\section{Observations, Analysis and Conclusion}
\subsection{Ablation Study}
% \subsection{General Ablations on Mixed Long-CoT Fine-tuning and UniGRPO}
We present quantitative ablation results of our \method across different training stages: Mixed Long-CoT fine-tuning and UniGRPO. All results generated follow the sampling process in \cref{app:evaluation-details}. As shown in the \cref{tab:ablation}, after Stage 1, our model still lags behind most baselines. In Stage 2, Mixed Long-CoT fine-tuning substantially enhances the model’s reasoning capabilities, particularly in mathematical and geometric domains. In Stage 3, UniGRPO further improves performance, allowing the model to achieve results comparable to state-of-the-art methods across various tasks, including mathematical reasoning, geometric problem-solving, and image generation benchmarks such as CLIP Score and ImageReward. These results demonstrate that UniGRPO effectively boosts both the model’s understanding/reasoning and generative capabilities.



\subsection{Design Choices of UniGRPO}
\label{app:ablation}
\paragraph{Effect of General Masking Strategy}

\begin{wrapfigure}{r}{0.42\textwidth}
    \captionsetup{font=small} 
    \vspace{-4mm}
  \begin{center}
    \includegraphics[width=0.42\textwidth]{fig/reward_trend.pdf}
  \end{center}
  \caption{Comparison of different masking strategies on GSM8K reward trends during training. }
  \vspace{-4mm}
  \label{fig:reward_trend}
\end{wrapfigure}




To evaluate the impact of our proposed masking strategy, we first conduct a comparative analysis with d1~\citep{zhao2025d1scalingreasoningdiffusion} within our reinforcement learning framework. Given the substantial computational cost associated with large-scale ablation studies, we perform these experiments on the GSM8K dataset, utilizing 8 A100 GPUs. Both the original d1 methodology and our UniGRPO approach are applied to this dataset, starting from the same pre-trained checkpoint of our \method. We present the reward trends during training in \cref{fig:reward_trend}. As shown in the figure, our method consistently achieves higher reward values during training, aligning well with our theoretical analysis. In contrast to d1, UniGRPO removes masking from the question and applies partial masking to the answer rather than masking it entirely. This results in input sequences that retain partial noise, encouraging the model to learn across multiple denoising timesteps. Consequently, this better leverages the intrinsic characteristics of diffusion models and improves the overall learning capacity.




\paragraph{Effect of Uniformly Random masking}
\begin{wrapfigure}{r}{0.42\textwidth}
    \captionsetup{font=small} 
  \begin{center}
    \includegraphics[width=0.42\textwidth]{fig/random.pdf}
  \end{center}
  \vspace{-4mm}
  \caption{Comparison of different random masking strategies on GSM8K reward trends during training. }
  \vspace{-4mm}
  \label{fig:uniform_masking_reward}
\end{wrapfigure}



In place of fully random masking across iterations, we adopt a uniformly random masking strategy for the answer portion. Specifically, we first sample a random starting timestep, and then uniformly generate the remaining denoising timesteps across the full diffusion timesteps (set to 1000 in our experiments). For instance, given a randomly selected starting timestep of 100 and a total of 5 training iterations, the remaining timesteps are uniformly spaced and set to 300, 500, 700, and 900. This design ensures a more consistent coverage of the diffusion process while retaining randomness.
We illustrate the training reward trends resulting from this structured masking strategy in \cref{fig:uniform_masking_reward}. As shown, the baseline approach with fully random timestep selection tends to introduce instability during training, leading to more frequent reward fluctuations and requiring a greater number of steps to converge. In contrast, our uniformly spaced sampling strategy effectively approximates the behavior of Monte Carlo averaging in log-likelihood estimation, resulting in improved stability and faster convergence.

\begin{figure}[ht]
    \centering
    \vspace{-0.0in}
    \includegraphics[width=1\linewidth]{fig/synergy.pdf}
    \caption{Qualitative Illustration of Synergy Across Modalities.}
    \vspace{-0.0in}
    \label{fig:synergy}
\end{figure}

\begin{table}[ht]
    \centering
    \caption{Ablations on Mixed Long-CoT fine-tuning and UniGRPO}
    \resizebox{\linewidth}{!}{
    \begin{tabular}{lcccccc}
        \toprule
         Model & GSM8K & MATH500 & GeoQA & CLEVR  &  CLIP Score & ImageReward\\
         \midrule
         \method After Stage 1& 17.4 &4.2  & 8.3  & 10.3  & 23.1 & 0.69\\
         + Mixed Long-CoT Finetuning & 65.2 & 26.5 & 15.9 & 27.5 & 29.4 & 0.84\\
         \rowcolor{mmada_color}+ UniGRPO (\method)&  73.4& 36.0 & 21.0 & 34.5 &32.5 & 1.15 \\
         \bottomrule
    \end{tabular}}
    
    \label{tab:ablation}
\end{table}

\newpage
\subsection{Synergy Across Various Tasks}


\begin{wrapfigure}{r}{0.42\textwidth}
    \captionsetup{font=small} 
  \vspace{-4mm}
  \begin{center}
    \includegraphics[width=0.42\textwidth]{fig/mmada_draw1.pdf}
  \end{center}
  \vspace{-5mm}
  \caption{Key Performance Metrics Across Three Tasks. }
  \label{fig:synergy_quanti}
  \vspace{-4mm}
\end{wrapfigure}

Throughout the joint training process, we observe a clear synergy across the three task categories—text generation, multimodal understanding, and image generation. As shown in \cref{fig:synergy_quanti}, all key performance metrics exhibit consistent improvements during Stage 2 (training steps 120K–200K), reflecting the mutually beneficial nature of our unified training framework. This synergy is also evident qualitatively: as illustrated in \cref{fig:synergy}, the model's responses—both textual and visual—become increasingly complex and coherent. Specifically, textual outputs grow more informative and logically structured, while visual understanding yields more precise and grounded descriptions. Consequently, for the same prompt, the generated images become more accurate, detailed, and better aligned with the given instructions, demonstrating the effectiveness of joint optimization in enhancing cross-modal alignment and compositionality.



% \paragraph{Self-Refinement}
\subsection{Sampling Efficiency}
\begin{table}[ht]
\centering
% \vspace{-4mm}
\caption{
Generation performance of \method under different denoising steps. 
\textbf{*} Metrics: CLIP Score for image generation and multimodal understanding, MMLU accuracy for text generation.
}
\label{tab:denoising_steps}
\resizebox{0.5\textwidth}{!}{ 
\begin{tabular}{lccc}
\toprule
\textbf{Task} & \textbf{Denoising Steps} & \textbf{Metrics*} \\
\midrule
\multirow{3}{*}{Image Generation} 
    & 1024 & 32.8 \\
    & 50   & 32.0 \\
    & 15   & 31.7\\
\midrule
\multirow{3}{*}{Multimodal Understanding} 
    & 1024 & 35.5 \\
    & 512  & 36.1 \\
    & 256  & 35.4 \\
\midrule
\multirow{3}{*}{Text Generation} 
    & 1024 & 66.9 \\
     & 512 & 66.3 \\
    & 256  & 65.7 \\
\bottomrule
\end{tabular}}
% \vspace{-4mm}
\end{table} 
We identify \textbf{sampling efficiency} as a key advantage of diffusion models over autoregressive (AR) approaches. Unlike AR models, which generate tokens sequentially, diffusion models enable parallel token generation within each denoising step, substantially reducing the number of forward passes required. To quantify this advantage, we evaluate the performance of \method{} under varying numbers of denoising steps.
In our setup, generation begins from 1024 \texttt{[MASK]} tokens, allowing up to 1024 denoising steps—corresponding to a $512 \times 512$ resolution image. As shown in \cref{tab:denoising_steps}, image generation maintains strong performance even with as few as 15 or 50 steps. For text and multimodal tasks, coherent outputs can be achieved with just a quarter or half of the full steps. These results underscore the efficiency potential of diffusion-based language models and suggest that future advances in sampling techniques or higher-order solvers could further enhance their speed and quality.
% \newpage
\subsection{Task Extension}
\begin{wrapfigure}{r}{0.42\textwidth}
    \captionsetup{font=small} 
  \vspace{-0mm}
  \begin{center}
    \includegraphics[width=0.42\textwidth]{fig/inpaiting.pdf}
  \end{center}
  \vspace{-2mm}
  \caption{Inpainting Task Extension.}
  \label{fig:inpainting}
  \vspace{-4mm}
\end{wrapfigure}
A notable advantage of diffusion-based models is their natural ability to perform \textit{inpainting} and \textit{extrapolation} without requiring additional fine-tuning. This stems from the fact that these tasks can be formulated as masked token prediction problems, which are inherently integrated into the training objective of diffusion models. While prior work such as Show-o demonstrates this property only in the context of image generation, \method extends it further to multimodal understanding and text generation.
As illustrated in Figure~\ref{fig:inpainting}, our model supports inpainting across three modalities: (i) predicting missing spans in text sequences, (ii) completing answers in visual question answering given an image and partial input, and (iii) performing image inpainting conditioned on incomplete visual prompts. These examples showcase the flexibility and generalization capabilities of our unified diffusion architecture across diverse generation and reasoning tasks.

\section{Related Work}

\paragraph{Multimodal Large Language Models for Multimodal Understanding}


Recent developments in large language models (LLMs) such as Gemini-2.0 \citep{team2023gemini}, o1-preview \citep{o1}, and DeepSeek-R1 \citep{guo2025deepseek} have improved the evolution of multimodal large language models (MLLMs) \citep{li2024multimodal,mllm_survey,mllm_hallu}. Early efforts in this domain, including LLaVA \citep{llava}, MiniGPT-4 \citep{minigpt4}, and InstructBLIP \citep{instructblip}, showcased impressive capabilities in multimodal understanding. These studies advanced the integration of LLMs into multimodal contexts by projecting features from pre-trained modality-specific encoders, such as CLIP \citep{clip}, into the input space of LLMs, thereby facilitating multimodal understanding and reasoning within a unified transformer.
Many efforts have been made for MLLMs regarding vision encoders, alignment adapters, and curated datasets \citep{bai2023qwen,mckinzie2024mm1,qwen_vl,wang2024qwen2}, and most of them follow an autoregressive generation paradigm that has been proven effective for text generation in LLMs. However, they are usually not capable of performing textual and multimodal reasoning concurrently.
In this work, \method develops diffusion foundation models to fill this gap.  

\paragraph{Diffusion Models and Autoregressive Models for Visual Generation}
A large number of diffusion models~\citep{rombach2022high,ramesh2022hierarchical,dalle2,sdxl,pixart,nichol2021glide,raphael,yang2024mastering,zhang2025itercomp,chen2024overview,tian2025diffusion,yuan2023reward,guo2024gradient,chen2023score,wang2024rectified,yang2023improving,yang2024structure,tian2024videotetris} have demonstrated notable success in visual generation. In addition to the typical denoising diffusion process on the continuous space, a series of frameworks, such as D3PM \citep{d3pm} and VQ-Diffusion \citep{gu2022vector}, adopt discrete diffusion modeling \citep{d3pm,ARDM} for visual generation. Specifically, the image is denoted as a sequence of discrete tokens using pretrained image tokenizers~\citep{pml2Book,vqgan, magvitv2,gu2024rethinking}. In tthe raining stage, the model is optimized to recover the original values of a portion of these tokens that are randomly masked.
Transformer series~\citep{attention, raffel2020exploring, gpt1,gpt3,llama} has demonstrated significant capabilities in autoregressive modeling for NLP tasks. Many approaches~\citep{parmar2018image, taming,ravuri2019classification,chen2020generative,kondratyuk2023videopoet} try to apply the autoregressive modeling to perform visual generation by modeling semantic dependency within visual details. 
For example, 
LlamaGen \citep{llamagen} employs Llama architectures \citep{llama} and refines codebook design to enhance the performance of discrete tokenizers in class-conditional image generation.  
VAR \citep{tian2024visual} replaces the ``next-token prediction'' paradigm with ``next-scale prediction'' by designing a multi-scale image tokenizer.
However, existing autoregressive methods still lag behind diffusion methods in terms of visual generation capabilities.
In this work, \method train diffusion models to model textual and visual contents, and infer efficiently with AR or Semi-AR sampling.

\paragraph{Unified Vision-Language Foundation Models} Recently, numerous studies \citep{seed-x,wu2023next,CoDI,x-vila,dreamllm,jointly} have focused on developing unified multimodal foundation models that excel in both understanding and generation. Approaches such as SEED-X \citep{seed-x} and DreamLLM \citep{dreamllm}, along with others \citep{vlgpt,DBLP:journals/corr/abs-2307-05222,DBLP:journals/corr/abs-2312-13286,team2024chameleon, lwm, wu2024vila, sun2023emu, anil2023gemini, wu2024janus, chen2025janus}, represent all modalities as a series of tokens and employ a unified transformer architecture to train the entire system end-to-end. For example, Emu3 \citep{wang2024emu3} trains a single transformer from scratch using a mixture of multimodal tokenized sequences, optimized solely with next-token prediction.
While these unified autoregressive models show promise, they can struggle with visual generation tasks. Transfusion \citep{zhou2024transfusion} and Show-o \citep{xie2024show} employ autoregressive modeling for text generation and diffusion modeling for visual generation. Nevertheless, they mainly focus on pretraining strategies. Exploring effective post-training designs is still lacking for existing unified multimodal foundation models. 

\section{Conclusion}

This work introduces a unified diffusion foundation model, namely \method, that integrates textual reasoning, multimodal understanding, and generation within a single probabilistic framework. To the best of our knowledge, \method is the first to systematically explore the design space of diffusion-based foundation models, proposing novel post-training strategies. Extensive experiments across diverse vision-language tasks demonstrate that \method is comparable to or even better than specialized models, highlighting the potential of diffusion models as a next-generation foundation paradigm for multimodal intelligence.  However, \method also has limitations due to its current model size (8B parameters), and we will use a larger model size for better performance. 

\bibliographystyle{unsrt}
% \bibliographystyle{plainnat}
\bibliography{main}

\clearpage

\beginappendix




\section{Preliminaries of Discrete Diffusion, PPO and GRPO}

\subsection{Discrete Diffusion and Mask Token Prediction} \label{app-pre-discrete}

Discrete denoising diffusion models have emerged as a powerful paradigm for modeling discrete data representations, particularly in visual and textual domains. Recent works such as VQ-Diffusion~\citep{vqdiffusion}, MaskGIT~\cite{chang2022maskgit}, and Muse~\cite{chang2023muse} demonstrate their effectiveness in generating high-quality discrete tokens. Building on these advancements, Show-o~\cite{xie2024show} unifies discrete diffusion and mask token prediction into a single framework, enabling joint optimization of noise corruption and semantic recovery. Below, we formalize the key components of this discrete diffusion process.

\subsubsection{Forward and Reverse Diffusion Processes}

The \textit{forward} diffusion process progressively corrupts the initial data $\bm{x}_0$ (e.g., a sequence of visual/textual tokens) into noisy latent variables $\bm{x}_1, ..., \bm{x}_T$ via a fixed Markov chain $q(\bm{x}_t|\bm{x}_{t-1})$. At each step, this process stochastically perturbs tokens—such as replacing them with uniform noise or introducing a special $[\text{MASK}]$ token. The \textit{reverse} process, parameterized by a learned model $p_\theta$, reconstructs $\bm{x}_0$ from $\bm{x}_T$ by sequentially sampling $q(\bm{x}_{t-1}|\bm{x}_t, \bm{x}_0)$.

To formalize this, consider a single token $x^i_0$ at position $i$ in $\bm{x}_0$, where $x^i_0 \in \{1, 2, ..., K\}$ corresponds to an entry in a codebook. For simplicity, we omit the index $i$ in subsequent derivations. The transition probabilities between consecutive tokens are defined by a matrix $\bm{Q}_t \in \mathbb{R}^{K \times K}$, with $[\bm{Q}_t]_{mn} = q(x_t = m | x_{t-1} = n)$. The forward Markov process for the entire token sequence is then expressed as:
\begin{equation}
    q(x_t | x_{t-1}) = \bm{v}^\top(x_t) \bm{Q}_t \bm{v}(x_{t-1}),
    \label{eq:markov}
\end{equation}
where $\bm{v}(x)$ is a one-hot vector of length $K$ (with 1 at position $x$). The categorical distribution over $x_t$ is derived from $\bm{Q}_t \bm{v}(x_{t-1})$.

A critical property of the Markov chain allows us to marginalize intermediate steps and compute the direct transition from $\bm{x}_0$ to $\bm{x}_t$:
\begin{equation}
    q(x_t | x_0) = \bm{v}^\top(x_t) \overline{\bm{Q}}_t \bm{v}(x_0), \quad \text{with } \overline{\bm{Q}}_t = \bm{Q}_T \bm{Q}_{T-1} \cdots \bm{Q}_1.
    \label{eq:forward_chain}
\end{equation}

Furthermore, the posterior distribution conditioned on $\bm{x}_0$ is analytically tractable:
\begin{equation}
    \begin{split}
        q(x_{t-1} | x_t, x_0) = \frac{q(x_t | x_{t-1}, x_0) q(x_{t-1} | x_0)}{q(x_t | x_0)} 
        = \frac{
            \left(\bm{v}^\top(x_t) \bm{Q}_t \bm{v}(x_{t-1})\right)
            \left(\bm{v}^\top(x_{t-1}) \overline{\bm{Q}}_{t-1} \bm{v}(x_0)\right)
        }{
            \bm{v}^\top(x_t) \overline{\bm{Q}}_t \bm{v}(x_0)
        }.
    \end{split}
    \label{eq:posterior}
\end{equation}
This property is essential for deriving the variational lower bound of the diffusion process.

\subsubsection{Transition Matrix Design and Masking Strategy}

The design of the transition matrix $\bm{Q}_t$ is pivotal to the success of discrete diffusion models. Early works~\cite{hoogeboom2021argmax,austin2021structured} propose injecting uniform noise into the categorical distribution, leading to:
\begin{equation}
    \bm{Q}_t =
    \begin{bmatrix}
        \alpha_t + \beta_t & \beta_t & \cdots & \beta_t \\
        \beta_t & \alpha_t + \beta_t & \cdots & \beta_t \\
        \vdots & \vdots & \ddots & \vdots \\
        \beta_t & \beta_t & \cdots & \alpha_t + \beta_t
    \end{bmatrix},
\end{equation}
where $\alpha_t \in [0, 1]$ controls the retention probability, and $\beta_t = (1 - \alpha_t)/K$ ensures uniform diffusion across all $K$ categories. However, this approach often introduces abrupt semantic changes due to aggressive replacement of tokens with unrelated categories.

To address this limitation, Show-o adopts a \textbf{mask-and-replace} strategy inspired by masked language modeling. A special $[\text{MASK}]$ token is introduced, expanding the token space to $K+1$ states. The transition matrix is redefined as:
\begin{equation}
    \bm{Q}_t =
    \begin{bmatrix}
        \alpha_t + \beta_t & \beta_t & \cdots & \beta_t & 0 \\
        \beta_t & \alpha_t + \beta_t & \cdots & \beta_t & 0 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        \beta_t & \beta_t & \cdots & \alpha_t + \beta_t & 0 \\
        \gamma_t & \gamma_t & \cdots & \gamma_t & 1
    \end{bmatrix},
    \label{eq:mask_transit}
\end{equation}
where:
- Ordinary tokens have a probability $\alpha_t$ to remain unchanged, $\beta_t$ to be uniformly diffused, and $\gamma_t$ to be replaced by $[\text{MASK}]$.
- The $[\text{MASK}]$ token retains its state with probability 1.

This design explicitly signals corrupted positions during the forward process, enabling the reverse network to focus on reconstructing masked regions. The parameters satisfy $\alpha_t + K\beta_t + \gamma_t = 1$, ensuring proper normalization.

\subsubsection{Variational Objective and Loss Simplification}

Following the D3PM framework~\cite{austin2021structured}, the variational lower bound for the discrete diffusion process is derived as:
\begin{equation}
    \mathbb{E}_{q(\mathbf{x}_0)}[\log p_\theta(\mathbf{x}_0)] \geq -\mathcal{L}_\text{ELBO}(\mathbf{x}_0, \theta) \geq \sum_{t=1}^T \mathbb{E}_{q(\mathbf{x}_0, \mathbf{x}_t)}[\log p_\theta(\mathbf{x}_0 | \mathbf{x}_t)] + C.
\end{equation}
To simplify training, Show-o leverages the mask token prediction objective. Specifically, the model learns a neural network $p_\theta$ to reconstruct masked tokens in $\mathbf{x}_0$ from the noised $\mathbf{x}_t$, following the methodology of MaskGIT~\cite{chang2022maskgit}. This approach avoids explicit modeling of the full posterior and instead focuses on recovering only the corrupted regions, significantly reducing computational complexity.


\subsection{Proximal Policy Optimization and Group Relative Policy Optimization}
\label{app-pre-ppo-grpo}
Proximal Policy Optimization (PPO)~\cite{schulman2017proximal} is a widely adopted reinforcement learning algorithm that balances policy updates with stability through a clipped surrogate objective. The core idea of PPO lies in constraining policy changes to a proximal region around the previous policy, thereby preventing large, destabilizing updates. This is achieved by introducing a clipping mechanism that bounds the importance sampling ratio during optimization. Specifically, PPO maximizes the following objective:
\begin{equation}
\begin{aligned}
\mathcal{J}_\text{PPO}(\theta) = \mathbb{E}_{(q,a)\sim \mathcal{D}, o_{\le t}\sim\pi_{\theta_{\text{old}}}(\cdot\mid q)} 
\Bigg[ 
\min \Bigg( \frac{\pi_{\theta}(o_t\mid q,o_{<t})}{\pi_{\theta_{\text{old}}}(o_t\mid q,o_{<t})} \hat{A}_t,  
\ \text{clip} \Bigg( \frac{\pi_{\theta}(o_t\mid q,o_{<t})}{\pi_{\theta_{\text{old}}}(o_t\mid q,o_{<t})}, 1 - \varepsilon, 1 + \varepsilon \Bigg) \hat{A}_t \Bigg) \Bigg],
\label{eq:ppoloss}
\end{aligned}
\end{equation}
where $(q,a)$ denotes a question-answer pair sampled from the dataset $\mathcal{D}$, $\varepsilon$ is the clipping threshold, and $\hat{A}_t$ is the estimated advantage at time step $t$. The advantage $\hat{A}_t$ is typically computed using Generalized Advantage Estimation (GAE)~\cite{schulman2015high}, which combines multiple temporal differences with discounting and mixing parameters $(\gamma,\lambda)$:
\begin{equation}
\begin{aligned}
    &\hat{A}_t^{\text{GAE}(\gamma,\lambda)} = \sum_{l=0}^{\infty} (\gamma\lambda)^l \delta_{t+l}, \quad \text{with } \delta_l = R_l + \gamma V(s_{l+1}) - V(s_l).
\end{aligned}
\end{equation}
This formulation ensures robustness against high-variance estimates while maintaining compatibility with off-policy data.

In contrast, Group Relative Policy Optimization (GRPO)~\citep{deepseekmath} introduces two key innovations to address limitations in PPO's value function dependency and global advantage estimation. First, GRPO eliminates the explicit modeling of the value function $V(s)$ and instead computes advantages in a group-relative manner. For each $(q,a)$ pair, the behavior policy $\pi_{\theta_\text{old}}$ generates $G$ responses $\{o_i\}_{i=1}^G$, and the advantage of each response is normalized relative to its group:
\begin{equation}
\label{eq:adv}
\hat{A}_{i,t} = \frac{r_i - \text{mean}(\{R_i\}_{i=1}^G)}{\text{std}(\{R_i\}_{i=1}^G)},
\end{equation}
where $r_i$ represents the raw reward for response $o_i$. This design enables GRPO to focus on relative performance within a local context, reducing sensitivity to absolute reward scales.

Second, GRPO extends PPO's clipped objective by incorporating an explicit KL divergence penalty term between the current policy $\pi_\theta$ and a reference policy $\pi_\text{ref}$. The final objective is defined as:
\begin{equation}
\begin{aligned}
\mathcal{J}_\text{GRPO}(\theta)& = \mathbb{E}_{(q,a)\sim \mathcal{D}, \{o_i\}_{i=1}^G\sim \pi_{\theta_\text{old}}(\cdot\mid q)} \\
&\Bigg[ \frac{1}{G}\sum_{i=1}^{G} \frac{1}{|o_i|}\sum_{t=1}^{|o_i|} \Bigg( 
\min \Big( r_{i,t}(\theta) \hat{A}_{i,t},  
\ \text{clip} \Big( r_{i,t}(\theta), 1 - \varepsilon, 1 + \varepsilon \Big) \hat{A}_{i,t} \Big)
- \beta D_{\text{KL}}(\pi_{\theta} || \pi_{\text{ref}}) 
\Bigg) \Bigg],
\label{eq:grpoloss}
\end{aligned}
\end{equation}
where $r_{i,t}(\theta) = \frac{\pi_{\theta}(o_{i,t} \mid q, o_{i,<t})}{\pi_{\theta_{\text{old}}}(o_{i,t} \mid q,o_{i,<t})}$ is the importance sampling ratio at time $t$. Notably, GRPO computes the loss at the sample level: for each sequence, the per-token loss is averaged first, followed by averaging across all sequences in the group. This hierarchical aggregation enhances stability in multi-response settings.






\section{Details of UniGRPO}
\label{app:grpo}

In this section, we detail our UniGRPO training process and illustrate the differences compared to LLaDA and  d1 (diff-GRPO)~\citep{zhao2025d1scalingreasoningdiffusion}. Notably, LLaDA does not incorporate a reinforcement learning procedure, yet it includes an algorithm for computing log probabilities. Diff-GRPO introduced the first GRPO-style RL algorithm for diffusion language models and proposed an alternative method to compute log probabilities for a given question-answer pair. We first outline the methodologies of these two prior works:

\paragraph{(1) LLaDA} While LLaDA does not employ an RL process, it offers a method to estimate the log probability of a given pair $(q,a)$ using Monte Carlo simulation. For a given answer, N mask ratios are randomly sampled from the interval $(0,1)$ (where $N=128$ in the original work). Subsequently, a batch of $N$ inputs is constructed. Each input instance consists of the original question tokens concatenated with the answer, where a portion of the answer tokens (corresponding to the sampled mask ratio) is replaced by \texttt{[MASK]} tokens. A model forward pass is performed on this batch, and the logits corresponding to the masked regions are collected and averaged to obtain the log probability $\log p(q,a)$. The principal drawback of this method is its significant computational overhead, rendering it impractical for on-policy RL algorithms.

\paragraph{(2) d1 (diff-GRPO)} The d1 framework~\citep{zhao2025d1scalingreasoningdiffusion} introduced the first GRPO-style RL algorithm for diffusion language models. To overcome the efficiency issues of LLaDA, d1 employs a novel masking strategy. In each iteration, only a single forward pass is performed, eschewing Monte Carlo simulation. The input is constructed by applying a random mask to the question tokens and completely masking all answer tokens. The stochasticity is intended to be achieved by selecting different tokens for masking across different iterations, even with the same mask ratio.
While this approach in d1 enables GRPO training, we identify potential limitations:
\begin{itemize}
    \item \textbf{Question Masking:} The random masking of question tokens is of unclear practical significance. In typical question-answering scenarios, the question is always fully observed during both training and inference. We argue that such random masking of questions serves primarily to introduce stochasticity without direct relevance to the task's practical application.
    \item \textbf{Answer Masking Strategy:} By consistently masking the entire answer, the model is effectively trained only on the initial denoising step (i.e., predicting the full sequence from a fully masked state). We contend that this approach provides insufficient learning depth, potentially causing the model to behave more like a single-step predictive AR model rather than leveraging the multi-step denoising capabilities inherent to diffusion models. This underutilizes the diffusion model's unique strengths.
\end{itemize}

As introduced in \cref{sec:unigrpo}, our UniGRPO addresses the masking issues with the following key modifications:
\begin{enumerate}
    \item \textbf{Unmasked Questions:} In UniGRPO, all question tokens remain unmasked during training. This aligns the training process with the inference conditions and practical use-cases where the question is always fully provided.
    \item \textbf{Iteratively Varied Answer Masking:} We apply a random mask ratio to the answer tokens. Crucially, this mask ratio, denoted $\rho_a$, varies uniformly with the training iteration $\mu$ (e.g., $\rho_a \sim U(0,1)$ sampled anew or cycled through discrete steps each iteration). This strategy maintains stochasticity while exposing the model to diverse denoising stages—from nearly fully masked to almost fully denoised answers. This enables UniGRPO to learn from multi-step denoising, aligning with standard diffusion model training and leveraging their full generative potential across multiple steps.
\end{enumerate}



At each iteration, we sample a mask ratio $r_\mu$ uniformly from a predefined range, apply this ratio to the answer tokens, and perform a single forward pass using the unmasked question and masked answer. We then compute token-level log-likelihoods in the masked regions, apply the GRPO objective, and update the policy accordingly.











\clearpage
\section{Qualitative Comparisons (with Reasoning CoT)}
\label{app-qualitative-cot}

\begin{tcolorbox}[title = {Qualitative Comparison of World Knowledge-Aware Text-to-Image Generation (1)}, breakable, fontupper=\small, fonttitle=\small]

\textbf{Prompt: }The largest terrestrial carnivore from the Arctic.

\noindent\rule{\linewidth}{0.5pt}

\textbf{Other models: }

\noindent % 防止段落缩进，确保 minipage 从行首开始
\vspace{\medskipamount}
\begin{minipage}[t]{0.31\linewidth} % 第一个小区块，[t] 表示顶部对齐
    \centering % 内部内容居中
    Show-o
    
    \vspace{0.5ex} % 文字和图片之间的小间距，可选
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-1-showo.png} % 图片宽度充满这个 minipage
\end{minipage}\hfill % \hfill 会在区块之间产生弹性空白，使它们均匀分布
\begin{minipage}[t]{0.31\linewidth} % 第二个小区块
    \centering
    Emu3
    
    \vspace{0.5ex}
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-1-emu3.png}
\end{minipage}\hfill
\begin{minipage}[t]{0.31\linewidth} % 第三个小区块
    \centering
    Janus Pro 7B
    
    \vspace{0.5ex}
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-1-janus.png}
\end{minipage}
\vspace{\medskipamount}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }

{\scriptsize\ttfamily\obeylines
\textbf{\textcolor{blue}{<think>}}
largest terrestrial carnivore from the Arctic is the Polar Bear.
\textbf{\textcolor{blue}{</think>}}
A majestic \textbf{\textcolor{mydarkgreen}{polar bear}} stands on a snowy landscape, its fur glowing against the icy backdrop.
}

\vspace{\medskipamount}
\centering\includegraphics[width=0.2\linewidth]{fig/appx_images/t2i-1-mmada.png}
\vspace{\medskipamount}

\end{tcolorbox}

\begin{tcolorbox}[title = {Qualitative Comparison of World Knowledge-Aware Text-to-Image Generation (2)}, breakable, fontupper=\small, fonttitle=\small]

\textbf{Prompt: }A famous landmark building in Europe.

\noindent\rule{\linewidth}{0.5pt}

\textbf{Other models: }

\noindent % 防止段落缩进，确保 minipage 从行首开始
\vspace{\medskipamount}
\begin{minipage}[t]{0.31\linewidth} % 第一个小区块，[t] 表示顶部对齐
    \centering % 内部内容居中
    Show-o
    
    \vspace{0.5ex} % 文字和图片之间的小间距，可选
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-2-showo.png} % 图片宽度充满这个 minipage
\end{minipage}\hfill % \hfill 会在区块之间产生弹性空白，使它们均匀分布
\begin{minipage}[t]{0.31\linewidth} % 第二个小区块
    \centering
    Emu3
    
    \vspace{0.5ex}
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-2-emu3.png}
\end{minipage}\hfill
\begin{minipage}[t]{0.31\linewidth} % 第三个小区块
    \centering
    Janus Pro 7B
    
    \vspace{0.5ex}
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-2-janus.png}
\end{minipage}
\vspace{\medskipamount}

\noindent\rule{\linewidth}{0.5 pt}

\textbf{MMaDA: }

{\scriptsize\ttfamily\obeylines
\textbf{\textcolor{blue}{<think>}}
A famous landmark building in Europe is likely the Eiffel Tower.
\textbf{\textcolor{blue}{</think>}}
\textbf{\textcolor{mydarkgreen}{The Eiffel Tower}} is hollow, with a cloudy and muted sky.
}

\vspace{\medskipamount}
\centering\includegraphics[width=0.2\linewidth]{fig/appx_images/t2i-2-mmada.png}
\vspace{\medskipamount}

\end{tcolorbox}

\begin{tcolorbox}[title = {Qualitative Comparison of World Knowledge-Aware Text-to-Image Generation (3)}, breakable, fontupper=\small, fonttitle=\small]

\textbf{Prompt: }A traditional strong spirit from Russia.

\noindent\rule{\linewidth}{0.5pt}

\textbf{Other models: }

\noindent % 防止段落缩进，确保 minipage 从行首开始
\vspace{\medskipamount}
\begin{minipage}[t]{0.31\linewidth} % 第一个小区块，[t] 表示顶部对齐
    \centering % 内部内容居中
    Show-o
    
    \vspace{0.5ex} % 文字和图片之间的小间距，可选
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-3-showo.png} % 图片宽度充满这个 minipage
\end{minipage}\hfill % \hfill 会在区块之间产生弹性空白，使它们均匀分布
\begin{minipage}[t]{0.31\linewidth} % 第二个小区块
    \centering
    Emu3
    
    \vspace{0.5ex}
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-3-emu3.png}
\end{minipage}\hfill
\begin{minipage}[t]{0.31\linewidth} % 第三个小区块
    \centering
    Janus Pro 7B
    
    \vspace{0.5ex}
    \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-3-janus.png}
\end{minipage}
\vspace{\medskipamount}

\noindent\rule{\linewidth}{0.5 pt}

\textbf{MMaDA: }

{\scriptsize\ttfamily\obeylines
\textbf{\textcolor{blue}{<think>}}
The user is describing a traditional strong spirit from Russia. A well-known spirit that fits this description is vodka.
\textbf{\textcolor{blue}{</think>}}
An silver bottle of \textbf{\textcolor{mydarkgreen}{vodka}} sits next to a decanter and a halved lime.
}

\vspace{\medskipamount}
\centering\includegraphics[width=0.2\linewidth]{fig/appx_images/t2i-3-mmada.png}
\vspace{\medskipamount}

\end{tcolorbox}

% \begin{tcolorbox}[title = {Qualitative Comparison of World Knowledge-Aware Text-to-Image Generation (4)}, breakable, fontupper=\small, fonttitle=\small]

% \textbf{Prompt: }A famous statue gifted to the United States by France.

% \noindent\rule{\linewidth}{0.5pt}

% \textbf{Other models: }

% \noindent % 防止段落缩进，确保 minipage 从行首开始
% \vspace{\medskipamount}
% \begin{minipage}[t]{0.31\linewidth} % 第一个小区块，[t] 表示顶部对齐
%     \centering % 内部内容居中
%     Show-o
    
%     \vspace{0.5ex} % 文字和图片之间的小间距，可选
%     \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-4-showo.png} % 图片宽度充满这个 minipage
% \end{minipage}\hfill % \hfill 会在区块之间产生弹性空白，使它们均匀分布
% \begin{minipage}[t]{0.31\linewidth} % 第二个小区块
%     \centering
%     Emu3
    
%     \vspace{0.5ex}
%     \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-4-emu3.png}
% \end{minipage}\hfill
% \begin{minipage}[t]{0.31\linewidth} % 第三个小区块
%     \centering
%     Janus Pro 7B
    
%     \vspace{0.5ex}
%     \includegraphics[width=0.5\linewidth]{fig/appx_images/t2i-4-janus.png}
% \end{minipage}
% \vspace{\medskipamount}

% \noindent\rule{\linewidth}{0.5 pt}

% \textbf{MMaDA: }

% {\scriptsize\ttfamily\obeylines
% \textbf{\textcolor{blue}{<think>}}
% The description refers to a well-known landmark that was given from France to the United States. This is known as the Statue of Liberty.
% \textbf{\textcolor{blue}{</think>}}
% A low-angle shot of \textbf{\textcolor{mydarkgreen}{the Statue of Liberty}} against a blue sky with a few wispy clouds.
% }

% \vspace{\medskipamount}
% \centering\includegraphics[width=0.2\linewidth]{fig/appx_images/t2i-4-mmada.png}
% \vspace{\medskipamount}

% \end{tcolorbox}

\begin{tcolorbox}[title = {Qualitative Comparison of Textual Reasoning (1)}, breakable, fontupper=\small, fonttitle=\small]

\textbf{Question: }How many plums will balance one pear, given that 3 apples and one pear weigh as much as 10 plums, and one apple and 6 plums balance one pear? Assume that fruits of the same kind have the same weight.

\noindent\rule{\linewidth}{0.5pt}

\textbf{Show-o: }

\textit{< Show-o does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Emu3: }

\textit{< Emu3 does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Janus Pro 7B: }

\textit{< Janus Pro 7B does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }
{\scriptsize\ttfamily

\textbf{\textcolor{blue}{<think>}}

Okay, let's see. I need to figure out how many plums are needed to balance the weight of a single pear. The problem gives me some information. Let me try to parse it step by step.

First, the problem says: 3 apples together with a pear have the same weight as 10 plums. Then, the combination of one apple and 6 plums weighs the same as a pear. I need to find out how many plums are needed to equal the weight of a single pear. Since all fruits are equal in weight, probably the same.

Let me start by writing this down. Let's assign variables. Let's let \( A \) represent the weight of an apple, \( R \) as the weight of a pear, and \( M \) as the weight of a plum. So, based on the first statement given:

3A + R = 10M

Right, because 3 apples plus a pear equals 10 plums.

Second, saying one apple and 6 plums weigh the same as a pear. So,
A + 6M = R

That's because all apples and plums are equal. So,

A = R - 6M

So, now we have two equations:

1) 3A + R = 10M

2) A + 6M = R

Our goal is to find out how many plums are needed to balance one pear. That is, we need to find the value of \( n \) in \( R = nM \).

From equation (2): \( A + 6M = R \implies A = R - 6M \).

Now substitute this value of \( A \) into equation (1):

3A + R = 10M  
3(R - 6M) + R = 10M  
3R - 18M + R = 10M  
4R - 18M = 10M  
4R = 28M  
R = 7M

So one pear weighs as much as 7 plums.

Let's check with equation (2):  
A + 6M = R  
A = R - 6M  
But if \( R = 7M \), then \( A = 7M - 6M = M \).

So an apple weighs as much as 1 plum, which works when we check the first equation:  
3A + R = 10M  
3(M) + 7M = 10M  
3M + 7M = 10M

So both equations are satisfied.

Therefore, the answer is 7 plums.

\textbf{\textcolor{blue}{</think>}}\\

To determine how many plums are to balance a single pear, we set up the equations based on the given conditions:

1. Let \( A \) represent the weight of an apple, \( R \) represent the weight of a pear, and \( M \) the weight of a plum.

2. The equations are:

   - Equation 1: \( 3A + R = 10M \)
   
   - Equation 2: \( A + 6M = R \) (1 apple + 6 plums equals a pear)

**Step 1: Solve for \( A \) in terms of \( R \) and \( M \)**  

From Equation 2: \( A = R - 6M \)

**Step 2: Substitute \( A \) into Equation 1**  

\( 3(R - 6M) + R = 10M \)  

\( 3R - 18M + R = 10M \)  

\( 4R - 18M = 10M \)  

\( 4R = 28M \)  

\( R = 7M \)


**Final Answer:**  

\textbf{\textcolor{mydarkgreen}{\(\boxed{7}\)}} plums are needed to balance the weight of a single pear.
}

\end{tcolorbox}

\newpage

\begin{tcolorbox}[title = {Qualitative Comparison of Textual Reasoning (2)}, breakable, fontupper=\small, fonttitle=\small]

\textbf{Question: }How many positive divisors does the integer \( n \) possess, where \( n = 7^{3} \times 11^{2} \times 13^{4} \)?

\noindent\rule{\linewidth}{0.5pt}

\textbf{Show-o: }

\textit{< Show-o does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Emu3: }

\textit{< Emu3 does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Janus Pro 7B: }

\textit{< Janus Pro 7B does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }
{\scriptsize\ttfamily

\textbf{\textcolor{blue}{<think>}}

Okay, so I need to figure out how many positive divisors the integer n has, where n is given as 7 cubed times 11 squared times 13 to the fourth power. Hmm, let me think. I remember that the number of divisors of a number is related to its prime factorization exponents. Let me try to recall the general formula here.

First, right that if a number is expressed in a product of primes, like a number in the form \( n = p_1^a \times p_2^b \times \ldots \times p_k^c \), then the number of positive divisors is (a + 1)(b + 1)\dots (c + 1). Right? Because for each exponent in the prime factorization, you add 1, then multiply them all together for the total number of divisors.

Let me verify that. Take an example: if n is \( 2^1 \times 3^1 \), then the positive divisors are 1, 2, 3, 6. Using the formula, (1+1)*(1+1) = 2*2 = 4, which checks out. Another example: like \( 2^2 \times 3^1 \). Then the divisors are 1, 2, 4, 3, 6, 12. That makes 6 divisors. The formula gives (2+1)*(1+1) = 3*2 = 6, so that's correct. Another example: \( 2^2 \times 5^2 \). Then divisors count is (2+1)*(2+1) = 3*3=9. So the formula seems correct.

So, applying this formula to n in the given problem: n is \( 7^3 \times 11^2 \times 13^4 \). Here, the exponents are 3, 2, and 4. So, applying the number of divisors formula should be (3 + 1)(2 + 1)(4 + 1). Let's compute that step by step.

First, calculate each exponent plus one: 3 + 1 is 4. 2 + 1 is 3. 4 + 1 is 5. Then, multiplying those results together: 4 * 3 * 5. Let me check that. 4 times 3 is 12, then 12 times 5 is 60. So the number of positive divisors is 60.

Wait, but let me make sure I didn't make a mistake in the multiplication. Let me verify the steps again. 4*3 is 12, 12*5 is 60. Yes, that seems correct.

But just to be thorough, let me write the original number n as \( 7^3 \times 11^2 \times 13^4 \). The prime factors are 7, 11, 13. Each raised to the exponents 3, 2, and 4, respectively. Since they are all prime, the formula for divisors uses the exponents as factors plus 1. Then multiplying the results: 4*3*5=60.

Is there any possible way this could be wrong? Well, maybe if the exponents were different, but the original expression is already in its exponential form. Let me check if I misread. For example, if one of the exponents was not what is stated, but the problem writes $7^3$ times $11^2$ times $13^4$, so the exponents are 3, 2, 4, that's correct.

Alternatively, could there be a mistake in the multiplication? Let me check 4*3*5 again. 4*3 is 12, 12*5 is 60. Correct. So 60 is the correct answer.

Wait, just to be thorough, let's list the combinations: any divisor is \( 7^a \times 11^b \times 13^c \), where \( 0 \leq a \leq 3 \), \( 0 \leq b \leq 2 \), and \( 0 \leq c \leq 4 \). The number of choices for a is 4, for b is 3, for c is 5, so 4*3*5 = 60 possible positive divisors. Perfect.

I don't see any mistakes here. The answer is 60.

\textbf{\textcolor{blue}{</think>}}\\

**Solution:**  

The integer \( n \) is given as \( n = 7^3 \times 11^2 \times 13^4 \). To determine the number of positive divisors of \( n \), we use the formula for the number of divisors: if a number is in the form \( p_1^{a} \times p_2^{b} \times \dots \times p_k^{c} \), then it has \( (a + 1)(b + 1)\dots(c + 1) \) divisors. 

**Step-by-Step Solution:**

1. **Identify the Exponent:**

   - \( 7^3 \): Exponent = 3
   
   - \( 11^2 \): Exponent = 2
   
   - \( 13^4 \): Exponent = 4
   

2. **Apply the Divisor Formula:**

   - For each prime exponent, \( (a + 1) \), and then multiply:
   
     - For exponent 3: \( 3 + 1 = 4 \)
     
     - For exponent 2: \( 2 + 1 = 3 \)
     
     - For exponent 4: \( 4 + 1 = 5 \)
     
   - Multiply the results: \( 4 \times 3 \times 5 = 60 \)

**Answer:**  
The integer \( n \) has \textbf{\textcolor{mydarkgreen}{\(\boxed{60}\)}} positive divisors.
}

\end{tcolorbox}

\clearpage

\begin{tcolorbox}[title = {Qualitative Comparison of Textual Reasoning (3)}, breakable, fontupper=\small, fonttitle=\small]

\textbf{Question: }A "Hope Number" is a natural number with an odd number of positive divisors. Find the largest "Hope Number" within 1000.

\noindent\rule{\linewidth}{0.5pt}

\textbf{Show-o: }

\textit{< Show-o does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Emu3: }

\textit{< Emu3 does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Janus Pro 7B: }

\textit{< Janus Pro 7B does not support this task >}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }
{\scriptsize\ttfamily

\textbf{\textcolor{blue}{<think>}}

Okay, so I need to find the largest "Hope Number" which is a natural number with an odd number of positive divisors. Let me start by recalling what a Hope Number is. The question says a number has an odd number of divisors. Hmm, that means a number has an odd number of different divisors.

First, I remember that numbers have divisors related to their prime factorization. If a number has the prime factorization \( n = p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \), the number of divisors is \( (a_1+1)(a_2+1)\cdots(a_k+1) \). This product is odd if and only if every \( a_i+1 \) is odd, i.e., if every exponent \( a_i \) is even. That means the number itself is a perfect square (since each exponent is even).

To check: for example, take 9. Its prime factorization is \(3^2\). Its divisors are 1, 3, 9. That's three divisors, which is odd. Another example, 16. Prime factorization is \(2^4\). Its divisors are 1, 2, 4, 8, 16. That's five divisors, which is odd. How about 15? Its prime factorization is \(3 \times 5\), which is not a perfect square. Its divisors are 1, 3, 5, 15—that's four divisors, which is even. Therefore, only perfect squares have an odd number of divisors.

Therefore, all Hope Numbers less than 1000 are perfect squares.

Therefore, the problem reduces to finding the largest perfect square less than or equal to 1000. Let's compute that.

The square root of 1000 is approximately 31.62. So we need to check the largest integer \( n \) such that \( n^2 \leq 1000 \). Let's check 31.

\( 31^2 = 961 \). \( 32^2 = 1024 \), which is greater than 1000.

So $31^2 = 961$ is the largest perfect square within 1000. Therefore, 961 is the largest Hope Number within 1000.

But wait, let me confirm that 961 is indeed a Hope Number. Let's count its divisors. Its prime factorization is \(31^2\), and 31 is prime. The exponents are all even, so the number of divisors is \(2+1=3\): 1, 31, 961, which is odd. That checks out.

And to be thorough, is there a perfect square between 961 and 1000? \( 32^2 = 1024 > 1000 \), so there are none.

Therefore, the largest Hope Number within 1000 is 961.

\textbf{\textcolor{blue}{</think>}}\\

The largest "Hope Number" within 1000 is **961**.

**Step-by-Step Explanation:**

1. **Understanding Hope Property:** A Hope Number has an odd number of positive divisors. A number has an odd number of divisors **if and only if it is a perfect square** (because in the divisor formula, all exponents must be even so all \(a_i+1\) are odd, and their product is odd).

2. **Largest Hope Number:** Compute perfect squares $\leq$ 1000. \(31^2=961\) and \(32^2=1024\). Since \(32^2 > 1000\), the largest one is \(31^2=961\).

3. **Verification:** Any non-square number below 1000 has an even number of divisors; only perfect squares have an odd number of divisors.

**Answer:** \textbf{\textcolor{mydarkgreen}{961}}
}

\end{tcolorbox}

% \newpage

% \begin{tcolorbox}[title = {Qualitative Comparison of Textual Reasoning (4)}, breakable, fontupper=\small, fonttitle=\small]

% \textbf{Question: }What is the result of \(\frac{(20-16) \times (12+8)}{4}\)?

% \noindent\rule{\linewidth}{0.5pt}

% \textbf{Show-o: }

% \textit{< Show-o does not support this task >}

% \noindent\rule{\linewidth}{0.5pt}

% \textbf{Emu3: }

% \textit{< Emu3 does not support this task >}

% \noindent\rule{\linewidth}{0.5pt}

% \textbf{Janus Pro 7B: }

% \textit{< Janus Pro 7B does not support this task >}

% \noindent\rule{\linewidth}{0.5pt}

% \textbf{MMaDA: }
% {\scriptsize\ttfamily

% \textbf{\textcolor{blue}{<think>}}

% Okay, let me try to figure out the result of this expression: \(\frac{(20 - 16) \times (12 + 8)}{4}\). Hmm, let's break this down step by step. First, I need to make sure I apply the order of operations correctly. Right, order of operations: parentheses first, then multiplication and division from left to right, then addition and subtraction from left to right. Let me write the expression again to visualize it better: \(\frac{(20 - 16) \times (12 + 8)}{4}\).

% So, the parentheses are within the numerator, which is part of the fraction, and the denominator is 4. According to PEMDAS, I should handle the parentheses first before doing the multiplication and division. Let me look at the parentheses part: (20 - 16) and (12 + 8). Both of those are inside parentheses, so I should compute each part separately.

% Starting with the first parentheses: 20 - 16. That's straightforward. 20 minus 16 equals 4. Okay, so that part simplifies to 4.

% Next, the second part inside the parentheses: 12 + 8. That's addition, so 12 plus 8 is 20. So now, the numerator becomes \(4 \times 20\).

% So now, the numerator is \(4 \times 20\). Let me do that multiplication. 4 times 20 is 80. So the numerator simplifies to 80.

% Now the entire expression is \(\frac{80}{4}\). Dividing that number by 4. Let me check that division. 4 goes into 80 twenty times (since 4 × 20 is 80), so 80 ÷ 4 = 20. So the result is 20.

% Wait a second, let me verify that again to make sure I didn't make a mistake. Let's go through each step again.

% Original expression: \(\frac{(20 - 16) \times (12 + 8)}{4}\).

% First parentheses: 20 - 16 = 4. Check.

% Second parentheses: 12 + 8 = 20. Check.

% Multiply those two: 4 × 20 = 80. Check.

% Divide by 4: 80 ÷ 4 = 20. Yep, that seems right.

% But wait, maybe there's another way to approach this. Sometimes with order of operations, parentheses are easier to catch. Let's see. The original expression is \(\frac{(20 - 16) \times (12 + 8)}{4}\). The parentheses in the numerator give us multiplication, and the denominator is 4. You could write it as \(\frac{4 \times 20}{4}\). Simplifying, \(4 \times 20 = 80\), then 80 divided by 4 is 20. Alternatively, you could factor out the 4 in the numerator and cancel with the denominator, but in this case, the result is still 20. So that's a good way. Either way, multiplication and division are done from left to right.

% Alternatively, let's check if there's any possibility that the entire numerator is being divided. Let's see. The entire expression is \(\frac{4 \times 20}{4}\), so 4 × 20 is 80, then 80 divided by 4 is 20. So that's another way to see it, which is reassuring. So that seems consistent.

% Another way to check: look at it by using properties of operations. But in this case, since the expression is written with multiplication and division, the order of operations tells us to do all calculations in the numerator first, then divide by the denominator. All steps lead to 20.

% Another check: let me verify the multiplication. 4 × 20 = 80. Then 80 divided by 4 is 20. So no errors in the operations.

% Hmm, I think that's solid. The answer should be 20. Let me just make sure there's no alternative interpretation. Let's write the original expression again:

% Original expression: \(\frac{(20-16) \times (12+8)}{4}\).

% The numerator is 4 × 20 = 80, denominator is 4. So 80 divided by 4 is 20. Yep, exactly, that's correct.

% Therefore, I'm confident that the answer is 20.

% \textbf{\textcolor{blue}{</think>}}\\



% The result of \(\frac{(20-16) \times (12+8)}{4}\) is calculated as follows:

% 1. **Simplify the parentheses inside the numerator:**

%    - \(20 - 16 = 4\)

%    - \(12 + 8 = 20\)

% 2. **Multiply the simplified results:**

%    - \(4 \times 20 = 80\)

% 3. **Divide by 4:**

%    - \(\frac{80}{4} = 20\)

% **Final Answer:** \textbf{\textcolor{mydarkgreen}{\(\boxed{20}\)}}
% }

% \end{tcolorbox}


\newpage
\begin{tcolorbox}[title = {Qualitative Comparison of Multimodal Reasoning (1)}, breakable, fontupper=\small, fonttitle=\small]

\vspace{\medskipamount}
{\includegraphics[width=0.25\linewidth]{fig/appx_images/mmu_2.png}}
\vspace{\medskipamount}

\textbf{Question: }In the given figure, a square ABCD is inscribed in a circle with center O. Point P is located on side CD. What is the value of angle APB?

\noindent\rule{\linewidth}{0.5pt}

\textbf{Show-o: }

{\scriptsize\ttfamily The angle APB is \textbf{\textcolor{red}{90 degrees}}.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Emu3: }

{\scriptsize\ttfamily The question seems to be a mix of letters and numbers, \textbf{\textcolor{red}{not a coherent question or statement}}. If you meant "is" or "are", the correct statement would be "The measure of angle APB is" or "The measure of angle APB is". If you meant "is" or "are", the correct statement would be "The measure of angle APB is" or "The measure of angle APB is".}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Janus Pro 7B: }

{\scriptsize\ttfamily To find the measure of angle APB, we can use the properties of the square and the circle.

1. **Square ABCD**: The square is circumscribed around the circle, meaning all its sides are tangent to the circle. This implies that the diagonal of the square is equal to the diameter of the circle.

2. **Circle O**: The center of the circle is also the center of the square.

3. **Diagonals of a Square**: The diagonals of a square bisect each other and are equal in length.

4. **Angle APB**: Since ABCD is a square, angle APB is a right angle (90 degrees).

Therefore, the measure of angle APB is \textbf{\textcolor{red}{90 degrees}}.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }
{\scriptsize\ttfamily

\textbf{\textcolor{blue}{<think>}}

To determine the measure of angle APB, we start by noting that square ABCD is circumscribed within circle O. This means that the circle is the circumcircle of the square, and the center of the circle is the same as the center of the square. Let's denote the center of the circle as O.

Since ABCD is a square, the diagonals AC and BD are equal in length and bisect each other at right angles. This means that angle AOB is 90 degrees because the diagonals of a square are equal and bisect each other at right angles.

Now, consider triangle APB. Since P lies on side CD, we can use the property of the circle that the angle subtended by a chord at the center is twice the angle subtended at any point on the circumference on the same side. In this case, AP and BP are chords of the circle meeting at point P on CD (which is opposite AB). Points A and B are at adjacent vertices, and P is on CD.

The key point is that when P moves along CD, angle APB is fixed, as AB is a fixed chord and P moves along the arc opposite AB. For a square inscribed in a circle, the arc AB subtends a right angle at any point on the circle opposite AB (i.e., on arc CD). Therefore, for any point P on CD, angle APB is always 45°. This follows because $\angle APB$ is half the degree of the arc AB, which is 90°, so:

\[
\angle APB = \frac{1}{2} \times 90^\circ = 45^\circ
\]

Thus, the measure of angle APB is \(\boxed{45^\circ}\).

\textbf{\textcolor{blue}{</think>}}\\

The measure of angle APB is \textbf{\textcolor{mydarkgreen}{\(\boxed{45^\circ}\)}}.
}

\end{tcolorbox}




\begin{tcolorbox}[title = {Qualitative Comparison of Multimodal Reasoning (2)}, breakable, fontupper=\small, fonttitle=\small]

\vspace{\medskipamount}
{\includegraphics[width=0.3\linewidth]{fig/appx_images/mmu_4.jpg}}
\vspace{\medskipamount}

\textbf{Question: }What number of large things are the same material as the purple cylinder?

\noindent\rule{\linewidth}{0.5pt}

\textbf{Show-o: }

{\scriptsize\ttfamily There are \textbf{\textcolor{red}{two}} large things that are the same material as the purple cylinder.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Emu3: }

{\scriptsize\ttfamily \textbf{\textcolor{red}{2}}}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Janus Pro 7B: }

{\scriptsize\ttfamily There is one large object that shares the same material as the purple cylinder, which is the large green cylinder.

Answer: \textbf{\textcolor{mydarkgreen}{1}}.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }
{\scriptsize\ttfamily

\textbf{\textcolor{blue}{<think>}}

To determine the number of items in the image, let's analyze the objects present:

1. The large green cylinder is made of a material that appears to be rubber.

2. The large yellow cube is made of a material that appears to be metal.

3. The small gray cube is made of a material that appears to be rubber.

4. The small cyan sphere is made of a material that appears to be metal.

5. The small cyan cylinder is made of a material that appears to be metal.

6. The small purple cylinder is made of a material that appears to be rubber.

The green cylinder is the only large object made of a material that is the same as the small purple cylinder. Therefore, there is 1 large object made of the same material as the purple cylinder.

\textbf{\textcolor{blue}{</think>}}\\

The number of large things that are the same material as the purple cylinder is \textbf{\textcolor{mydarkgreen}{1}}.
}

\end{tcolorbox}


\begin{tcolorbox}[title = {Qualitative Comparison of Multimodal Reasoning (4)}, breakable, fontupper=\small, fonttitle=\small]

\vspace{\medskipamount}
{\includegraphics[width=0.3\linewidth]{fig/appx_images/mmu_6.png}}
\vspace{\medskipamount}

\textbf{Question: }Divide 24 apples equally among the people in the picture, how many apples can each person get?

\noindent\rule{\linewidth}{0.5pt}

\textbf{Show-o: }

{\scriptsize\ttfamily In the image, there are four people, each with a different number of apples. The person with the most apples has \textbf{\textcolor{red}{the most apples}}.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Emu3: }

{\scriptsize\ttfamily \textbf{\textcolor{red}{24}}.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{Janus Pro 7B: }

{\scriptsize\ttfamily To divide 24 apples equally among the four people in the picture, you would divide 24 by 4.

24 / 4 = 6

Each person can get \textbf{\textcolor{mydarkgreen}{6}} apples.}

\noindent\rule{\linewidth}{0.5pt}

\textbf{MMaDA: }
{\scriptsize\ttfamily

\textbf{\textcolor{blue}{<think>}}

To determine how many apples each person can get, we need to divide the total number of apples by the number of people.

\(24 \div 4 = 6\)

So, each person can get 6 apples.

\textbf{\textcolor{blue}{</think>}}\\

Each person can get \textbf{\textcolor{mydarkgreen}{6}} apples.
}

\end{tcolorbox}

% \clearpage
\section{Qualitative Examples (without Reasoning CoT)}
\label{app-qualitative-general}
\method can also efficiently handle general and simple tasks without generating CoT.


\begin{tcolorbox}[title = {Qualitative Example of Text-to-Image Generation}, breakable, fontupper=\small, fonttitle=\small]

\includegraphics[width=.95\linewidth]{fig/appx_images/general_t2i/general_t2i_merge.pdf}

\end{tcolorbox}

\begin{tcolorbox}[title = {Qualitative Example of Multimodal Understanding (1)}, breakable, fontupper=\small, fonttitle=\small]

\begin{minipage}[c]{0.35\textwidth}
\vspace{0pt}
\includegraphics[width=\linewidth]{fig/appx_images/general_mmu/dolphins.png}

\textbf{Prompt: }Please describe this image in detail.
\end{minipage}
\hfill
\begin{minipage}[c]{0.61\textwidth}
\vspace{0pt}
\method:

The image captures a serene and picturesque scene of a group of dolphins swimming in the ocean during sunset. The sky is painted with hues of orange, yellow, and pink, transitioning into a deep blue as it meets the horizon. The sun is setting on the horizon, casting a warm glow that reflects off the water's surface, creating a shimmering effect. The dolphins are positioned in the foreground, with their bodies partially submerged in the water, their heads and upper bodies visible above the surface. The water appears calm, with gentle ripples around the dolphins, adding to the tranquil atmosphere. The overall composition of the image conveys a sense of peace and natural beauty.
\end{minipage}

\end{tcolorbox}

\begin{tcolorbox}[title = {Qualitative Example of Multimodal Understanding (2)}, breakable, fontupper=\small, fonttitle=\small]

\begin{minipage}[c]{0.35\textwidth}
\vspace{0pt}
\includegraphics[width=\linewidth]{fig/appx_images/general_mmu/flowerfield.jpg}

\textbf{Prompt: }Please describe this image in detail.
\end{minipage}
\hfill
\begin{minipage}[c]{0.61\textwidth}
\vspace{0pt}
\method:

The image is a vibrant and colorful landscape painting that captures the essence of a sunset over a field of yellow flowers. The sky is a stunning blend of warm, soft hues, transitioning from deep purples and blues at the top to warm oranges and yellows near the horizon. The sun is positioned low on the horizon, casting a golden glow that illuminates the sky and the field below. The field of yellow flowers, likely sunflowers, stretches across the foreground, creating a sea of color that contrasts beautifully with the cooler tones of the sky. The painting is executed with broad, expressive brushstrokes, giving it a dynamic and lively feel. The overall composition is balanced and harmonious, capturing the tranquility and beauty of the natural world.
\end{minipage}

\end{tcolorbox}



\clearpage