Results for
The formula comes from @yuruyurau. (https://x.com/yuruyurau)
digital life 1
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:2e4;
x = mod(i, 100);
y = floor(i./100);
k = x./4 - 12.5;
e = y./9 + 5;
o = vecnorm([k; e])./9;
while true
t = t + pi/90;
q = x + 99 + tan(1./k) + o.*k.*(cos(e.*9)./4 + cos(y./2)).*sin(o.*4 - t);
c = o.*e./30 - t./8;
SHdl.XData = (q.*0.7.*sin(c)) + 9.*cos(y./19 + t) + 200;
SHdl.YData = 200 + (q./2.*cos(c));
drawnow
end
digital life 2
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:1e4;
x = i;
y = i./235;
e = y./8 - 13;
while true
t = t + pi/240;
k = (4 + sin(y.*2 - t).*3).*cos(x./29);
d = vecnorm([k; e]);
q = 3.*sin(k.*2) + 0.3./k + sin(y./25).*k.*(9 + 4.*sin(e.*9 - d.*3 + t.*2));
SHdl.XData = q + 30.*cos(d - t) + 200;
SHdl.YData = 620 - q.*sin(d - t) - d.*39;
drawnow
end
digital life 3
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 1, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:1e4;
x = mod(i, 200);
y = i./43;
k = 5.*cos(x./14).*cos(y./30);
e = y./8 - 13;
d = (k.^2 + e.^2)./59 + 4;
a = atan2(k, e);
while true
t = t + pi/20;
q = 60 - 3.*sin(a.*e) + k.*(3 + 4./d.*sin(d.^2 - t.*2));
c = d./2 + e./99 - t./18;
SHdl.XData = q.*sin(c) + 200;
SHdl.YData = (q + d.*9).*cos(c) + 200;
drawnow; pause(1e-2)
end
digital life 4
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 1, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:4e4;
x = mod(i, 200);
y = i./200;
k = x./8 - 12.5;
e = y./8 - 12.5;
o = (k.^2 + e.^2)./169;
d = .5 + 5.*cos(o);
while true
t = t + pi/120;
SHdl.XData = x + d.*k.*sin(d.*2 + o + t) + e.*cos(e + t) + 100;
SHdl.YData = y./4 - o.*135 + d.*6.*cos(d.*3 + o.*9 + t) + 275;
SHdl.CData = ((d.*sin(k).*sin(t.*4 + e)).^2).'.*[1,1,1];
drawnow;
end
digital life 5
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 1, 'filled','o','w',...
'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:1e4;
x = mod(i, 200);
y = i./55;
k = 9.*cos(x./8);
e = y./8 - 12.5;
while true
t = t + pi/120;
d = (k.^2 + e.^2)./99 + sin(t)./6 + .5;
q = 99 - e.*sin(atan2(k, e).*7)./d + k.*(3 + cos(d.^2 - t).*2);
c = d./2 + e./69 - t./16;
SHdl.XData = q.*sin(c) + 200;
SHdl.YData = (q + 19.*d).*cos(c) + 200;
drawnow;
end
digital life 6
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 1:1e4;
y = i./790;
k = y; idx = y < 5;
k(idx) = 6 + sin(bitxor(floor(y(idx)), 1)).*6;
k(~idx) = 4 + cos(y(~idx));
while true
t = t + pi/90;
d = sqrt((k.*cos(i + t./4)).^2 + (y/3-13).^2);
q = y.*k.*cos(i + t./4)./5.*(2 + sin(d.*2 + y - t.*4));
c = d./3 - t./2 + mod(i, 2);
SHdl.XData = q + 90.*cos(c) + 200;
SHdl.YData = 400 - (q.*sin(c) + d.*29 - 170);
drawnow; pause(1e-2)
end
digital life 7
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 1:1e4;
y = i./345;
x = y; idx = y < 11;
x(idx) = 6 + sin(bitxor(floor(x(idx)), 8))*6;
x(~idx) = x(~idx)./5 + cos(x(~idx)./2);
e = y./7 - 13;
while true
t = t + pi/120;
k = x.*cos(i - t./4);
d = sqrt(k.^2 + e.^2) + sin(e./4 + t)./2;
q = y.*k./d.*(3 + sin(d.*2 + y./2 - t.*4));
c = d./2 + 1 - t./2;
SHdl.XData = q + 60.*cos(c) + 200;
SHdl.YData = 400 - (q.*sin(c) + d.*29 - 170);
drawnow; pause(5e-3)
end
digital life 8
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl{6} = [];
for j = 1:6
SHdl{j} = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.3);
end
t = 0;
i = 1:2e4;
k = mod(i, 25) - 12;
e = i./800; m = 200;
theta = pi/3;
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
while true
t = t + pi/240;
d = 7.*cos(sqrt(k.^2 + e.^2)./3 + t./2);
XY = [k.*4 + d.*k.*sin(d + e./9 + t);
e.*2 - d.*9 - d.*9.*cos(d + t)];
for j = 1:6
XY = R*XY;
SHdl{j}.XData = XY(1,:) + m;
SHdl{j}.YData = XY(2,:) + m;
end
drawnow;
end
digital life 9
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl{14} = [];
for j = 1:14
SHdl{j} = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.1);
end
t = 0;
i = 1:2e4;
k = mod(i, 50) - 25;
e = i./1100; m = 200;
theta = pi/7;
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
while true
t = t + pi/240;
d = 5.*cos(sqrt(k.^2 + e.^2) - t + mod(i, 2));
XY = [k + k.*d./6.*sin(d + e./3 + t);
90 + e.*d - e./d.*2.*cos(d + t)];
for j = 1:14
XY = R*XY;
SHdl{j}.XData = XY(1,:) + m;
SHdl{j}.YData = XY(2,:) + m;
end
drawnow;
end
Having tackled a given problem is not the end of the game, and the fun is far from over. Thanks to the test suite in place, we can continue tweaking our solutions ("refactoring") so that it still passes the tests while improving its ranking with regard to "Cody size".
Although reducing the Cody size does not necessarily mean a solution will perform more efficiently nor be more readable (quite the contrary, actually…), it is a fun way to delve into the intricacies of MATLAB code and maybe win a Cody Leader badge!
I am not talking about just basic hacks. The size constraint urges us to find an “out-of-the box” way of solving a problem, a way of thinking creatively, of finding other means to achieve a desired computation result, that uses less variables, that is less cumbersome, or that is more refined.
The past few days have taught me several useful tricks that I would like to share with anyone wishing to reduce the solution size of their Cody submission. Happy to learn about other tricks you may know of, please share!
- Use this File Exchange submission to get the size of your solution: https://fr.mathworks.com/matlabcentral/fileexchange/34754-calculate-size
- Use existing MATLAB functions that may already perform the desired calculations but that you might have overlooked (as I did with histcount and digraph).
- Use vectorization amply. It’s what make the MATLAB language so concise and elegant!
- Before creating a matrix of replicated values, check if your operation requires it. See Compatible Array Sizes for Basic Operations. For example, you can write x == y with x being a column vector and y a row vector, thereby obtaining a 2 by 2-matrix.
- Try writing out for loops instead of vectorization. Sometimes it’s actually smaller from a Cody point of view.
- Avoid nested functions and subfunctions. Try anonymous functions if used in several places. (By all means, DO write nested functions and subfunctions in real life!)
- Avoid variable assignments. If you declare variables, look for ones you can use in multiples places for multiple purposes. If you have a variable used only in one place, replace it with its expression where you need it. (Do not do this in real life!)
- Instead of variable assignments, write hardcoded constants. (Do not do this in real life!)
- Instead of indexed assignments, look for a way to use multiplying or exponentiating with logical indexes. (For example, compare Solution 14896297 and Solution 14897502 for Problem 61069. Clueless - Lord Ned in the Game Room with the Technical Computing Language).
- Replace x == 0 with ~x if x is a numeric scalar or vector or matrix that doesn’t contain any NaN (the latter is smaller in size by 1)
- Instead of x == -1, see if x < 0 works (smaller in size by 1).
- Instead of [1 2], write 1:2 (smaller in size by 1).
- “sum(sum(x))” is actually smaller than “sum(x, 1:2)”
- Instead of initialising a matrix of 2s with 2 * ones(m,n), write repmat(2,m,n) (smaller in size by 1).
- If you have a matrix x and wish to initialize a matrix of 1s, instead of ones(size(x)), write x .^ 0 (works as long as x doesn’t contain any NaN) (smaller in size by 2).
- Unfortunately, x ^-Inf doesn’t provide any reduction compared to zeros(size(x)), and it doesn’t work when x contains 0 or 1.
- Beware of Operator Precedence and avoid unnecessary parenthesis (special thanks to @Stefan Abendroth for bringing that to my attention ;)) :
- Instead of x * (y .^ 2), write x * y .^2 (smaller in size by 1).
- Instead of x > (y == z), write y == z < x (smaller in size by 1).
18. Ask help from other solvers: ideas coming from a new pair of eyes can bring unexpected improvements!
That’s all I can see for now.
Having applied all those tips made me realise that a concise yet powerful code, devoid of the superfluous, also has a beauty of its own kind that we can appreciate.
Yet we do not arrive at those minimalist solutions directly, but through several iterations, thanks to the presence of tests that allow us to not worry about breaking anything, and therefore try out sometimes audacious ideas.
That's why I think the main interest lies in that it prompts to think of our solutions differently, thereby opening ways to better understand the problem statement at hand and the inner workings of the possible solutions.
Hope you’ll find it fun and useful!
P.S.: Solvers, please come help us reduce even more the size of the leading solution for Problem 61069. Clueless - Lord Ned in the Game Room with the Technical Computing Language!
The challenge:
You are given a string of lowercase letters 'a' to 'z'.
Each character represents a base-26 digit:
- 'a' = 0
1. Understand the Base-26 Conversion Process:
Let the input be s = 'aloha'.
Convert each character to a digit:
digits = double(s) - double('a');
This works because:
double('a') = 97
double('b') = 98
So:
double('a') - 97 = 0
double('l') - 97 = 11
double('o') - 97 = 14
double('h') - 97 = 7
double('a') - 97 = 0
Now you have:
[0 11 14 7 0]
2. Interpret as Base-26:
For a number with n digits:
d1 d2 d3 ... dn
Value = d1*26^(n-1) + d2*26^(n-2) + ... + dn*26^0
So for 'aloha' (5 chars):
0*26^4 + 11*26^3 + 14*26^2 + 7*26^1 + 0*26^0
MATLAB can compute this automatically.
3. Avoid loops — Use MATLAB vectorization:
You can compute the weighted sum using dot
digits = double(s) - 'a';
powers = 26.^(length(s)-1:-1:0);
result = dot(digits, powers);
This is clean, short, and vectorized.
4.Test with the examples:
char2num26('funfunfun')
→ 1208856210289
char2num26('matlab')
→ 142917893
char2num26('nasa')
→ 228956
Hi Creative Coders!
I've been working my way through the problem set (and enjoying all the references), but the final puzzle has me stumped. I've managed to get 16/20 of the test cases to the right answer, and the rest remain very unsolvable for my current algorithm. I know there's some kind of leap of logic I'm missing, but can't figure out quite what it is. Can any of you help?
What I've Done So Far
My current algorithm looks a bit like this. The code is doing its best to embody spaghetti at the moment, so I'll refrain from posting the whole thing to spare you all from trying to follow my thought processes.
Step 1: Go through all the turns and fill out tables of 'definitely', 'maybe', and 'clue' based on the information provided in a single run through the turns. This means that the case mentioned in the problem where information from future turns affecting previous turns does not matter yet. 'Definitely' information is for when I know a card must belong to a certain player (or to no-one). 'Maybe' starts off with all players in all cells, and when a player is found to not be able to have a card, their number is removed from the cell. Think of Sudoku notes where someone has helpfully gone through the grid and put every single possible number in each cell. 'Clue' contains information about which cards players were hinted about.
Example from test case 1:
definitelyTable =
6×3 table
G1 G2 G3
____________ ____________ ____________
{[ 0]} {0×0 double} {0×0 double}
{0×0 double} {[ -1]} {[ 1]}
{0×0 double} {[ 6]} {[ 0]}
{[ 3]} {[ 4]} {0×0 double}
{0×0 double} {[ 0]} {0×0 double}
{[ 5]} {[ 3]} {0×0 double}
maybeTable =
6×3 table
G1 G2 G3
_________ _______ _______
{[ 0]} {[2 5]} {[1 2]}
{[ 4]} {[ 0]} {[ 0]}
{[2 4 6]} {[ 0]} {[ 0]}
{[ 0]} {[ 0]} {[1 4]}
{[ 1 4]} {[ 0]} {[ 1]}
{[ 0]} {[ 0]} {[2 4]}
clueTable =
6×3 table
G1 G2 G3
____________ ____________ ____________
{0×0 double} {[ 5 6]} {[ 2 4]}
{[ 4 6]} {[ 4 6]} {0×0 double}
{[ 2 6]} {[ 5 6]} {0×0 double}
{0×0 double} {[ 4]} {[ 4 5 6]}
{[ 4]} {0×0 double} {[ 1 4 6]}
{[ 2 5]} {0×0 double} {[ 2 4 5 6]}
(-1 indicates the card is in the envelope. 0 indicates the card is commonly known.)
Step 2: While a solution has not yet been found, loop through all the turns again. This is the part where future turn info can now be fed back into previous turns, and where my sticky test cases loop forever. I've coded up each of the implementation tips from the problem statement for this stage.
Where It All Comes Undone
The problem is, for certain test cases (e.g., case 5), I reach a point where going through all turns yields no new information. I either end up with an either-or scenario, where the potential culprit card is one of two choices, or with so little information it doesn't look like there is anywhere left to turn.
I solved some of the either-or cases by adding a snippet that assumes one of the values and then tries to solve the problem based on that new information. If it can't solve it, then it tries the other option and goes round again. Unfortunately, however, this results in an infinite flip-flop for some cases as neither guess resolves the puzzle.
Essentially guessing the solution and following through to a logical inconsistency for however many combinations of players and cards sounds a) inefficient and b) not the way this was intended to be solved. Does anyone have any hints that might get me on track to solve this mystery?
Thank you to everyone who attended the workshop A Hands-On Introduction to Reinforcement Learning! Now that you all have had some time to digest the content, I wanted to create a thread where you could ask any further questions, share insights, or discuss how you're applying the concepts to your work. Please feel free to share your thoughts in the thread below! And for your reference, I have attached a PDF version of the workshop presentation slides to this post.
If you were interested in joining the RL workshop but weren't able to attend live (maybe because you were in one of our other fantastic workshops instead!), you can find the workshop hands-on material in this shared MATLAB Drive folder. To access the exercises, simply download the MATLAB Project Archive (.mlproj) file or copy it to your MATLAB Drive, extract the files, and open the project (.prj). Each exercise has its own live script (.mlx file) which contains all the instructions and individual steps for each exercise. Happy (reinforcement) learning!
Is it possible to get the slides from the Hands-On-Workshops?
I can't find them in the proceedings. I'm particularly interested in the Reinforcement Learning workshop, but unfortunately I couldn't participate.
Thanks in advance!
To track the current leader after each match, you can use cumulative scores. First, calculate the cumulative sum for each player across the matches. Then, after eaayer with the highest score.
Hint: Use cumsum(S, 1) to get cumulative scores along the rows (matches). Loop through each row to keep track of the leader. If multiple players tie, pick the lowest index.
Example:
If S = [5 3 4; 2 6 2; 3 5 7], after match 3, the cumulative scores are [10 14 13]. Player 2 leads with 14 hilbs.
This method keeps your code clean and avoids repeatedly summing rows.
Great material, examples and skillfully guided. And, of course, very useful.
Thanks!
Congratulations to all the Cool Coders who have completed the problem set. I hope you weren't too cool to enjoy the silliness I put into the problems.
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.)
Fittingly for a Creative Coder, @Vasilis Bellos clearly enjoyed the silliness I put into the problems. 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.)
Extracting the digits of a number will be useful to solve many Cody problems.
Instead of iteratively dividing by 10 and taking the remainder, the digits of a number can be easily extracted using String operations.
%Extract the digits of N
N = 1234;
d = num2str(N)-'0';
d =
1 2 3 4
Instead of looping with if-statements, use logical indexing:
A(A < 0) = 0;
One line, no loops, full clarity.
Whenever a problem repeats in cycles (like indexing or angles), mod() keeps your logic clean:
idx = mod(i-1, n) + 1;
No if-else chaos!
Hi, what’s the best way to learn MATLAB, Simulink, and Simscape? Do you recommend a learning path? I work in the Electrical & Electronics area for automotive systems.
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.
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.
Don’t miss out on two incredible keynotes that will shape the future of engineering and innovation:
1️⃣What’s New in MATLAB and Simulink in 2025
Get an inside look at the latest features designed to supercharge your workflows:
- A redesigned MATLAB desktop with customizable sidebars, light/dark themes, and new panels for coding tasks
- MATLAB Copilot – your AI-powered assistant for learning, idea generation, and productivity
- Simulink upgrades, including an enhanced quick insert tool, auto-straightening signal lines, and new methods for Python integration
- New options to deploy AI models on Qualcomm and Infineon hardware
2️⃣Accelerating Software-Defined Vehicles with Model-Based Design
See how MathWorks + NXP are transforming embedded system development for next-gen vehicles:
- Vehicle electrification example powered by MATLAB, Simulink, and NXP tools
- End-to-end workflow: modeling → validation → code generation → hardware deployment → real-time cloud monitoring
📅 When: November 13
💡Why Join? Stay ahead with cutting-edge tools, workflows, and insights from industry leaders.
The Cody Contest 2025 is underway, and it includes a super creative problem group which many of us have found fascinating. The central theme of the problems, expertly curated by @Matt Tearle, humorously revolves around the whims of the capricious dictator Lord Ned, as he goes out of his way to complicate the lives of his subjects and visitors alike. We cannot judge whether or not there's any truth to the rumors behind all the inside jokes, but it's obvious that the team had a lot of fun creating these; and we had even more fun solving them.
Today I want to showcase a way of graphically solving and visualizing one of those problems which I found very elegant, The Bridges of Nedsburg.
To briefly reiterate the problem, the number of islands and the arrangement of bridges of the city of Nedsburg are constantly changing. Lord Ned has decided to take advantage of this by charging visitors with an increasingly expensive n-bridge pass which allows them to cross up to n bridges in one journey. Given the Connectivity Matrix C, we are tasked with calculating the minimum n needed so that there is a path from every island to every other island in n steps or fewer.
Matt kindly provided us with some useful bit of math in the description detailing how to calculate the way to get from one island to another in an number of m steps. However, he has also hidden an alternative path to the solution in plain sight, in one of the graphs he provided. This involves the extremely useful and versatile class digraph, representing directed graphs, which have directional edges connecting the nodes. Here's some further great documentation and other cool resources on the topic for those who are interested in learning more about it:
Let's start using this class to explore a graphical solution to Lord Ned's conundrum. We will use the unit tests included in the problem to visualize the solution. We can retrieve the connectivity matrix for each case using the following function:
function C = getConnectivityMatrix(unit_test)
% Number of islands and bridge arrangement
switch unit_test
case 1
m = 3; idx = [3;4;8];
case 2
m = 3; idx = [3;4;7;8];
case 3
m = 4; idx = [2;7;8;10;13];
case 4
m = 4; idx = [4;5;7;8;9;14];
case 5
m = 5; idx = [5;8;11;12;14;18;22;23];
case 6
m = 5; idx = [2;5;8;14;20;21;24];
case 7
m = 6; idx = [3;4;7;11;18;23;24;26;30;32];
case 8
m = 6; idx = [3;11;12;13;18;19;28;32];
case 9
m = 7; idx = [3;4;6;8;13;14;20;21;23;31;36;47];
case 10
m = 7; idx = [4;11;13;14;19;22;23;26;28;30;34;35;37;38;45];
case 11
m = 8; idx = [2;4;5;6;8;12;13;17;27;39;44;48;54;58;60;62];
case 12
m = 8; idx = [3;9;12;20;24;29;30;31;33;44;48;50;53;54;58];
case 13
m = 9; idx = [8;9;10;14;15;22;25;26;29;33;36;42;44;47;48;50;53;54;55;67;80];
case 14
m = 9; idx = [8;10;22;32;37;40;43;45;47;53;56;57;62;64;69;70;73;77;79];
case 15
m = 10; idx = [2;5;8;13;16;20;24;27;28;36;43;49;53;62;71;75;77;83;86;87;95];
case 16
m = 10; idx = [4;9;14;21;22;35;37;38;44;47;50;51;53;55;59;61;63;66;69;76;77;84;85;86;90;97];
end
C = zeros(m);
C(idx) = 1;
end
The case in the example refers to unit test case 2.
unit_test = 2;
C = getConnectivityMatrix(unit_test);
disp(C)
D = digraph(C);
figure
p = plot(D,'LineWidth',1.5,'ArrowSize',10);
This is the same as the graph provided in the example. Another very useful method of digraph is shortestpath. This allows us to calculate the path and distance from one single node to another. For example:
% Path and distance from node 1 to node 2
[path12,dist12] = shortestpath(D,1,2);
fprintf('The shortest path from island %d to island %d is: %s. The minimum number of steps is: n = %d\n', 1, 2, join(string(path12), ' -> '),dist12)
% Path and distance from node 2 to node 1
[path21,dist21] = shortestpath(D,2,1);
fprintf('The shortest path from island %d to island %d is: %s. The minimum number of steps is: n = %d\n', 2, 1, join(string(path21), ' -> '),dist21)
figure
p = plot(D,'LineWidth',1.5,'ArrowSize',10);
highlight(p,path12,'EdgeColor','r','NodeColor','r','LineWidth',2)
highlight(p,path21,'EdgeColor',[0 0.8 0],'LineWidth',2)
But that's not all! digraph can also provide us with a matrix of the distances d, i.e. the steps needed to travel from island i to island j, where i and j are the rows and columns of d respectively. This is accomplished by using its distances method. The distance matrix can be visualized as:
d = distances(D);
figure
% Using pcolor w/ appending matrix workaround for convenience
pcolor([d,d(:,end);d(end,:),d(end,end)])
% Alternatively you can use imagesc(d), but you'll have to recreate the grid manually
axis square
set(gca,'YDir','reverse','XTick',[],'YTick',[])
[X,Y] = meshgrid(1:height(d));
text(X(:)+0.5,Y(:)+0.5,string(d(:)),'FontSize',11)
colormap(interp1(linspace(0,1,4), [1 1 1; 0.7 0.9 1; 0.6 0.7 1; 1 0.3 0.3], linspace(0,1,8)))
clim([-0.5 7+0.5])
This confirms what we saw before, i.e. you need 1 step to go from island 1 to island 2, but 2 steps for vice versa. It also confirms that the minimum number of steps n that you need to buy the pass for is 2 (which also occurs for traveling from island 3 to island 2). As it's not the point of the post to give the full solution to the problem but rather present the graphical way of visualizing it I will not include the code of how to calculate this, but I'm sure that by now it's reduced to a trivial problem which you have already figured out how to solve.
That being said, now that we have the distance matrix, let's continue with the visualizations. First, let's plot the corresponding paths for each of these combinations:
figure
tiledlayout(size(C,1),size(C,2),'TileSpacing','tight','Padding','tight');
for i = 1:size(C,1)
for j = 1:size(C,2)
nexttile
p = plot(D,'ArrowSize',10);
highlight(p,shortestpath(D,i,j),'EdgeColor','r','NodeColor','r','LineWidth',2)
lims = axis;
text(lims(1)+diff(lims(1:2))*0.05,lims(3)+diff(lims(3:4))*0.9,sprintf('n = %d',d(i,j)))
end
end
This allows us to go from the distance matrix to visualizing the paths and number of steps for each corresponding case. Things are rather simple for this 3-island example case, but evil Lord Ned is just getting started. Let's now try to solve the problem for all provided unit test cases:
% Cell array of connectivity matrices
C = arrayfun(@getConnectivityMatrix,1:16,'UniformOutput',false);
% Cell array of corresponding digraph objects
D = cellfun(@digraph,C,'UniformOutput',false);
% Cell array of corresponding distance matrices
d = cellfun(@distances,D,'UniformOutput',false);
% id of solutions: Provided as is to avoid handing out the code to the full solution
id = [2, 2, 9, 3, 4, 6, 16, 4, 44, 43, 33, 34, 7, 18, 39, 2];
First, let's plot the distance matrix for each case:
figure
tiledlayout('flow','TileSpacing','compact','Padding','compact');
% Vary this to plot different combinations of cases
plot_cases = 1:numel(C);
for i = plot_cases
nexttile
pcolor([d{i},d{i}(:,end);d{i}(end,:),d{i}(end,end)])
axis square
set(gca,'YDir','reverse','XTick',[],'YTick',[])
title(sprintf('Case %d',i),'FontWeight','normal','FontSize',8)
end
c = colorbar('Ticks',0:7,'TickLength',0,'Limits',[-0.5 7+0.5],'FontSize',8);
c.Layout.Tile = 'East';
c.Label.String = 'Number of Steps';
c.Label.FontSize = 8;
colormap(interp1(linspace(0,1,4), [1 1 1; 0.7 0.9 1; 0.6 0.7 1; 1 0.3 0.3], linspace(0,1,8)))
clim(findobj(gcf,'type','axes'),[-0.5 7+0.5])
We immediately notice some inconsistencies, perhaps to be expected of the eccentric and cunning dictator. Things are pretty simple for the configurations with a small number of islands, but the minimum number of steps n can increase sharply and disproportionally to the additional number of islands. Cases 8 and 9 specifically have a particularly large n (relative to their grid dimensions), and case 14 has the largest n, almost double that of case 16 despite the fact that the latter has one extra island.
To visualize how this is possible, let's plot the path corresponding to the largest n for each case (though note that there might be multiple possible paths for each case):
figure
tiledlayout('flow','TileSpacing','tight','Padding','tight');
for i = plot_cases
nexttile
% Changing the layout to circular so we can better visualize the paths
p = plot(D{i},'ArrowSize',10,'Layout','Circle');
% Alternatively we could use the XData and YData properties if the positions of the islands were provided
axis([-1.5 1.5 -1.5 1.75])
[row,col] = ind2sub(size(d{i}),id(i));
highlight(p,shortestpath(D{i},row,col),'EdgeColor','r','NodeColor','r','LineWidth',2)
lims = axis;
text(lims(1)+diff(lims(1:2))*0.05,lims(3)+diff(lims(3:4))*0.9,sprintf('n = %d',d{i}(row,col)))
end
And busted! Unraveled! Exposed! Lord Ned has clearly been taking advantages of the tectonic forces by instructing his corrupt civil engineer lackeys to design the bridges to purposely force the visitors to go around in circles in order to drain them of their precious savings. In particular, for cases 8 and 9, he would have them go through every single island just to get from one island to another, whereas for case 14 they would have to visit 8 of the 9 islands just to get to their destination. If that's not diabolical then I don't know what is!
Ned jokes aside, I hope you enjoyed this contest just as much as I did, and that you found this article useful. I look forward to seeing more creative problems and solutions in the future.
Pure Matlab
82%
Simulink
18%
11 votes
