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Implement Integer.sqrt for large integers #3872

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Merged
merged 1 commit into from
May 23, 2025
+11 −2
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Implement Integer.sqrt for large integers #3872

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1 change: 0 additions & 1 deletion spec/tags/core/integer/sqrt_tags.txt
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12 changes: 11 additions & 1 deletion src/main/ruby/truffleruby/core/integer.rb
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Original file line number Diff line number Diff line change
Expand Up @@ -350,7 +350,17 @@ def self.try_convert(obj)
def self.sqrt(n)
n = Primitive.rb_to_int(n)
raise Math::DomainError if n.negative?
Math.sqrt(n).floor
return Math.sqrt(n).floor if n < 0xfffffffffffff

shift = n.bit_length / 4
x = sqrt(n >> (2 * shift)) << shift
x = (x + n / x) / 2
xx = x * x
while xx > n
xx -= 2 * x - 1
x -= 1
end
x
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Wondering how the constant 0xfffffffffffff was chosen. Is it based on benchmarking results?

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Larger is better as long as Math.sqrt(n).floor returns correct value.
0xfffffffffffff is 52bit integer and double has 53bit precision.

If n is larger than 0xfffffffffffff == 0x4000000**2-1, Math.sqrt(n).floor might not return correct value.
Integer.sqrt(0x4000001**2-1) should return 0x4000001 - 1
Math.sqrt(0x4000001**2-1).floor is 0x4000001

end

private def upto_internal(val)
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