A VLM answers every question at temperature 0 with reasoning enabled. Its answer is compared with the official ground truth by the verifier described below; matches go to GeomVerse_RS_think and GeomVerse_RS_nothink, non-matches go to GeomVerse_rejected with the ground truth preserved.

The question field is byte-identical to the prompt the model actually received, including any appended instruction, so the stored training text and the inference-time input cannot drift apart.

How a match is decided

Every rule below is stated with its exact constants and a runnable reference implementation, so any accept/reject decision in this dataset can be recomputed. All snippets assume:

def levenshtein(a, b):
    if a == b: return 0
    if not a: return len(b)
    if not b: return len(a)
    prev = list(range(len(b) + 1))
    for i, ca in enumerate(a, 1):
        cur = [i] + [0] * len(b)
        for j, cb in enumerate(b, 1):
            cur[j] = min(prev[j] + 1, cur[j - 1] + 1, prev[j - 1] + (ca != cb))
        prev = cur
    return prev[len(b)]

`anls` — Average Normalized Levenshtein Similarity

The official DocVQA/InfoVQA metric. Both strings are lowercased and stripped, the normalized Levenshtein distance NL = lev(pred, gt) / max(len(pred), len(gt)) is computed, and the score is 1 - NL when NL < 0.5, otherwise 0. A pair is accepted when the score is above zero and both guards below pass.

def anls_pair(pred, gt, tau=0.5):
    p, q = pred.lower().strip(), gt.lower().strip()
    if not p and not q: return 1.0
    m = max(len(p), len(q))
    if m == 0: return 0.0
    nl = levenshtein(p, q) / m
    return (1.0 - nl) if nl < tau else 0.0

Multiple choice. When the question enumerates its own lettered options, the only correct target is the letter, so a single-letter reference is also matched against the first standalone letter in the prediction — C, (C), The answer is C and **Answer: C** all match C.

Two guards that override a positive score

Numeric guard. ANLS was designed for OCR text and is far too permissive on numbers: 99 degrees vs 80 degrees differs by two characters out of ten and scores 0.8. When both sides state numbers and those numbers disagree, the pair is rejected regardless of string similarity. Thousand separators are removed first, so 1,329.87 and 1329.87 are the same number.

_DIGITS = re.compile(r"\d+(?:\.\d+)?")
_THOUS  = re.compile(r"(?<=\d),(?=\d\d\d\b)")

def num_conflict(a, b):
    a, b = _THOUS.sub("", a or ""), _THOUS.sub("", b or "")
    na, nb = _DIGITS.findall(a), _DIGITS.findall(b)
    if not na or not nb:
        return False
    fa, fb = [float(x) for x in na], [float(x) for x in nb]
    if len(fa) == 1 and len(fb) == 1:
        return abs(fa[0] - fb[0]) > 1e-6 * max(1.0, abs(fb[0]))
    return set(na) != set(nb)

Edit budget. ANLS's fixed tau=0.5 lets a 7-character answer differ by 3 characters, which accepts SBOXCAR against a reference of boxcar at a score of 0.857. That threshold is meant for evaluating a model, where OCR noise is expected; selecting TRAINING data needs precision, because an accepted row is shipped as ground truth. The tolerated number of character edits scales with length, and is measured after stripping case, punctuation, whitespace and a leading article — otherwise stop. vs stop would burn the whole budget on a period.

reference length	edits allowed
≤ 8	0 (must be exact)
9–16	1
17–32	2
> 32	length / 12

_ART = re.compile(r"^(?:a|an|the)\s+")

def budget_norm(x):
    return re.sub(r"[^a-z0-9]", "", _ART.sub("", (x or "").strip().lower()))

def edit_budget(gt):
    n = len(gt or "")
    return 0 if n <= 8 else 1 if n <= 16 else 2 if n <= 32 else max(3, n // 12)

def within_budget(pred, gt):
    p, g = budget_norm(pred), budget_norm(gt)
    return p == g or levenshtein(p, g) <= edit_budget(g)

Shared pre-processing (applies to every verifier below)

Both sides are expanded into a small set of candidate strings and the best-scoring pair decides. The prediction contributes: the raw string; its first line; its first paragraph; the text after the last Answer: marker; the span between Answer: and Rationale:; the stated final answer of a worked solution (recognised from \boxed{…}, Answer:, the final answer is …, the answer is …, so the answer is …, 故选X); and a markdown-stripped variant. The reference contributes the same forms.

When the reference is a wrapper around the real answer (The answer is Red), the raw string is replaced by the extracted answer rather than kept beside it: the wrapper text is shared by every wrong answer too, so The answer is Blue would otherwise score 0.78 against it.

import re

def final_answer(gt):
    """Stated answer of a worked solution, or None."""
    pats = [r"\\boxed\{([^{}]{1,80})\}",
            r"(?is)\banswer\s*[:：]\s*(.{1,80}?)\s*$",
            r"(?is)\b(?:the\s+)?final\s+answer\s+is\s*[:：]?\s*(.{1,80}?)\s*[.。]?\s*$",
            r"(?is)\b(?:the\s+)?answer\s+is\s*[:：]?\s*(.{1,80}?)\s*[.。]?\s*$",
            r"(?s)故\s*选\s*[:：]?\s*\(?([A-H])\)?"]
    for p in pats:
        ms = list(re.finditer(p, (gt or "").strip()))
        if ms:
            v = re.sub(r"\s+", " ", ms[-1].group(1).strip().strip("$ \t.。，,;:"))
            if v and len(v) <= 80:
                return v
    return None

Second pass: model-as-judge

String verifiers cannot judge free-form answers at all, and they under-credit answers that are correct but differently phrased. Every item the verifier rejected is therefore re-examined by the same VLM in a text-only pass that sees the question, the reference and the candidate — no image, no re-generation, it only grades text that was already produced.

The judge is instructed to compare against the reference and never to solve the question itself, so an answer is not promoted merely because the model believes it is true. It answers CORRECT, WRONG, or UNSURE; numbers must agree, and UNSURE is reserved for references that are themselves unusable or ambiguous. UNSURE is never counted as correct — only a CORRECT verdict moves a row into the accepted splits. WRONG and UNSURE both remain in the rejected split, so an unresolved case is never silently promoted to a pass.

Calibration: run over items the string verifier had already accepted, the judge agreed on 59/60. Spot-checking its overturns shows it discriminates rather than rubber-stamps — it keeps \frac{L}{n} rejected against a reference of \frac{L}{n-1}, and returns UNSURE when the model argues no finite answer exists against a numeric reference.

What `answer` holds

In the accepted splits answer is the model's own verified output — the text that passed the verifier — so it agrees with the reasoning in think. The original reference annotation is kept beside it as org_answer. Train on answer; use org_answer when you need the source label.

The two differ more often than one would guess: the verifier accepts on meaning, not on characters, so casing, spacing and formatting routinely differ (bridgestone vs Bridgestone, ( n x m ) vs (n x m)). Storing the reference in the answer slot would pair a model's reasoning with someone else's wording.

In the rejected split the roles are different: answer is the official ground truth and wrong_vlm is what the model said instead.

Schema

image — HF Image() — renders directly in the dataset viewer
qa — list of {question, answer, note}
clean_meta — per-row provenance and the cleaning policy that admitted it
image_sha256 — content hash of the image

Examples

Three rows taken straight from this split. Images are the original files as stored in the dataset, and long fields are shown in full.

Example 1

field	value
`question`	Given the lengths of sides AB and BC in the right triangle BCD, and knowing the area of the triangle, how can you determine the length of side CD? Use the given information and the properties of right triangles to solve this problem. End your solution with the final answer inside \boxed{}.
`answer`	To determine the length of side CD, we can use the formula for the area of a right triangle, which is given by: [ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ] In this case, the area of triangle BCD is 82, and the lengths of sides AB and BC are 6 and 8.68 respectively. We can use these values to find the length of side CD. First, we need to identify which side is the base and which is the height. Since AB is perpendicular to BC, AB is the height and BC is the base. Plugging in the known values: [ 82 = \frac{1}{2} \times 8.68 \times \text{CD} ] Solving for CD: [ 82 = 4.34 \times \text{CD} ] [ \text{CD} = \frac{82}{4.34} ] [ \text{CD} \approx 18.89 ] Therefore, the length of side CD is approximately 18.89 units.
`note`	long-form: excluded from rejection sampling

Example 2

field	value
`question`	Given the area of the parallelogram ABCD is 114 and the lengths of sides AB and AD are 20 and 15 respectively, calculate the height of the parallelogram from vertex A to side BC. Round your computations to 2 decimal places. End your solution with the final answer inside \boxed{}.
`answer`	To find the height of the parallelogram from vertex A to side BC, we can use the formula for the area of a parallelogram, which is given by: [ \text{Area} = \text{base} \times \text{height} ] In this case, the base is side AD, which is 15 units, and the area is 114 square units. Let's denote the height from vertex A to side BC as ( h ). [ 114 = 15 \times h ] Solving for ( h ): [ h = \frac{114}{15} = 7.6 ] Therefore, the height of the parallelogram from vertex A to side BC is 7.6 units.
`note`	long-form: excluded from rejection sampling

Example 3

field	value
`question`	If the gray shape is a rectangle where a semi-circle has been removed from one side of it, and the lime shape is a rectangle where a semi-circle has been removed from one side of it, and the perimeter of the lime shape is 88, determine the ratio of the areas of the gray shape to the lime shape. Assume $\pi=3.14$. Round computations to 2 decimal places. End your solution with the final answer inside \boxed{}.
`answer`	To determine the ratio of the areas of the gray shape to the lime shape, we need to follow these steps: 1. Determine the dimensions of the lime shape: - The diameter of the semi-circle in the lime shape is equal to the side of the rectangle with length 15. - The shape has two sides with equal but unknown lengths, one side with length 15, and one semi-circle arc with diameter 15. - The perimeter of the lime shape is given by the formula: $2 \times \text{Unknown Side} + 15 + \frac{15 \times \pi}{2}$. - Given that the perimeter is 88, we can set up the equation: $2 \times \text{Unknown Side} + 15 + \frac{15 \times 3.14}{2} = 88$. - Solving for the unknown side: $2 \times \text{Unknown Side} = 88 - 15 - \frac{15 \times 3.14}{2} = 88 - 15 - 23.55 = 49.45$. - Therefore, the length of the side marked with "$a$" is $\frac{49.45}{2} = 24.73$. 2. Compute the area of the gray shape: - The gray shape is a rectangle with dimensions 24.73 and 14. - The area of the rectangle is calculated as: $24.73 \times 14 = 346.22$. - The area of the semi-circle removed from the rectangle is calculated as: $\frac{3.14 \times 14^2}{8} = \frac{3.14 \times 196}{8} = \frac{615.44}{8} = 76.93$. - Therefore, the area of the gray shape is the area of the rectangle minus the area of the semi-circle: $346.22 - 76.93 = 269.29$. 3. Compute the area of the lime shape: - The lime shape is a rectangle with dimensions 24.73 and 15. - The area of the rectangle is calculated as: $24.73 \times 15 = 371.95$. - The area of the semi-circle removed from the rectangle is calculated as: $\frac{3.14 \times 15^2}{8} = \frac{3.14 \times 225}{8} = \frac{706.5}{8} = 88.31$. - Therefore, the area of the lime shape is the area of the rectangle minus the area of the semi-circle: $371.95 - 88.31 = 283.64$. 4. Determine the ratio of the areas: - The ratio of the areas of the gray shape to the lime shape is: $\frac{269.29}{283.64} \approx 0.95$. Final Answer: The ratio of the areas of the gray shape to the lime shape is approximately 0.95
`note`	long-form: excluded from rejection sampling

Cleaning applied before rejection sampling

Provenance check. Each QA item records where it came from; items whose recorded source does not resolve to the image they are attached to are dropped.
Task purity. The pool's InfinityMM slice assigns samples to a family by image match rather than by dataset origin, which pulls unrelated general-VQA questions into benchmark families. Families where that content measured off-task keep only their genuine benchmark QA.
Multiple-choice normalisation. Where options had been flattened into prose, they are re-rendered as a lettered block and the reference is rewritten to the bare letter, so the task is well posed and the answer is unambiguous.

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GeomVerse — GeomVerse_longform

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