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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 everyone!
I’m Kishen Mahadevan, Senior Product Manager at MathWorks, where I focus on controls and deep learning. I’m excited to be speaking at MATLAB EXPO this year!
In one of my sessions, I’ll share how AI-based reduced order models (ROMs) are transforming engineering workflows—using battery fast charging as an example—making it easier to reuse high-fidelity models for real-time control and deployment.
I’d love to have you join the conversation at the EXPO and right here in the community!
Feel free to drop any questions or thoughts ahead of the event.
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.
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
Hi Everyone,
Require some guidance and pointers on model ee_pmlsm_drive please.
Its regarding a PMSM linear machine model, with a cascade (position and speed) outer loop and a current controller inner loop (Id and Iq currents).
The current Simulink model uses a low voltage DC supply (48v) and uses a step input to the system. My system uses a HV source (400v) and the input is sinusoidal position, with an operating frequency range of 0 to 20hz.
I have used the original model, re-created my own (to learn Simulink/Simscape) with a HV battery source (400v) and used machine parameters that match the application (peer reviewed publication).
As my power electronics background is limited, my background is mechanical, i am unable to tune the inner and outer loops, and am unsure in what order to tune. My project is to use the linear PMSM to drive (motoring) a linear piston for combustion and also use the linear piston to drive the PMSM (generating).
Using the built in PID tuner for the outer speed and position loops i am encountering an error (plant cannot be linearised). I am using a simple 1 hz sine wave to simulate the reference position set-point. With all PID values set to default i have no dq currents, however i have idq ref from the outer velocity controller loop.
Any help with resources and guidance as to how to tune the loops for my updated parameters on this model would be great.
Experimenting with Agentic AI
44%
I am an AI skeptic
0%
AI is banned at work
11%
I am happy with Conversational AI
44%
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.
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);
Pure Matlab
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Simulink
18%
11 votes
Don’t miss out on two incredible keynotes that will shape the future of engineering and innovation:
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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! 🚀
In the FAQs, I saw the procedure to download the "mobile background", is the the same thing as an award? If yes, good, else how can we get an award and what are the available ones?
iaabdulhameed@knu.ac.kr
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
It’s exciting to dive into a new dataset full of unfamiliar variables but it can also be overwhelming if you’re not sure where to start. Recently, I discovered some new interactive features in MATLAB live scripts that make it much easier to get an overview of your data. With just a few clicks, you can display sparklines and summary statistics using table variables, sort and filter variables, and even have MATLAB generate the corresponding code for reproducibility.
The Graphics and App Building blog published an article that walks through these features showing how to explore, clean, and analyze data—all without writing any code.
If you’re interested in streamlining your exploratory data analysis or want to see what’s new in live scripts, you might find it helpful:
If you’ve tried these features or have your own tips for quick data exploration in MATLAB, I’d love to hear your thoughts!
Submit your questions about this work in the comment section below.
isequal() is your best friend for Cody! It compares arrays perfectly without rounding errors — much safer than == for matrix outputs.
I am Prof Ansar Interested in coding challenge taker inmatlab
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