Results for
If you haven't solved the problem yet, below hints guide how the algorithm should be implemented and clarify subtle rules that are easy to miss.
1. Shield is ONLY defended in HOME matches of the CURRENT holder - Even if a team beats the Shield holder in an away match, that does NOT count as a Shield defense.
2. A team defends the Shield ONLY when:
> They currently hold it.
> They are home team in that match
3. Shield transfer happens ONLY if the HOLDER plays a home match AND loses - A team may lose an away match — no effect.
4. The output ALWAYS includes the initial holder as the first row.
5. Defenses count resets for each new holder. - Every holder accumulates their own count until they lose it at home.
6. Match numbers are 1-indexed in the input, but “0” is used for initial state - The first real match is Match 1, but the output starts with Match 0.
7. Output row is created ONLY WHEN SHIELD CHANGES HANDS - This is an important hidden detail. A new row is appended, When the current holder loses a home match → Shield taken by visitor. If no loss at home occurs after that → no new row until next change.
8. The last holder’s defense count goes until the season ends - Even if they lose away later.
9. If a holder never gets a home match, defenses = 0.
10. In case the holder loses their very first home match → defenses = 0.
11. Shield changes only on HOME LOSS, not on a draw.
I hope above hints will help you in solving the problem.
Thanks and Regards,
Dev
Many MATLAB Cody problems involve solving congruences, modular inverses, Diophantine equations, or simplifying ratios under constraints. A powerful tool for these tasks is the Extended Euclidean Algorithm (EEA), which not only computes the greatest common divisor, gcd(a,b), but also provides integers x and y such that: a*x + b*y = gcd(a,b) - which is Bezout's identity.
Use of the Extended Euclidean Algorithm is very using in solving many different types of MATLAB Cody problems such as:
- Computing modular inverses safely, even for very large numbers
- Solving linear Diophantine equations
- Simplifing fractions or finding nteger coefficients without using symbolic tools
- Avoiding loops (EEA can be implemented recursively)
Below is a recursive implementation of the EEA.
function [g,x,y] = egcd(a,b)
% a*x + b*y = g [gcd(a,b)]
if b == 0
g = a; x = 1; y = 0;
else
[g, x1, y1] = egcd(b, mod(a,b));
x = y1;
y = x1 - floor(a/b)*y1;
end
end
Problem:
Given integers a and m, return the modular inverse of a (mod m).
If the inverse does not exist, return -1.
function inv = modInverse(a,m)
[g,x,~] = egcd(a,m);
if g ~= 1 % inverse doesn't exist
inv = -1;
else
inv = mod(x,m); % Bézout coefficient gives the inverse
end
end
%find the modular inverse of 19 (mod 5)
inv=modInverse(19,5)
Congratulations to all the Relentless Coders who have completed the problem set. I hope you weren't too busy relentlessly solving problems to enjoy the silliness I put into them.
If you've solved the whole problem set, don't forget to help out your teammates with suggestions, tips, tricks, etc. But also, just for fun, I'm curious to see which of my many in-jokes and nerdy references you noticed. Many of the problems were inspired by things in the real world, then ported over into the chaotic fantasy world of Nedland.
I guess I'll start with the obvious real-world reference: @Ned Gulley (I make no comment about his role as insane despot in any universe, real or otherwise.)
Hi Everyone!
As this is the most difficult question in problem group "Cody Contest 2025". To solve this problem, It is very important to understand all the hidden clues in the problem statement. Because everything is not directly visible.
For those who tried the problem, but were not able to solve. You might have missed any of the below hints -
- “The other players do not get to see which card has been shown, but they do know which three cards were asked for and that the player asked had one of them.” - Even when the card identity isn’t revealed (result = 0), you still gain partial knowledge — the asked player must have at least one of those three cards, meaning you can mark other players as not having all three simultaneously.
- "If it is your turn, you know the exact identity of that card" - You only know the exact shown card when result = 1, 2, or 3 — and it must be your turn. If someone else asked (even if you know result = 0), you don’t know which one was shown. So the meaning of result depends on whose turn it was, which is implicit — MATLAB code must assume that turns alternate 1→m→1, so your turn index is determined by (t-1) mod m + 1 == pnum.
- "Any leftover cards are placed face-up so that all players can see them" - These cards (commoncards) are not in anyone’s hand and cannot be in the envelope. So they’re not just visible — they’re logical constraints to eliminate from deduction.
- “It may be possible to determine the solution from less information than is given, but the information given will always be sufficient.”
- "Turn order is implied, not given explicitly" - Players take turns in order (1 to m, and back to 1).
On considering all the clues and constraints in the question, you will definitely be able to card for each category present in envelope.
I hope above clues will be useful for you.
Thank you, wishing you the success!
Regards,
Dev
When solving Cody problems, sometimes your solution takes too long — especially if you’re recomputing large arrays or iterative sequences every time your function is called.
The Cody work area resets between separate runs of your code, but within one Cody test suite, your function may be called multiple times in a single session.
This is where persistent variables come in handy.
A persistent variable keeps its value between function calls, but only while MATLAB is still running your function suite.
This means:
- You can cache results to avoid recomputation.
- You can accumulate data across multiple calls.
- But it resets when Cody or MATLAB restarts.
Suppose you’re asked to find the n-th Fibonacci number efficiently — Cody may time out if you use recursion naively. Here’s how to use persistent to store computed values:
function f = fibPersistent(n)
import java.math.BigInteger
persistent F
if isempty(F)
F=[BigInteger('0'),BigInteger('1')];
for k=3:10000
F(k)=F(k-1).add(F(k-2));
end
end
% Extend the stored sequence only if needed
while length(F) <= n
F(end+1)=F(end).add(F(end-1));
end
f = char(F(n+1).toString); % since F(1) is really F(0)
end
%calling function 100 times
K=arrayfun(@(x)fibPersistent(x),randi(10000,1,100),'UniformOutput',false);
K(100)
The fzero function can handle extremely messy equations — even those mixing exponentials, trigonometric, and logarithmic terms — provided the function is continuous near the root and you give a reasonable starting point or interval.
It’s ideal for cases like:
- Solving energy balance equations
- Finding intersection points of nonlinear models
- Determining parameters from experimental data
Example: Solving for Equilibrium Temperature in a Heat Radiation-Conduction Model
Suppose a spacecraft component exchanges heat via conduction and radiation with its environment. At steady state, the power generated internally equals the heat lost:
Given constants:
= 25 W- k = 0.5 W/K
- ϵ = 0.8
- σ = 5.67e−8 W/m²K⁴
- A = 0.1 m²
= 250 K
Find the steady-state temperature, T.
% Given constants
Qgen = 25;
k = 0.5;
eps = 0.8;
sigma = 5.67e-8;
A = 0.1;
Tinf = 250;
% Define the energy balance equation (set equal to zero)
f = @(T) Qgen - (k*(T - Tinf) + eps*sigma*A*(T.^4 - Tinf^4));
% Plot for a sense of where the root lies before implementing
fplot(f, [250 300]); grid on
xlabel('Temperature (K)'); ylabel('f(T)')
title('Energy Balance: Root corresponds to steady-state temperature')
% Use fzero with an interval that brackets the root
T_eq = fzero(f, [250 300]);
fprintf('Steady-state temperature: %.2f K\n', T_eq);
I set my 3D matrix up with the players in the 3rd dimension. I set up the matrix with: 1) player does not hold the card (-1), player holds the card (1), and unknown holding the card (0). I moved through the turns (-1 and 1) that are fixed first. Then cycled through the conditional turns (0) while checking the cards of each player using the hints provided until it was solved. The key for me in solving several of the tests (11, 17, and 19) was looking at the 1's and 0's being held by each player.
sum(cardState==1,3);%any zeros in this 2D matrix indicate possible cards in the solution
sum(cardState==0,3)>0;%the ones in this 2D matrix indicate the only unknown positions
sum(cardState==1,3)|sum(cardState==0,3)>0;%oring the two together could provide valuable information
Some MATLAB Cody problems prohibit loops (for, while) or conditionals (if, switch, while), forcing creative solutions.
One elegant trick is to use nested functions and recursion to achieve the same logic — while staying within the rules.
Example: Recursive Summation Without Loops or Conditionals
Suppose loops and conditionals are banned, but you need to compute the sum of numbers from 1 to n. This is a simple example and obvisously n*(n+1)/2 would be preferred.
function s = sumRecursive(n)
zero=@(x)0;
s = helper(n); % call nested recursive function
function out = helper(k)
L={zero,@helper};
out = k+L{(k>0)+1}(k-1);
end
end
sumRecursive(10)
- The helper function calls itself until the base case is reached.
- Logical indexing into a cell array (k>0) act as an 'if' replacement.
- MATLAB allows nested functions to share variables and functions (zero), so you can keep state across calls.
Tips:
- Replace 'if' with logical indexing into a cell array.
- Replace for/while with recursion.
- Nested functions are local and can access outer variables, avoiding global state.
Many MATLAB Cody problems involve recognizing integer sequences.
If a sequence looks familiar but you can’t quite place it, the On-Line Encyclopedia of Integer Sequences (OEIS) can be your best friend.
OEIS will often identify the sequence, provide a formula, recurrence relation, or even direct MATLAB-compatible pseudocode.
Example: Recognizing a Cody Sequence
Suppose you encounter this sequence in a Cody problem:
1, 1, 2, 3, 5, 8, 13, 21, ...
Entering it on OEIS yields A000045 – The Fibonacci Numbers, defined by:
F(n) = F(n-1) + F(n-2), with F(1)=1, F(2)=1
You can then directly implement it in MATLAB:
function F = fibSeq(n)
F = zeros(1,n);
F(1:2) = 1;
for k = 3:n
F(k) = F(k-1) + F(k-2);
end
end
fibSeq(15)
When solving MATLAB Cody problems involving very large integers (e.g., factorials, Fibonacci numbers, or modular arithmetic), you might exceed MATLAB’s built-in numeric limits.
To overcome this, you can use Java’s java.math.BigInteger directly within MATLAB — it’s fast, exact, and often accepted by Cody if you convert the final result to a numeric or string form.
Below is an example of using it to find large factorials.
function s = bigFactorial(n)
import java.math.BigInteger
f = BigInteger('1');
for k = 2:n
f = f.multiply(BigInteger(num2str(k)));
end
s = char(f.toString); % Return as string to avoid overflow
end
bigFactorial(100)
Hey Relentless Coders! 😎
Let’s get to know each other. Drop a quick intro below and meet your teammates! This is your chance to meet teammates, find coding buddies, and build connections that make the contest more fun and rewarding!
You can share:
- Your name or nickname
- Where you’re from
- Your favorite coding topic or language
- What you’re most excited about in the contest
Let’s make Team Relentless Coders an awesome community—jump in and say hi! 🚀
Welcome to the Cody Contest 2025 and the Relentless Coders team channel! 🎉
You never give up. When a problem gets tough, you dig in deeper. This is your space to connect with like-minded coders, share insights, and help your team win. To make sure everyone has a great experience, please keep these tips in mind:
- Follow the Community Guidelines: Take a moment to review our community standards. Posts that don’t follow these guidelines may be flagged by moderators or community members.
- Ask Questions About Cody Problems: When asking for help, show your work! Include your code, error messages, and any details needed to reproduce your results. This helps others provide useful, targeted answers.
- Share Tips & Tricks: Knowledge sharing is key to success. When posting tips or solutions, explain how and why your approach works so others can learn your problem-solving methods.
- Provide Feedback: We value your feedback! Use this channel to report issues or share creative ideas to make the contest even better.
Have fun and enjoy the challenge! We hope you’ll learn new MATLAB skills, make great connections, and win amazing prizes! 🚀
作ったコードは公開して使ってもらいましょう！ということでその方法をブログで紹介します。
GitHub や File Exchange で公開しているコードがあれば、ぜひこのスレで教えてください！
ブログで紹介している大まかな３ステップをここにまとめます。
1. GitHub でコードを公開・開発する
- GitHub 上でのリポジトリ公開はコミュニティ形成にもつながります。
- R2025a 以降は MATLAB の Markdown サポートも強化されており、README.md を充実させると理解や導入が促進されます。
2. File Exchange に展開（GitHub と連携して自動同期）
- File Exchangeで公開することで MATLAB 内から検索・インストールが可能になります。
- GitHub と File Exchange の連携設定により、GitHub の更新を自動的に File Exchange に反映させることも可能です。
3. 「Open in MATLAB Online」ボタンやリンクを追加
- GitHub リポジトリに「Open in MATLAB Online」リンクやボタンを埋め込むことで、ブラウザ上でコードを試せます。
群馬産業技術センター様をお招きし、製造現場での異常検知の取り組みについてご紹介いただくオンラインセミナーを開催します。
実際の開発事例を通して、MATLABを使った「教師なし」異常検知の進め方や、予知保全に役立つ最新機能もご紹介します。
✅ 異常検知・予知保全に興味がある方
✅ データ活用を何から始めればいいか迷っている方
✅ 実際の現場事例を知りたい方
ぜひお気軽にご参加ください！
Simulinkモデルを生成AIで自動的に作成できたら便利だと思いませんか？
QiitaのSacredTubesさんは、このアイデアを実験的に試みた記事を公開しています。
その方法は、まず生成AIでVerilogコードを作成し、それをSimulinkに取り込んでモデル化するというものです。（ここではHDL Coderというツールボックスの機能が使われました：importhdl）
まだ実用段階には至っていませんが、モデルベース開発（MBD）と生成AIの可能性を探る上で、非常に興味深い試みです。
生成AIの限界と可能性を考えるきっかけとして、一読の価値があります。
---
もし「Simulink Copilot」のような生成AIツールが登場するとしたら、
どんな機能があったら嬉しいと思いますか？
- 自然言語でブロック図を生成？
- 既存モデルの自動ドキュメント化？
- シミュレーション結果の要約と解釈？
皆さんのアイデアをぜひシェアしてください！
- 昨日までちゃんと動いていたのに・・
- ヘルプページ通りに書いているのに・・
MATLAB 関数がエラーを出すようになることありますよね（？）そんな時にみなさんがまず確認するもの、何かありますか？教えてください！
例えば
which -all plot
をコマンドウィンドウで実行して、もともと MATLAB で定義されている plot 関数（MATLAB のインストールフォルダにある plot 関数）がちゃんと頭に出てくるかどうか確認します。
キーと値の組み合わせでデータを格納できるディクショナリ。R2022bでdictionaryコマンドが登場し、最近のバージョンではreaddictionaryとwritedictionaryでJSONファイルからの読み込み・書き込みにも対応しました。
私はMIDIデータからピアノの演奏動画を作るプログラムで、ディクショナリを使いました。音のノート番号をキーにして、patchで白と黒で鍵盤を塗りつぶしたmatlab.graphics.Graphicsデータ型を値にしたディクショナリで保存して、MIDIで鳴らされた音のノート番号からlookupでグラフのオブジェクトを取得し、FaceColorを変更してハイライトするというもの。
コード例
%% MIDIデータの.matファイルを読み取ってピアノを描画するサンプル
fig = figure('Position', [34 328 1626 524]);
ax = axes;
whiteKeyY = [0 0 150 150];
whiteKeyColor = [1 1 1];
blackKeyY = [50 50 150 150];
blackKeyColor = [0.1 0.1 0.1];
edgeColor = [0 0 0];
% ディクショナリの定義
d = configureDictionary("double", "matlab.graphics.Graphics");
% 白鍵を描画
for n = 1:9
pos = 23*7*(n-1);
d = insert(d, 21 + (n-1)*12, patch([pos+5 pos+28 pos+28 pos+5],whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 21 + (n-1)*12));
d = insert(d, 23 + (n-1)*12, patch([pos+28 pos+51 pos+51 pos+28], whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 23 + (n-1)*12));
d = insert(d, 24 + (n-1)*12, patch([pos+51 pos+74 pos+74 pos+51], whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 24 + (n-1)*12));
if n < 9
d = insert(d, 26 + (n-1)*12, patch([pos+74 pos+97 pos+97 pos+74], whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 26 + (n-1)*12));
d = insert(d, 28 + (n-1)*12, patch([pos+97 pos+120 pos+120 pos+97], whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 28 + (n-1)*12));
d = insert(d, 29 + (n-1)*12, patch([pos+120 pos+143 pos+143 pos+120], whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 29 + (n-1)*12));
d = insert(d, 31 + (n-1)*12, patch([pos+143 pos+166 pos+166 pos+143], whiteKeyY, whiteKeyColor, 'EdgeColor', edgeColor, 'UserData', 31 + (n-1)*12));
end
end
% 黒鍵を描画。白鍵の上になるようにループを分けています
for n = 1:9
pos = 23*7*(n-1);
d = insert(d, 22 + (n-1)*12, patch([pos+23 pos+33 pos+33 pos+23], blackKeyY, blackKeyColor, 'EdgeColor', [0 0 0], 'UserData', 22 + (n-1)*12));
if n < 9
d = insert(d, 25 + (n-1)*12, patch([pos+69 pos+79 pos+79 pos+69], blackKeyY, blackKeyColor, 'EdgeColor', [0 0 0], 'UserData', 25 + (n-1)*12));
d = insert(d, 27 + (n-1)*12, patch([pos+92 pos+102 pos+102 pos+92], blackKeyY, blackKeyColor, 'EdgeColor', [0 0 0], 'UserData', 27 + (n-1)*12));
d = insert(d, 30 + (n-1)*12, patch([pos+138 pos+148 pos+148 pos+138], blackKeyY, blackKeyColor, 'EdgeColor', [0 0 0], 'UserData', 30 + (n-1)*12));
d = insert(d, 32 + (n-1)*12, patch([pos+161 pos+171 pos+171 pos+161], blackKeyY, blackKeyColor, 'EdgeColor', [0 0 0], 'UserData', 32 + (n-1)*12));
end
end
xticklabels({})
yticklabels({})
xlim([5 1362])
drawnow
%% MIDI音源の.matファイルを読み込み
matData = load('fur-elise.mat');
msg = matData.receivedMessages;
eventTimes = [msg.Timestamp] - msg(1).Timestamp;
n = 1;
numNotes = 0;
lastNote = 0;
highlightedCircles = cell(1, 127);
% 音が鳴った鍵盤だけハイライトする
tic
while toc < max(eventTimes)
if toc > eventTimes(n)
thisMsg = msg(n);
if thisMsg.Type == "NoteOn"
numNotes = numNotes + 1;
lastNote = thisMsg.Note;
thisPatch = lookup(d, thisMsg.Note);
thisPatch.FaceColor = '#CCFFCC';
drawnow
elseif thisMsg.Type == "NoteOff"
numNotes = 0;
thisPatch = lookup(d, thisMsg.Note);
[~, ~, wOrB] = calcNotePos(thisMsg.Note);
if wOrB == "w"
thisPatch.FaceColor = 'white';
else
thisPatch.FaceColor = 'black';
end
drawnow
end
n = n+1;
end
end
%% サブ関数
function [pianoPos, centerPos, wOrB] = calcNotePos(note)
tempVar = idivide(int64(note), int64(12)); % 12で割った商
pos = 23*7*(tempVar-1);
switch mod(note, 12)
case 0 % C
pianoPos = pos + 62.5;
centerPos = 30;
wOrB = "w";
case 2 % D
pianoPos = pos + 85.5;
centerPos = 30;
wOrB = "w";
case 4 % E
pianoPos = pos + 108.5;
centerPos = 30;
wOrB = "w";
case 5 % F
pianoPos = pos + 131.5;
centerPos = 30;
wOrB = "w";
case 7 % G
pianoPos = pos + 154.5;
centerPos = 30;
wOrB = "w";
case 9 % A
pianoPos = pos + 177.5;
centerPos = 30;
wOrB = "w";
case 11 % B
pianoPos = pos + 200.5;
centerPos = 30;
wOrB = "w";
case 1 % C#
pianoPos = pos + 69;
centerPos = 100;
wOrB = "b";
case 3 % D#
pianoPos = pos + 92;
centerPos = 100;
wOrB = "b";
case 6 % F#
pianoPos = pos + 138;
centerPos = 100;
wOrB = "b";
case 8 % G#
pianoPos = pos + 161;
centerPos = 100;
wOrB = "b";
case 10 % A#
pianoPos = pos + 184;
centerPos = 100;
wOrB = "b";
end
end
皆さんはディクショナリを使ってますか？ もし使っていたら、どういう活用をしているか、聞かせてください！
どの方法を使う事が多いですか？他によく使う方法があれば教えてくださいー。
方法①
Livescript 上で for ループ内で描画を編集させて描いた動画は「アニメーションのエクスポート」から動画ファイルに出力するのが一番簡単ですね。再生速度やら細かい設定ができない点は要注意。
方法②
exportgraphics 関数で "Append" オプション指定で実現できるようになった（R2022a から）のでこれも便利ですね。
N = 100;
x = linspace(0,4*pi,N);
y = sin(x);
filename = 'animation_sample.gif'; % Specify the output file name
if exist(filename,'file')
delete(filename)
end
h = animatedline;
axis([0,4*pi,-1,1]) % x軸の表示範囲を固定
for k = 1:length(x)
addpoints(h,x(k),y(k)); % ループでデータを追加
exportgraphics(gca,filename,"Append",true)
end
方法③
R2021b 以前のバージョンだとこんな感じ。
各ループで画面キャプチャして、imwrite で動画ファイルにフレーム追加していくイメージです。"DelayTime" を使って細かい指定ができるので、必要に応じて今でも利用します。
for k = 1:length(x)
addpoints(h,x(k),y(k)); % ループでデータを追加
drawnow % グラフアップデート
frame = getframe(gcf); % Figure 画面をムービーフレーム（構造体）としてキャプチャ
tmp = frame2im(frame); % 画像に変更
[A,map] = rgb2ind(tmp,256); % RGB -> インデックス画像に
if k == 1 % 新規 gif ファイル作成
imwrite(A,map,filename,'gif','LoopCount',Inf,'DelayTime',0.2);
else % 以降、画像をアペンド
imwrite(A,map,filename,'gif','WriteMode','append','DelayTime',0.2);
end
end
これからは生成AIでコードを１から書くという事が減ってくるのかと思いますが，皆さんがMATLABのコードを書く時に意識しているご自身のルールのようなものがあれば教えてください．
MATLAB言語は柔軟に書けますが，自然と個人個人のルールというものが出来上がってきているのでは，と思います．
私はParameter, Valueペアの引数がある関数はそれぞれのペアを新しい行に書く，というのをよくやります．
h = plot(x, y, "ro-", ...
"LineWidth", 2, ...
"MarkerSize", 10, ...
"MarkerFaceColor", "g");
Parameter=Valueでも同じです．
h = plot(x, y, "ro-", ...
LineWidth = 2, ...
MarkerSize = 10, ...
MarkerFaceColor = "g");
また，一時期は "=" を揃えることもやってました（今はやってませんが）．
h = plot(x, y, "ro-", ...
LineWidth = 2, ...
MarkerSize = 10, ...
MarkerFaceColor = "g");
皆さんにはどのようなルールがありますか？
