v2_1_10 - Show latest stable - 0 notes - Class: Range
bsearch() public

By using binary search, finds a value in range which meets the given condition in O(log n) where n is the size of the range.

You can use this method in two use cases: a find-minimum mode and a find-any mode. In either case, the elements of the range must be monotone (or sorted) with respect to the block.

In find-minimum mode (this is a good choice for typical use case), the block must return true or false, and there must be a value x so that:

  • the block returns false for any value which is less than x, and

  • the block returns true for any value which is greater than or equal to i.

If x is within the range, this method returns the value x. Otherwise, it returns nil.

ary = [0, 4, 7, 10, 12]
(0...ary.size).bsearch {|i| ary[i] >= 4 } #=> 1
(0...ary.size).bsearch {|i| ary[i] >= 6 } #=> 2
(0...ary.size).bsearch {|i| ary[i] >= 8 } #=> 3
(0...ary.size).bsearch {|i| ary[i] >= 100 } #=> nil

(0.0...Float::INFINITY).bsearch {|x| Math.log(x) >= 0 } #=> 1.0

In find-any mode (this behaves like libc’s bsearch(3)), the block must return a number, and there must be two values x and y (x <= y) so that:

  • the block returns a positive number for v if v < x,

  • the block returns zero for v if x <= v < y, and

  • the block returns a negative number for v if y <= v.

This method returns any value which is within the intersection of the given range and x…y (if any). If there is no value that satisfies the condition, it returns nil.

ary = [0, 100, 100, 100, 200]
(0..4).bsearch {|i| 100 - ary[i] } #=> 1, 2 or 3
(0..4).bsearch {|i| 300 - ary[i] } #=> nil
(0..4).bsearch {|i|  50 - ary[i] } #=> nil

You must not mix the two modes at a time; the block must always return either true/false, or always return a number. It is undefined which value is actually picked up at each iteration.

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