debian-mirror-gitlab/lib/gitlab/ci/pipeline_duration.rb

module Gitlab
  module Ci
    # # Introduction - total running time
    #
    # The problem this module is trying to solve is finding the total running
    # time amongst all the jobs, excluding retries and pending (queue) time.
    # We could reduce this problem down to finding the union of periods.
    #
    # So each job would be represented as a `Period`, which consists of
    # `Period#first` as when the job started and `Period#last` as when the
    # job was finished. A simple example here would be:
    #
    # * A (1, 3)
    # * B (2, 4)
    # * C (6, 7)
    #
    # Here A begins from 1, and ends to 3. B begins from 2, and ends to 4.
    # C begins from 6, and ends to 7. Visually it could be viewed as:
    #
    #     0  1  2  3  4  5  6  7
    #        AAAAAAA
    #           BBBBBBB
    #                       CCCC
    #
    # The union of A, B, and C would be (1, 4) and (6, 7), therefore the
    # total running time should be:
    #
    #     (4 - 1) + (7 - 6) => 4
    #
    # # The Algorithm
    #
    # The algorithm used here for union would be described as follow.
    # First we make sure that all periods are sorted by `Period#first`.
    # Then we try to merge periods by iterating through the first period
    # to the last period. The goal would be merging all overlapped periods
    # so that in the end all the periods are discrete. When all periods
    # are discrete, we're free to just sum all the periods to get real
    # running time.
    #
    # Here we begin from A, and compare it to B. We could find that
    # before A ends, B already started. That is `B.first <= A.last`
    # that is `2 <= 3` which means A and B are overlapping!
    #
    # When we found that two periods are overlapping, we would need to merge
    # them into a new period and disregard the old periods. To make a new
    # period, we take `A.first` as the new first because remember? we sorted
    # them, so `A.first` must be smaller or equal to `B.first`. And we take
    # `[A.last, B.last].max` as the new last because we want whoever ended
    # later. This could be broken into two cases:
    #
    #     0  1  2  3  4
    #        AAAAAAA
    #           BBBBBBB
    #
    # Or:
    #
    #     0  1  2  3  4
    #        AAAAAAAAAA
    #           BBBB
    #
    # So that we need to take whoever ends later. Back to our example,
    # after merging and discard A and B it could be visually viewed as:
    #
    #     0  1  2  3  4  5  6  7
    #        DDDDDDDDDD
    #                       CCCC
    #
    # Now we could go on and compare the newly created D and the old C.
    # We could figure out that D and C are not overlapping by checking
    # `C.first <= D.last` is `false`. Therefore we need to keep both C
    # and D. The example would end here because there are no more jobs.
    #
    # After having the union of all periods, we just need to sum the length
    # of all periods to get total time.
    #
    #     (4 - 1) + (7 - 6) => 4
    #
    # That is 4 is the answer in the example.
    module PipelineDuration
      extend self

      Period = Struct.new(:first, :last) do
        def duration
          last - first
        end
      end

      def from_pipeline(pipeline)
        status = %w[success failed running canceled]
        builds = pipeline.builds.latest
          .where(status: status).where.not(started_at: nil).order(:started_at)

        from_builds(builds)
      end

      def from_builds(builds)
        now = Time.now

        periods = builds.map do |b|
          Period.new(b.started_at, b.finished_at || now)
        end

        from_periods(periods)
      end

      # periods should be sorted by `first`
      def from_periods(periods)
        process_duration(process_periods(periods))
      end

      private

      def process_periods(periods)
        return periods if periods.empty?

        periods.drop(1).inject([periods.first]) do |result, current|
          previous = result.last

          if overlap?(previous, current)
            result[-1] = merge(previous, current)
            result
          else
            result << current
          end
        end
      end

      def overlap?(previous, current)
        current.first <= previous.last
      end

      def merge(previous, current)
        Period.new(previous.first, [previous.last, current.last].max)
      end

      def process_duration(periods)
        periods.sum(&:duration)
      end
    end
  end
end