Find the number of outputs of a value.

Hi,

I need help finding the number of outputs a function has. This function finds pythagorean triples given a maxumum value of the hypotenuse. Addditionally, it negates all outputs that are not positive integers. The output of the code looks like:

Input the maximum length of a triangle: 10

Triangle:

Side A = 3

Side B = 4

Hypotenuse = 5

x = 3 %x is the value of Side A

Triangle:

Side A = 6

Side B = 8

Hypotenuse = 10

x = 6

I need to find how many x values the function has produced which should be 2 for this. I could also solve my problem by numbering the triangle by n+1 so the first triangle would be 1, the second 2, and so forth.

Thanks

[Adam Danz]

It's unclear what the output of the function is. Does the function print those lines to the command window or is there a numeric output that you used to create those lines? If the function is printing lines to the command window, those lines are not really 'outputs'. What does the function look like and can you edit the function?

[Zach Adams]

First off, I apologize for my lack of knowledge with MatLab (i just started). I attached a pdf of the function. The function does print those lines to the command window when I run the function. Additionally, it runs the "main program script" which takes a user input for an integer value and sets it equal to C.

[Zach Adams]

I just realized the if conditions are not correct. The only difference would be that the A, B, and C values would all be positive integers that are greater than 0 with C being the greatest and A^2 + B^2 = C^2

[Adam Danz]

No need to appologize.

There are some areas of concern in your code (not listed in any order, below).

1. There is an input C and an output C. It's generally a bad idea to use the same variable name for an input and an output unless the input is literally just being sent as an output (which isn't the case in your code).
2. The input C is being overwritten at least twice. First, in the initial for-loop and second, within the 2nd for-loop (see block of code below). The value of C becomes meaningless.
3. MainProgramFunction appears to be a script. Although scripts are allowed to be called from functions, it's generally a bad idea because the code becomes much less traceable, contained, and controlled. The whole point of functions is reproducibility and containment but when a script takes over and alters/adds variables, you lose those benefits.

My matlab is installing the r2019b update#1 so I can't run any code right now but that's my feedback at first glance.

% Simplified version of your code (lines removed) showing the problem 
% of variable C being overwritten 

% 1) C is an input and an output without the same value
%               v                          v
function [A, B, C] = triplefindingfunction(C)

% 2) The input C is overwritten as a loop variable 
%   v       v
for C = 1:1:C

for m = round(sqrt(C/2)):round(sqrt(C))
 A = m^2 - n^2;
 B = 2*m*n;
 
 %3) The value of C changes each iteration but C is already a loop variable 
 % v
   C = m^2 + n^2;

To answer your question about counting the number of outputs, I see your function contains 3 outputs [A,B,C] and I think you'd like to know the number of values in each of those outputs. The way your code is written now, you'll only have 1 value per output because you're not saving any of the data on each iteration of the for-loops.

[Zach Adams]

Thank you for leading me into the right direction! I am going to attempt to rewrite the code so I do not repeat variable names.

[Adam Danz]

Good, I think that's the right course of action. Feel free to submit the updated code here as a comment. By the way, it's easier if you just copy-paste the code as a comment and then format it using the [ > ] button rather than attaching it as a PDF.

[Zach Adams]

function [A, B, hypotenuse] = M(C);
%This calculates the pythagorean triples given the length of the
%hypotenuse.
H;
disp([1:1:C]);
for max = 1:1:C
    for    m = round(sqrt(max/2)):round(sqrt(max))
        m == sqrt(max/2) && m == sqrt(max);
        if m == floor(m);
            n = sqrt(max - m^2);
        end
        A = m^2 -n^2;
        B = 2*m*n;
        hypotenuse = m^2 + n^2;
        if A > B;
            temp_A=A;
            A = B;
            B = temp_A;
        end
        for hypotenuse > A && hypotenuse > B && hypotenuse^2 == B^2+A^2 &&  A >= 0 && A == round(A) &&  B >= 0 && B == round(B);
            
            fprintf('Side A:')
            A
            fprintf('Side B:')
            B
            fprintf('Hypotenuse:')
            hypotenuse
        end
    end
end

This is my new code, H asks for a value of C from the user. I now need to know how many values of A (it could also be B or C) were printed in order to find how many pythagorean triples were made.

Answer 1

[Adam Danz]

@Zach Adams, first, more feedback. Please take some time to digest that. Second, I've improved your code without changing the algorithm and I've shown you how to determine the number of outputs.

Feedback on code

1) What is H and where does it come from? If H is supposed to be a variable, doesn't it cause an error? If H is a script or a function without any inputs, that needs improved for 2 reasons. a) "H" is a really bad function/file name. b) calling a script from a function defeats the advantages of functions - reproducibility and self-containment. If H is a function that merely asks for input you could move that code into his function. If H is a function that is used by other function, it's OK to keep it as it is but the name should really change. Functions should generally contain verbs or describe what it does. Example: getUserInput().

2) I see you have a variable named "max". You should avoid naming variables after common functions for two reasons. a) if you have a variable named max you will not be able to use the max() function anywhere in the same workspace and b) other people reading your code, including your future-self, will get confused because they are used to the "max" being a function. You can rename it to something like "maxC" if you'd like.

3) Use smart-indentation by selecting the entire code (ctrl + a) and then ctrl+i. I've corrected the code in your comment so that it has smart indentation. This makes it MUCH easier to read.

4) Your final for-loop is actually a conditional statement. Instead of "for hypotenuse > A &&..." use "if hypotenuse > A &&..."

5) 'M' is a very bad function name (same reasons as in point #1 above).

Cleaner version of your code

See inline comments.

function [A, B, hypotenuse] = findPythagTriplets(C)
%This calculates the pythagorean triples given the length of the hypotenuse.
% H;  % H isn't even used anywhere
disp(1:1:C);
A = []; 
B = []; 
hypotenuse = [];
for maxC = 1:1:C
    for m = round(sqrt(maxC/2)):round(sqrt(maxC))
       % m == sqrt(maxC/2) && m == sqrt(maxC);   %This line doesn't do anything at all
        if m == floor(m)
            n = sqrt(maxC - m^2);
        end
        A(end+1) = m^2 -n^2; %* The A(end+1) is generally bad practice but I didn't want to make major changes to your code
        B(end+1) = 2*m*n;    % this too
        hypotenuse(end+1) = m^2 + n^2; %this too
        if A(end) > B(end)
            temp_A=A(end);
            A(end) = B(end);
            B(end) = temp_A;
        end
        if hypotenuse(end) > A(end) && hypotenuse(end) > B(end) && hypotenuse(end)^2 == B(end)^2+A(end)^2 &&  ...
                A(end) >= 0 && A(end) == round(A(end)) &&  B(end) >= 0 && B(end) == round(B(end))
            
            fprintf('Side A:')
            disp(A(end))
            fprintf('Side B:')
            disp(B(end))
            fprintf('Hypotenuse:')
            disp(hypotenuse(end))
        else
            A(end) = []; 
            B(end) = []; 
            hypotenuse(end) = []; 
        end
    end
end

Determine number of values in the output

To determine the number of triangles, just call the function with the output and count the number of values in the output variables.

[A, B, hypotenuse] = findPythagTriplets(10)
numberOfTriangles = numel(A)
% numberOfTriangles =
%      2

[Zach Adams]

Thank you. Especially for sticking with me because I definitely learned a lot through this and I appreciate that.

[Adam Danz]

Thanks for the feedback! I was at the same stage several years ago so I know how ya feel.

Answer 2

[Sulaymon Eshkabilov]

Hi,

Here is your code improved a bit and bugs removed:

function [A, B, hypotenuse] = M(C)
%This calculates the pythagorean triples given the length of the
%hypotenuse.

disp([1:1:C]);
for max = 1:1:C
   for    m = round(sqrt(max/2)):round(sqrt(max))
       m == sqrt(max/2) && m == sqrt(max);
      if m == floor(m)
         n = sqrt(max - m^2);
      end
  A = m^2 -n^2;
  B = 2*m*n;
  hypotenuse = m^2 + n^2;
if A > B
   temp_A=A;
   A = B;
   B = temp_A;
end
if hypotenuse>A && hypotenuse > B && hypotenuse^2 == B^2+A^2 &&  A >= 0 && A == round(A) &&  B >= 0 && B == round(B)
  
  fprintf('Side A  = %f \n', A) 
  fprintf('Side B = %f  \n', B)
  fprintf('Hypotenuse  = %f \n', hypotenuse)
   end
 end
end

Good luck.

[Zach Adams]

Thank you for cleaning it up, but i was wondering if there is a way to determine how many times the if statement displays a value for the triangles. This would help me find how many pythagorean triples were created.