Special pythagorean triplet: Difference between revisions
→{{header|ALGOL 68}}: Show the number of iterations reuired and don't stop after the first (only) solution has been found.
(→{{header|ALGOL 68}}: Make better use of Euclid's formula to reduce the possible candidates.) |
(→{{header|ALGOL 68}}: Show the number of iterations reuired and don't stop after the first (only) solution has been found.) |
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=={{header|ALGOL 68}}==
Uses Euclid's formula, as in the XPL0 sample but also uses the fact that M and N must be factors of half the triangle's perimeter to reduce the number of candidate M's to check. A loop is not needed to find N once a candidate M has been found.
<br>Does not stop after the solution has been found, thus verifying there is only one solution.
<lang algol68>BEGIN # find the product of the of the Pythagorian triplet a, b, c where: #
# a + b + c = 1000, a2 + b2 = c2, a < b < c #
Line 13 ⟶ 14:
INT half perimeter = perimeter OVER 2;
INT max factor := half perimeter;
FOR m WHILE m < max factor DO
# using Euclid's formula: #
# a = m^2 - n^2, b = 2mn, c = m^2 + n^2 for some integer m, n, so #
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INT b := 2 * m * n;
INT c = m2 + n2;
IF ( a + b + c )
# have found the
▲ ELSE
IF b < a THEN INT t = a; a := b; b := t FI;
print( ( "a = ", whole( a, 0 ), ", b = ", whole( b, 0 ), ", c = ", whole( c, 0 ) ) );
print( ( "; a * b * c = ", whole( a * b * c, 0 ), newline ) )
FI;
▲ FI
FI
OD;
print( ( whole( count, 0 ), " iterations", newline ) )
END</lang>
{{out}}
<pre>
a = 200, b = 375, c = 425; a * b * c = 31875000
24 iterations
</pre>
Note that if we stopped after finding the solution, it would be 20 iterations.
=={{header|ALGOL W}}==
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