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! 🚀
Automating Parameter Identifiability Analysis in SimBiology
Is it possible to develop a MATLAB Live Script that automates a series of SimBiology model fits to obtain likelihood profiles? The goal is to fit a kinetic model to experimental data while systematically fixing the value of one kinetic constant (e.g., k1) and leaving the others unrestricted.
The script would perform the following:
Use a pre-configured SimBiology project where the best fit to the experimental data has already been established (including dependent/independent variables, covariates, the error model, and optimization settings).
Iterate over a defined sequence of fixed values for a chosen parameter.
For each fixed value, run the estimation to optimize the remaining parameters.
Record the resulting Sum of Squared Errors (SSE) for each run.
The final output would be a likelihood profile—a plot of SSE versus the fixed parameter value (e.g., k1)—to assess the practical identifiability of each model parameter.
I want to observe the time (Tmax) to reach maximum drug concentration (Cmax) in my model. I have set up the OBSERVABLES as follows (figure1): Cmax = max(Blood.lL15); Tmax_LT = time(Conc_lL15_LT_nm == max(Conc_lL15_LT_nm)); Tmax_Tm = time(Conc_lL15_Tumor_nm == max(Conc_lL15_Tumor_nm)); After running the Sobol indices program for global sensitivity analysis, with inputs being some parameters and their ranges, the output for Cmax works, but there are some prompts, as shown in figure2. Additionally, when outputting Tmax, the program does not run successfully and reports some errors, as shown in figure2. How can I resolve the errors when outputting Tmax?
Hi All,
I'm currently verifying a global sensitivity analysis done in SimBiology and I'm a touch confused. This analysis was run with every parameter and compartment volume in the model. To my understanding the fraction of unexplained variance is 1 - the sum of the first order variances, therefore if the model dynamics are dominated by interparameter effects you might see a higher fraction of unexplained variance. In this analysis however, as the attached figure shows (with input at t=20 minutes), the most sensitive four parameters seem to sum, in first order sensitivities to roughly one at each time point and the total order sensitivies appear nearly identical. So how is the fraction of unexplained variance near one?
Thank you for your help!
Hi to everyone!
To simplify the explanation and the problem, I simulated the kinetics of an irreversible first-order reaction, A -> B. I implemented it in two independent compartments, R and P. I simulated the effect of a dilution in R by doubling at t= 0,1 the R volume. I programmed in P that, at t = 0.1, the instantaneous concentration of A and B would be reduced by half. I am sending an attach with the implementation of these simulations in the Simbiology interface.
When the simulations of the two compartments are plotted, it can be seen that the responses are not equal. That is, from t = 0.1 s, the reaction follow an exponential function in R with half of the initial amplitude and half of the initial value of k1. That is, the relaxation time is doubled. Meanwhile, in P, from t = 0.1, the reaction follows exponential kinetics with half the amplitude value but maintaining the initial value of k = 10. Without a doubt, the correct simulation is the latter (compartment P) where only the effect is observed in the amplitude and not in the relaxation time. Could you tell me what the error is that makes these kinetics that should be equal not be?
Thank you in advance!
Luis B.
Hi All,
I've been producing a QSP model of glucose homeostasis for a while now for my PhD project, recently I've been able to expand it to larger time series, i.e. 2 days of data rather than a singular injection or a singular meal. My problem is as follows: If I put 75g of glucose into my stomach glucose species any later than (exactly) 8.5 hours I get an integration tolerance error. Curiosly, I can put 25g of glucose in at any time up to 15.9 hours, then any later an error. I have disabled all connections to my glucose absorption chain, i.e. stomach -> duodenum -> jenenum -> ileum -> removal, to isolate the cause of this. I had initially thought it may be because I mechanistically model liver glycogen and that does deplete over time, but I've tested enough to show that that does nothing. My next test is to isolate the glucose absorption chain into a seperate model and see if the issue persists but I'm completely baffled!
These are the equations, to my eye there's no reason why there would be such a sharp glucose quantity/time dependence, they all begin at a value of 0:
d(Gs)/dt = -(kw*(1-Gd^14/(Igd^14+Gd^14))*Gs) #Stomach glucose
d(Gd)/dt = (kw*(1-Gd^14/(Igd^14+Gd^14))*Gs) - (kdj*Gd) #Duodenal Glucose
d(Gj)/dt = (kdj*Gd) - (kji*Gj) #Jejunal Glucose
d(Gi)/dt = (kji*Gj) - (kic*Gi) #Ileal Glucose
(The sigmoidicity of gastric emptying slowing term (^14) was parameterised off of paracetamol absorption data and appears to be correct!)
Thank you for your help, best regards,
Dan
Pre-Edit: I changed the run time to 30 hours and now I can't use the 75g input any later than 7.9 hours not 8.5 hours anymore!
Edit: This is how it appears at all times prior to it failing for 75g:
Join us for a hands-on workshop focussing on Global Sensitivity Analysis (GSA) for QSP using SimBiology. This complimentary session aims to give you theoretical background and hands-on experience with GSA methods.
🗓️ Event Details: June 25, 1-5pm CEST @PAGE 2024, Rome
🔖 Registration: Free, but space is limited. Early registration is recommended to ensure your place in this workshop.
For more details and to register, please visit https://www.mathworks.com/company/events/seminars/global-sensitivity-analysis-with-simbiology-4361208.html
Hi All,
I'm trying to produce some nice figures from the graphical user interface and have a set of local sensitivity analyses that I'd like to combine.
I have two inputs that vary the sensitivies of my system and would like to plot them on top of each other like:
Is there either a way to do this natively in simbiology (when you try and use 'keep results from each run' it plots them both as a time series) or to export the sensitivity data to the normal matlab programatic UI where I can combine them by hand?
Thank you for your help!
Hi All,
I'm trying to plot my observables in the analyzer and running into far more problems than expected. I have two species in my model that are invovled, blood.Insulin (pM) and blood.Glucose (mM), all I want to do is plot the ratio of these two (blood.Insulin/blood.Glucose (dimensionless)) along with my other species in Model Simulation, to compare it to the same ratio from my data.
First, there doesn't seem to be a way to directly add an observable to the logged states in 'Simulate Model', so I've tried to used 'Calculate Observable' based on the data from my last run (IVGTT.LastRun.results) but it says that units are required when unit conversion is enabled, but it should be dimensionless!
My next idea would be to make a non-constant parameter with a repeated assignment, but I feel like I should be able to do this without resorting to that?
Any help or ideas would be appreciated. Thank you, best regards,
Dan
