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O45
O45
Last activity 2018-10-23

Best solution to cody problems

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?
James
James
Last activity 2015-6-11

Issue with Cody Challenge Badges?

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?

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?
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"
http://www.mathworks.com/matlabcentral/cody/problems/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
Jan
Jan
Last activity 2013-8-9

Magic strings and numbers in Matlab

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
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):
whats
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
Adam Kaas
Adam Kaas
Last activity 2012-6-6

Coolest things MATLAB can do

As I'm becoming more and more familiar with MATLAB, I'm starting to fall in love with it. I was wondering what are the coolest things that you all know MATLAB can do? As for me so far, the auto-code generation into another language is the coolest thing.
It is not uncommon for students to be assigned questions which they are required to complete "without using any built-in functions". There is not a great deal that can be programmed in MATLAB without using any built-in functions, but a little can be done -- but what, exactly is possible?
What a "built-in function" is, exactly, is open to interpretation. In the below, I refer instead to "publicly visible routines". Keywords (see below) are not publicly visible routines (they are "statements" or components of statements.) Any documented operation or call that invokes a MATLAB-supplied .m or .p or mex file or built-in library to do its work is a publicly visible routines. If you can use documented methods override the normal meaning of a statement or expression in practice by supplying alternate code, then the code probably involves publicly visible routines. If the language design is such that you could use documented methods to override the normal meaning of a statement or expression in theory (such as the behavior of adding two double, the code for which is in practice bundled into an internal MATLAB library), then I would still consider that a call to a publicly visible routine.
A MATLAB-supplied routine that is not documented, which is used for internal MATLAB purposes, could perhaps be held not to be a publicly visible routine, but it certainly would still be a "built-in function".
I exclude from the list any routine which there is no direct way to access, and is only used for internal purposes, such as the memory allocation routines.
This is what I have come up with:
  • the names defined as "keywords" do not in themselves involve function calls to publicly visible routines. These keywords currently include 'break', 'case', 'catch', 'classdef', 'continue', 'else', 'elseif', 'end', 'for', 'function', 'global', 'if', 'otherwise', 'parfor', 'persistent', 'return', 'spmd', 'switch', 'try', 'while'. There is no functional form of any of these: for example, one cannot use global(s) to declare the name contained in the variable "s" to be global. (However, you can define an "end" method; https://www.mathworks.com/help/matlab/matlab_oop/object-end-indexing.html )
  • scalar numeric double precision real-valued constants are handled at parse time, including unary plus and unary minus in front of them
  • scalar numeric double precision constants followed immediately by "i" or "j" create a complex-value constant at parse time, including unary plus and unary minus in front of them
  • whether a complete complex constant with real and imaginary part is handled at parse time is unknown
  • literal character vectors and string objects are handled at parse time
  • in sufficiently new versions, int64() and uint64() around an integer constant is handled at parse time. This was a change from previous versions which handled it at run time (after the integer had been converted to double precision...)
  • whether any other casts such as uint16() or logical() are now handled at parse time is unknown
  • assignment of a compete variable (no indexing, no substructure references, etc.) to a plain variable (no indexing, no substructure references, etc.) does not involve any function calls to publicly visible routines (unless I have overlooked a case involving objects)
  • "if" or "while" applied to a scalar logical constant or to a scalar logical variable does not involve any function calls to publicly visible routines. However, it is not known whether there is any method to construct a logical value without calling a MATLAB routine: "true" and "false" are MATLAB routines, not constants, and logical() of a numeric constant might be handled at run time
  • "for" in which the range is named as a scalar constant or scalar variable do not involve any function calls to publicly visible routines; for example, "for K = 5"
  • defining an anonymous function does not involve any function calls to publicly visible routines
I may have overlooked something due to shortage of chocolate in my bloodstream.
The language described above is not Turing complete, and is not "sufficiently powerful" for the purposes of the Church-Rosser Theorem of general-purpose computability. It is also not possible to do any arithmetic in it, as arithmetic must be reducible to the Peano Postulates, and those require at the very least the ability to compare a value for equality with 0, which in MATLAB would require a call to the MATLAB routine "eq".
We all know that MATLAB is probably the best software for engineering purposes, I think it's a little expensive unless you have it for free on your school or place you work, please share your opinion about MATLAB cost, including toolboxes, student versions... is it that expensive?