 'number' divisible by 'divisor' then sets 'status'
# of false and break up the loop.
if number % divisor == 0:
status = False
break
# precondition
assert isinstance(status,bool), "'status' must been from type bool"
return status
# ------------------------------------------
def sieveEr(N):
"""
input: positive integer 'N' > 2
returns a list of prime numbers from 2 up to N.
This function implements the algorithm called
sieve of erathostenes.
"""
# precondition
assert isinstance(N,int) and (N > 2), "'N' must been an int and > 2"
# beginList: conatins all natural numbers from 2 upt to N
beginList = [x for x in range(2,N+1)]
ans = [] # this list will be returns.
# actual sieve of erathostenes
for i in range(len(beginList)):
for j in range(i+1,len(beginList)):
if (beginList[i] != 0) and \
(beginList[j] % beginList[i] == 0):
beginList[j] = 0
# filters actual prime numbers.
ans = [x for x in beginList if x != 0]
# precondition
assert isinstance(ans,list), "'ans' must been from type list"
return ans
# --------------------------------
def getPrimeNumbers(N):
"""
input: positive integer 'N' > 2
returns a list of prime numbers from 2 up to N (inclusive)
This function is more efficient as function 'sieveEr(...)'
"""
# precondition
assert isinstance(N,int) and (N > 2), "'N' must been an int and > 2"
ans = []
# iterates over all numbers between 2 up to N+1
# if a number is prime then appends to list 'ans'
for number in range(2,N+1):
if isPrime(number):
ans.append(number)
# precondition
assert isinstance(ans,list), "'ans' must been from type list"
return ans
# -----------------------------------------
def primeFactorization(number):
"""
input: positive integer 'number'
returns a list of the prime number factors of 'number'
"""
# precondition
assert isinstance(number,int) and number >= 0, \
"'number' must been an int and >= 0"
ans = [] # this list will be returns of the function.
# potential prime number factors.
factor = 2
quotient = number
if number == 0 or number == 1:
ans.append(number)
# if 'number' not prime then builds the prime factorization of 'number'
elif not isPrime(number):
while (quotient != 1):
if isPrime(factor) and (quotient % factor == 0):
ans.append(factor)
quotient /= factor
else:
factor += 1
else:
ans.append(number)
# precondition
assert isinstance(ans,list), "'ans' must been from type list"
return ans
# -----------------------------------------
def greatestPrimeFactor(number):
"""
input: positive integer 'number' >= 0
returns the greatest prime number factor of 'number'
"""
# precondition
assert isinstance(number,int) and (number >= 0), \
"'number' bust been an int and >= 0"
ans = 0
# prime factorization of 'number'
primeFactors = primeFactorization(number)
ans = max(primeFactors)
# precondition
assert isinstance(ans,int), "'ans' must been from type int"
return ans
# ----------------------------------------------
def smallestPrimeFactor(number):
"""
input: integer 'number' >= 0
returns the smallest prime number factor of 'number'
"""
# precondition
assert isinstance(number,int) and (number >= 0), \
"'number' bust been an int and >= 0"
ans = 0
# prime factorization of 'number'
primeFactors = primeFactorization(number)
ans = min(primeFactors)
# precondition
assert isinstance(ans,int), "'ans' must been from type int"
return ans
# ----------------------
def isEven(number):
"""
input: integer 'number'
returns true if 'number' is even, otherwise false.
"""
# precondition
assert isinstance(number, int), "'number' must been an int"
assert isinstance(number % 2 == 0, bool), "compare bust been from type bool"
return number % 2 == 0
# ------------------------
def isOdd(number):
"""
input: integer 'number'
returns true if 'number' is odd, otherwise false.
"""
# precondition
assert isinstance(number, int), "'number' must been an int"
assert isinstance(number % 2 != 0, bool), "compare bust been from type bool"
return number % 2 != 0
# ------------------------
def goldbach(number):
"""
Goldbach's assumption
input: a even positive integer 'number' > 2
returns a list of two prime numbers whose sum is equal to 'number'
"""
# precondition
assert isinstance(number,int) and (number > 2) and isEven(number), \
"'number' must been an int, even and > 2"
ans = [] # this list will returned
# creates a list of prime numbers between 2 up to 'number'
primeNumbers = getPrimeNumbers(number)
lenPN = len(primeNumbers)
# run variable for while-loops.
i = 0
j = None
# exit variable. for break up the loops
loop = True
while (i < lenPN and loop):
j = i+1
while (j < lenPN and loop):
if primeNumbers[i] + primeNumbers[j] == number:
loop = False
ans.append(primeNumbers[i])
ans.append(primeNumbers[j])
j += 1
i += 1
# precondition
assert isinstance(ans,list) and (len(ans) == 2) and \
(ans[0] + ans[1] == number) and isPrime(ans[0]) and isPrime(ans[1]), \
"'ans' must contains two primes. And sum of elements must been eq 'number'"
return ans
# ----------------------------------------------
def gcd(number1,number2):
"""
Greatest common divisor
input: two positive integer 'number1' and 'number2'
returns the greatest common divisor of 'number1' and 'number2'
"""
# precondition
assert isinstance(number1,int) and isinstance(number2,int) \
and (number1 >= 0) and (number2 >= 0), \
"'number1' and 'number2' must been positive integer."
rest = 0
while number2 != 0:
rest = number1 % number2
number1 = number2
number2 = rest
# precondition
assert isinstance(number1,int) and (number1 >= 0), \
"'number' must been from type int and positive"
return number1
# ----------------------------------------------------
def kgV(number1, number2):
"""
Least common multiple
input: two positive integer 'number1' and 'number2'
returns the least common multiple of 'number1' and 'number2'
"""
# precondition
assert isinstance(number1,int) and isinstance(number2,int) \
and (number1 >= 1) and (number2 >= 1), \
"'number1' and 'number2' must been positive integer."
ans = 1 # actual answer that will be return.
# for kgV (x,1)
if number1 > 1 and number2 > 1:
# builds the prime factorization of 'number1' and 'number2'
primeFac1 = primeFactorization(number1)
primeFac2 = primeFactorization(number2)
elif number1 == 1 or number2 == 1:
primeFac1 = []
primeFac2 = []
ans = max(number1,number2)
count1 = 0
count2 = 0
done = [] # captured numbers int both 'primeFac1' and 'primeFac2'
# iterates through primeFac1
for n in primeFac1:
if n not in done:
if n in primeFac2:
count1 = primeFac1.count(n)
count2 = primeFac2.count(n)
for i in range(max(count1,count2)):
ans *= n
else:
count1 = primeFac1.count(n)
for i in range(count1):
ans *= n
done.append(n)
# iterates through primeFac2
for n in primeFac2:
if n not in done:
count2 = primeFac2.count(n)
for i in range(count2):
ans *= n
done.append(n)
# precondition
assert isinstance(ans,int) and (ans >= 0), \
"'ans' must been from type int and positive"
return ans
# ----------------------------------
def getPrime(n):
"""
Gets the n-th prime number.
input: positive integer 'n' >= 0
returns the n-th prime number, beginning at index 0
"""
# precondition
assert isinstance(n,int) and (n >= 0), "'number' must been a positive int"
index = 0
ans = 2 # this variable holds the answer
while index < n:
index += 1
ans += 1 # counts to the next number
# if ans not prime then
# runs to the next prime number.
while not isPrime(ans):
ans += 1
# precondition
assert isinstance(ans,int) and isPrime(ans), \
"'ans' must been a prime number and from type int"
return ans
# ---------------------------------------------------
def getPrimesBetween(pNumber1, pNumber2):
"""
input: prime numbers 'pNumber1' and 'pNumber2'
pNumber1 < pNumber2
returns a list of all prime numbers between 'pNumber1' (exclusiv)
and 'pNumber2' (exclusiv)
"""
# precondition
assert isPrime(pNumber1) and isPrime(pNumber2) and (pNumber1 < pNumber2), \
"The arguments must been prime numbers and 'pNumber1' < 'pNumber2'"
number = pNumber1 + 1 # jump to the next number
ans = [] # this list will be returns.
# if number is not prime then
# fetch the next prime number.
while not isPrime(number):
number += 1
while number < pNumber2:
ans.append(number)
number += 1
# fetch the next prime number.
while not isPrime(number):
number += 1
# precondition
assert isinstance(ans,list) and ans[0] != pNumber1 \
and ans[len(ans)-1] != pNumber2, \
"'ans' must been a list without the arguments"
# 'ans' contains not 'pNumber1' and 'pNumber2' !
return ans
# ----------------------------------------------------
def getDivisors(n):
"""
input: positive integer 'n' >= 1
returns all divisors of n (inclusive 1 and 'n')
"""
# precondition
assert isinstance(n,int) and (n >= 1), "'n' must been int and >= 1"
ans = [] # will be returned.
for divisor in range(1,n+1):
if n % divisor == 0:
ans.append(divisor)
#precondition
assert ans[0] == 1 and ans[len(ans)-1] == n, \
"Error in function getDivisiors(...)"
return ans
# ----------------------------------------------------
def isPerfectNumber(number):
"""
input: positive integer 'number' > 1
returns true if 'number' is a perfect number otherwise false.
"""
# precondition
assert isinstance(number,int) and (number > 1), \
"'number' must been an int and >= 1"
divisors = getDivisors(number)
# precondition
assert isinstance(divisors,list) and(divisors[0] == 1) and \
(divisors[len(divisors)-1] == number), \
"Error in help-function getDivisiors(...)"
# summed all divisors up to 'number' (exclusive), hence [:-1]
return sum(divisors[:-1]) == number
# ------------------------------------------------------------
def simplifyFraction(numerator, denominator):
"""
input: two integer 'numerator' and 'denominator'
assumes: 'denominator' != 0
returns: a tuple with simplify numerator and denominator.
"""
# precondition
assert isinstance(numerator, int) and isinstance(denominator,int) \
and (denominator != 0), \
"The arguments must been from type int and 'denominator' != 0"
# build the greatest common divisor of numerator and denominator.
gcdOfFraction = gcd(abs(numerator), abs(denominator))
# precondition
assert isinstance(gcdOfFraction, int) and (numerator % gcdOfFraction == 0) \
and (denominator % gcdOfFraction == 0), \
"Error in function gcd(...,...)"
return (numerator // gcdOfFraction, denominator // gcdOfFraction)
# -----------------------------------------------------------------
def factorial(n):
"""
input: positive integer 'n'
returns the factorial of 'n' (n!)
"""
# precondition
assert isinstance(n,int) and (n >= 0), "'n' must been a int and >= 0"
ans = 1 # this will be return.
for factor in range(1,n+1):
ans *= factor
return ans
# -------------------------------------------------------------------
def fib(n):
"""
input: positive integer 'n'
returns the n-th fibonacci term , indexing by 0
"""
# precondition
assert isinstance(n, int) and (n >= 0), "'n' must been an int and >= 0"
tmp = 0
fib1 = 1
ans = 1 # this will be return
for i in range(n-1):
tmp = ans
ans += fib1
fib1 = tmp
return ans
| [] | [] | [] |
archives/1098994933_python.zip | other/sierpinski_triangle.py | #!/usr/bin/python
# encoding=utf8
'''Author Anurag Kumar | anuragkumarak95@gmail.com | git/anuragkumarak95
Simple example of Fractal generation using recursive function.
What is Sierpinski Triangle?
>>The Sierpinski triangle (also with the original orthography Sierpinski), also called the Sierpinski gasket or the Sierpinski Sieve,
is a fractal and attractive fixed set with the overall shape of an equilateral triangle, subdivided recursively into smaller
equilateral triangles. Originally constructed as a curve, this is one of the basic examples of self-similar sets, i.e.,
it is a mathematically generated pattern that can be reproducible at any magnification or reduction. It is named after
the Polish mathematician Wacław Sierpinski, but appeared as a decorative pattern many centuries prior to the work of Sierpinski.
Requirements(pip):
- turtle
Python:
- 2.6
Usage:
- $python sierpinski_triangle.py <int:depth_for_fractal>
Credits: This code was written by editing the code from http://www.riannetrujillo.com/blog/python-fractal/
'''
import turtle
import sys
PROGNAME = 'Sierpinski Triangle'
points = [[-175,-125],[0,175],[175,-125]] #size of triangle
def getMid(p1,p2):
return ( (p1[0]+p2[0]) / 2, (p1[1] + p2[1]) / 2) #find midpoint
def triangle(points,depth):
myPen.up()
myPen.goto(points[0][0],points[0][1])
myPen.down()
myPen.goto(points[1][0],points[1][1])
myPen.goto(points[2][0],points[2][1])
myPen.goto(points[0][0],points[0][1])
if depth>0:
triangle([points[0],
getMid(points[0], points[1]),
getMid(points[0], points[2])],
depth-1)
triangle([points[1],
getMid(points[0], points[1]),
getMid(points[1], points[2])],
depth-1)
triangle([points[2],
getMid(points[2], points[1]),
getMid(points[0], points[2])],
depth-1)
if __name__ == '__main__':
if len(sys.argv) !=2:
raise ValueError('right format for using this script: '
'$python fractals.py <int:depth_for_fractal>')
myPen = turtle.Turtle()
myPen.ht()
myPen.speed(5)
myPen.pencolor('red')
triangle(points,int(sys.argv[1]))
| [] | [] | [] |
archives/1098994933_python.zip | other/tower_of_hanoi.py | def moveTower(height, fromPole, toPole, withPole):
'''
>>> moveTower(3, 'A', 'B', 'C')
moving disk from A to B
moving disk from A to C
moving disk from B to C
moving disk from A to B
moving disk from C to A
moving disk from C to B
moving disk from A to B
'''
if height >= 1:
moveTower(height-1, fromPole, withPole, toPole)
moveDisk(fromPole, toPole)
moveTower(height-1, withPole, toPole, fromPole)
def moveDisk(fp,tp):
print('moving disk from', fp, 'to', tp)
def main():
height = int(input('Height of hanoi: ').strip())
moveTower(height, 'A', 'B', 'C')
if __name__ == '__main__':
main()
| [] | [] | [] |
archives/1098994933_python.zip | other/two_sum.py | """
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
"""
def twoSum(nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
chk_map = {}
for index, val in enumerate(nums):
compl = target - val
if compl in chk_map:
indices = [chk_map[compl], index]
print(indices)
return [indices]
else:
chk_map[val] = index
return False
| [] | [] | [] |
archives/1098994933_python.zip | other/word_patterns.py | import pprint, time
def getWordPattern(word):
word = word.upper()
nextNum = 0
letterNums = {}
wordPattern = []
for letter in word:
if letter not in letterNums:
letterNums[letter] = str(nextNum)
nextNum += 1
wordPattern.append(letterNums[letter])
return '.'.join(wordPattern)
def main():
startTime = time.time()
allPatterns = {}
with open('Dictionary.txt') as fo:
wordList = fo.read().split('\n')
for word in wordList:
pattern = getWordPattern(word)
if pattern not in allPatterns:
allPatterns[pattern] = [word]
else:
allPatterns[pattern].append(word)
with open('Word Patterns.txt', 'w') as fo:
fo.write(pprint.pformat(allPatterns))
totalTime = round(time.time() - startTime, 2)
print(('Done! [', totalTime, 'seconds ]'))
if __name__ == '__main__':
main()
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_01/sol1.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>>> solution(4)
3
>>> solution(10)
23
>>> solution(600)
83700
>>> solution(-7)
0
"""
return sum([e for e in range(3, n) if e % 3 == 0 or e % 5 == 0])
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol2.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>>> solution(4)
3
>>> solution(10)
23
>>> solution(600)
83700
"""
sum = 0
terms = (n - 1) // 3
sum += ((terms) * (6 + (terms - 1) * 3)) // 2 # sum of an A.P.
terms = (n - 1) // 5
sum += ((terms) * (10 + (terms - 1) * 5)) // 2
terms = (n - 1) // 15
sum -= ((terms) * (30 + (terms - 1) * 15)) // 2
return sum
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol3.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""
This solution is based on the pattern that the successive numbers in the
series follow: 0+3,+2,+1,+3,+1,+2,+3.
Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>>> solution(4)
3
>>> solution(10)
23
>>> solution(600)
83700
"""
sum = 0
num = 0
while 1:
num += 3
if num >= n:
break
sum += num
num += 2
if num >= n:
break
sum += num
num += 1
if num >= n:
break
sum += num
num += 3
if num >= n:
break
sum += num
num += 1
if num >= n:
break
sum += num
num += 2
if num >= n:
break
sum += num
num += 3
if num >= n:
break
sum += num
return sum
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol4.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>>> solution(4)
3
>>> solution(10)
23
>>> solution(600)
83700
"""
xmulti = []
zmulti = []
z = 3
x = 5
temp = 1
while True:
result = z * temp
if result < n:
zmulti.append(result)
temp += 1
else:
temp = 1
break
while True:
result = x * temp
if result < n:
xmulti.append(result)
temp += 1
else:
break
collection = list(set(xmulti + zmulti))
return sum(collection)
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol5.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
"""A straightforward pythonic solution using list comprehension"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>>> solution(4)
3
>>> solution(10)
23
>>> solution(600)
83700
"""
return sum([i for i in range(n) if i % 3 == 0 or i % 5 == 0])
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol6.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>>> solution(4)
3
>>> solution(10)
23
>>> solution(600)
83700
"""
a = 3
result = 0
while a < n:
if a % 3 == 0 or a % 5 == 0:
result += a
elif a % 15 == 0:
result -= a
a += 1
return result
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_02/sol1.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous two
terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=10, we have {2,8}, sum is
10.
"""
def solution(n):
"""Returns the sum of all fibonacci sequence even elements that are lower
or equals to n.
>>> solution(10)
10
>>> solution(15)
10
>>> solution(2)
2
>>> solution(1)
0
>>> solution(34)
44
"""
i = 1
j = 2
sum = 0
while j <= n:
if j % 2 == 0:
sum += j
i, j = j, i + j
return sum
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/sol2.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous two
terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=10, we have {2,8}, sum is
10.
"""
def solution(n):
"""Returns the sum of all fibonacci sequence even elements that are lower
or equals to n.
>>> solution(10)
[2, 8]
>>> solution(15)
[2, 8]
>>> solution(2)
[2]
>>> solution(1)
[]
>>> solution(34)
[2, 8, 34]
"""
ls = []
a, b = 0, 1
while b <= n:
if b % 2 == 0:
ls.append(b)
a, b = b, a + b
return ls
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/sol3.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous
two terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=10, we have {2,8}, sum is
10.
"""
def solution(n):
"""Returns the sum of all fibonacci sequence even elements that are lower
or equals to n.
>>> solution(10)
10
>>> solution(15)
10
>>> solution(2)
2
>>> solution(1)
0
>>> solution(34)
44
"""
if n <= 1:
return 0
a = 0
b = 2
count = 0
while 4 * b + a <= n:
a, b = b, 4 * b + a
count += a
return count + b
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/sol4.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous two
terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=10, we have {2,8}, sum is
10.
"""
import math
from decimal import Decimal, getcontext
def solution(n):
"""Returns the sum of all fibonacci sequence even elements that are lower
or equals to n.
>>> solution(10)
10
>>> solution(15)
10
>>> solution(2)
2
>>> solution(1)
0
>>> solution(34)
44
>>> solution(3.4)
2
>>> solution(0)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution(-17)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution([])
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
>>> solution("asd")
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
"""
try:
n = int(n)
except (TypeError, ValueError) as e:
raise TypeError("Parameter n must be int or passive of cast to int.")
if n <= 0:
raise ValueError("Parameter n must be greater or equal to one.")
getcontext().prec = 100
phi = (Decimal(5) ** Decimal(0.5) + 1) / Decimal(2)
index = (math.floor(math.log(n * (phi + 2), phi) - 1) // 3) * 3 + 2
num = Decimal(round(phi ** Decimal(index + 1))) / (phi + 2)
sum = num // 2
return int(sum)
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_03/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_03/sol1.py | """
Problem:
The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor
of a given number N?
e.g. for 10, largest prime factor = 5. For 17, largest prime factor = 17.
"""
import math
def isprime(no):
if no == 2:
return True
elif no % 2 == 0:
return False
sq = int(math.sqrt(no)) + 1
for i in range(3, sq, 2):
if no % i == 0:
return False
return True
def solution(n):
"""Returns the largest prime factor of a given number n.
>>> solution(13195)
29
>>> solution(10)
5
>>> solution(17)
17
>>> solution(3.4)
3
>>> solution(0)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution(-17)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution([])
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
>>> solution("asd")
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
"""
try:
n = int(n)
except (TypeError, ValueError) as e:
raise TypeError("Parameter n must be int or passive of cast to int.")
if n <= 0:
raise ValueError("Parameter n must be greater or equal to one.")
maxNumber = 0
if isprime(n):
return n
else:
while n % 2 == 0:
n = n / 2
if isprime(n):
return int(n)
else:
n1 = int(math.sqrt(n)) + 1
for i in range(3, n1, 2):
if n % i == 0:
if isprime(n / i):
maxNumber = n / i
break
elif isprime(i):
maxNumber = i
return maxNumber
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_03/sol2.py | """
Problem:
The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor
of a given number N?
e.g. for 10, largest prime factor = 5. For 17, largest prime factor = 17.
"""
def solution(n):
"""Returns the largest prime factor of a given number n.
>>> solution(13195)
29
>>> solution(10)
5
>>> solution(17)
17
>>> solution(3.4)
3
>>> solution(0)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution(-17)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution([])
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
>>> solution("asd")
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
"""
try:
n = int(n)
except (TypeError, ValueError) as e:
raise TypeError("Parameter n must be int or passive of cast to int.")
if n <= 0:
raise ValueError("Parameter n must be greater or equal to one.")
prime = 1
i = 2
while i * i <= n:
while n % i == 0:
prime = i
n //= i
i += 1
if n > 1:
prime = n
return int(prime)
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_04/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_04/sol1.py | """
Problem:
A palindromic number reads the same both ways. The largest palindrome made from
the product of two 2-digit numbers is 9009 = 91 x 99.
Find the largest palindrome made from the product of two 3-digit numbers which
is less than N.
"""
def solution(n):
"""Returns the largest palindrome made from the product of two 3-digit
numbers which is less than n.
>>> solution(20000)
19591
>>> solution(30000)
29992
>>> solution(40000)
39893
"""
# fetchs the next number
for number in range(n - 1, 10000, -1):
# converts number into string.
strNumber = str(number)
# checks whether 'strNumber' is a palindrome.
if strNumber == strNumber[::-1]:
divisor = 999
# if 'number' is a product of two 3-digit numbers
# then number is the answer otherwise fetch next number.
while divisor != 99:
if (number % divisor == 0) and (
len(str(int(number / divisor))) == 3
):
return number
divisor -= 1
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_04/sol2.py | """
Problem:
A palindromic number reads the same both ways. The largest palindrome made from
the product of two 2-digit numbers is 9009 = 91 x 99.
Find the largest palindrome made from the product of two 3-digit numbers which
is less than N.
"""
def solution(n):
"""Returns the largest palindrome made from the product of two 3-digit
numbers which is less than n.
>>> solution(20000)
19591
>>> solution(30000)
29992
>>> solution(40000)
39893
"""
answer = 0
for i in range(999, 99, -1): # 3 digit nimbers range from 999 down to 100
for j in range(999, 99, -1):
t = str(i * j)
if t == t[::-1] and i * j < n:
answer = max(answer, i * j)
return answer
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_05/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_05/sol1.py | """
Problem:
2520 is the smallest number that can be divided by each of the numbers from 1
to 10 without any remainder.
What is the smallest positive number that is evenly divisible(divisible with no
remainder) by all of the numbers from 1 to N?
"""
def solution(n):
"""Returns the smallest positive number that is evenly divisible(divisible
with no remainder) by all of the numbers from 1 to n.
>>> solution(10)
2520
>>> solution(15)
360360
>>> solution(20)
232792560
>>> solution(22)
232792560
>>> solution(3.4)
6
>>> solution(0)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution(-17)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution([])
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
>>> solution("asd")
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
"""
try:
n = int(n)
except (TypeError, ValueError) as e:
raise TypeError("Parameter n must be int or passive of cast to int.")
if n <= 0:
raise ValueError("Parameter n must be greater or equal to one.")
i = 0
while 1:
i += n * (n - 1)
nfound = 0
for j in range(2, n):
if i % j != 0:
nfound = 1
break
if nfound == 0:
if i == 0:
i = 1
return i
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_05/sol2.py | """
Problem:
2520 is the smallest number that can be divided by each of the numbers from 1
to 10 without any remainder.
What is the smallest positive number that is evenly divisible(divisible with no
remainder) by all of the numbers from 1 to N?
"""
""" Euclidean GCD Algorithm """
def gcd(x, y):
return x if y == 0 else gcd(y, x % y)
""" Using the property lcm*gcd of two numbers = product of them """
def lcm(x, y):
return (x * y) // gcd(x, y)
def solution(n):
"""Returns the smallest positive number that is evenly divisible(divisible
with no remainder) by all of the numbers from 1 to n.
>>> solution(10)
2520
>>> solution(15)
360360
>>> solution(20)
232792560
>>> solution(22)
232792560
"""
g = 1
for i in range(1, n + 1):
g = lcm(g, i)
return g
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_06/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_06/sol1.py | # -*- coding: utf-8 -*-
"""
Problem:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural
numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first N natural
numbers and the square of the sum.
"""
def solution(n):
"""Returns the difference between the sum of the squares of the first n
natural numbers and the square of the sum.
>>> solution(10)
2640
>>> solution(15)
13160
>>> solution(20)
41230
>>> solution(50)
1582700
"""
suma = 0
sumb = 0
for i in range(1, n + 1):
suma += i ** 2
sumb += i
sum = sumb ** 2 - suma
return sum
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_06/sol2.py | # -*- coding: utf-8 -*-
"""
Problem:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural
numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first N natural
numbers and the square of the sum.
"""
def solution(n):
"""Returns the difference between the sum of the squares of the first n
natural numbers and the square of the sum.
>>> solution(10)
2640
>>> solution(15)
13160
>>> solution(20)
41230
>>> solution(50)
1582700
"""
suma = n * (n + 1) / 2
suma **= 2
sumb = n * (n + 1) * (2 * n + 1) / 6
return int(suma - sumb)
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_06/sol3.py | # -*- coding: utf-8 -*-
"""
Problem:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural
numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first N natural
numbers and the square of the sum.
"""
import math
def solution(n):
"""Returns the difference between the sum of the squares of the first n
natural numbers and the square of the sum.
>>> solution(10)
2640
>>> solution(15)
13160
>>> solution(20)
41230
>>> solution(50)
1582700
"""
sum_of_squares = sum([i * i for i in range(1, n + 1)])
square_of_sum = int(math.pow(sum(range(1, n + 1)), 2))
return square_of_sum - sum_of_squares
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_07/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_07/sol1.py | # -*- coding: utf-8 -*-
"""
By listing the first six prime numbers:
2, 3, 5, 7, 11, and 13
We can see that the 6th prime is 13. What is the Nth prime number?
"""
from math import sqrt
def isprime(n):
if n == 2:
return True
elif n % 2 == 0:
return False
else:
sq = int(sqrt(n)) + 1
for i in range(3, sq, 2):
if n % i == 0:
return False
return True
def solution(n):
"""Returns the n-th prime number.
>>> solution(6)
13
>>> solution(1)
2
>>> solution(3)
5
>>> solution(20)
71
>>> solution(50)
229
>>> solution(100)
541
"""
i = 0
j = 1
while i != n and j < 3:
j += 1
if isprime(j):
i += 1
while i != n:
j += 2
if isprime(j):
i += 1
return j
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_07/sol2.py | # -*- coding: utf-8 -*-
"""
By listing the first six prime numbers:
2, 3, 5, 7, 11, and 13
We can see that the 6th prime is 13. What is the Nth prime number?
"""
def isprime(number):
for i in range(2, int(number ** 0.5) + 1):
if number % i == 0:
return False
return True
def solution(n):
"""Returns the n-th prime number.
>>> solution(6)
13
>>> solution(1)
2
>>> solution(3)
5
>>> solution(20)
71
>>> solution(50)
229
>>> solution(100)
541
>>> solution(3.4)
5
>>> solution(0)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution(-17)
Traceback (most recent call last):
...
ValueError: Parameter n must be greater or equal to one.
>>> solution([])
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
>>> solution("asd")
Traceback (most recent call last):
...
TypeError: Parameter n must be int or passive of cast to int.
"""
try:
n = int(n)
except (TypeError, ValueError) as e:
raise TypeError("Parameter n must be int or passive of cast to int.")
if n <= 0:
raise ValueError("Parameter n must be greater or equal to one.")
primes = []
num = 2
while len(primes) < n:
if isprime(num):
primes.append(num)
num += 1
else:
num += 1
return primes[len(primes) - 1]
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_07/sol3.py | # -*- coding: utf-8 -*-
"""
By listing the first six prime numbers:
2, 3, 5, 7, 11, and 13
We can see that the 6th prime is 13. What is the Nth prime number?
"""
import math
import itertools
def primeCheck(number):
if number % 2 == 0 and number > 2:
return False
return all(number % i for i in range(3, int(math.sqrt(number)) + 1, 2))
def prime_generator():
num = 2
while True:
if primeCheck(num):
yield num
num += 1
def solution(n):
"""Returns the n-th prime number.
>>> solution(6)
13
>>> solution(1)
2
>>> solution(3)
5
>>> solution(20)
71
>>> solution(50)
229
>>> solution(100)
541
"""
return next(itertools.islice(prime_generator(), n - 1, n))
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_08/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_08/sol1.py | # -*- coding: utf-8 -*-
"""
The four adjacent digits in the 1000-digit number that have the greatest
product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the
greatest product. What is the value of this product?
"""
import sys
N = """73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450"""
def solution(n):
"""Find the thirteen adjacent digits in the 1000-digit number n that have
the greatest product and returns it.
>>> solution(N)
23514624000
"""
LargestProduct = -sys.maxsize - 1
for i in range(len(n) - 12):
product = 1
for j in range(13):
product *= int(n[i + j])
if product > LargestProduct:
LargestProduct = product
return LargestProduct
if __name__ == "__main__":
print(solution(N))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_08/sol2.py | # -*- coding: utf-8 -*-
"""
The four adjacent digits in the 1000-digit number that have the greatest
product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the
greatest product. What is the value of this product?
"""
from functools import reduce
N = (
"73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450"
)
def solution(n):
"""Find the thirteen adjacent digits in the 1000-digit number n that have
the greatest product and returns it.
>>> solution(N)
23514624000
"""
return max(
[
reduce(lambda x, y: int(x) * int(y), n[i : i + 13])
for i in range(len(n) - 12)
]
)
if __name__ == "__main__":
print(solution(str(N)))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_09/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_09/sol1.py | """
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def solution():
"""
Returns the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a < b < c
2. a**2 + b**2 = c**2
3. a + b + c = 1000
# The code below has been commented due to slow execution affecting Travis.
# >>> solution()
# 31875000
"""
for a in range(300):
for b in range(400):
for c in range(500):
if a < b < c:
if (a ** 2) + (b ** 2) == (c ** 2):
if (a + b + c) == 1000:
return a * b * c
if __name__ == "__main__":
print("Please Wait...")
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_09/sol2.py | """
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def solution(n):
"""
Return the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a < b < c
2. a**2 + b**2 = c**2
3. a + b + c = 1000
>>> solution(1000)
31875000
"""
product = -1
d = 0
for a in range(1, n // 3):
"""Solving the two equations a**2+b**2=c**2 and a+b+c=N eliminating c
"""
b = (n * n - 2 * a * n) // (2 * n - 2 * a)
c = n - a - b
if c * c == (a * a + b * b):
d = a * b * c
if d >= product:
product = d
return product
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_09/sol3.py | """
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def solution():
"""
Returns the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a**2 + b**2 = c**2
2. a + b + c = 1000
# The code below has been commented due to slow execution affecting Travis.
# >>> solution()
# 31875000
"""
return [
a * b * c
for a in range(1, 999)
for b in range(a, 999)
for c in range(b, 999)
if (a * a + b * b == c * c) and (a + b + c == 1000)
][0]
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_10/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_10/sol1.py | """
Problem Statement:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
from math import sqrt
def is_prime(n):
for i in range(2, int(sqrt(n)) + 1):
if n % i == 0:
return False
return True
def sum_of_primes(n):
if n > 2:
sumOfPrimes = 2
else:
return 0
for i in range(3, n, 2):
if is_prime(i):
sumOfPrimes += i
return sumOfPrimes
def solution(n):
"""Returns the sum of all the primes below n.
# The code below has been commented due to slow execution affecting Travis.
# >>> solution(2000000)
# 142913828922
>>> solution(1000)
76127
>>> solution(5000)
1548136
>>> solution(10000)
5736396
>>> solution(7)
10
"""
return sum_of_primes(n)
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_10/sol2.py | """
Problem Statement:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
import math
from itertools import takewhile
def primeCheck(number):
if number % 2 == 0 and number > 2:
return False
return all(number % i for i in range(3, int(math.sqrt(number)) + 1, 2))
def prime_generator():
num = 2
while True:
if primeCheck(num):
yield num
num += 1
def solution(n):
"""Returns the sum of all the primes below n.
# The code below has been commented due to slow execution affecting Travis.
# >>> solution(2000000)
# 142913828922
>>> solution(1000)
76127
>>> solution(5000)
1548136
>>> solution(10000)
5736396
>>> solution(7)
10
"""
return sum(takewhile(lambda x: x < n, prime_generator()))
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_10/sol3.py | """
https://projecteuler.net/problem=10
Problem Statement:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million using Sieve_of_Eratosthenes:
The sieve of Eratosthenes is one of the most efficient ways to find all primes
smaller than n when n is smaller than 10 million. Only for positive numbers.
"""
def prime_sum(n: int) -> int:
""" Returns the sum of all the primes below n.
>>> prime_sum(2_000_000)
142913828922
>>> prime_sum(1_000)
76127
>>> prime_sum(5_000)
1548136
>>> prime_sum(10_000)
5736396
>>> prime_sum(7)
10
>>> prime_sum(7.1) # doctest: +ELLIPSIS
Traceback (most recent call last):
...
TypeError: 'float' object cannot be interpreted as an integer
>>> prime_sum(-7) # doctest: +ELLIPSIS
Traceback (most recent call last):
...
IndexError: list assignment index out of range
>>> prime_sum("seven") # doctest: +ELLIPSIS
Traceback (most recent call last):
...
TypeError: can only concatenate str (not "int") to str
"""
list_ = [0 for i in range(n + 1)]
list_[0] = 1
list_[1] = 1
for i in range(2, int(n ** 0.5) + 1):
if list_[i] == 0:
for j in range(i * i, n + 1, i):
list_[j] = 1
s = 0
for i in range(n):
if list_[i] == 0:
s += i
return s
if __name__ == "__main__":
# import doctest
# doctest.testmod()
print(prime_sum(int(input().strip())))
| [
"int"
] | [
374
] | [
377
] |
archives/1098994933_python.zip | project_euler/problem_11/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_11/sol1.py | """
What is the greatest product of four adjacent numbers (horizontally,
vertically, or diagonally) in this 20x20 array?
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
"""
import os
def largest_product(grid):
nColumns = len(grid[0])
nRows = len(grid)
largest = 0
lrDiagProduct = 0
rlDiagProduct = 0
# Check vertically, horizontally, diagonally at the same time (only works
# for nxn grid)
for i in range(nColumns):
for j in range(nRows - 3):
vertProduct = (
grid[j][i] * grid[j + 1][i] * grid[j + 2][i] * grid[j + 3][i]
)
horzProduct = (
grid[i][j] * grid[i][j + 1] * grid[i][j + 2] * grid[i][j + 3]
)
# Left-to-right diagonal (\) product
if i < nColumns - 3:
lrDiagProduct = (
grid[i][j]
* grid[i + 1][j + 1]
* grid[i + 2][j + 2]
* grid[i + 3][j + 3]
)
# Right-to-left diagonal(/) product
if i > 2:
rlDiagProduct = (
grid[i][j]
* grid[i - 1][j + 1]
* grid[i - 2][j + 2]
* grid[i - 3][j + 3]
)
maxProduct = max(
vertProduct, horzProduct, lrDiagProduct, rlDiagProduct
)
if maxProduct > largest:
largest = maxProduct
return largest
def solution():
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution()
70600674
"""
grid = []
with open(os.path.dirname(__file__) + "/grid.txt") as file:
for line in file:
grid.append(line.strip("\n").split(" "))
grid = [[int(i) for i in grid[j]] for j in range(len(grid))]
return largest_product(grid)
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_11/sol2.py | """
What is the greatest product of four adjacent numbers (horizontally,
vertically, or diagonally) in this 20x20 array?
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
"""
import os
def solution():
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution()
70600674
"""
with open(os.path.dirname(__file__) + "/grid.txt") as f:
l = []
for i in range(20):
l.append([int(x) for x in f.readline().split()])
maximum = 0
# right
for i in range(20):
for j in range(17):
temp = l[i][j] * l[i][j + 1] * l[i][j + 2] * l[i][j + 3]
if temp > maximum:
maximum = temp
# down
for i in range(17):
for j in range(20):
temp = l[i][j] * l[i + 1][j] * l[i + 2][j] * l[i + 3][j]
if temp > maximum:
maximum = temp
# diagonal 1
for i in range(17):
for j in range(17):
temp = (
l[i][j]
* l[i + 1][j + 1]
* l[i + 2][j + 2]
* l[i + 3][j + 3]
)
if temp > maximum:
maximum = temp
# diagonal 2
for i in range(17):
for j in range(3, 20):
temp = (
l[i][j]
* l[i + 1][j - 1]
* l[i + 2][j - 2]
* l[i + 3][j - 3]
)
if temp > maximum:
maximum = temp
return maximum
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_12/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_12/sol1.py | """
Highly divisible triangular numbers
Problem 12
The sequence of triangle numbers is generated by adding the natural numbers. So
the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten
terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors?
"""
from math import sqrt
def count_divisors(n):
nDivisors = 0
for i in range(1, int(sqrt(n)) + 1):
if n % i == 0:
nDivisors += 2
# check if n is perfect square
if n ** 0.5 == int(n ** 0.5):
nDivisors -= 1
return nDivisors
def solution():
"""Returns the value of the first triangle number to have over five hundred
divisors.
# The code below has been commented due to slow execution affecting Travis.
# >>> solution()
# 76576500
"""
tNum = 1
i = 1
while True:
i += 1
tNum += i
if count_divisors(tNum) > 500:
break
return tNum
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_12/sol2.py | """
Highly divisible triangular numbers
Problem 12
The sequence of triangle numbers is generated by adding the natural numbers. So
the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten
terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors?
"""
def triangle_number_generator():
for n in range(1, 1000000):
yield n * (n + 1) // 2
def count_divisors(n):
return sum(
[2 for i in range(1, int(n ** 0.5) + 1) if n % i == 0 and i * i != n]
)
def solution():
"""Returns the value of the first triangle number to have over five hundred
divisors.
# The code below has been commented due to slow execution affecting Travis.
# >>> solution()
# 76576500
"""
return next(
i for i in triangle_number_generator() if count_divisors(i) > 500
)
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_13/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_13/sol1.py | """
Problem Statement:
Work out the first ten digits of the sum of the following one-hundred 50-digit
numbers.
"""
def solution(array):
"""Returns the first ten digits of the sum of the array elements.
>>> import os
>>> sum = 0
>>> array = []
>>> with open(os.path.dirname(__file__) + "/num.txt","r") as f:
... for line in f:
... array.append(int(line))
...
>>> solution(array)
'5537376230'
"""
return str(sum(array))[:10]
if __name__ == "__main__":
n = int(input().strip())
array = []
for i in range(n):
array.append(int(input().strip()))
print(solution(array))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_14/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_14/sol1.py | # -*- coding: utf-8 -*-
"""
Problem Statement:
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains
10 terms. Although it has not been proved yet (Collatz Problem), it is thought
that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
"""
def solution(n):
"""Returns the number under n that generates the longest sequence using the
formula:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
# The code below has been commented due to slow execution affecting Travis.
# >>> solution(1000000)
# {'counter': 525, 'largest_number': 837799}
>>> solution(200)
{'counter': 125, 'largest_number': 171}
>>> solution(5000)
{'counter': 238, 'largest_number': 3711}
>>> solution(15000)
{'counter': 276, 'largest_number': 13255}
"""
largest_number = 0
pre_counter = 0
for input1 in range(n):
counter = 1
number = input1
while number > 1:
if number % 2 == 0:
number /= 2
counter += 1
else:
number = (3 * number) + 1
counter += 1
if counter > pre_counter:
largest_number = input1
pre_counter = counter
return {"counter": pre_counter, "largest_number": largest_number}
if __name__ == "__main__":
result = solution(int(input().strip()))
print(
(
"Largest Number:",
result["largest_number"],
"->",
result["counter"],
"digits",
)
)
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_14/sol2.py | # -*- coding: utf-8 -*-
"""
Collatz conjecture: start with any positive integer n. Next term obtained from
the previous term as follows:
If the previous term is even, the next term is one half the previous term.
If the previous term is odd, the next term is 3 times the previous term plus 1.
The conjecture states the sequence will always reach 1 regardless of starting
n.
Problem Statement:
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains
10 terms. Although it has not been proved yet (Collatz Problem), it is thought
that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
"""
def collatz_sequence(n):
"""Returns the Collatz sequence for n."""
sequence = [n]
while n != 1:
if n % 2 == 0:
n //= 2
else:
n = 3 * n + 1
sequence.append(n)
return sequence
def solution(n):
"""Returns the number under n that generates the longest Collatz sequence.
# The code below has been commented due to slow execution affecting Travis.
# >>> solution(1000000)
# {'counter': 525, 'largest_number': 837799}
>>> solution(200)
{'counter': 125, 'largest_number': 171}
>>> solution(5000)
{'counter': 238, 'largest_number': 3711}
>>> solution(15000)
{'counter': 276, 'largest_number': 13255}
"""
result = max([(len(collatz_sequence(i)), i) for i in range(1, n)])
return {"counter": result[0], "largest_number": result[1]}
if __name__ == "__main__":
result = solution(int(input().strip()))
print(
"Longest Collatz sequence under one million is %d with length %d"
% (result["largest_number"], result["counter"])
)
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_15/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_15/sol1.py | """
Starting in the top left corner of a 2×2 grid, and only being able to move to
the right and down, there are exactly 6 routes to the bottom right corner.
How many such routes are there through a 20×20 grid?
"""
from math import factorial
def lattice_paths(n):
"""
Returns the number of paths possible in a n x n grid starting at top left
corner going to bottom right corner and being able to move right and down
only.
bruno@bruno-laptop:~/git/Python/project_euler/problem_15$ python3 sol1.py 50
1.008913445455642e+29
bruno@bruno-laptop:~/git/Python/project_euler/problem_15$ python3 sol1.py 25
126410606437752.0
bruno@bruno-laptop:~/git/Python/project_euler/problem_15$ python3 sol1.py 23
8233430727600.0
bruno@bruno-laptop:~/git/Python/project_euler/problem_15$ python3 sol1.py 15
155117520.0
bruno@bruno-laptop:~/git/Python/project_euler/problem_15$ python3 sol1.py 1
2.0
>>> lattice_paths(25)
126410606437752
>>> lattice_paths(23)
8233430727600
>>> lattice_paths(20)
137846528820
>>> lattice_paths(15)
155117520
>>> lattice_paths(1)
2
"""
n = (
2 * n
) # middle entry of odd rows starting at row 3 is the solution for n = 1,
# 2, 3,...
k = n / 2
return int(factorial(n) / (factorial(k) * factorial(n - k)))
if __name__ == "__main__":
import sys
if len(sys.argv) == 1:
print(lattice_paths(20))
else:
try:
n = int(sys.argv[1])
print(lattice_paths(n))
except ValueError:
print("Invalid entry - please enter a number.")
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_16/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_16/sol1.py | """
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
"""
def solution(power):
"""Returns the sum of the digits of the number 2^power.
>>> solution(1000)
1366
>>> solution(50)
76
>>> solution(20)
31
>>> solution(15)
26
"""
num = 2 ** power
string_num = str(num)
list_num = list(string_num)
sum_of_num = 0
for i in list_num:
sum_of_num += int(i)
return sum_of_num
if __name__ == "__main__":
power = int(input("Enter the power of 2: ").strip())
print("2 ^ ", power, " = ", 2 ** power)
result = solution(power)
print("Sum of the digits is: ", result)
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_16/sol2.py | """
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
"""
def solution(power):
"""Returns the sum of the digits of the number 2^power.
>>> solution(1000)
1366
>>> solution(50)
76
>>> solution(20)
31
>>> solution(15)
26
"""
n = 2 ** power
r = 0
while n:
r, n = r + n % 10, n // 10
return r
if __name__ == "__main__":
print(solution(int(str(input()).strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_17/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_17/sol1.py | """
Number letter counts
Problem 17
If the numbers 1 to 5 are written out in words: one, two, three, four, five,
then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in
words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20
letters. The use of "and" when writing out numbers is in compliance withBritish
usage.
"""
def solution(n):
"""Returns the number of letters used to write all numbers from 1 to n.
where n is lower or equals to 1000.
>>> solution(1000)
21124
>>> solution(5)
19
"""
# number of letters in zero, one, two, ..., nineteen (0 for zero since it's
# never said aloud)
ones_counts = [0, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8]
# number of letters in twenty, thirty, ..., ninety (0 for numbers less than
# 20 due to inconsistency in teens)
tens_counts = [0, 0, 6, 6, 5, 5, 5, 7, 6, 6]
count = 0
for i in range(1, n + 1):
if i < 1000:
if i >= 100:
# add number of letters for "n hundred"
count += ones_counts[i // 100] + 7
if i % 100 != 0:
# add number of letters for "and" if number is not multiple
# of 100
count += 3
if 0 < i % 100 < 20:
# add number of letters for one, two, three, ..., nineteen
# (could be combined with below if not for inconsistency in
# teens)
count += ones_counts[i % 100]
else:
# add number of letters for twenty, twenty one, ..., ninety
# nine
count += ones_counts[i % 10]
count += tens_counts[(i % 100 - i % 10) // 10]
else:
count += ones_counts[i // 1000] + 8
return count
if __name__ == "__main__":
print(solution(int(input().strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_19/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_19/sol1.py | """
Counting Sundays
Problem 19
You are given the following information, but you may prefer to do some research
for yourself.
1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century
unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century
(1 Jan 1901 to 31 Dec 2000)?
"""
def solution():
"""Returns the number of mondays that fall on the first of the month during
the twentieth century (1 Jan 1901 to 31 Dec 2000)?
>>> solution()
171
"""
days_per_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
day = 6
month = 1
year = 1901
sundays = 0
while year < 2001:
day += 7
if (year % 4 == 0 and not year % 100 == 0) or (year % 400 == 0):
if day > days_per_month[month - 1] and month != 2:
month += 1
day = day - days_per_month[month - 2]
elif day > 29 and month == 2:
month += 1
day = day - 29
else:
if day > days_per_month[month - 1]:
month += 1
day = day - days_per_month[month - 2]
if month > 12:
year += 1
month = 1
if year < 2001 and day == 1:
sundays += 1
return sundays
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_20/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_20/sol1.py | """
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
"""
def factorial(n):
fact = 1
for i in range(1, n + 1):
fact *= i
return fact
def split_and_add(number):
"""Split number digits and add them."""
sum_of_digits = 0
while number > 0:
last_digit = number % 10
sum_of_digits += last_digit
number = number // 10 # Removing the last_digit from the given number
return sum_of_digits
def solution(n):
"""Returns the sum of the digits in the number 100!
>>> solution(100)
648
>>> solution(50)
216
>>> solution(10)
27
>>> solution(5)
3
>>> solution(3)
6
>>> solution(2)
2
>>> solution(1)
1
"""
f = factorial(n)
result = split_and_add(f)
return result
if __name__ == "__main__":
print(solution(int(input("Enter the Number: ").strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_20/sol2.py | """
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
"""
from math import factorial
def solution(n):
"""Returns the sum of the digits in the number 100!
>>> solution(100)
648
>>> solution(50)
216
>>> solution(10)
27
>>> solution(5)
3
>>> solution(3)
6
>>> solution(2)
2
>>> solution(1)
1
"""
return sum([int(x) for x in str(factorial(n))])
if __name__ == "__main__":
print(solution(int(input("Enter the Number: ").strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_21/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_21/sol1.py | # -.- coding: latin-1 -.-
from math import sqrt
"""
Amicable Numbers
Problem 21
Let d(n) be defined as the sum of proper divisors of n (numbers less than n
which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and
each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55
and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and
142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
"""
def sum_of_divisors(n):
total = 0
for i in range(1, int(sqrt(n) + 1)):
if n % i == 0 and i != sqrt(n):
total += i + n // i
elif i == sqrt(n):
total += i
return total - n
def solution(n):
"""Returns the sum of all the amicable numbers under n.
>>> solution(10000)
31626
>>> solution(5000)
8442
>>> solution(1000)
504
>>> solution(100)
0
>>> solution(50)
0
"""
total = sum(
[
i
for i in range(1, n)
if sum_of_divisors(sum_of_divisors(i)) == i
and sum_of_divisors(i) != i
]
)
return total
if __name__ == "__main__":
print(solution(int(str(input()).strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_22/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_22/sol1.py | # -*- coding: latin-1 -*-
"""
Name scores
Problem 22
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file
containing over five-thousand first names, begin by sorting it into
alphabetical order. Then working out the alphabetical value for each name,
multiply this value by its alphabetical position in the list to obtain a name
score.
For example, when the list is sorted into alphabetical order, COLIN, which is
worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would
obtain a score of 938 × 53 = 49714.
What is the total of all the name scores in the file?
"""
import os
def solution():
"""Returns the total of all the name scores in the file.
>>> solution()
871198282
"""
with open(os.path.dirname(__file__) + "/p022_names.txt") as file:
names = str(file.readlines()[0])
names = names.replace('"', "").split(",")
names.sort()
name_score = 0
total_score = 0
for i, name in enumerate(names):
for letter in name:
name_score += ord(letter) - 64
total_score += (i + 1) * name_score
name_score = 0
return total_score
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_22/sol2.py | # -*- coding: latin-1 -*-
"""
Name scores
Problem 22
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file
containing over five-thousand first names, begin by sorting it into
alphabetical order. Then working out the alphabetical value for each name,
multiply this value by its alphabetical position in the list to obtain a name
score.
For example, when the list is sorted into alphabetical order, COLIN, which is
worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would
obtain a score of 938 × 53 = 49714.
What is the total of all the name scores in the file?
"""
import os
def solution():
"""Returns the total of all the name scores in the file.
>>> solution()
871198282
"""
total_sum = 0
temp_sum = 0
with open(os.path.dirname(__file__) + "/p022_names.txt") as file:
name = str(file.readlines()[0])
name = name.replace('"', "").split(",")
name.sort()
for i in range(len(name)):
for j in name[i]:
temp_sum += ord(j) - ord("A") + 1
total_sum += (i + 1) * temp_sum
temp_sum = 0
return total_sum
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_234/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_234/sol1.py | """
https://projecteuler.net/problem=234
For an integer n ≥ 4, we define the lower prime square root of n, denoted by
lps(n), as the largest prime ≤ √n and the upper prime square root of n, ups(n),
as the smallest prime ≥ √n.
So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. Let us
call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n) divides n,
but not both.
The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8,
10 and 12. 15 is not semidivisible because it is a multiple of both lps(15) = 3
and ups(15) = 5. As a further example, the sum of the 92 semidivisible numbers
up to 1000 is 34825.
What is the sum of all semidivisible numbers not exceeding 999966663333 ?
"""
def fib(a, b, n):
if n==1:
return a
elif n==2:
return b
elif n==3:
return str(a)+str(b)
temp = 0
for x in range(2,n):
c=str(a) + str(b)
temp = b
b = c
a = temp
return c
def solution(n):
"""Returns the sum of all semidivisible numbers not exceeding n."""
semidivisible = []
for x in range(n):
l=[i for i in input().split()]
c2=1
while(1):
if len(fib(l[0],l[1],c2))<int(l[2]):
c2+=1
else:
break
semidivisible.append(fib(l[0],l[1],c2+1)[int(l[2])-1])
return semidivisible
if __name__ == "__main__":
for i in solution(int(str(input()).strip())):
print(i)
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_24/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_24/sol1.py | """
A permutation is an ordered arrangement of objects. For example, 3124 is one
possible permutation of the digits 1, 2, 3 and 4. If all of the permutations
are listed numerically or alphabetically, we call it lexicographic order. The
lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5,
6, 7, 8 and 9?
"""
from itertools import permutations
def solution():
"""Returns the millionth lexicographic permutation of the digits 0, 1, 2,
3, 4, 5, 6, 7, 8 and 9.
>>> solution()
'2783915460'
"""
result = list(map("".join, permutations("0123456789")))
return result[999999]
if __name__ == "__main__":
print(solution())
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_25/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_25/sol1.py | # -*- coding: utf-8 -*-
"""
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000
digits?
"""
def fibonacci(n):
if n == 1 or type(n) is not int:
return 0
elif n == 2:
return 1
else:
sequence = [0, 1]
for i in range(2, n + 1):
sequence.append(sequence[i - 1] + sequence[i - 2])
return sequence[n]
def fibonacci_digits_index(n):
digits = 0
index = 2
while digits < n:
index += 1
digits = len(str(fibonacci(index)))
return index
def solution(n):
"""Returns the index of the first term in the Fibonacci sequence to contain
n digits.
>>> solution(1000)
4782
>>> solution(100)
476
>>> solution(50)
237
>>> solution(3)
12
"""
return fibonacci_digits_index(n)
if __name__ == "__main__":
print(solution(int(str(input()).strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_25/sol2.py | # -*- coding: utf-8 -*-
"""
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000
digits?
"""
def fibonacci_generator():
a, b = 0, 1
while True:
a, b = b, a + b
yield b
def solution(n):
"""Returns the index of the first term in the Fibonacci sequence to contain
n digits.
>>> solution(1000)
4782
>>> solution(100)
476
>>> solution(50)
237
>>> solution(3)
12
"""
answer = 1
gen = fibonacci_generator()
while len(str(next(gen))) < n:
answer += 1
return answer + 1
if __name__ == "__main__":
print(solution(int(str(input()).strip())))
| [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_28/__init__.py | [] | [] | [] |
|
archives/1098994933_python.zip | project_euler/problem_28/sol1.py | """
Starting with the number 1 and moving to the right in a clockwise direction a 5
by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed
in the same way?
"""
from math import ceil
def diagonal_sum(n):
"""Returns the sum of the numbers on the diagonals in a n by n spiral
formed in the same way.
>>> diagonal_sum(1001)
669171001
>>> diagonal_sum(500)
82959497
>>> diagonal_sum(100)
651897
>>> diagonal_sum(50)
79697
>>> diagonal_sum(10)
537
"""
total = 1
for i in range(1, int(ceil(n / 2.0))):
odd = 2 * i + 1
even = 2 * i
total = total + 4 * odd ** 2 - 6 * even
return total
if __name__ == "__main__":
import sys
if len(sys.argv) == 1:
print(diagonal_sum(1001))
else:
try:
n = int(sys.argv[1])
print(diagonal_sum(n))
except ValueError:
print("Invalid entry - please enter a number")
| [] | [] | [] |