- You need to return true/false - not as string
- You compare values to be bigger or equal -> [1 1 1] is not increasing, but your solution return true
- Your solution wont work with single values, because x(i+1) does not exist

# Problem 10 of MATLAB cody challenge

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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

### Accepted Answer

Dennis
on 30 Jul 2018

There are a few problems with your solution:

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

Guillaume
on 30 Jul 2018

Edited: 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.

### More Answers (2)

Paolo
on 28 Jul 2018

You can use:

all(diff(x)>0)

##### 4 Comments

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
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!

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
on 2 Feb 2020

Edited: 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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