very easy

Nice one!

Adding comment for testing badges!!

What exactly the error is

function tf = is_it_odd(n)

if
rem(n,2)==0
tf = 'false';
else
tf='true';
end
end
end
where is the problem

whyyyyy

fun problem, however, smallest solutions are broken for all problems as they use ! & regexp

great problem

what is wrong here?

help aha?

come on really? whats wrong with this?

You need to assign your answer to the output variable tf.

...

I think there is a problem with the Cody server since yesterday.

Apologies for the down time. It should be working now

You actually don't need the Boolean assignments, technically a 0 and a 1 as the output will be good enough

so easy

I have tried many times this one. It runs fine but I am not able to understand why is it showing as wrong solution.
Can anybody help on this?

Remove the single quotes in true/false. It might work. ☺

function tf = is_it_odd(n)
if mod(n,2)==0
tf = 'false';
else
tf = 'ture';
end

why assertion failed. it works in matlab.

Thanks for your answer, you had a typo in true, also you should remove ' ' from false and true

very easy

good

easy

good job

good

good job

gj

good job

too demanding

good job

my solution was right and i tested it i m not sure what is going on
function [tf] = is_it_odd(n)
if rem(n,2)>0
tf='true';
else
tf='false';
end
end

my answer correct

good job

How can be a smaller size? tf = rem(n,2)

good job

I don't think that simply using tf=rem(n,2) should be allowed, its implicit casting which isn't a good thing to depend on.

Had cofused mod with % in python

good job

What is the problem here?

good job

this work on matlab but not here, why?

good job

My code works but it fails the test suite despite it giving the correct output. What's the deal?

true, not 'true', false, not 'false'

good job

Assertation failed.

Code works perfectly on my computer.
Must be a bug with this particular problem.

The output should be logical,and it should't be char type:
function tf = is_it_odd(n)
x = n/2;
y = floor(x);
if x == y
tf = false;
else
tf = true;
end
end

good job

good solution

Does anyone know why this doesn't answer the problem? When worked out in matlab it works just fine.

1. do not use "input(...)", "n" is defined variable and input doesn't return anything. 2. "=" is used for assignments, not for comparisions. This one returns syntax error in line 3 (if ...), use "==" instead. 3. Output should be logical, not string. Use "false" instead of "'false'" etc. 4. Return the output via variable specified for it in first line of the function, in this case it is "tf". Do not use "fprintf" for that. 5. Read The Friendly Manual. Just type "doc" in Command Window - there are tons of documentation, tutorials.

good job

This solution works offline?

good job

Nice work on a creative solution that has the "smallest size" per Cody's grading criteria. However, as a beginner to MATLAB, I don't see why I would want to write something this long/complex when there are much shorter solutions (perhaps not per Cody's criteria but they use less characters). I guess it's all about winning the contest.

want to how can be size 9

9...............how can you do that......

I tried this in my pc, all results appear as it is in your ans_correct.

good job

what is wrong with the solution?

You're returning the string "true" or "false" which the test suite is trying to compare to a boolean variable true or false. Since the string does not equal the boolean value, assert fails.

good job

% i mean that: test = is_it_odd; :D

good job

Why is function "odd" not working?

good job

Is this code wrong?
tf=floor(n/2)*2==n

yup

good job

How could i get this any shorter?

good job

Why is this solution not working?? works on my matlab 2014 version!!

I think the ' ' characters might be the problem.

good job

it should use mod.

good job

realized my true/false assignments were off. They should be reverse...if mod(n,2) ~= 0; tf=true; else tf = false; end

good job

can anyone tell me better solution than this?

Well done brother :)

function tf = is_it_odd(n)
mod(n,2);
end

This is called cheating....Even this code is failed

for 21 size. little editing :)
function ans = is_it_odd(n)
if mod(n,2)==0; false;
else
true
end

bugger

I think the assertion is flawed, because

isequal(is_it_odd(2),true) == true == 1

returns

ans = 1

Here, false and 0 (as double) mean same.
Also true and 1 (as double) mean same.

what the matter,such an easy problem prevent me from continuing solving the following problems,so why,l can't accept the format of the solver function,it is so stupid,it is not flexible,l can't stand

this returns "1" & not "true", but the validation for this cody problem is spurious since isequal(true,1) is actually true (when it should really throw an error).

Why doesn't " if mod(n,2)==0, tf = 'false', else tf = 'true', end " work?

it should be tf = false

Seems to me that the "best" answer would be tf = logical(mod(n,2)). But I guess the description does not explicitly require logicals, so maybe I am reading too much into the problem. Using logical makes the solution 17 compared to 13 without.

by calling logical function you add extra code.

I'am bad!

I don't think you know what odd/even means.
odd = 1 3 5 ...
even = 2 4 6 ...

I know this isn't correct, only submitted it because I couldn't figure out how it was possible to solve the problem with code size 12, and wanted to see the solution, only to find out that the 'leading solution' was wrong.

Does it work if n=3 or greater??
Pretty confusing!!

Ofcourse not, the problem with cody is that you can see the unit test.

But the good thing about Cody is that you can update the test suite. Thanks for the comments, and please keep flagging bad tests so we can fix them.

This doesn't return a logical 'true' or 'false'.

Persistent server error trying to evaluate this solution...

Hooray for ambiguous classing requirements.

How about this for the "best" solution:

x_odd = x(is_it_odd(x)); % :/

The answer should be logical(mod(n,2)).