Results for
The toughest problem in the Cody Contest 2025 is Clueless - Lord Ned in the Game Room. Thank you Matt Tearle for such as wonderful problem. We can approach this clueless(!) tough problem systematically.
Initialize knowledge Matrix
Based on the hints provided in the problem description, we can initialize a knowledge matrix of size n*3 by m+1. The rows of the knowledge matrix represent the different cards and the columns represent the players. In the knowledge matrix, the first n rows represent category 1 cards, the next n rows, category 2 and the next category 3. We can initialize this matrix with zeros. On the go, once we know that a player holds the card, we can make that entry as 1 and if a player doesn't have the card, we can make that entry as -1.
yourcards processing
These are cards received by us.
- In the knowledge matrix, mark the entries as 1 for the cards received. These entries will be the some elements along the column pnum of the knowledge matrix.
- Mark all other entries along the column pnum as -1, as we don't receive other cards.
- Mark all other entries along the rows corresponding to the received cards as -1, as other players cannot receive the cards that are with us.
commoncards processing
These are the common cards kept open.
- In the knowledge matrix, mark the entries as 1 for the common cards. These entries will be some elements along the column (m+1) of the knowledge matrix.
- Mark all other entries along the column (m+1) as -1, as other cards are not common.
- Mark all other entries along the rows corresponding to the common cards as -1, as other players cannot receive the cards that are common.
Result -1 processing
In the turns input matrix, the result (5th column) value -1 means, the corresponding player doesn't have the 3 cards asked.
- Find all the rows with result as -1.
- For those corresponding players (1st element in each row of turns matrix), mark -1 entries in the knowledge matrix for those 3 absent cards.
pnum turns processing
These are our turns, so we get definite answers for the asked cards. Make sure to traverse only the rows corresponding to our turn.
- The results with -1 are already processed in the previous step.
- The results other than -1 means, that particular card is present with the asked player. So mark the entry as 1 for the corresponding player in the knowledge matrix.
- Mark all other entries along the row corresponding to step 2 as -1, as other players cannot receive this card.
Result 0 processing
So far, in the yourcards processing, commoncards processing, result -1 processing and pnum turns processing, we had very straightforward definite knowledge about the presence/absence of the card with a player. This step onwards, the tricky part of the problem begins.
result 0 means, any one (or more) of the asked cards are present with the asked player. We don't know exactly which card.
- For the asked player, if we have a definite no answer (-1 value in the knowledge matrix) for any two of the three asked cards, then we are sure about the card that is present with the player.
- Mark the entry as 1 for the definitely known card for the corresponding player in the knowledge matrix.
- Mark all other entries along the row corresponding to step 2 as -1, as other players cannot receive this card.
Cards per Player processing
Based on the number of cards present in the yourcards, we know the ncards, the number of cards per player.
Check along each column of the knowledge matrix, that is for each player.
- If the number of ones (definitely present cards) is equal to ncards, we can make all other entries along the column as -1, as this player cannot have any other card.
- If the sum of number of ones (definitely present cards) and the number of zeros (unknown cards) is equal to ncards, we can (i) mark the zero entries as one, as the unknown cards have become definitely present cards, (ii) mark all other entries along the column as -1, as other players cannot have any other card.
Category-wise cards checking
For each category, we must get a definite card to be present in the envelope.
- In each category (For every group of n rows of knowledge matrix), check for a row with all -1s. That is a card which is definitely not present with any of the players. Then this card will surely be present in the envelope. Add it to the output.
- If we could not find an all -1 row, then in that category, check each row for a 1 to be present. Note down the rows which doesn't have a 1. Those cards' players are still unknown. If we have only one such row (unknown card), then it must be in the envelope, as from each category one card is present in the envelope. Add it to the output.
- For the card identified in Step 2, mark all the entries along that row in the knowledge matrix as -1, as this card doesn't belong to any player.
Looping Over
In our so far steps, we could note that, the knowledge matrix got updated even after "Result 0 processing" step. This updation in the knowledge matrix may help the "Result 0 processing" step, if we perform it again. So, we can loop over the steps, "Result 0 processing", "Cards per Player processing" and "Category-wise cards checking" again. This ensures that, we will get the desired number of envelop cards (three in our case) as output.
Experimenting with Agentic AI
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9 votes
Instead of growing arrays inside a loop, preallocate with zeros(), ones(), or nan(). It avoids memory fragmentation and speeds up Cody solutions.
A = zeros(1,1000);
Cody often hides subtle hints in example outputs — like data shape, rounding, or format. Matching those exactly saves you a lot of debugging time.
isequal() is your best friend for Cody! It compares arrays perfectly without rounding errors — much safer than == for matrix outputs.
When Cody hides test cases, test your function with random small inputs first. If it works for many edge cases, it will almost always pass the grader.
I realized that using vectorized logic instead of nested loops makes Cody problems run much faster and cleaner. Functions like any(), all(), and logical indexing can replace multiple for-loops easily !
Hi cool guys,
I hope you are coding so cool!
FYI, in Problem 61065. Convert Hexavigesimal to Decimal in Cody Contest 2025 there's a small issue with the text:
[ ... For example, the text ‘aloha’ would correspond to a vector of values [0 11 14 7 0], thus representing the base-26 value 202982 = 11*263 + 14*262 + 7*26 ...]
The bold section should be:
202982 = 11*26^3 + 14*26^2 + 7*26
Run MATLAB using AI applications by leveraging MCP. This MCP server for MATLAB supports a wide range of coding agents like Claude Code and Visual Studio Code.
Check it out and share your experiences below. Have fun!
GitHub repo: https://github.com/matlab/matlab-mcp-core-server
Yann Debray's blog post: https://blogs.mathworks.com/deep-learning/2025/11/03/releasing-the-matlab-mcp-core-server-on-github/
Pick a team, solve Cody problems, and share your best tips and tricks. Whether you’re a beginner or a seasoned MATLAB user, you’ll have fun learning, connecting with others, and competing for amazing prizes, including MathWorks swags, Amazon gift cards, and virtual badges.
How to Participate
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- Bonus Round: Two top players from each team will be invited to a fun code-along event
Contest Timeline
- Main Round: Nov 10 – Dec 7, 2025
- Bonus Round: Dec 8 – Dec 19, 2025
Prizes (updated 11/19)
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Hello MATLAB Central community,
My name is Yann. And I love MATLAB. I also love Python ... 🐍 (I know, not the place for that).
I recently decided to go down the rabbit hole of AI. So I started benchmarking deep learning frameworks on basic examples. Here is a recording of my experiment:
Happy to engage in the debate. What do you think?
Large Language Models (LLMs) with MATLAB was updated again today to support the newly released OpenAI models GPT-5, GPT-5 mini, GPT-5 nano, GPT-5 chat, o3, and o4-mini. When you create an openAIChat object, set the ModelName name-value argument to "gpt-5", "gpt-5-mini", "gpt-5-nano", "gpt-5-chat-latest", "o4-mini", or "o3".
This is version 4.4.0 of this free MATLAB add-on that lets you interact with LLMs on MATLAB. The release notes are at Release v4.4.0: Support for GPT-5, o3, o4-mini · matlab-deep-learning/llms-with-matlab
The Graphics and App Building Blog just launched its first article on R2025a features, authored by Chris Portal, the director of engineering for the MATLAB graphics and app building teams.
Over the next few months, we'll publish a series of articles that showcase our updated graphics system, introduce new tools and features, and provide valuable references enriched by the perspectives of those involved in their development.
To stay updated, you can subscribe to the blog (look for the option in the upper left corner of the blog page). We also encourage you to join the conversation—your comments and questions under each article help shape the discussion and guide future content.
Large Languge model with MATLAB, a free add-on that lets you access LLMs from OpenAI, Azure, amd Ollama (to use local models) on MATLAB, has been updated to support OpenAI GPT-4.1, GPT-4.1 mini, and GPT-4.1 nano.
According to OpenAI, "These models outperform GPT‑4o and GPT‑4o mini across the board, with major gains in coding and instruction following. They also have larger context windows—supporting up to 1 million tokens of context—and are able to better use that context with improved long-context comprehension."
What would you build with the latest update?
Provide insightful answers
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Provide label-AI answer
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Provide answer by both AI and human
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Do not use AI for answers
46%
Give a button "chat with copilot"
10%
use AI to draft better qustions
5%
1561 votes
%% 清理环境
close all; clear; clc;
%% 模拟时间序列
t = linspace(0,12,200); % 时间从 0 到 12，分 200 个点
% 下面构造一些模拟的"峰状"数据，用于演示
% 你可以根据需要替换成自己的真实数据
rng(0); % 固定随机种子，方便复现
baseIntensity = -20; % 强度基线（z 轴的最低值）
numSamples = 5; % 样本数量
yOffsets = linspace(20,140,numSamples); % 不同样本在 y 轴上的偏移
colors = [ ...
0.8 0.2 0.2; % 红
0.2 0.8 0.2; % 绿
0.2 0.2 0.8; % 蓝
0.9 0.7 0.2; % 金黄
0.6 0.4 0.7]; % 紫
% 构造一些带多个峰的模拟数据
dataMatrix = zeros(numSamples, length(t));
for i = 1:numSamples
% 随机峰参数
peakPositions = randperm(length(t),3); % 三个峰位置
intensities = zeros(size(t));
for pk = 1:3
center = peakPositions(pk);
width = 10 + 10*rand; % 峰宽
height = 100 + 50*rand; % 峰高
% 高斯峰
intensities = intensities + height*exp(-((1:length(t))-center).^2/(2*width^2));
end
% 再加一些小随机扰动
intensities = intensities + 10*randn(size(t));
dataMatrix(i,:) = intensities;
end
%% 开始绘图
figure('Color','w','Position',[100 100 800 600],'Theme','light');
hold on; box on; grid on;
for i = 1:numSamples
% 构造 fill3 的多边形顶点
xPatch = [t, fliplr(t)];
yPatch = [yOffsets(i)*ones(size(t)), fliplr(yOffsets(i)*ones(size(t)))];
zPatch = [dataMatrix(i,:), baseIntensity*ones(size(t))];
% 使用 fill3 填充面积
hFill = fill3(xPatch, yPatch, zPatch, colors(i,:));
set(hFill,'FaceAlpha',0.8,'EdgeColor','none'); % 调整透明度、去除边框
% 在每条曲线尾部标注 Sample i
text(t(end)+0.3, yOffsets(i), dataMatrix(i,end), ...
['Sample ' num2str(i)], 'FontSize',10, ...
'HorizontalAlignment','left','VerticalAlignment','middle');
end
%% 坐标轴与视角设置
xlim([0 12]);
ylim([0 160]);
zlim([-20 350]);
xlabel('Time (sec)','FontWeight','bold');
ylabel('Frequency (Hz)','FontWeight','bold');
zlabel('Intensity','FontWeight','bold');
% 设置刻度（根据需要微调）
set(gca,'XTick',0:2:12, ...
'YTick',0:40:160, ...
'ZTick',-20:40:200);
% 设置视角（az = 水平旋转，el = 垂直旋转）
view([211 21]);
% 让三维坐标轴在后方
set(gca,'Projection','perspective');
% 如果想去掉默认的坐标轴线，也可以尝试
% set(gca,'BoxStyle','full','LineWidth',1.2);
%% 可选：在后方添加一个浅色网格平面 (示例)
% 这个与题图右上方的网格类似
[Xplane,Yplane] = meshgrid([0 12],[0 160]);
Zplane = baseIntensity*ones(size(Xplane)); % 在 Z = -20 处画一个竖直面的框
surf(Xplane, Yplane, Zplane, ...
'FaceColor',[0.95 0.95 0.9], ...
'EdgeColor','k','FaceAlpha',0.3);
%% 进一步美化（可根据需求调整）
title('3D Stacked Plot Example','FontSize',12);
constantplane("x",12,FaceColor=rand(1,3),FaceAlpha=0.5);
constantplane("y",0,FaceColor=rand(1,3),FaceAlpha=0.5);
constantplane("z",-19,FaceColor=rand(1,3),FaceAlpha=0.5);
hold off;
Have fun! Enjoy yourself!
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