# Is this working right

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Hey guys. I am trying to do this Assignment and i am pretty sure its correct but it keeps saing its now when i run the emulator that checks it.

Here is the assignemnt:

Write a function called sort3 that takes a 3-element vector as its sole arguments. It uses if-statements, possibly nested, to return the three elements of the vector as three scalar output arguments in nondecreasing order, i.e., the first output argument equals the smallest element of the input vector and the last output argument equals the largest element. NOTE: Your function may not use any built-in functions, e.g., sort, min, max, median, etc.

my code

function [a,b,c] = sort3(v)

if v(1) >= v(2) && v(2) >= v(3)

a = v(1); b = v(2); c = v(3);

elseif v(1) >= v(3) && v(3) >= v(2)

a = v(1); b = v(3); c = v(2);

elseif v(2) >= v(1) && v(1) >= v(3)

a = v(2); b = v(1); c = v(3);

elseif v(2) >= v(3) && v(3) >= v(1)

a = v(2); b = v(3); c = v(1);

elseif v(3) >= v(1) && v(1) >= v(2)

a = v(3); b = v(1); c = v(2);

elseif v(3) >= v(2) && v(2) >= v(1)

a = v(3); b = v(2); c = v(1);

end

end

### Answers (8)

Jan
on 13 Mar 2017

Edited: Jan
on 26 Jul 2018

This can be written shorter with less chances for typos:

function [a, b, c] = sort3(v)

a = v(1);

b = v(2);

c = v(3);

if a > b, [a, b] = swap(a, b); end

if b > c, [b, c] = swap(b, c); end

if a > b, [a, b] = swap(a, b); end

end

function [b, a] = swap(a, b) % empty function body

end

Prefer one command per line in real code. I've move the if block to single lines only to emphasize the pattern optically here.

##### 0 Comments

Adam
on 13 Mar 2017

##### 0 Comments

Steven Lord
on 13 Mar 2017

One way to increase the confidence that your code is doing what you expect is to pass in test cases for which you know the correct answer and check that you get the output you expected. Adam's test case of [1 2 3] is one example. Others include:

- [2 3 1] (the input is not sorted already)
- [1 2 1] (duplicate elements)
- [3 3 3] (all the elements are duplicates)
- [-1 0 -2] (negative numbers and zero)
- [exp(1) pi sqrt(2)] (non-integer values)

##### 0 Comments

Walter Fanka
on 23 Oct 2017

Edited: Walter Roberson
on 17 Oct 2020

function [s1,s2,s3] = sort3(v_3)

a = v_3(1); b = v_3(2); c = v_3(3);

if a <= b && a <= c s1 = a; if b <= c s2 = b; s3 = c; else s2 = c; s3 = b; end end

if b < a && b <= c s1 = b; if a <= c s2 = a; s3 = c; else s2 = c; s3 = a; end end

if c < a && c < b s1 = c; if a < b s2 = a; s3 = b; else s2 = b; s3 = a; end end

end

##### 1 Comment

Jan
on 24 Oct 2017

Something must be unnecessary here: If the first two conditions "a <= b && a <= c" and "b < a && b <= c" have been rejected before, it cannot be required to check for the third "c < a && c < b": If this third condition would be false, s1, s2, s3 would be undefined and the function would crash. So either this third condition is not needed, or the code fails for some input.

Do you see the bunch of warnings in the editor? Insert commas to calm down the MLint code checker:

if a <= b && a <= c, s1 = a; if b <= c, s2 = b; s3 = c; else, s2 = c; s3 = b; end, end

% ^ ^ ^ ^

Or better use line breaks:

if a <= b && a <= c

s1 = a;

if b <= c

s2 = b;

s3 = c;

else

s2 = c;

s3 = b;

end

end

I know, I have used multiple command per line also, but this was a bad example.

Vignesh M
on 4 May 2018

Edited: Walter Roberson
on 17 Oct 2020

The question says to return them in 'non-decreasing' order. Your function is for decreasing order.

function [u1,u2,u3] = sort3(v)

if v(1) <= v(2) && v(2) <= v(3);

u1=v(1);u2=v(2);u3=v(3);

elseif v(1) <= v(3) && v(3) <= v(2);

u1=v(1);u2=v(3);u3=v(2);

elseif v(2) <= v(1) && v(1) <= v(3);

u1=v(2);u2=v(1);u3=v(3);

elseif v(2) <= v(3) && v(3) <= v(1);

u1=v(2);u2=v(3);u3=v(1);

elseif v(3) <= v(1) && v(1) <= v(2);

u1=v(3);u2=v(1);u3=v(2);

else v(3) <= v(2) && v(2) <= v(1);

u1=v(3);u2=v(2);u3=v(1);

end

##### 2 Comments

Heirleking
on 25 Jul 2018

Jan
on 26 Jul 2018

Edited: Jan
on 18 Oct 2020

@David Silva: There are no commas, but semicolons. They are not required. If you are in doubt, remove them and see what happens.

As described in a comment above, commas help, if you append the body of the if branch to the same line:

if a==b disp('match') end % ?!? Confusing!

if a==b, disp('match'); end % Better! Less confusion.

if a==b % Best! This is purely clear

disp('match')

end

Mohamed Ahmed Khedr
on 14 Oct 2018

Your code working in a good way if you just make the following edit>>> change the arrangement of outputs

function [c,b,a] = sort3(v)

##### 0 Comments

zehra ülgen
on 17 Oct 2020

Edited: Jan
on 18 Oct 2020

function A = sort3(x,y,z)

if x < y && x < z % x is the smallest one

if y < z

A = [x y z];

else

A = [x z y];

end

elseif y < x && y < z % y is the smallest one

if x < z

A = [y x z];

else

A = [y z x];

end

elseif z < x && z < y % z is the smallest one

if x < y

A = [z x y];

else

A = [z y x];

end

end

##### 2 Comments

Jan
on 18 Oct 2020

The original question has a vector as input and wants 3 scalars as output.

Your function fails if two inputs are equal.

zehra ülgen
on 18 Oct 2020

Sorry, you right! I couldn't realize that these questions are not same.

"Write a function called sort3 that takes three unequal scalar arguments (the function does not have to check the format of the input or the inequality of the arguments). It uses if-statements, possibly nested, to return the three values of these arguments in a single row vector in increasing order, i.e., element one of the output vector equals the smallest input argument and element three of the output vector equals the largest input argument. NOTE: The function may not use the built-in function sort."

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