Commit 4df2c7b1 authored by Baber's avatar Baber
Browse files

add math500

parent 2793ceb6
# MATH
ℹ️ This is the 0-shot variant! reproducing https://huggingface.co/datasets/meta-llama/Llama-3.1-8B-Instruct-evals/viewer/Llama-3.1-8B-Instruct-evals__math__details?row=0
## Paper
Measuring Mathematical Problem Solving With the MATH Dataset
https://arxiv.org/abs/2103.03874
Many intellectual endeavors require mathematical problem solving, but this skill remains beyond the capabilities of computers. To measure this ability in machine learning models, we introduce MATH, a new dataset of 12,500 challenging competition mathematics problems. Each problem in MATH has a full step-by-step solution which can be used to teach models to generate answer derivations and explanations.
NOTE: The few-shot and the generated answer extraction is based on the [Minerva](https://arxiv.org/abs/2206.14858) and exact match equivalence is calculated using the `sympy` library. This requires additional dependencies, which can be installed via the `lm-eval[math]` extra.
Homepage: https://github.com/hendrycks/math
## Citation
```
@article{hendrycksmath2021,
title={Measuring Mathematical Problem Solving With the MATH Dataset},
author={Dan Hendrycks and Collin Burns and Saurav Kadavath and Akul Arora and Steven Basart and Eric Tang and Dawn Song and Jacob Steinhardt},
journal={NeurIPS},
year={2021}
}
@misc{2206.14858,
Author = {Aitor Lewkowycz and Anders Andreassen and David Dohan and Ethan Dyer and Henryk Michalewski and Vinay Ramasesh and Ambrose Slone and Cem Anil and Imanol Schlag and Theo Gutman-Solo and Yuhuai Wu and Behnam Neyshabur and Guy Gur-Ari and Vedant Misra},
Title = {Solving Quantitative Reasoning Problems with Language Models},
Year = {2022},
Eprint = {arXiv:2206.14858},
}
```
### Groups and Tasks
[//]: # (#### Groups)
[//]: # ()
[//]: # (- `llama_math`)
#### Tasks
- `llama_math_algebra`
- `llama_math_counting_and_prob`
- `llama_math_geometry`
- `llama_math_intermediate_algebra`
- `llama_math_num_theory`
- `llama_math_prealgebra`
- `llama_math_precalc`
### Checklist
The checklist is the following:
For adding novel benchmarks/datasets to the library:
* [x] Is the task an existing benchmark in the literature?
* [x] Have you referenced the original paper that introduced the task?
* [x] If yes, does the original paper provide a reference implementation? If so, have you checked against the reference implementation and documented how to run such a test?
* The implementation in the original paper is one where the model is first fine-tuned on the data. They do have a few-shot evaluation for GPT-3, however the few-shot context used here is sourced from [Lewkowycz et al](https://arxiv.org/abs/2206.14858). The achieved accuracy on Llama-2 models is comparable to that provided in the paper, though not identical.
If other tasks on this dataset are already supported:
* [x] Is the "Main" variant of this task clearly denoted?
* [x] Have you provided a short sentence in a README on what each new variant adds / evaluates?
* [x] Have you noted which, if any, published evaluation setups are matched by this variant?
### Variant Wishlist
- [ ] zero-shot variant
task: math500
dataset_path: EleutherAI/hendrycks_math
process_docs: !function utils.process_docs
output_type: generate_until
training_split: train
test_split: test
doc_to_text: "Solve the following math problem efficiently and clearly:\n\n- For simple problems (2 steps or fewer):\nProvide a concise solution with minimal explanation.\n\n- For complex problems (3 steps or more):\nUse this step-by-step format:\n\n## Step 1: [Concise description]\n[Brief explanation and calculations]\n\n## Step 2: [Concise description]\n[Brief explanation and calculations]\n\n...\n\nRegardless of the approach, always conclude with:\n\nTherefore, the final answer is: $\\\\boxed{answer}$. I hope it is correct.\n\nWhere [answer] is just the final number or expression that solves the problem.\n\nProblem: {{ problem }}"
process_results: !function utils.process_results
doc_to_target: "{{answer if few_shot is undefined else solution}}"
generation_kwargs:
# until:
# - "Problem:"
max_gen_toks: 5120
do_sample: false
temperature: 0
metric_list:
- metric: exact_match
aggregation: mean
higher_is_better: true
num_fewshot: 0
metadata:
version: 1.0
dataset_kwargs:
trust_remote_code: true
import re
import signal
from typing import Dict, List, Optional
import datasets
from lm_eval.utils import eval_logger
try:
import sympy
from sympy.parsing.latex import parse_latex
except ModuleNotFoundError:
raise ModuleNotFoundError(
"`sympy` is required for generating translation task prompt templates. \
please install sympy via pip install lm-eval[math] or pip install -e .[math]",
)
template = "Solve the following math problem efficiently and clearly:\n\n- For simple problems (2 steps or fewer):\nProvide a concise solution with minimal explanation.\n\n- For complex problems (3 steps or more):\nUse this step-by-step format:\n\n## Step 1: [Concise description]\n[Brief explanation and calculations]\n\n## Step 2: [Concise description]\n[Brief explanation and calculations]\n\n...\n\nRegardless of the approach, always conclude with:\n\nTherefore, the final answer is: $\\\\boxed{answer}$. I hope it is correct.\n\nWhere [answer] is just the final number or expression that solves the problem.\n\nProblem: {{ problem }}"
# taken from
# https://github.com/wellecks/lm-evaluation-harness/blob/master/lm_eval/tasks/minerva_math.py
# def doc_to_text(doc: dict) -> str:
# return "Problem:" + "\n" + doc["problem"] + "\n\n" + "Solution:"
def process_docs(dataset: datasets.Dataset) -> datasets.Dataset:
def _process_doc(doc: dict) -> dict:
out_doc = {
"problem": doc["problem"],
"solution": doc["solution"],
"answer": normalize_final_answer(doc["answer"]),
}
if getattr(doc, "few_shot", None) is not None:
out_doc["few_shot"] = True
return out_doc
return dataset.map(_process_doc)
# def list_fewshot_samples() -> list[dict]:
# return [
# {
# "problem": "Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}",
# "solution": "The expressions inside each square root must be non-negative. Therefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\\boxed{[2,5)}$.\nFinal Answer: The final answer is $[2,5)$. I hope it is correct.",
# "few_shot": "1",
# },
# {
# "problem": "If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$",
# "solution": "We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\nFinal Answer: The final answer is $24$. I hope it is correct.",
# "few_shot": "1",
# },
# {
# "problem": "Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?",
# "solution": "If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$:\n\\begin{align*}\n30n&=480\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align*}\nFinal Answer: The final answer is $16$. I hope it is correct.",
# "few_shot": "1",
# },
# {
# "problem": "If the system of equations\n\n\\begin{align*}\n6x-4y&=a,\\\n6y-9x &=b.\n\\end{align*}has a solution $(x, y)$ where $x$ and $y$ are both nonzero,\nfind $\\frac{a}{b},$ assuming $b$ is nonzero.",
# "solution": "If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\nFinal Answer: The final answer is $-\\frac{2}{3}$. I hope it is correct.",
# "few_shot": "1",
# },
# ]
def process_results(doc: dict, results: List[str]) -> Dict[str, int]:
candidates = results[0]
answer = normalize_final_answer(remove_boxed(last_boxed_only_string(candidates)))
if is_equiv(answer, doc["answer"]):
retval = 1
else:
retval = 0
results = {
"exact_match": retval,
}
return results
def last_boxed_only_string(string: str) -> Optional[str]:
idx = string.rfind("\\boxed")
if "\\boxed " in string:
return "\\boxed " + string.split("\\boxed ")[-1].split("$")[0]
if idx < 0:
idx = string.rfind("\\fbox")
if idx < 0:
return None
i = idx
right_brace_idx = None
num_left_braces_open = 0
while i < len(string):
if string[i] == "{":
num_left_braces_open += 1
if string[i] == "}":
num_left_braces_open -= 1
if num_left_braces_open == 0:
right_brace_idx = i
break
i += 1
if right_brace_idx is None:
retval = None
else:
retval = string[idx : right_brace_idx + 1]
return retval
def remove_boxed(s: str) -> str:
if not s:
return ""
if "\\boxed " in s:
left = "\\boxed "
assert s[: len(left)] == left
return s[len(left) :]
left = "\\boxed{"
assert s[: len(left)] == left
assert s[-1] == "}"
return s[len(left) : -1]
class timeout:
def __init__(self, seconds=1, error_message="Timeout"):
self.seconds = seconds
self.error_message = error_message
def handle_timeout(self, signum, frame):
raise TimeoutError(self.error_message)
def __enter__(self):
signal.signal(signal.SIGALRM, self.handle_timeout)
signal.alarm(self.seconds)
def __exit__(self, type, value, traceback):
signal.alarm(0)
def is_equiv(x1: str, x2: str) -> bool:
"""
x1 and x2 are normalized latex string
"""
try:
with timeout(seconds=5):
try:
parsed_x1 = parse_latex(x1)
parsed_x2 = parse_latex(x2)
except (
sympy.parsing.latex.errors.LaTeXParsingError,
sympy.SympifyError,
TypeError,
):
eval_logger.debug(f"couldn't parse one of {x1} or {x2}")
return False
try:
diff = parsed_x1 - parsed_x2
except TypeError:
eval_logger.debug(f"couldn't subtract {x1} and {x2}")
return False
try:
if sympy.simplify(diff) == 0:
return True
else:
return False
except ValueError:
eval_logger.debug(
f"Had some trouble simplifying when comparing {x1} and {x2}"
)
except TimeoutError:
eval_logger.debug(f"Timed out comparing {x1} and {x2}")
return False
except ImportError as e:
eval_logger.error(e)
raise
except Exception as e:
eval_logger.debug(f"Failed comparing {x1} and {x2} with {e}")
return False
def get_unnormalized_answer(text: str) -> str:
INVALID_ANSWER = "[invalidanswer]"
end_seq = "I hope it is correct."
text += end_seq
match = re.search(
r"Final Answer: The final answer is(.*?). I hope it is correct.",
text,
)
if match:
return match.group(1).strip()
else:
return INVALID_ANSWER
SUBSTITUTIONS = [
("an ", ""),
("a ", ""),
(".$", "$"),
("\\$", ""),
(r"\ ", ""),
(" ", ""),
("mbox", "text"),
(",\\text{and}", ","),
("\\text{and}", ","),
("\\text{m}", "\\text{}"),
]
REMOVED_EXPRESSIONS = [
"square",
"ways",
"integers",
"dollars",
"mph",
"inches",
"ft",
"hours",
"km",
"units",
"\\ldots",
"sue",
"points",
"feet",
"minutes",
"digits",
"cents",
"degrees",
"cm",
"gm",
"pounds",
"meters",
"meals",
"edges",
"students",
"childrentickets",
"multiples",
"\\text{s}",
"\\text{.}",
"\\text{\ns}",
"\\text{}^2",
"\\text{}^3",
"\\text{\n}",
"\\text{}",
r"\mathrm{th}",
r"^\circ",
r"^{\circ}",
r"\;",
r",\!",
"{,}",
'"',
"\\dots",
]
def normalize_final_answer(final_answer: Optional[str]) -> str:
"""
Normalize a final answer to a quantitative reasoning question.
Copied character for character from appendix D of Lewkowycz et al. (2022)
"""
if not final_answer:
return ""
final_answer = final_answer.split("=")[-1]
for before, after in SUBSTITUTIONS:
final_answer = final_answer.replace(before, after)
for expr in REMOVED_EXPRESSIONS:
final_answer = final_answer.replace(expr, "")
# Extract answer that is in LaTeX math, is bold,
# is surrounded by a box, etc.
final_answer = re.sub(r"(.*?)(\$)(.*?)(\$)(.*)", "$\\3$", final_answer)
final_answer = re.sub(r"(\\text\{)(.*?)(\})", "\\2", final_answer)
final_answer = re.sub(r"(\\textbf\{)(.*?)(\})", "\\2", final_answer)
final_answer = re.sub(r"(\\overline\{)(.*?)(\})", "\\2", final_answer)
final_answer = re.sub(r"(\\boxed\{)(.*)(\})", "\\2", final_answer)
# Normalize shorthand TeX:
# \fracab -> \frac{a}{b}
# \frac{abc}{bef} -> \frac{abc}{bef}
# \fracabc -> \frac{a}{b}c
# \sqrta -> \sqrt{a}
# \sqrtab -> sqrt{a}b
final_answer = re.sub(r"(frac)([^{])(.)", "frac{\\2}{\\3}", final_answer)
final_answer = re.sub(r"(sqrt)([^{])", "sqrt{\\2}", final_answer)
final_answer = final_answer.replace("$", "")
# Normalize 100,000 -> 100000
if final_answer.replace(",", "").isdigit():
final_answer = final_answer.replace(",", "")
return final_answer
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