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Problem 10 of MATLAB cody challenge

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NAVNEET NAYAN
NAVNEET NAYAN on 28 Jul 2018
Edited: Guillaume on 2 Feb 2020
I was trying to solve this question in cody challenge: Problem 10. Determine whether a vector is monotonically increasing. I tried following code:
i=1;
while i<length(x)
if x(i)<=x(i+1)
tf='true';
else
tf='false';
break;
end
%
i=i+1;
end
When I am running this piece of code on MATLAB editor everything is Ok. But when I am submitting this, incorrect answer results. Format to make a function for this problem is given as:
function tf = mono_increase(x)
tf = false;
end
Can anyone sort it out?

  0 Comments

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Accepted Answer

Dennis
Dennis on 30 Jul 2018
There are a few problems with your solution:
  1. You need to return true/false - not as string
  2. You compare values to be bigger or equal -> [1 1 1] is not increasing, but your solution return true
  3. Your solution wont work with single values, because x(i+1) does not exist
A working solution based on your approach might look like this:
if length(x)==1
tf=true;
else
i=1;
while i<length(x)
if x(i)<x(i+1)
tf=true;
else
tf=false;
break;
end
%
i=i+1;
end
end

  3 Comments

NAVNEET NAYAN
NAVNEET NAYAN on 30 Jul 2018
Thanks...I was making the mistake what you have pointed out in above 3 points.
Guillaume
Guillaume on 30 Jul 2018
A proper implementation in matlab would do this without a loop (e.g. see Paolo's answer). If you were to implement this with a loop, the following would be cleaner:
function tf = mono_increase(x)
tf = true;
for idx = 2:numel(x)
if x(idx) <= x(idx-1)
tf = false;
break; %for a better cody score, omit this line. for a better implementation, keep it!
end
end
end
Any loop implementation will score very poorly on cody.
NAVNEET NAYAN
NAVNEET NAYAN on 31 Jul 2018
thanks Guillaume for the worthy suggestions.

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More Answers (2)

Paolo
Paolo on 28 Jul 2018
You can use:
all(diff(x)>0)

  4 Comments

Show 1 older comment
Paolo
Paolo on 30 Jul 2018
Ah, that's interesting :)
function tf = mono_increase(x)
tf = all(diff(x)>0)
end
gives size 16. Whereas
function tf = mono_increase(x)
tf = issorted(x, 'strictascend')
end
gives size 13.
How exactly is the size computed? And how can the leading solution be size 9?
Guillaume
Guillaume on 30 Jul 2018
Most of the very low scoring solutions no longer work. The two 9 overwrite assert.m with a system command. The 10s use regexp with a dynamic regular expression to execute code and bypass assignment to the function with the so-called ans trick.
For better or for worse, the score of a cody solution only depends on the number of nodes in the parse tree of the code. You can see the parse tree of code with the undocumented mtree class.
t = mtree('tf = all(diff(x)>0)');
t.rawdump
t = mtree('tf = issorted(x, ''strictascend'')');
t.rawdump
%also
%t = mtree('-filename', 'nameoffile.m')
There are plenty of tricks to get low score. Command form, if possible, cost less than function form. A char array regardless of its length only cost one so stuffing code in text helps. Compare:
str2num 1+2+3+4 %command form cheaper than function for
mtree('str2num 1+2+3+4')
1+2+3+4
mtree('1+2+3+4')
Omitting assignment using ans also helps:
%score 11:
function ans = mono_increase(x)
issorted(x, 'strictascend'); %no assignment so result is assigned to ans
end
As demonstrated above, low scoring solutions are rarely good matlab code and are often inefficient. They are optimised for the scoring system, not for the speed of execution or readability.
Paolo
Paolo on 1 Aug 2018
Thanks for the detailed answer Guillaume.
Very interesting indeed, those are some cool tricks. Wouldn't have thought people had put all this effort into hacking cody!

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Sriram Nayak
Sriram Nayak on 2 Feb 2020
i=1;
while i<length(x)
if x(i)<=x(i+1)
tf='true';
else
tf='false';
break;
end
%
i=i+1;
end

  1 Comment

Guillaume
Guillaume on 2 Feb 2020
I'm afraid this is is not going to work. The char array `true` and the logical value true are not the same at all.
In term of cody score
i = 1;
while i < endbound
%do something
i = i + 1;
end
is going to score you very badly against the equivalent and much simpler for loop:
for i = 1:endbound
%dosomething
end

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