Problem 23. Finding Perfect Squares

too difficult with only documentation

The last if is just a way to tell the function to say false, as I was getting desperate :) but the logical indexing works!

You should have stated upfront that if statements are not allowed.

These bans are sometimes used, I believe, to keep people from submitting "cheat" solutions that simply check which test case they're being fed and produce the corresponding output.

As you observe, however, this is often overzealous in that in many problems, case distinctions arise naturally in the solution. (Not to mention that the letters "if" may well appear together in a solution that doesn't even use if statements.)

And I agree, if such restrictions are imposed they should be communicated clearly.

nice!

This code works only if the vector is ordered in an ascending order.
If we have for example [4 3 2] it will not work, need a sqrt check then. Or even worse, if the vector is not ordered at all!

tricky

very challenging problem!!

I don't understand why #4 if false. Can someone help me?

Test case #4 is true, not false, because the vector contains a perfect square and its square root.
Your function is not passing because it overwrites the check variable for each vector element, only actually checking the last value to return the true/false variable. It happened to pass the other four cases because their answers (either false or true) happen to correlate with your check for the last (or only) element in the vector.

GOOD

a = [20:30];
assert(isequal(isItSquared(a),false))
Assertion failed.
It should be true because in the array there is 25.
a = [20:30]
a = 20 21 22 23 24 25 26 27 28 29 30

The question asks that whether a perfect square and it's root both are present in the array or not. You have to check for both and not one only.

Hi,
I verified this code with the given test suite. but the system is not accepting the code as it is.

thanks

TR

What's wrong with this code? Runs perfectly on my computer.

I didn't see the leading code,so I put my code here to give some inspiration.
Looking forward to a shorter one
b=~isempty(intersect(a,a.^2));

whoops, don't know how I managed to confuse .^ with .* for a bit there ---I was like, "What's WRONG?"

nice

You cheater!

This type of cheat/hack answers are no longer allowed.

Good question

0 and 1 elements in a will show the bug in your code

nice one-liner

a vector 'a' with elements 0 or 1 will show the bug in your code

cool!

Wow! This took me a couple of days to solve, but I did it and I’m very proud! :D

Hi. A vector with ones or zeroes as v = [3, 1] will show the bug in your code.

Lol

Test case : a = [20:30] contains a square. but its expected error is false. It should be fixed

hello,
where do you find a perfect square in [20:30]?

Hi Jann B,
25 is a perfect square, no?

Very interesting solution!

How is the output true if the vector contains only one number? People posting such meaningless solutions: Please have some common sense!

If that one number is 1, then the vector contains both a number and its square.

easy easy

That's better...

a vector as v = [3,2,7,0,5] will show you your code's bug

a little smaller now

Nice ☺

Works

just kidding...

very easy

I want to make this code shorter!!

Case2,a = [20:30],fails.
Others pass.
Why???

in 2 test suite in a=[20:30] , contains 5^2 =25 so it has to be true not false

But there seems to be a better choice. Why forgetting ismember?

Test 2 in the test suite is wrong!
25 is clearly a perfect square.

the problem is to see if a number is the square of another number in the vector. Although 25 is a perfect square, 5 is not included in the vector

Though it passes the test suite, I feel like I could make this a better way but I am drawing a blank. Any feedback would be appreciated.

good

if a = [20:30] the output should be true as 25 is a perfect square, while in the test case output is given to be false.

25 is not a square of any other members of the group. If the group included 5 it would be true.

hur dur durrr durr. sqrt(25)==5, not included in givens

Yeah, I did it

My solution returns false for [20:30]; why not for [6 10 12 14 101]?

Instead of a(1):a(end) you should just write a.

Where did I do wrong？

The desired results are not the character strings 'true' and 'false' but the logical values true and false.

hint ; any and ismember.

submitted tested solution but still showing error......
L 17 (C 1-3): Parse error at END: usage might be invalid MATLAB syntax.

interesting solution... I like it :)

I disagree. Almost every problem has a solution wrapped in this function. It's cheating as it hides the complexity of the code from the scoring algorithm. That's why "eval" is not allowed ...

This is the correctly executing in matlab

I wonder...if you cheat, why even bother to actually solve the problem?

I have not heard of ismember before, thanks.

As primitive as my solutions are, I learn just as much from reading other peoples as doing the problem itself. Great setup mathworks :)

Code works on Matlab, but not on your test suite?

Srtr2func, classic way:)a new thing learned.

a = [20:30] 25 is in that array and is perfect square. Why should it return false?

because 5 isn't in the array. the perfect square and it's root should both be in the array.

test case 3 should be 'false'

it states 'one of the other' numbers, so for a = [1] it should be false

Assertion 2. is not correct. vector 'a' return with 6th element as 25 & thts a perfect square.. So it must be true!!