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