how I delete combination rows that have the same numbers from matrix and only keeping one of the combinations?

Question

 for a=1:50; %numbers 1 through 50
    for b=1:50;
        c=sqrt(a^2+b^2);

        if c<=50&c(rem(c,1)==0);%if display only if c<=50 and c=c/1 has remainder of 0

           pyth=[a,b,c];%pythagorean matrix

           disp(pyth)
        else c(rem(c,1)~=0);%if remainder doesn't equal to 0, omit output

        end      
    end   
end

answer=
        3     4     5
        4     3     5
        5    12    13
        6     8    10
        7    24    25
        8     6    10
        8    15    17
        9    12    15
        9    40    41
       10    24    26
       12     5    13
       12     9    15
       12    16    20
       12    35    37
       14    48    50
       15     8    17
       15    20    25
       15    36    39
       16    12    20
       16    30    34
       18    24    30
       20    15    25
       20    21    29
       21    20    29
       21    28    35
       24     7    25
       24    10    26
       24    18    30
       24    32    40
       27    36    45
       28    21    35
       30    16    34
       30    40    50
       32    24    40
       35    12    37
       36    15    39
       36    27    45
       40     9    41
       40    30    50
       48    14    50

This problem involves the Pythagorean theorem but we cannot use the built in function so I had to write one myself. The problem is for example columns 1 & 2 from the first two rows have the same numbers. How do I code it so it only deletes one of the rows if the columns 1 and 2 have the same number combination? I've tried unique function but it doesn't really delete the combinations. I have read about deleting duplicates from previous posts but those have confused me even more. Any help on how to go about this problem will help me immensely!

Thank you

Answer 1

welcome to StackOverflow.

The problem in your code seems to be, that pyth only contains 3 values, [a, b, c]. The unique() funcion used in the next line has no effect in that case, because only one row is contained in pyth. another issue is, that the values idx and out are calculated in each loop cycle. This should be placed after the loops. An example code could look like this:

pyth = zeros(0,3);

for a=1:50
    for b=1:50
        c = sqrt(a^2 + b^2);  
        if c<=50 && rem(c,1)==0
            abc_sorted = sort([a,b,c]);
            pyth = [pyth; abc_sorted];

        end
    end
end

% do final sorting outside of the loop
[~,idx]  = unique(pyth, 'rows', 'stable');
out = pyth(idx,:);
disp(out)

a few other tips for writing MATLAB code:

• You do not need to end for or if/else stements with a semicolon
• else statements cover any other case not included before, so they do not need a condition.

Some performance reommendations:

• Due to the symmetry of a and b (a^2 + b^2 = b^2 + a^2) the b loop could be constrained to for b=1:a, which would roughly save you half of the loop cycles.
• if you use && for contencation of scalar values, the second part is not evaluated, if the first part already fails (source).

Regards,

Chris

Answer 2

You can also linearize your algorithm (but we're still using bruteforce):

[X,Y] = meshgrid(1:50,1:50);                             %generate all the combination
C     = (X(:).^2+Y(:).^2).^0.5;                          %sums of two square for every combination
ind   = find(rem(C,1)==0 & C<=50);                       %get the index
res   = unique([sort([X(ind),Y(ind)],2),C(ind)],'rows'); %check for uniqueness

Now you could really optimized your algorithm using math, you should read this question. It will be useful if n>>50.