Datasets:
task_id stringclasses 410
values | source stringclasses 11
values | family stringclasses 2
values | effort stringclasses 5
values | sample int64 0 0 | model stringclasses 1
value | provider stringclasses 1
value | served_by stringclasses 1
value | content stringlengths 9 2.42k ⌀ | reasoning stringlengths 28 91k | finish_reason stringclasses 2
values | latency_s float64 0.39 64.7 | usage dict | ts timestamp[s]date 2026-10-22 06:29:50 2026-10-22 06:46:57 | correct bool 2
classes | grade_reason stringclasses 10
values | pred stringclasses 217
values | public_pass bool 2
classes |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
aime24-60 | aime24 | math | high | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{204} | We need solve problem. Need parse. Walk distance 9 km. Speed s km/h, total time 4 hours including t minutes in coffee shop. Speed s+2 km/h, total time 2h24m including t. Need total time when speed s+1/2 km/h, including t minutes. Need output minutes presumably.
Let's solve carefully.
Let distance d=9 km. Walking time... | stop | 1.31 | {
"prompt_tokens": 196,
"completion_tokens": 665,
"reasoning_tokens": 648
} | 2026-10-22T06:43:57 | true | exact | 204 | null |
aime24-60 | aime24 | math | low | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{204} | Let's parse problem.
Walk length 9 km. Speed s km/h, total time 4 hours including t minutes coffee. Speed s+2, total time 2 hours 24 minutes including t minutes. Find total minutes when speed s+1/2 including t.
Let t minutes. Convert to hours maybe.
Walking time = distance/speed. Total time = walk time + t (in hours... | stop | 1.18 | {
"prompt_tokens": 195,
"completion_tokens": 568,
"reasoning_tokens": 551
} | 2026-10-22T06:36:37 | true | exact | 204 | null |
aime24-60 | aime24 | math | max | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | We set up equations using total time = walking time + coffee time.
Let \(s\) be in km/h and \(t\) in minutes.
\[
240 = \frac{540}{s}+t,\qquad
144 = \frac{540}{s+2}+t.
\]
Subtracting:
\[
96 = 540\left(\frac1s-\frac1{s+2}\right)
= \frac{1080}{s(s+2)}.
\]
So
\[
s(s+2)=\frac{1080}{96}=\frac{45}{4}.
\]
Thus
\[
4s^2+... | We need solve problem. Need parse. Every morning Aya goes for 9-km walk and stops at coffee shop afterwards. When she walks constant speed s km/h, walk takes her 4 hours, including t minutes spent in coffee shop. When walks s+2 km/h, takes 2 hours 24 minutes, including t minutes. Suppose speed s+1/2 km/h. Find number m... | stop | 1.44 | {
"prompt_tokens": 196,
"completion_tokens": 940,
"reasoning_tokens": 646
} | 2026-10-22T06:30:41 | true | exact | 204 | null |
aime24-60 | aime24 | math | medium | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{204} | Let me set up equations.
Walk distance = 9 km.
Case 1: speed s km/h. Walking time = 9/s hours. Plus coffee t minutes = t/60 hours. Total = 4 hours.
So 9/s + t/60 = 4.
Case 2: speed s+2. Walking time = 9/(s+2). Total = 2 hours 24 min = 2.4 hours = 12/5 hours.
So 9/(s+2) + t/60 = 12/5.
Subtract: 9/s - 9/(s+2) = 4 - 1... | stop | 1 | {
"prompt_tokens": 195,
"completion_tokens": 575,
"reasoning_tokens": 558
} | 2026-10-22T06:40:48 | true | exact | 204 | null |
aime24-60 | aime24 | math | xhigh | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{204} | We need solve problem. Need parse carefully. Every morning Aya goes for 9-km-long walk and stops at coffee shop afterwards. "When she walks at constant speed s km/h, walk takes her 4 hours, including t minutes spent in coffee shop." "When she walks s+2 km/h, walk takes her 2 hours 24 minutes, including t minutes..." Su... | stop | 1.13 | {
"prompt_tokens": 196,
"completion_tokens": 676,
"reasoning_tokens": 659
} | 2026-10-22T06:37:47 | true | exact | 204 | null |
aime24-61 | aime24 | math | high | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | We place \(B=(0,0)\), \(C=(9,0)\). Then \(A=(\frac13,\frac{4\sqrt{14}}3)\).
The tangents at \(B\) and \(C\) meet at
\[
D=\left(\frac92,-\frac{27\sqrt{14}}{11}\right).
\]
We compute
\[
DB=\frac{225}{22},\qquad DA=\frac{325}{22}.
\]
By power of a point,
\[
DA\cdot DP=DB^2,
\]
so
\[
DP=\frac{DB^2}{DA}
=\frac{2025}{286}.
... | We need solve geometry. Need AP length? Triangle ABC with sides AB=5, BC=9, AC=10 inscribed in circle ω. Tangents at B,C meet D. AD intersects circle at P. Need AP. Ambiguity: chord? line AD intersects circle at A and P; AP could be chord length between A and P (possibly minor). Need value m/n. Need compute.
Let's par... | stop | 5.93 | {
"prompt_tokens": 169,
"completion_tokens": 4198,
"reasoning_tokens": 3932
} | 2026-10-22T06:45:28 | true | exact | 113 | null |
aime24-61 | aime24 | math | low | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{113} | We need solve geometry. Need final boxed m+n. Let's solve carefully.
Problem: Triangle ABC inscribed in circle ω. Tangents at B and C intersect at D. Line AD intersects circle at P (besides A? likely second intersection). Given side lengths AB=5, BC=9, AC=10. Find AP length? AP can be m/n. Need m+n.
Need compute. Let... | stop | 5.05 | {
"prompt_tokens": 168,
"completion_tokens": 3367,
"reasoning_tokens": 3350
} | 2026-10-22T06:37:06 | true | exact | 113 | null |
aime24-61 | aime24 | math | max | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{113} | We need solve geometry. Need final answer boxed. Let's parse.
Triangle ABC inscribed in circle ω. Tangents at B and C intersect at D. Line AD intersects circle at P (besides? A on circle, line through A and D intersects circle at A and P). Given side lengths AB=5, BC=9, AC=10. Need AP length. "AP can be written form m... | stop | 10.91 | {
"prompt_tokens": 169,
"completion_tokens": 6453,
"reasoning_tokens": 6437
} | 2026-10-22T06:36:27 | true | exact | 113 | null |
aime24-61 | aime24 | math | medium | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | We place \(B=(0,0)\), \(C=(9,0)\). From \(AB=5\), \(AC=10\), we get \(A=\left(\frac13,\frac{4\sqrt{14}}{3}\right)\).
The circumcircle is
\[
x^2+y^2-9x+vy=0,\qquad v=-\frac{33\sqrt{14}}{28}.
\]
The tangents at \(B\) and \(C\) meet at
\[
D=\left(\frac92,-\frac{27\sqrt{14}}{11}\right).
\]
Let \(w=D-A\). Then
\[
w=\left(\... | We need solve geometry. Need final answer m+n. Need carefully compute AP.
Problem: triangle ABC in circle omega. Tangents at B,C meet at D. AD intersects circle at P (besides A? likely second intersection). Given side lengths AB=5, BC=9, AC=10. Need AP length. Need express rational m/n? AP maybe rational? Need compute... | stop | 10.38 | {
"prompt_tokens": 168,
"completion_tokens": 6825,
"reasoning_tokens": 6500
} | 2026-10-22T06:35:27 | true | exact | 113 | null |
aime24-61 | aime24 | math | xhigh | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | We place \(B=(0,0)\) and \(C=(9,0)\). From \(AB=5,\ AC=10\), we get \(A=\left(\frac13,\frac{4\sqrt{14}}3\right)\).
The circumcircle is
\[
x^2+y^2-9x+ky=0,\qquad k=-\frac{33}{2\sqrt{14}}.
\]
The tangents at \(B,C\) meet at
\[
D=\left(\frac92,-\frac{27\sqrt{14}}{11}\right).
\]
Thus
\[
\vec{AD}=\left(\frac{25}{6},-\frac{... | We need solve geometry. Need final answer boxed. Let's parse problem.
Triangle ABC inscribed in circle omega. Tangents at B and C intersect at D. Line AD intersects circle again at P (besides A? "AD intersect ω at P" likely second intersection; A is on circle and line AD, so P is other point). Given side lengths AB=5,... | stop | 7.34 | {
"prompt_tokens": 169,
"completion_tokens": 5051,
"reasoning_tokens": 4722
} | 2026-10-22T06:44:08 | true | exact | 113 | null |
aime24-62 | aime24 | math | high | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | Let the blue vertices form a subset \(B\subseteq \mathbb Z_8\). We need a rotation by \(k\) positions such that no blue vertex lands on a blue vertex, i.e.
\[
(B+k)\cap B=\varnothing.
\]
If \(|B|>4\), this is impossible since two disjoint translates would contain \(2|B|>8\) vertices. Thus all colorings with at most \(... | We need solve combinatorics probability. Need parse problem carefully.
Regular octagon vertices independently colored red/blue with equal probability. 2^8 colorings. Event: "the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices." Need probability ... | stop | 12.92 | {
"prompt_tokens": 139,
"completion_tokens": 6794,
"reasoning_tokens": 6414
} | 2026-10-22T06:33:57 | true | exact | 371 | null |
aime24-62 | aime24 | math | low | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{373} | We need solve combinatorics probability. Need parse problem carefully.
Regular octagon vertices independently colored red/blue prob 1/2. We consider random 2-coloring of 8 positions. Condition: "the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertice... | stop | 8.65 | {
"prompt_tokens": 138,
"completion_tokens": 3905,
"reasoning_tokens": 3887
} | 2026-10-22T06:43:19 | false | mismatch | 373 | null |
aime24-62 | aime24 | math | max | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | The blue vertices form a random subset \(B\subseteq \mathbb Z_8\). We need existence of a rotation \(k\) such that \(B+k\subseteq B^c\), i.e. \((B+k)\cap B=\varnothing\).
- If \(|B|=0,1,2,3\), this is always possible.
- If \(|B|\ge 5\), impossible since \(2|B|>8\).
- If \(|B|=4\), need \(B+k=B^c\). Counting such 4-sub... | We need solve combinatorics probability. Need parse problem carefully. Regular octagon vertices independently colored red/blue equiprobable. Total 2^8 colorings. Condition: "octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices" Need interpret. We have... | stop | 14.18 | {
"prompt_tokens": 139,
"completion_tokens": 7127,
"reasoning_tokens": 6888
} | 2026-10-22T06:44:52 | true | exact | 371 | null |
aime24-62 | aime24 | math | medium | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | \boxed{371} | We need solve combinatorics probability. Need parse problem carefully.
Regular octagon vertices independently colored red/blue equal prob. Total colorings 2^8. We want probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices.
Interp... | stop | 9.26 | {
"prompt_tokens": 138,
"completion_tokens": 4663,
"reasoning_tokens": 4643
} | 2026-10-22T06:33:46 | true | exact | 371 | null |
aime24-62 | aime24 | math | xhigh | 0 | reflection/Beam-1.0-501B-A23B:baseten | hf | baseten | The random coloring is determined by the subset \(B\) of blue vertices. The octagon can be rotated so all blue vertices land on originally red positions iff there is a rotation \(k\) such that \(B+k\) is disjoint from \(B\). This happens iff the set of pairwise cyclic distances between blue vertices misses at least one... | We need solve combinatorics probability. Need parse carefully. Regular octagon vertices independently red/blue prob 1/2 each. We have coloring. Condition: "the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices" Need probability over random coloring... | stop | 27.87 | {
"prompt_tokens": 139,
"completion_tokens": 14012,
"reasoning_tokens": 13843
} | 2026-10-22T06:40:53 | true | exact | 371 | null |
End of preview. Expand in Data Studio
BeamDial v0.1: what each reasoning level buys on Beam
2,050 graded runs: 410 tasks with known answers, each run at all five reasoning_effort levels (low, medium, high, xhigh, max) on Reflection's Beam 1.0 501B A23B.
| Level | Accuracy | 95% interval | Mean completion tokens | vs max |
|---|---|---|---|---|
| low | 96.3% | 94.1 to 97.8% | 914 | 55% |
| medium | 96.1% | 93.8 to 97.6% | 1,236 | 75% |
| high | 97.3% | 95.3 to 98.5% | 1,471 | 89% |
| xhigh | 97.3% | 95.3 to 98.5% | 1,603 | 97% |
| max | 96.6% | 94.4 to 98.0% | 1,657 | 100% |
Finding: on standard math and code, all five levels are a statistical tie on accuracy. low uses 55% of max's tokens. The dial does change thinking (on competition math, max reasons 2.2× longer than low); it just doesn't buy more right answers on these tasks.
What's inside
| Config / file | Rows | What it is |
|---|---|---|
runs (data/runs_graded.jsonl) |
2,050 | One row per (task, level): Beam's answer, its reasoning, token usage, latency, finish reason, and the grade |
tasks (data/tasks.jsonl) |
410 | The prompts and gold answers (or test code) |
data/flagged_tasks.json |
6 | Tasks with broken reference answers or tests, with the reason for each |
results/summary.json |
Every number in the report, including replayed routing policies | |
results/BeamDial-results-v0.1.xlsx |
Spreadsheet with live formulas | |
figures/ |
Launch figures (PNG, 3200×1800) |
Key fields in runs: task_id, source, family, effort, content, reasoning, finish_reason, latency_s, usage.prompt_tokens, usage.completion_tokens, usage.reasoning_tokens, correct, grade_reason, pred, public_pass, served_by.
How it was made
- Tasks: GSM8K (40), MATH levels 1 to 5 (160), AMC 2023 (40), AIME 2024 (30), OlympiadBench numeric (40), HumanEval (50), MBPP test split (50). Sampled with a fixed seed.
- Model and serving:
reflection/Beam-1.0-501B-A23Bthrough Hugging Face Inference Providers, pinned to Baseten. Temperature 0.7,max_completion_tokens32,000, one sample per task per level, October 22, 2026. - Grading: math via math-verify on the last
\boxed{}answer; code by running the benchmark's own tests in an isolated process with a timeout.public_passrecords only the test a developer could see (the MBPP example or the HumanEval docstring examples). - Code: everything to reproduce, re-grade and re-chart is at https://github.com/CoderRahul01/beamdial
Known limits
- One sample per cell at temperature 0.7, so per-task results are noisy. Compare levels using the intervals above.
- Math and Python only. Agent, tool-calling and long-context tasks are coming in v0.2.
- One provider (Baseten). Other servers may treat
reasoning_effortdifferently. - 6 tasks are flagged as broken benchmark items (one wrong GSM8K reference, five underspecified MBPP tests). Excluding them,
lowscores 97.8%. - These benchmarks are old and public, so Beam has likely seen many of them in training. That can make
lowlook better than it would on new problems. - Median latency was about 1.1 s at every level. Means differ only because of a long tail of hard prompts.
- The "re-ask at high when low thinks long" policy in
results/summary.jsonuses a threshold picked on this data. With the threshold picked on held-out folds it scores 97.2%, a tie with always usinghigh. - An early version of our grader under-counted math answers (math-verify cannot run in worker threads). This release uses the fixed grader. The fix and regression tests are in the repo.
License
The prompts and reference answers come from the public benchmarks listed above and keep their original licenses and terms. Beam's outputs, our grades, annotations and code are released under Apache 2.0. Reflection's model card permits use of Beam outputs, including for training other models.
Citation
@misc{beamdial2026,
title = {BeamDial: what each reasoning level buys on Beam},
author = {Rahul Pandey},
year = {2026},
url = {https://huggingface.co/datasets/PengoSword123/beamdial-v0.1}
}
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