Easy-to-Hard Generalization Models
Collection
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OPRMs (Outcome & Process Reward Models) are trained to predict the correctness of each step on the position of "\n\n", as well as the correctness of the whole solution on the position of "<eos>".
Usage:
import torch
from transformers import AutoTokenizer, AutoModelForCausalLM
model_name = "ScalableMath/llemma-7b-oprm-prm800k-level-1to3-hf"
model = AutoModelForCausalLM.from_pretrained(model_name, torch_dtype=torch.bfloat16, device_map="auto")
tokenizer = AutoTokenizer.from_pretrained("EleutherAI/llemma_7b")
qa_example = """# Question
Convert the point $(0,3)$ in rectangular coordinates to polar coordinates. Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
# Solution
To convert from rectangular to polar coordinates, I need to use the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}(y/x).$
In this case, $x = 0$ and $y = 3,$ so I can plug them into the formulas.
For $r,$ I get $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.$
For $\theta,$ I get $\theta = \tan^{-1}(3/0).$
This is undefined, since the tangent function is not defined at $0.$
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 $90^\circ.$
Therefore, I can choose any angle in the range $(0,\pi/2)$ as the value of $\theta.$
I will choose $\theta = \pi/2,$ since it is the simplest and most natural choice.
Therefore, the polar coordinates of the point $(0,3)$ are $(3,\pi/2).$
# Answer
(3,\pi/2)"""
begin_solution_tokens = tokenizer.encode("\n\n# Solution", add_special_tokens=False)[1:]
scoring_tokens = tokenizer.encode("\n\n", add_special_tokens=False)[1:]
eos_token = tokenizer.eos_token_id
input_ids = tokenizer.encode(qa_example)
begin_solution_flag = False
candidate_positions = []
for start_idx in range(len(input_ids)):
if tuple(input_ids[start_idx:start_idx+len(begin_solution_tokens)]) == tuple(begin_solution_tokens):
begin_solution_flag = True
if begin_solution_flag and tuple(input_ids[start_idx:start_idx+len(scoring_tokens)]) == tuple(scoring_tokens):
candidate_positions.append(start_idx)
if input_ids[start_idx] == eos_token:
candidate_positions.append(start_idx)
break
# maybe delete the first and the second to last candidate_positions
# because they are "\n\n" after "# Solution" and after "# Answer"
del candidate_positions[0]
del candidate_positions[-2]
input_tensor = torch.tensor([input_ids])
candidate_positions = torch.tensor(candidate_positions)
with torch.no_grad():
logits = model(input_tensor).logits
scores =logits.mean(dim=-1)
step_scores = scores[0][candidate_positions]
step_probs = torch.sigmoid(step_scores)
print(step_probs)
# tensor([0.8093, 0.9566, 0.9872, 0.9890, 0.9797, 0.3090, 0.8044, 0.7677, 0.8105, 0.5247])