Create README.md
Browse files
README.md
ADDED
@@ -0,0 +1,82 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
ORMs are trained to predict the correctness of the whole solution on the position of "\<eos\>".
|
2 |
+
But they are actually trained to forcast the correctness of the whole solution on each token (i.e., token-level loss).
|
3 |
+
|
4 |
+
Usage:
|
5 |
+
|
6 |
+
```python
|
7 |
+
import torch
|
8 |
+
from transformers import AutoTokenizer, AutoModelForCausalLM
|
9 |
+
|
10 |
+
model_name = "ScalableMath/llemma-7b-orm-prm800k-level-1to3-hf"
|
11 |
+
model = AutoModelForCausalLM.from_pretrained(model_name, torch_dtype=torch.bfloat16, device_map="auto")
|
12 |
+
|
13 |
+
tokenizer = AutoTokenizer.from_pretrained("EleutherAI/llemma_7b")
|
14 |
+
|
15 |
+
qa_example = """# Question
|
16 |
+
|
17 |
+
Convert the point $(0,3)$ in rectangular coordinates to polar coordinates. Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
|
18 |
+
|
19 |
+
# Solution
|
20 |
+
|
21 |
+
To convert from rectangular to polar coordinates, I need to use the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}(y/x).$
|
22 |
+
|
23 |
+
In this case, $x = 0$ and $y = 3,$ so I can plug them into the formulas.
|
24 |
+
|
25 |
+
For $r,$ I get $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.$
|
26 |
+
|
27 |
+
For $\theta,$ I get $\theta = \tan^{-1}(3/0).$
|
28 |
+
|
29 |
+
This is undefined, since the tangent function is not defined at $0.$
|
30 |
+
|
31 |
+
However, I can use the fact that the point $(0,3)$ lies on the positive $y$-axis, which has an angle of $\pi/2$ radians or $90^\circ.$
|
32 |
+
|
33 |
+
Therefore, I can choose any angle in the range $(0,\pi/2)$ as the value of $\theta.$
|
34 |
+
|
35 |
+
I will choose $\theta = \pi/2,$ since it is the simplest and most natural choice.
|
36 |
+
|
37 |
+
Therefore, the polar coordinates of the point $(0,3)$ are $(3,\pi/2).$
|
38 |
+
|
39 |
+
# Answer
|
40 |
+
|
41 |
+
(3,\pi/2)"""
|
42 |
+
|
43 |
+
begin_solution_tokens = tokenizer.encode("\n\n# Solution", add_special_tokens=False)[1:]
|
44 |
+
scoring_tokens = tokenizer.encode("\n\n", add_special_tokens=False)[1:]
|
45 |
+
eos_token = tokenizer.eos_token_id
|
46 |
+
|
47 |
+
input_ids = tokenizer.encode(qa_example)
|
48 |
+
|
49 |
+
begin_solution_flag = False
|
50 |
+
|
51 |
+
candidate_positions = []
|
52 |
+
|
53 |
+
for start_idx in range(len(input_ids)):
|
54 |
+
if tuple(input_ids[start_idx:start_idx+len(begin_solution_tokens)]) == tuple(begin_solution_tokens):
|
55 |
+
begin_solution_flag = True
|
56 |
+
|
57 |
+
if begin_solution_flag and tuple(input_ids[start_idx:start_idx+len(scoring_tokens)]) == tuple(scoring_tokens):
|
58 |
+
candidate_positions.append(start_idx)
|
59 |
+
|
60 |
+
if input_ids[start_idx] == eos_token:
|
61 |
+
candidate_positions.append(start_idx)
|
62 |
+
break
|
63 |
+
|
64 |
+
# maybe delete the first and the second to last candidate_positions
|
65 |
+
# because they are "\n\n" after "# Solution" and after "# Answer"
|
66 |
+
del candidate_positions[0]
|
67 |
+
del candidate_positions[-2]
|
68 |
+
|
69 |
+
input_tensor = torch.tensor([input_ids])
|
70 |
+
candidate_positions = torch.tensor(candidate_positions)
|
71 |
+
|
72 |
+
with torch.no_grad():
|
73 |
+
logits = model(input_tensor).logits
|
74 |
+
scores =logits.mean(dim=-1)
|
75 |
+
step_scores = scores[0][candidate_positions]
|
76 |
+
step_probs = torch.sigmoid(step_scores)
|
77 |
+
|
78 |
+
print(step_probs)
|
79 |
+
|
80 |
+
# only the last logprob is orm's output
|
81 |
+
# tensor([0.4531, 0.3882, 0.3748, 0.4785, 0.4087, 0.3166, 0.3040, 0.2295, 0.2628, 0.2568])
|
82 |
+
```
|