Problem 10. Determine whether a vector is monotonically increasing

simpler than it looks

tough problem

good problem

good problem

nice

Nice problem. You have to take care of the inequalities.

cool problem

Eh okay. I dislike this problem.
First, I used a "for loop", but the results were often not logical/ reliable. I than used

a = 1 : 1 : length(x) - 1;
if all( x(a) < x(a + 1))
tf = true;
end

and it worked. But why can't I use a "for loop"?! I feel like I did not learn anything from this

Compare the number in current index with the number in next index. If it is less than or equal to zero return false.

I truely hate this

I have got some good news for you - These solutions are not valid anymore!

nice

I don't understand your solution?

Hi Hidd, that is logical. The leading solution system is a bit broken, there are a few ways around it. That is why people are calling attention to fixing it. But as far as I know it's existed for 2+ years.

L=length(x);
if L==1
tf=true; % adding a condition for single element
return %we must close the case for faster operation
end
for ii = 1:(L-1)
if x(ii+1) > x(ii)
tf= true ;
else
tf=false ;
break
end
end

end

kinda cheated but test values were stupid

niceeeeeeeeee

in the test suite, test number 3 tests for vector [0 0 0 0 0] if it is monotically increasing, but in the assertion it claims it is false, meaning it is not monotically increasing, which i dont think is true.
the assertion should check it as true.
this is from wikipedia:
[ A function is called monotonically increasing if for all x and y such that x <= y one has f(x)<=f(y) ]
the (issorted) function should be enough to check for monotically increasing.

-0--

How is/isn't x = [0]; monotonically increasing vector when there is no other element to compare?

*Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return false otherwise.*

Good pblm ..try with length(unique(x))==length(x) and all(diff(x))>=0)).tis shd be the hint for the pblm

Thanks for the hint

nice question

you are using MATLAB in a C++ way lol

nice solve!

One easily understandable size 15 solution is (use rot13 to read it): shapgvba gs = zbab_vapernfr(k) gs = vfrdhny(k, havdhr(k)); raq

Did it in one line...but took me a while to figure it out. Matlab n00b

Nice one ☺

is that guy real :'D

Can anybody helpe me? What's wrongs with this?

What's wrong? I tried it on the matlab desktop,it's okay.

same problem here

sort(x) will always return a increasing vector.
So all(diff(sort(x)))>0 is always true, if x has just unique elements.

try all((diff(x))>0)

great answer

(^-^)v

function tf = mono_increase(x)
tf = false;

if x(1,1)>=y(1, length(x))
fprintf('false \n');
elseif x(1,1)==0
fprintf('true\n');
else
fprintf('false\n');
end
end

%this is compiling in matlab with all inputs

Could someone help me? I don't know why it's wrong.

You are returning the character strings 'true' and 'false' instead of the logical values true and false (so just remove the quotes).

This solution code must have sth wrong. When i run this code on my matlab, it will give the right answer however when I run it on this web, it doesn't work

true and false are booleans. 'true' and 'false' are strings. It's not the same thing.

Can't improve more. Any help (trick) is appreciated...

the sign function is not needed

Nice problem

I agree

I don't understand why this solution is wrong, can I get some help please?

This code returns the strings 'true' and 'false' instead of the logical values: true and false.

Where did I do wrong？Please help me！

You fail test 2, since the vector x has length 1, you never evaluate anything in your loop. This can be fixed by initializing tf before the loop.

Look into recursive solution instead of this loop.

Why are you incrementing the solution with i=i+1? The for loop automatically increments from 1 to n-1. You are also only really testing the last two values in the vector. Your function returns false for each pair in vector but overwrites tf = false with true when it reaches the last two entries. You need to break out of the loop if you detect one false.

This doesn't solve the problem...

lots of things dont.

cheater.

sahi hai...

This will return true for vectors like [1 2 2 3] that are sorted but not monotonically increasing. Try changing the last if to
if length(x) ~= length(unique(x))

Mattias is right!

this is the solution right???

Nice work, champ.

Bruh

where is the problem here? it gives coreect result in matlab

tf should be TRUE or FALSE, not a string

I dont see any problem in matlab. what is the problem here ?
thanks

Can anybody explain me, why this solution is smaler than the version where x = diff(x); is saved befor?

I mean:
x = diff(x);
tf = all(sign(x) > 0) || isempty(x);

all(diff(x)>0)

you can reduce your score by 6 points by removing the logic after ||.its taken care by 'all'

why is the requested answer for x = [0 0 0 0 0] false? constant sequences are both, monotonically increasing and decreasing.

You are right, your solution is good. In the problem describson should be "strictly monotonically increasing".

The leading solution is wonderful!!!

By the looks of this...
I have a long way to go.

Impressive.

This is simultaneously abusive and beautiful.

good question

don't know how I missed "isequal" the first time around... it's right there in the test suite. DUH.

why this solution is not right? In the fist case, it returns 16, isn't it equal true?

Why does regexp work like this? Is it a bug? If so, could the Cody team please fix this so that players can't use it to score easy points?

This is just gaming the test set for a limited program with an artificially low score...

so I did it in 1 single line but my size is too big :((

Here's my one-liner (similar to yours):

tf = ~sum(x(2:end) - x(1:end-1) <= 0)

you did good job

assertion 2 of the test suite is monotonically increasing (x = [0])? Don't you have to have at least two numbers?

Nice shot!!!

Didn't think about using all , well played.

Good function of unique!

Nice :)!

excellent

Really an awesome problem yaar!!

assert(monotonic increasing == nondecreasing,'check the definition')

I think test 2 is wrong. Is x=0 monotonically increasing? I think not.

i dont know why this is not working .

OK, I had to test the leader's work myself, and I get it now. If any values repeat, unique()'s answer is shorter and therefore not equal. Since unique() sorts the answer in ascending order, only a strictly increasing function has both terms equal.

can I say: WOW

Clever although slow

Why doesn't this fail on test 2? Why should diff(0) > 0 be true?

The weirdness is due to this: all([]) is true.