Issue #19075 has been updated by Dan0042 (Daniel DeLorme). sawa (Tsuyoshi Sawada) wrote in #note-11:
`bsearch_rindex`
I love this one, how it has a direct parallel in the way the condition must be inverted when using `index` vs `rindex`. ```ruby a = [5,6,6,6,7] a.index{ |e| e >= 6 } #=>1 a.rindex{ |e| e <= 6 } #=>3 (inverted condition) a.bsearch_index{ |e| e >= 6 } #=>1 a.bsearch_rindex{ |e| e <= 6 } #=>3 (inverted condition just like rindex and bsearch(target: :last) implementation above) ``` ---------------------------------------- Feature #19075: Binary searching for the last element https://bugs.ruby-lang.org/issues/19075#change-104468 * Author: kyanagi (Kouhei Yanagita) * Status: Open * Priority: Normal ---------------------------------------- My latest proposal is https://bugs.ruby-lang.org/issues/19075#note-6. I will leave the initial proposal below. --- PR: https://github.com/ruby/ruby/pull/6611 (I'm going to talk about `Array` here, but the same argument can be made for `Range`. If `Array#bsearch_last` is acceptable, I will work also for `Range`.) Ruby's bsearch returns the first element which satisfies the given block. ```ruby # Search the first element greater than 18 array = [10, 15, 20, 25] array.bsearch { |x| x > 18 } # => 20 ``` If we want the last element, we need to invert the condition and step backward. ```ruby # Search the last element less than 18 array = [10, 15, 20, 25] index = array.bsearch_index { |x| !(x < 18) } array[index-1] # => 15 ``` Of course, we need to consider `nil` and the boundary. ```ruby # Search the last element less than 100 index = array.bsearch_index { |x| !(x < 100) } # => nil if index.nil? array.last # => 25 else array[index-1] end ``` ```ruby # Search the last element less than 0 index = array.bsearch_index { |x| !(x < 0) } # => 0 if index.nil? array.last elsif index == 0 nil else array[index-1] end ``` This is where mistakes can easily be made, so I propose `Array#bsearch_last` and `Array#bsearch_last_index`. `Array#bsearch_last` returns the last element which satisfies the given block. `Array#bsearch` requires that all false-evaluating elements precede all true-evaluating elements. As is clear from the meaning of the method, conversely to `bsearch`, `bsearch_last` requires that all true-evaluating elements precede all false-evaluating elements. (If `bsearch_last` is acceptable, the name "find-minimum mode" should be changed.) ```ruby array = [10, 15, 20, 25] array.bsearch_last { |x| x < 18 } # => 15 array.bsearch_last { |x| x < 100 } # => 25 array.bsearch_last { |x| x < 0 } # => nil ``` There are several possible options for find-any mode. (1) `bsearch_last` does not support find-any mode. A block for `bsearch_last` must return `true`, `false` or `nil`. ``` [1, 2, 3].bsearch_last { 0 } # => TypeError ``` My pull request tentatively includes this implementation. (2) `bsearch_last` supports find-any mode and it behaves like `bsearch`. `bsearch` with find-any mode returns an element, for which the block returns zero. If multiple elements satisfy the condition, it is not determined which of them will be returned. It is conceivable that `bsearch_last` behaves in the same way as `bsearch`. ``` # current behavior # It is not specified whether `:b`, `:c`, or `:d` is returned. [[1,:a], [2, :b], [2, :c], [2, :d], [3, :e]].bsearch { |a, b| 2 <=> a } # => [2, :c] ``` (3) `bsearch_last` supports find-any mode and returns the last element. Make `bsearch` return the first element. Change the behavior of `bsearch` to return the first element for which the block returns zero. `bsearch_last` returns the last element for which the block returns zero. ``` # Change it like this: [[1,:a], [2, :b], [2, :c], [2, :d], [3, :e]].bsearch { |a, b| 2 <=> a } # => [2, :b] [[1,:a], [2, :b], [2, :c], [2, :d], [3, :e]].bsearch_last { |a, b| 2 <=> a } # => [2, :d] ``` (If this option is adopted, the name "find-any mode" should be renamed.) -- https://bugs.ruby-lang.org/