This repository contains the OneThinker-8B model presented in: OneThinker: All-in-one Reasoning Model for Image and Video
Since OneThinker-8B shares the same architecture as Qwen3-VL-8B, it naturally supports easy and efficient inference.
For inference, please refer to:
Code: https://github.com/tulerfeng/OneThinker
We introduce OneThinker, an all-in-one multimodal reasoning generalist that is capable of thinking across a wide range of fundamental visual tasks within a single model.
We construct the large-scale OneThinker-600k multi-task training corpus and build OneThinker-SFT-340k with high-quality CoT annotations for cold-start SFT. Moreover, we propose EMA-GRPO, a new RL method that balances heterogeneous reward signals across diverse visual tasks, via simply tracking task-wise moving averages of reward std.
OneThinker demonstrates strong performance on 31 benchmarks across 10 fundamental vision tasks, while showing cross-task knowledge transfer and promising zero-shot generalization toward a unified multimodal reasoning generalist.
All code, models, and data are fully released.
Our model obtains significant performance gains after training based on Qwen3-VL-Instruct-8B across diverse visual tasks. For examle, OneThinker-8B reaches 70.6% accuracy on MMMU, 64.3% on MathVerse, 66.2% on VideoMMMU, 93.7 on Refcoco-testA, 54.9 J&F on ReasonVOS.
Besides, we also observe beneficial cross-task and cross-modality knowledge transfer, along with promising preliminary zero-shot generalization under unified training. This highlights the effectiveness and generalization ability of our unified training framework across diverse visual tasks.
Some reasoning examples are as follows.
Question:
As shown in the figure, AB is the diameter of ⊙O, and points C and D are on ⊙O. If ∠ABD = 50.0, then the degree of ∠BCD is () Choices: (A) 30° (B) 35° (C) 40° (D) 45°
We know that AB is the diameter of circle O. So, any angle inscribed in a semicircle is a right angle. That means ∠ACB should be 90° because triangle ACB is inscribed in the semicircle with AB as the diameter. Now, let's look at ∠ABD. The problem says ∠ABD is 50°. Let's think about quadrilateral or triangles here. Maybe we can use the inscribed angle theorem, which states that an inscribed angle is half the measure of its intercepted arc. First, let's consider arc AD or arc AB. Since AB is the diameter, the arc AB is 180°. ∠ABD is an inscribed angle that intercepts arc AD. So ∠ABD intercepts arc AD, so the measure of arc AD would be 2 × ∠ABD = 2 × 50° = 100°? Wait, maybe not. Let's check another angle. Alternatively, ∠BAD and ∠BCD might be related. Let's see. ∠BAD is an inscribed angle intercepting arc BD. ∠BCD is also an inscribed angle intercepting arc BD. So they might be equal or supplementary?
Wait, AB is the diameter, so ∠ADB is also a right angle (since angle inscribed in a semicircle). So triangle ADB is a right-angled triangle at D. In triangle ADB, ∠ABD is 50°, so ∠BAD = 90° - 50° = 40°.
Now, ∠BCD and ∠BAD are inscribed angles that intercept the same arc BD. Because ∠BAD intercepts arc BD, and ∠BCD also intercepts arc BD. So they should be equal. Therefore, ∠BCD = ∠BAD = 40°. Let's verify again.
AB is diameter, so ∠ACB = 90°, ∠ADB = 90°. In △ABD, ∠BAD = 90° - 50° = 40°. ∠BCD and ∠BAD both subtend arc BD, so they are equal. Hence, ∠BCD is 40°, which is option C.</think>
<answe>C</answer>
Question:
Given the bounding box [537,403,768,703] of the target object in the first frame, track this object in each frame.
<answer>{"boxes": {"1": [537, 403, 768, 703], "2": [537, 353, 797, 690], "3": [532, 323, 827, 687], "4": [519, 282, 850, 690], "5": [515, 260, 877, 694], "6": [510, 246, 887, 694], "7": [509, 246, 897, 703], "8": [509, 232, 947, 710], "9": [509, 210, 972, 732], "10": [482, 200, 989, 787], "11": [469, 196, 995, 856], "12": [459, 203, 995, 929], "13": [422, 221, 995, 987], "14": [397, 244, 995, 994], "15": [389, 250, 995, 994], "16": [382, 250, 995, 994], "17": [370, 250, 995, 994], "18": [364, 250, 995, 994], "19": [362, 250, 995, 994], "20": [354, 244, 995, 994], "21": [350, 239, 995, 994], "22": [347, 239, 995, 994], "23": [347, 244, 965, 979], "24": [350, 244, 945, 944], "25": [350, 244, 904, 910], "26": [350, 244, 845, 860], "27": [347, 244, 797, 779], "28": [347, 244, 772, 750], "29": [350, 250, 764, 742], "30": [350, 250, 757, 739], "31": [350, 250, 750, 739], "32": [350, 250, 750, 739]}}</answer>
Question:
Pleae segment the shorter one.
<answer>{"time": 0.0, "boxes": [426, 300, 528, 775], "positive_points": [[454, 450], [468, 550], [472, 650]], "negative_points": [[300, 500], [600, 500], [400, 200]]}</answer>
If you find our work helpful for your research, please consider citing our work.
@article{feng2025onethinker,
title={OneThinker: All-in-one Reasoning Model for Image and Video},
author={Feng, Kaituo and Zhang, Manyuan and Li, Hongyu and Fan, Kaixuan and Chen, Shuang and Jiang, Yilei and Zheng, Dian and Sun, Peiwen and Zhang, Yiyuan and Sun, Haoze and others},
journal={arXiv preprint arXiv:2512.03043},
year={2025}
}
9 commits
1 commits
This repository contains the OneThinker-8B model presented in: OneThinker: All-in-one Reasoning Model for Image and Video
Since OneThinker-8B shares the same architecture as Qwen3-VL-8B, it naturally supports easy and efficient inference.
For inference, please refer to:
Code: https://github.com/tulerfeng/OneThinker
We introduce OneThinker, an all-in-one multimodal reasoning generalist that is capable of thinking across a wide range of fundamental visual tasks within a single model.
We construct the large-scale OneThinker-600k multi-task training corpus and build OneThinker-SFT-340k with high-quality CoT annotations for cold-start SFT. Moreover, we propose EMA-GRPO, a new RL method that balances heterogeneous reward signals across diverse visual tasks, via simply tracking task-wise moving averages of reward std.
OneThinker demonstrates strong performance on 31 benchmarks across 10 fundamental vision tasks, while showing cross-task knowledge transfer and promising zero-shot generalization toward a unified multimodal reasoning generalist.
All code, models, and data are fully released.
Our model obtains significant performance gains after training based on Qwen3-VL-Instruct-8B across diverse visual tasks. For examle, OneThinker-8B reaches 70.6% accuracy on MMMU, 64.3% on MathVerse, 66.2% on VideoMMMU, 93.7 on Refcoco-testA, 54.9 J&F on ReasonVOS.
Besides, we also observe beneficial cross-task and cross-modality knowledge transfer, along with promising preliminary zero-shot generalization under unified training. This highlights the effectiveness and generalization ability of our unified training framework across diverse visual tasks.
Some reasoning examples are as follows.
Question:
As shown in the figure, AB is the diameter of ⊙O, and points C and D are on ⊙O. If ∠ABD = 50.0, then the degree of ∠BCD is () Choices: (A) 30° (B) 35° (C) 40° (D) 45°
We know that AB is the diameter of circle O. So, any angle inscribed in a semicircle is a right angle. That means ∠ACB should be 90° because triangle ACB is inscribed in the semicircle with AB as the diameter. Now, let's look at ∠ABD. The problem says ∠ABD is 50°. Let's think about quadrilateral or triangles here. Maybe we can use the inscribed angle theorem, which states that an inscribed angle is half the measure of its intercepted arc. First, let's consider arc AD or arc AB. Since AB is the diameter, the arc AB is 180°. ∠ABD is an inscribed angle that intercepts arc AD. So ∠ABD intercepts arc AD, so the measure of arc AD would be 2 × ∠ABD = 2 × 50° = 100°? Wait, maybe not. Let's check another angle. Alternatively, ∠BAD and ∠BCD might be related. Let's see. ∠BAD is an inscribed angle intercepting arc BD. ∠BCD is also an inscribed angle intercepting arc BD. So they might be equal or supplementary?
Wait, AB is the diameter, so ∠ADB is also a right angle (since angle inscribed in a semicircle). So triangle ADB is a right-angled triangle at D. In triangle ADB, ∠ABD is 50°, so ∠BAD = 90° - 50° = 40°.
Now, ∠BCD and ∠BAD are inscribed angles that intercept the same arc BD. Because ∠BAD intercepts arc BD, and ∠BCD also intercepts arc BD. So they should be equal. Therefore, ∠BCD = ∠BAD = 40°. Let's verify again.
AB is diameter, so ∠ACB = 90°, ∠ADB = 90°. In △ABD, ∠BAD = 90° - 50° = 40°. ∠BCD and ∠BAD both subtend arc BD, so they are equal. Hence, ∠BCD is 40°, which is option C.</think>
<answe>C</answer>
Question:
Given the bounding box [537,403,768,703] of the target object in the first frame, track this object in each frame.
<answer>{"boxes": {"1": [537, 403, 768, 703], "2": [537, 353, 797, 690], "3": [532, 323, 827, 687], "4": [519, 282, 850, 690], "5": [515, 260, 877, 694], "6": [510, 246, 887, 694], "7": [509, 246, 897, 703], "8": [509, 232, 947, 710], "9": [509, 210, 972, 732], "10": [482, 200, 989, 787], "11": [469, 196, 995, 856], "12": [459, 203, 995, 929], "13": [422, 221, 995, 987], "14": [397, 244, 995, 994], "15": [389, 250, 995, 994], "16": [382, 250, 995, 994], "17": [370, 250, 995, 994], "18": [364, 250, 995, 994], "19": [362, 250, 995, 994], "20": [354, 244, 995, 994], "21": [350, 239, 995, 994], "22": [347, 239, 995, 994], "23": [347, 244, 965, 979], "24": [350, 244, 945, 944], "25": [350, 244, 904, 910], "26": [350, 244, 845, 860], "27": [347, 244, 797, 779], "28": [347, 244, 772, 750], "29": [350, 250, 764, 742], "30": [350, 250, 757, 739], "31": [350, 250, 750, 739], "32": [350, 250, 750, 739]}}</answer>
Question:
Pleae segment the shorter one.
<answer>{"time": 0.0, "boxes": [426, 300, 528, 775], "positive_points": [[454, 450], [468, 550], [472, 650]], "negative_points": [[300, 500], [600, 500], [400, 200]]}</answer>
If you find our work helpful for your research, please consider citing our work.
@article{feng2025onethinker,
title={OneThinker: All-in-one Reasoning Model for Image and Video},
author={Feng, Kaituo and Zhang, Manyuan and Li, Hongyu and Fan, Kaixuan and Chen, Shuang and Jiang, Yilei and Zheng, Dian and Sun, Peiwen and Zhang, Yiyuan and Sun, Haoze and others},
journal={arXiv preprint arXiv:2512.03043},
year={2025}
}
9 commits
1 commits