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I started with Cody today and I found the file calculateSize.m https://de.mathworks.com/matlabcentral/fileexchange/34754-calculate-size which should determine the size of my solution. Unfortunately I do not get it to run. Can someone explain this to me or maybe explain how exactly the size of the solution is calculated?
Something seems to either have changed with webread/urlread behavior with R2016a or the Cody site has changed its authentication after it's September 2016 upgrade. Now, when ever I make a call using either webread (with weboptions) urlread using "My Cody" url, I get the html that corresponds to an unauthenticated user.
I've tried passing credentials using weboption and also cached the credentials using the built in Matlab browser - neither helped in seeing the authenticated version of this page.
Any help in understanding where the problem/solution lies would be helpful. Easily reproduced by running...
URL = 'https://www.mathworks.com/matlabcentral/cody/players/1261697-gregory/solved';
[Contents Status] = urlread(URL);
Compare this to the source pulled down when authenticated with Chrome (for example, search for the work "two"). FYI: this has worked for a couple of years up until now...
Inspired by Chad Greene's " MATLAB jokes or puns " thread, and in celebration of 15 years of the MathWorks Community site, does anyone out there want to share their poetic creativity? Limericks, haiku, sonnets... Go!
And to start off, my (slightly off-topic) submission on Chad's thread:
There was an old math guy called Cleve
who, while teaching, a pipe-dream conceived:
of a language so clean
you can say what you mean!
From our suffering we've all been relieved.
Hi, I am uploading the answers to cody problems. Although most of answers are correct but my size is bigger than the best answer. How can I view the best answer?
Apologies for putting this question here, but I'm not sure there's a specific forum for Cody-related questions.
I recently noticed that a new badge for "Magic Numbers Master" had been created and awarded. When I entered my profile to check that out, I noticed that I had received that badge, but lost the badge for "Cody Challenge Master." I thought that maybe my solution had been messed up because of a change in one of the test suites that my solution missed, as this has happened a number of times on other problems. According to the text underneath the badge icon, I've still solved all 96 of the questions in the Cody Challenge. All of the problems listed as part of that challenge are still marked as "Solved." Has anyone else run into this issue?
Are there any good Matlab jokes? I don't mean why or any other Easter eggs, I mean good jokes involving Matlab. Actually, that bar may be a bit too high. Any jokes, good or bad, let's hear 'em.
I created a solution for Cody Problem 1745 (Get me!) that is very simple and works perfectly on my computer; but in Cody, it results in the error:
Error: Undefined function or variable 'me'.
The test code is:
get = me();
y = rand(1,12345);
!rm now.m
!rm assert.m
assert(get == now)
My solution is:
classdef me
methods
function tf = eq(obj,~)
tf = true;
end
end
end
Why doesn't Cody accept this?
(P. S.) And why doesn't it accept function definitions like
function justDoIt
(which are appropriate for some problems and accepted by MATLAB) but requires at least one input and output?
if i want write fewer codes in cody,what should I pay attention to?
Can anyone explain how Cody calculates the size of a program?
I read the help, but couldn't understand. :(
Why cody does not allow 'eval', 'evalc','feval' etc functions in the solutions? What is alternative to these functions?
e.g. I need to use:
if f='3*x'
x=0:10;
y=eval(vectorise(f))
so that values of y can be calculated as 3.*x
bt cody does not let us to use functions like eval
What is the alternative?
I have been stuck in the Cody question: Problem 31. Remove all the words that end with "ain"
for the whole afternoon. My codes below passed the first two tests but failed in the third one, because my codes cannot separate "" and ain. Anyone could help?
function s2 = remAin(s1)
s1_cell = strread(s1, '%s');
[s1_cell{find(not(cellfun('isempty', regexp(s1_cell, '\w*ain\>'))) == 1)}] = deal(' ');
s2 = strjoin(s1_cell');
s2 = strrep(s2, sprintf('%c%c%c', 32,32,32), sprintf('%c%c', 32,32));
if s2(end) == ' ' && s2(end-1) == ' '
s2(end) = [];
end
end
Salam,
While I was playing with Cody, I found one interesting question that asked to write MATLAB function to create times-table (can be found here: http://bit.ly/1cWZGGM ). The question itself was easy, but I had problems to figure out how could someone solve this problem with compact code of size 10 !!
After I've checked the leading answer, it seems to be written in Chinese not in MATLAB :)
Here is the answer:
function m = timestables(n)
regexp '' '(?@A=repmat([1:n]'',1,n);m=A.*A'';)'
end
Could anyone translate this to me so that I can use it in my future attempts ? :P
Regards.
In a previous Q & A, Jan Simon pointed to Cody: Sum 1:2^n. The current leading solution to that problem has node-count (or more simply, "length") 10. Apparently, 10 is the minimal length (per the official length-function on File Exchange) of any function taking input & generating output:
function y = test_cody_solution(x)
y = x;
end
Per Cody instruction examples, additional computation within a function definition increases the solution length. For example, both of the following functions have length 12:
function y = test_cody_solution(x)
y = [x];
end
function y = test_cody_solution(x)
y = x+1;
end
My question is: what kinds of ninja-style coding idioms even exist in MATLAB which actually perform definite computation but at the same time do not increase the node-count above 10? I'm not able to imagine what could be going on in order for someone to solve a given non-trivial Cody puzzle in length 10 or 11? IOW, without respect to any particular Cody problem, could someone please give an example of a non-trivial function which somehow comes in at or just above the absolute lower bound? Any explanation of the magic would be appreciated as well.
Thanks, Brad
Some of Matlab's toolbox functions are affected by magic strings or magic numbers, which are strings or numbers with a deeper meaning besides the normal value. Both are considered as bad programming patters, because they provoke confusions, when the magic keys appear with the normal meaning by accident. See http://en.wikipedia.org/wiki/Anti-pattern
Example 1:
clear('myVariable')
clear('variables')
While the 1st clears the variable myVariable, the later clears all variables. Here 'variables' has a meta-meaning. The problem appears, when 'variables' is an existing variable:
a = 1;
variables = 2;
clear('variables')
disp(a) % >> 1
Only variables is cleared, which cannot be understood directly when its definition is 1000 lines before.
Example 2:
uicontrol('String', 'default')
This creates a button with the empty string '' instead of the expected 'default', because this is the magic string to invoke the default value get(0, 'DefaultUIControlString'). The same concerns properties of other graphic objects also, e.g. the 'name' property of figure or the string of uimenu. There is a workaround which allows the user to display 'default': Simply use '\default'. Unfortunately this is doubled magic, because in consequence it is impossible to display the string '\default'. Obviously a bad idea.
Example 3:
Graphic handles are doubles (although gobject of the new R2013a seems, like this is subject to changes? [EDITED: Yes, it changed with HG2 in R2014a]). But then a handle can be confused with data:
a = axes; % e.g. 0.0048828125
plot(a, 2, '+')
But you cannot draw the point [0.0048828125, 2] by this way, because the 1st input is considered as handle of the parent. Here all possible values of handles are magic. Collisions are very unlikely, but there is no way to avoid them reliably - as long as handles have the type double.
Question:
Which functions are concerned by magic values? What are the pitfalls and workarounds?
function b = most_change(a)
a(:,1)=a(:,1)*0.25;
a(:,2)=a(:,2)*0.1;
a(:,3)=a(:,3)*0.05;
a(:,4)=a(:,4)*0.01;
d=sum(a,2);
c=max(d);
for i=1:length(d)
if d(i)==c
b=i;
end
end
i got wa please explain idont understand
How do I can delete my account?
I am wondering what others use for those little short-cuts or niceties in MATLAB. I have in mind something you wrote or something somebody else wrote or an underused MW function.
Here are my two favorites.
This is a simple script I use. Here is the entire contents of CLC.m (yes, it is capitalized):
clear all,close all,clc
Very simple, but I use it all the time. Here is another one I use so often that I forget not every machine has it (though every machine should, IMO):
Here is an underused MW function that I occasionally employ when working on someone else's machine. The usual response is, "Wait, what did you just do?"
home
What are some of yours?
Given a tic tac toe board:
1 represents X
0 represents empty.
-1 represents O
It is X's move. If there is an immediate win possibility, choose a square for an immediate win. Otherwise return 0.
Return absolute index of the square of choice. If multiple square are valid, return them in order.
Example:
Input a = [ 1 0 1
-1 1 0
0 -1 -1]
Output wins is [4 8]
Can anyone explain it in detail?
I'm confused with the sentence I marked ans bold style.
Thanks a lot~~~
I've written a valid answer to the last Cody problem, but it is not even close to the best answer. I have no idea how they made this short answer. To unlock it I need to solve another Cody question, but there are none left... :(
Anybody know how to unlock the last question?
Hi,
i'm "solving" number 30 cody's problem.
I think to solve that whit sortrows function.
If I have a z vector:
j = sqrt(-1);
z = [-4 6 3+4*j 1+j 0];
my funtion is:
function z = complexSort(z)
z(2,:)=sqrt(real(z).^2+imag(z).^2);
z=sortrows(z',-2);
z=z(:,1);
end
End it return the result
z =
6.0000 6.0000
3.0000 - 4.0000i 5.0000
-4.0000 4.0000
1.0000 - 1.0000i 1.4142
0 0
The question is: why imagine part in input is positive e sortrows trasform it in negative?
best regards
Marco
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