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Programming-Idioms

  • JS
  • C++

Idiom #361 Test for perfect squarity

Set the boolean b to true if the integer n is a square number, false otherwise

E.g. 2025 → true

math
b := round(sqrt(n)) ** 2 = n;

Lazy: just square the square root to see if you get the original value again
b := number mod (round(sqrt(number)) = 0;

Basically the same as checking that squaring the result gives back the original, just in a different way (not necessarily faster).
import static java.lang.Math.sqrt;
boolean b = sqrt(n) % 1 == 0;
New implementation...
< > ⌨
steenslag 
b := round(sqrt(n)) ** 2 = n;
b := number mod (round(sqrt(number)) = 0;
boolean b = sqrt(n) % 1 == 0;
b = math.sqrt(n).is_integer()
class Integer
  def perfect_square?
    Integer.sqrt(self)**2 == self
  end
end

b = n.perfect_square?