2025年4月16日

利用圖像思考

OpenAI o3 和 o4-mini 在思路鏈中對圖像進行推理，代表了視覺感知的重大突破。

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OpenAI o3 和 o4-mini 是我們 o-系列中最新的視覺推理模型。我們的模型首次能夠在思路鏈中利用圖像思考，而不僅僅是「看到」它們。

與我們早期 OpenAI o1 模型相似，o3 和 o4-mini 經過訓練，可以在回答前進行更長時間的思考，並在回覆用戶之前使用較長的內部思路鏈。o3 和 o4-mini 進一步擴展了這項能力，使其能夠在思路鏈中利用圖像思考，本功能透過使用工具轉換用戶上傳的圖像來實現，允許對它們進行裁剪、放大和旋轉以及使用其他簡單的圖像處理技術。更重要的是，這些功能是原生的，無需依賴單獨的專用模型。

ChatGPT 增強的視覺智能能夠透過比以往更徹底、更準確、更可靠的圖像分析技術，幫助您解決更棘手的問題。它可以將進階推理與網頁搜尋和圖像處理等工具（自動縮放、裁剪、翻轉或增強圖像）無縫整合，即使是不完美的照片也能從中提取見解。例如，您可以上傳經濟學問題集的照片以獲取逐步解釋，或分享組建錯誤的螢幕截圖以快速獲得根本原因分析。

這種方法為測試時運算擴展提供了一個新的方向，無縫融合了視覺和文字推理，正如其在多模態基準測試中的頂尖表現所反映，標誌著我們向多模態推理邁出了重要一步。

視覺推理實踐

「利用圖像思考」可讓您更輕鬆地與 ChatGPT 互動。您可以透過拍照提問，無需擔心物件的位置（包括文字倒置或單張照片含多個物理問題的情況）。即使物件乍看之下並不明顯，視覺推理也能讓模型將其放大以更清晰地觀察。

所有示例均使用 OpenAI o3 完成。

What is written on the notebook?

對筆記本上的內容進行推理，歷時 20 秒

Analyzed image

Analyzed image

The user posted an image at /mnt/data/own_notebook.jpg and asked me to read what’s written on it. I’ll need to load the image so I can inspect the text. Once I view it, I realize the text is upside down, so I’ll rotate it so it’s readable.
From there, I can check what’s written and share my findings clearly with the user.

Analyzed image

The user shared an image found at /mnt/data/own_notebook.jpg and asked me to read the text on it. I’ll load the image and zoom in to see the writing clearly. It turns out the text is upside down, so I’ll rotate it for easy reading. Afterward, I’ll crop the lower part of the image to focus on the relevant area, and rotate the cropped section to ensure the text is right side up.

Analyzed image

Done

It says: “4th February – finish roadmap.”

我們最新的視覺推理模型與 Python 資料分析、網絡搜尋、圖像產生等其他工具協同工作，創造性地有效解決更複雜的問題，為用戶提供第一個多模態代理體驗。

solve this maze and plot a path with red-colored line

推理迷宮解決方案和路徑，歷時 1 分 44 秒

Analyzed image

The user provided an image path, but I’ve yet to load it. First, I’ll need to load the image from the given path for analysis. Once that’s done, I can detect the maze path, potentially marked by black lines, and plot a red line to showcase it. The image is relatively small—404x

I’m thinking the start could be at the top left missing border piece, or maybe near the left border. To help figure this out, I’ll look at the first row pixels to find any white areas near the border. Since the maze walls are black on white, I’ll try a thresholding approach. To invert it, I’ll use a threshold value of 200, which would set white to 255 and black closer to 0. I’ll apply this threshold using the inverse method.

Okay, I initially thought I could approximate the maze start near the top-left opening. I applied some thresholding to detect white pixels but encountered an issue—nothing changed. It seems the image might have high noise. Oddly, when inspecting the grayscale, all pixel values are zero, which suggests the image is completely black. This doesn’t match what we saw earlier with Matplotlib, so I’ll delve deeper into how PIL loaded the image and inspect the underlying pixel array.

Analyzed

Python

1print(imgcv.max(), imgcv.min(), imgcv.mean())

0 0 0.0

Analyzed

Python

1import numpy as np
2img_np = np.array(img)
3print(img_np.shape, img_np.dtype, img_np.max(), img_np.min())

(404, 404, 4) uint8 255 0

It looks like the lines are black with a transparent background, but when PIL loads the image, it’s showing a white background due to how it’s handling the alpha channel. I’ll try converting the image to RGB, but the pixel values are still showing all zeros. This is odd, since the black lines should be visible. Hmm, earlier the display had a white background. Let me check the pixel values at coordinate (200, 200) to get a clearer picture.

Analyzed

Python

1img_rgb = img.convert('RGB')
2imgcv = cv2.cvtColor(np.array(img_rgb), cv2.COLOR_RGB2GRAY)
3print(imgcv.max(), imgcv.min())

0 0

Analyzed

Python

1print(img_np[200,200])

[0 0 0 0]

The image has black lines with a transparent background, but it appears white due to how PIL handled the alpha channel. When displayed earlier, I saw white areas, which were actually transparent areas shown on a white grid (the Matplotlib background color). So, the lines are black with the transparent spaces representing the open path. I’ll need to treat these transparent areas as the open path when interpreting the maze. It’s essentially a black line maze on a clear canvas.
I see that the maze walls are black pixels, and the path is transparent with an alpha value of 0. I need to find the path from an open cell on one border to another on the opposite border. To do this, I’ll create a binary matrix where the passable cells have an alpha of 0 and are not part of a line border. Now, I’ll focus on processing the image channels to identify the path.

Analyzed

Python

1alpha = img_np[:,:,3]
2# passable pixel if alpha == 0 (transparent) else wall
3passable = alpha == 0
4# find openings along border
5h,w = passable.shape
6openings = []
7# top row
8for x in range(w):
9    if passable[0,x]:
10        openings.append((0,x))
11# bottom row
12for x in range(w):
13    if passable[h-1,x]:
14        openings.append((h-1,x))
15# left col
16for y in range(h):
17    if passable[y,0]:
18        openings.append((y,0))
19# right col
20for y in range(h):
21    if passable[y,w-1]:
22        openings.append((y,w-1))
23len(openings)

1616

The walls are black, and the path is transparent (alpha 0). I need to find a way from the open cell at one border to the open cell at the opposite border. I’ll create a binary matrix where the passable cells are transparent (alpha 0). However, there are a lot of transparent cells, and the paths might not be clear due to line gaps. I plan to treat the lines as about 3 pixels thick. I’ll try BFS/DFS to analyze the passable area from the starting point.
It seems the maze's border cells might connect to "air" outside the maze—this is the transparent area surrounding the maze. But the path is only open through legitimate openings in the border walls. Outside of these openings, the transparent background might merge with the outside area, making it harder to differentiate. I need to treat the inside walkway as enclosed by walls, ensuring it connects to the outside only at those openings, not through a continuous open background.
I’m thinking of dilating the walls by 3 pixels. This might cause openings smaller than 3 pixels to close, which could block the path. However, this might inadvertently seal parts of the interior walkway, leading to a disconnected path. So, I should try morphological closing—using a small kernel to dilate the walls—and then check for passable areas. This could help me manage the maze more effectively while keeping the path clear.

Analyzed

Python

1# find pattern of vertical walls
2# examine alpha channel row 0 lines to see thickness
3unique_rows = np.unique(np.sum(~passable, axis=1))
4unique_rows[:20]

array([ 0, 16, 20, 22, 24, 26, 28, 30, 32, 34, 94, 150, 178, 192, 206, 220, 234, 248, 276, 290])

Analyzed

Python

1import cv2
2# create wall mask (1 where wall)
3wall = ~passable
4# dilate walls by 3 px
5kernel = np.ones((3,3), np.uint8)
6wall_dil = cv2.dilate(wall.astype(np.uint8), kernel, iterations=3)
7pass_dil = wall_dil==0
8# label connected regions in pass_dil
9num_labels, labels = cv2.connectedComponents(pass_dil.astype(np.uint8))
10print(num_labels)
11

It might be easier to treat the entire transparent area outside the walls as open, and then use BFS starting from the top-left outside region. However, since the walkway is also transparent, using BFS would flood the inside walkway from the entrance openings. I think a better approach might be to switch to grid cells, considering the maze lines are about 4 pixels wide. Alternatively, I could use a shortest path algorithm on the pixel level, restricting movement to inside the maze’s bounding box to solve the path.
I’m considering dilating the walls by around 3 pixels. If there are openings smaller than that, they could seal, disconnecting the interior walkway from the outside. The interior walkway might also get blocked, causing the path to be cut off. To test this, I’ll try using a morphological operation where I dilate the walls with a small kernel and check for passable regions. If needed, I could increase the dilation to 5 pixels to see how it affects the maze structure.
One idea is to invert the walls and walkway

Analyzed

Python

1wall_dil = cv2.dilate(wall.astype(np.uint8), kernel, iterations=5)
2pass_dil = wall_dil==0
3num_labels, labels = cv2.connectedComponents(pass_dil.astype(np.uint8))
4print(num_labels)

I’m thinking of analyzing the top border, specifically between 0 and around 10 pixel rows. I could look for positions where the wall (alpha not zero) is absent and detect where the top border has an opening — essentially where all the rows for a specific x position have alpha equal to zero (transparent). That should allow me to identify where the walkway might begin at the top border.

Analyzed

Python

1# find opening columns on top border vertical line: for each x where all first maybe 10 rows walls present?
2open_cols_top = []
3for x in range(w):
4    if np.all(alpha[0:10,x] == 0):
5        open_cols_top.append(x)
6open_cols_top[:20], len(open_cols_top)

([0, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 403], 16)

I’m considering dilating the walls by 3 pixels, but if the openings are smaller, they might seal the walkway. This could block the path. I’ll try using morphological closing to see if dilating the walls with a small kernel helps. After dilation, the walkway might still be connected, so I’ll need to check border openings and spot any gaps on the rectangle's border. I’ll also inspect a few initial pixels to determine wall thickness, which seems to be about 4 pixels.

Analyzed

Python

1print(alpha[0,0:30])

[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]

Analyzed

Python

1print(alpha[1,0:50])

[ 0 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255 255]

The walls are drawn with black lines, and walkway openings are the gaps in these lines. I want to check the top border for missing walls, so I’ll inspect the first 10 rows for each column where the "alpha" value is zero, indicating no wall. It looks like there are 16 such columns, mostly near the edges. Some areas (like positions 195-208) probably represent real openings, while others (like positions 0 and 403) may be due to missing borders outside the maze. Now, I’ll check row 1 for walls.
I’ll check row 1, where alpha equals 255, indicating pixels that are part of the walls. From this, I realize that the exterior walls are

Analyzed

Python

1wall_rows = []
2for y in range(h):
3    if np.mean(alpha[y,:]>0) > 0.5:
4        wall_rows.append(y)
5# keep unique groups
6rows=[]
7prev=-10
8for y in wall_rows:
9    if y-prev>2: rows.append(y)
10    prev=y
11rows

[1, 17, 49, 97, 113, 145, 161, 177, 193, 209, 241, 257, 289, 305, 321, 337, 369, 401]

I checked alpha at row 1, and 255 indicates the presence of a wall. The border walls are likely located at row 1, row h-2, and certain columns. The outside region and walkway are alpha==0, but not within the border lines. The walkway seems enclosed by walls drawn at intervals, every 4 pixels. I’ll check horizontal wall positions, looking for rows with mostly 255-alpha values. I noticed a pattern, with gaps roughly every 16 pixels, though it’s not consistent due to vertical walls. I plan to compute the differences in these gaps.

Analyzed

Python

1diffs = [j-i for i,j in zip(rows[:-1],rows[1:])]
2diffs

[16, 32, 48, 16, 32, 16, 16, 16, 16, 32, 16, 32, 16, 16, 16, 32, 32]

I checked row 1 and observed that alpha 255 indicates a wall. The border walls may be at row 1 and row h-2, as well as certain columns. Outside the border area and the walkway both have alpha 0. It seems the walkway is enclosed by walls drawn at odd intervals. The grid cell width might be around 4 px. The horizontal wall positions appear roughly every 16 px, but the pattern isn’t uniform due to vertical walls. I’ll compute the differences for a clearer picture.
Okay, there are still many openings along row 1, but I think clusters around columns 195–208 might actually correspond to the real entrance. This span is about 14 px wide. The openings at 0 and 403 could be corners, so they might not be true entrances. It seems the real entrance is in the 195–208 range. Now, I’ll look at the left border to check for vertical openings, focusing on specific y values.

Analyzed

Python

1openings_top = [x for x in range(w) if alpha[1,x]==0]
2len(openings_top), openings_top[:20]

(16, [0, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 403])

Analyzed

Python

1openings_left = [y for y in range(h) if alpha[y,1]==0]
2openings_left[:20], len(openings_left)

([0, 403], 2)

I’m thinking of using morphological thinning to simplify the walkway into a one-pixel width, then running BFS to find a path. I’ll need to separate the walkway region from the outside. A flood fill from the border can help mark the outside, leaving the walkway region, but I need the walkway. A simpler plan is to target specific openings for entrances and exits. I can detect these gaps at row 1 in the border, where openings might appear as alpha[1,x]==0. Let’s check that approach.
I plan to use morphological thinning to narrow the walkway down to a one-pixel width, then run BFS to find the best path. To separate the walkway region from the outside, I’ll flood-fill around the border where alpha==0 to mark the outside. Then, I can address gaps in the walkway with morphological closing. Instead of filling all gaps, I’ll focus on entrance/exit openings, which I can detect by finding regions where alpha[1,x] == 0. Let’s check row 1 for these openings.
Looking at row 1, there are many potential openings, but the clusters between 195-208 (about 14 px wide) seem to be the real entrance. The row's openings might just be where there’s no wall. The left border also has vertical gaps, and I see that the top-left corner is open. The right and bottom borders also have openings, so I think the entrance and exit are at the top and bottom of the maze. With these connections, the walkway is accessible through these openings, so flood filling isn’t viable.
I’m trying to define the interior of the maze, focusing on the region between (2, h-3) for rows and (2, w-3) for columns. To set up the starting point, I’ll look for the first passable cell near the top opening, specifically looking between x=195 and x=208 in row 2. I’ll select the middle of that range, x=201, as my start point. For the end point, I’ll look for something similar near the bottom, around (h-3, 201).

Analyzed

Python

1openings_right = [y for y in range(h) if alpha[y,w-2]==0]
2openings_right[:20], len(openings_right)

([0, 403], 2)

Analyzed

Python

1openings_bottom = [x for x in range(w) if alpha[h-2,x]==0]
2openings_bottom[:20], len(openings_bottom)

([0, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 403], 16)

Analyzed

Python

1start=(2,201)
2end=(401,201)
3print(passable[start], passable[end])

True True

Analyzed

Python

1from collections import deque
2import numpy as np
3
4# Define mask of valid passable inside region
5valid = passable.copy()
6# exclude row0,row h-1, col0,col w-1
7valid[0,:]=False
8valid[h-1,:]=False
9valid[:,0]=False
10valid[:,w-1]=False
11
12start_pt=start
13end_pt=end
14
15prev = -np.ones((h,w,2), dtype=int)
16visited = np.zeros((h,w), dtype=bool)
17q=deque()
18q.append(start_pt)
19visited[start_pt]=True
20found=False
21dirs=[(1,0),(-1,0),(0,1),(0,-1)]
22while q:
23    y,x=q.popleft()
24    if (y,x)==end_pt:
25        found=True
26        break
27    for dy,dx in dirs:
28        ny,nx=y+dy,x+dx
29        if 0<=ny<h and 0<=nx<w and valid[ny,nx] and not visited[ny,nx]:
30            visited[ny,nx]=True
31            prev[ny,nx]=[y,x]
32            q.append((ny,nx))
33print(found)

True

I’ll start by selecting the starting and ending points as passable pixels near specific coordinates, using the average of the opening cluster for the start point. Then, I can compute the shortest path using BFS, restricting it to the interior region. After finding the path, I’ll overlay it on the original image by drawing a red line. But to ensure the red line stands out, I’ll convert the image to RGBA and then display it for a clear view.

Analyzed

Python

1# reconstruct path
2path=[]
3p=end_pt
4while p!=(-1,-1):
5    path.append(p)
6    y,x=p
7    prev_y, prev_x = prev[y,x]
8    if prev_y==-1:
9        break
10    p=(prev_y, prev_x)
11path = path[::-1]
12len(path)

1144

Analyzed

Python

1from PIL import ImageDraw
2img_rgba = img.copy()
3draw = ImageDraw.Draw(img_rgba)
4for y,x in path:
5    draw.point((x,y), fill=(255,0,0,255))
6# Save
7img_rgba.save('/mnt/data/maze_solved.png')

Analyzed image

Done

Matplotlib Chart

Here’s one possible solution path for your maze, traced in red.

Python

1from PIL import Image, ImageDraw
2import numpy as np
3import cv2
4import matplotlib.pyplot as plt
5from collections import deque
6# Load the original maze (walls are black, background is transparent)
7maze_path = '/mnt/data/d19c14d3-e182-4dba-99dc-647e4f725536.png'
8img = Image.open(maze_path)
9alpha = np.array(img)[:, :, 3]
10h, w = alpha.shape
11passable = alpha == 0  # transparent pixels are walkable
12# We determined the entrance & exit are roughly centered at the top & bottom openings
13start = (2, 201)
14end   = (h - 3, 201)
15# Breadth‑first search on walkable pixels (inside the border)
16valid = passable.copy()
17valid[0, :] = valid[-1, :] = valid[:, 0] = valid[:, -1] = False
18prev = -np.ones((h, w, 2), dtype=int)
19visited = np.zeros((h, w), bool)
20q = deque([start])
21visited[start] = True
22dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
23while q:
24    y, x = q.popleft()
25    if (y, x) == end:
26        break
27    for dy, dx in dirs:
28        ny, nx = y + dy, x + dx
29        if 0 <= ny < h and 0 <= nx < w and valid[ny, nx] and not visited[ny, nx]:
30            visited[ny, nx] = True
31            prev[ny, nx] = (y, x)
32            q.append((ny, nx))
33# Reconstruct the path
34path = []
35p = end
36while (p != (-1, -1)):
37    path.append(p)
38    py, px = p
39    p = tuple(prev[py, px])
40path.reverse()
41# Draw the solution in red
42solved = img.copy()
43draw = ImageDraw.Draw(solved)
44for y, x in path:
45    draw.point((x, y), fill=(255, 0, 0, 255))
46# Show the solved maze
47plt.figure(figsize=(6, 6))
48plt.imshow(solved)
49plt.axis('off')
50plt.show()
51# Save for download if the user wants it later
52solved_path = '/mnt/data/maze_solved.png'
53solved.save(solved_path)
54print(f"Solved maze saved to {solved_path}")
55

基準效能

為了突出視覺推理與我們過往的多模態模型相比的改進，我們在多項人類考試和機器學習基準上測試了 OpenAI o3 和 o4-mini。在我們測試的所有多模態任務中，這些新的視覺推理模型表現都明顯優於其前代模型。

所有模型都是在高「推理負載」設定下進行評估的，類似 ChatGPT 中所謂的「o4-mini-high」版本。

尤其是，利用圖像思考（無需依賴瀏覽）在我們評估的所有感知基準測試中都取得了顯著提升。我們的模型在 STEM 問答（MMMU、MathVista）、圖表解讀與推理 (CharXiv)、感知基礎能力 (VLMs are Blind) 以及視覺搜尋 (V*) 方面均創下最新技術水平表現。在 V* 測試中，我們的視覺推理方法達到了 95.7% 的準確率，大致上解決了基準測試中的問題。

限制和下一個工作

「利用圖像思考」目前存在以下限制：

推理鏈過長：模型可能會執行冗餘或不必要的工具呼叫和圖像處理步驟，導致思路鏈過長。
感知錯誤：模型仍然可能犯下基本的感知錯誤。即使工具呼叫正確地推進了推理過程，視覺上的誤解也可能導致最終答案出錯。
可靠性：模型在多次嘗試解決同一問題時，可能會採用不同的視覺推理過程，部分過程可能導致錯誤結果。

OpenAI o3 和 o4-mini 顯著提升了最先進的視覺推理能力，代表著我們朝著更廣泛的多模態推理邁出了重要一步。這些模型在視覺感知任務上表現出頂尖水平的準確度，使其能夠解決以往無法處理的問題。

我們正在不斷改進模型的圖像推理能力，使其更加簡潔、更少冗餘、更加可靠。我們很高興能夠繼續推進多模態推理領域的研究，並讓人們探索這些改進如何提升日常工作效率。

4 月 16 日更新：o3 在 Charxiv-r、Mathvista 和 vlmsareblind 上的測試結果已經更新，反映出原始評估時不存在的系統提示更改。

作者

OpenAI

Contributors

Aditya Ramesh, Aidan Clark, Aleksandra Spyra, Alex Tachard Passos, Alexander Kirillov, Ali Kalami, Amy McDonald Sandjideh, Andrei Gheorghe, Andrew Gibiansky, Andrew Tulloch, Angela Baek, Anubha Srivastava, Avital Oliver, Behrooz Ghorbani, Ben Leimberger, Borys Minaiev, Bowen Cheng, Brandon McKinzie, Carpus Chang, Cary Hudson, Casey Chu, Charlotte Cole, Chen Shen, Dan Roberts, Dana Palmie, Daniel Kappler, David Medina, Edmund Wong, Eric Mitchell, Eric Ning, Freddie Sulit, Haiming Bao, Haitang Hu, Hongyu Ren, Hyeonwoo Noh, Jakub Pachocki, James Betker, James Qin, Jamie Kiros, Jason Ai, Jerry Tworek, Jessica Liang, Ji Lin, Jiahui Yu, Jianfeng Wang, Joseph Mo, Kenji Hata, Kevin King, Kristian Georgiev, Kshitij Gupta, Lauren Yang, Li Jing, Lin Yang, Linden Li, Mark Chen, Martin Li, Max Schwarzer, Mia Glaese, Michael Malek, Minnia Feng, Nacho Soto, Nat McAleese, Niko Felix, Peter Faiman, Prafulla Dhariwal, Rajkumar Samuel, Rapha Gontijo Lopes, Ravi Teja Mullapudi, Reiichiro Nakano, Rennie Song, Ricky Xu, Sam Altman, Sean Fitzgerald, Shengjia Zhao, Shengli Hu, Shuchao Bi, Spencer Papay, Szi-chieh Yu, Wenda Zhou, Yang Lu, Yara Khakbaz, Yunxing Dai, Zhishuai Zhang

Here’s one possible solution path for your maze, traced in red.

Python

1from PIL import Image, ImageDraw
2import numpy as np
3import cv2
4import matplotlib.pyplot as plt
5from collections import deque
6# Load the original maze (walls are black, background is transparent)
7maze_path = '/mnt/data/d19c14d3-e182-4dba-99dc-647e4f725536.png'
8img = Image.open(maze_path)
9alpha = np.array(img)[:, :, 3]
10h, w = alpha.shape
11passable = alpha == 0  # transparent pixels are walkable
12# We determined the entrance & exit are roughly centered at the top & bottom openings
13start = (2, 201)
14end   = (h - 3, 201)
15# Breadth‑first search on walkable pixels (inside the border)
16valid = passable.copy()
17valid[0, :] = valid[-1, :] = valid[:, 0] = valid[:, -1] = False
18prev = -np.ones((h, w, 2), dtype=int)
19visited = np.zeros((h, w), bool)
20q = deque([start])
21visited[start] = True
22dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
23while q:
24    y, x = q.popleft()
25    if (y, x) == end:
26        break
27    for dy, dx in dirs:
28        ny, nx = y + dy, x + dx
29        if 0 <= ny < h and 0 <= nx < w and valid[ny, nx] and not visited[ny, nx]:
30            visited[ny, nx] = True
31            prev[ny, nx] = (y, x)
32            q.append((ny, nx))
33# Reconstruct the path
34path = []
35p = end
36while (p != (-1, -1)):
37    path.append(p)
38    py, px = p
39    p = tuple(prev[py, px])
40path.reverse()
41# Draw the solution in red
42solved = img.copy()
43draw = ImageDraw.Draw(solved)
44for y, x in path:
45    draw.point((x, y), fill=(255, 0, 0, 255))
46# Show the solved maze
47plt.figure(figsize=(6, 6))
48plt.imshow(solved)
49plt.axis('off')
50plt.show()
51# Save for download if the user wants it later
52solved_path = '/mnt/data/maze_solved.png'
53solved.save(solved_path)
54print(f"Solved maze saved to {solved_path}")
55