MATLAB: My function output is not what I was expecting, can someone have a look and suggest why this might be the case. Thanks

homework

The assignment asks to write a function to find unique isosceles triangles of integer sides and area.
A=sqrt(m*(m-a)(m-b)(m-a)
where
m = (a+a+b)/2 or P/2.
for example P=16 produces a=5, b=6 A=12
    clear;
    for P = 1:100
      [a,b,A] = IsoTrig(P);
      fprintf('P: %d, a: %d, b: %d, A: %d\n', P, a, b, A);
    end;
function [a,b,A] = IsoTrig(P)
  if P<0 || fix(P) ~=P;
    disp('error, input is not a non negative integer')
    return
  end
  n = 0 % variable to sum loop passes
  m=P/2;
  for ii=1:P;
    a=ii/2;
    b=P-(2*a);
    if b>= 2*a || b<=0;
      continue
    end
    for A=sqrt((m*(m-a)*(m-b)*(m-a)));
      if fix(A)== A && A ~= 0 && fix(a)==a && fix(b) == b && n == 0
        a=a; 
        b=b; 
        A=A; 
        n = n+1;
      else fix(A) ~= A || fix(a) ~= a || fix (b) ~= b || A == 0 || n == 1;   
        a = 0;
        b = 0;
        A = 0;
        continue
      end
      [P,a,b,A]
    end
  end
end
  • So we are talking about unique Heronian isosceles triangle.
*My code doesn't output what I want, the [P,a,b,A] gives me Heronian isosceles results but the function output itself is different.
*As can be seen my [P,a,b,A] call results is different to the fprintf results. The current program doesn't find any results to fprintf.
*I don't know how to isolate the unique triangles (where 1 heronian isosceles exists for given P) My code if working only reduces the output to 1 not per given P not find unique.
OUTPUT as stands
P: 101, a: 5.050000e+01, b: 0, A: 0
P: 102, a: 51, b: 0, A: 0
P: 103, a: 5.150000e+01, b: 0, A: 0
P: 104, a: 52, b: 0, A: 0
P: 105, a: 5.250000e+01, b: 0, A: 0
P: 106, a: 53, b: 0, A: 0
P: 107, a: 5.350000e+01, b: 0, A: 0
ans =
     108    30    48   432     <- from [P,a,b,A] call
P: 108, a: 54, b: 0, A: 0
P: 109, a: 5.450000e+01, b: 0, A: 0
P: 110, a: 55, b: 0, A: 0
P: 111, a: 5.550000e+01, b: 0, A: 0
ans =
     112    35    42   588
P: 112, a: 56, b: 0, A: 0
P: 113, a: 5.650000e+01, b: 0, A: 0
P: 114, a: 57, b: 0, A: 0
P: 115, a: 5.750000e+01, b: 0, A: 0
P: 116, a: 58, b: 0, A: 0
P: 117, a: 5.850000e+01, b: 0, A: 0
P: 118, a: 59, b: 0, A: 0
P: 119, a: 5.950000e+01, b: 0, A: 0
P: 120, a: 60, b: 0, A: 0
P: 121, a: 6.050000e+01, b: 0, A: 0
P: 122, a: 61, b: 0, A: 0
P: 123, a: 6.150000e+01, b: 0, A: 0
P: 124, a: 62, b: 0, A: 0
P: 125, a: 6.250000e+01, b: 0, A: 0
ans =
     126    35    56   588
P: 126, a: 63, b: 0, A: 0
P: 127, a: 6.350000e+01, b: 0, A: 0
ans =
     128    34    60   480
P: 128, a: 64, b: 0, A: 0
P: 129, a: 6.450000e+01, b: 0, A: 0
P: 130, a: 65, b: 0, A: 0
P: 131, a: 6.550000e+01, b: 0, A: 0
P: 132, a: 66, b: 0, A: 0
P: 133, a: 6.650000e+01, b: 0, A: 0
P: 134, a: 67, b: 0, A: 0
P: 135, a: 6.750000e+01, b: 0, A: 0
P: 136, a: 68, b: 0, A: 0
P: 137, a: 6.850000e+01, b: 0, A: 0
P: 138, a: 69, b: 0, A: 0
P: 139, a: 6.950000e+01, b: 0, A: 0
P: 140, a: 70, b: 0, A: 0
P: 141, a: 7.050000e+01, b: 0, A: 0
P: 142, a: 71, b: 0, A: 0
P: 143, a: 7.150000e+01, b: 0, A: 0
ans =
     144    37    70   420
P: 144, a: 72, b: 0, A: 0
P: 145, a: 7.250000e+01, b: 0, A: 0
P: 146, a: 73, b: 0, A: 0
P: 147, a: 7.350000e+01, b: 0, A: 0
P: 148, a: 74, b: 0, A: 0
P: 149, a: 7.450000e+01, b: 0, A: 0

Best Answer

Without looking at your code, I first did some pencil and paper work.
Heron's formula for an isosceles triangle, with sides of {a,a,b}, would reduce as:
m = (a + a + b)/2 = a + b/2
A = sqrt(m*(m-a)*(m-a)*)(m-b))
but m-a is just b/2. And m-b is a-b/2. Substitute, and do some quick algebra...
A = sqrt(a^2 - b^2/4) * b/2
This may be a bit simpler to work with than the original formula. It does point out a restriction on a, that a>=b/2. Of course, that should be obvious, since an isosceles triangle MUST have that property.
Next, if you condition your search on the triangle perimeter, then we see another piece of information to reduce the search. If P is an odd integer, then so must be b. Likewise, if P is even, then b is also.
Better, is to look at it from the perspective of a. a must be an integer, at least as large as P/4, and no larger than P/2. (Think about it! Check my logic.) So for ANY given integer perimeter P, the possible side lengths are:
a = ceil(P/4):ceil(P/2 - 1);
b = P - 2*a;
It is always a good idea to test things out, just to be confident you have it right. In fact, I'll write a couple of tiny helper functions to compute a and b for any perimeter.
afun = @( P) ceil(P/4):ceil(P/2 - 1);
bfun = @( P) P - 2*afun( P);
First, try an even perimeter.
P = 12;
[afun( P);bfun( P)]
ans =
     3     4     5
     6     4     2
The first row of that array is a, the second row is b.
Now try an odd value for the perimeter.
P = 13;
[afun( P);bfun( P)]
ans =
     4     5     6
     5     3     1
Next, we require that the area is also an integer.
Areafun = @( P) sqrt(afun( P).^2 - bfun( P).^2/4) .* bfun( P)/2;
But can we do just a bit more before we go into actual numbers? We said before that if P is an odd number, then so must be b. What was the area again?
A = sqrt(a^2 - b^2/4) * b/2
If b is odd, then a^2-b^2/4 must ALWAYS be non-integer, and so must be the area. So we can conclude that to have an integer area, an integral sided isosceles triangle must always have an even perimeter.
So a simple code might just loop over the even integers, starting with P=4, searching for integer values of Areafun.
afun = @( P) ceil(P/4):ceil(P/2 - 1);
bfun = @( P) P - 2*afun( P);
Areafun = @( P) sqrt(afun( P).^2 - bfun( P).^2/4) .* bfun( P)/2;
for P = 4:2:100
  A = Areafun( P);
  % identify triangles with an integer, non-zero area
  ind = find((A == round(A)) & (A ~= 0));
  if ~isempty(ind)
    a = afun( P);
    b = bfun( P);
    disp('Solution(s) found')
    P
    [a(ind);b(ind);A(ind)]
  end
end
Could we do even better? Perhaps by thinking more deeply about the problem, perhaps we might.
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