间隔和一组间隔之间的区别？

Question

我有一组不重叠、不相邻的间隔，例如。 [{10,15}，{30,35}，{20,25}]。它们没有排序，但我可以根据需要对其进行排序。

现在我得到了一些新的间隔，例如。 {5,32} 并希望生成一组新的区间来描述差异：这个新区间所涵盖的不在集合中的范围。在本例中，答案是：[{5,9}, {16,19}, {26,29}]。

计算这个的快速算法是什么？请注意，该集合通常包含 1 个，有时 2 个，很少有 3 个或更多项目，因此我想针对这种情况进行优化。

对于上下文，这里是最初从开始+结束数据的输入流创建集合的代码，我在其中进行合并：

type Interval struct {
    start int
    end   int
}

func (i *Interval) OverlapsOrAdjacent(j Interval) bool {
    return i.end+1 >= j.start && j.end+1 >= i.start
}

func (i *Interval) Merge(j Interval) bool {
    if !i.OverlapsOrAdjacent(j) {
        return false
    }
    if j.start < i.start {
        i.start = j.start
    }
    if j.end > i.end {
        i.end = j.end
    }
    return true
}

type Intervals []Interval

func (ivs Intervals) Len() int           { return len(ivs) }
func (ivs Intervals) Swap(i, j int)      { ivs[i], ivs[j] = ivs[j], ivs[i] }
func (ivs Intervals) Less(i, j int) bool { return ivs[i].start < ivs[j].start }

func (ivs Intervals) Merge(iv Interval) Intervals {
    ivs = append(ivs, iv)
    merged := make(Intervals, 0, len(ivs))
    for _, iv := range ivs {
        for i := 0; i < len(merged); {
            if iv.Merge(merged[i]) {
                merged = append(merged[:i], merged[i+1:]...)
            } else {
                i++
            }
        }
        merged = append(merged, iv)
    }
    return merged
}

func (ivs Intervals) MergeUsingSort(iv Interval) Intervals {
    ivs = append(ivs, iv)
    sort.Sort(ivs)
    merged := make(Intervals, 0, len(ivs))
    merged = append(merged, ivs[0])
    for i := 1; i < len(ivs); i++ {
        last := len(merged) - 1
        if !merged[last].Merge(ivs[i]) {
            merged = append(merged, ivs[i])
        }
    }
    return merged
}

func (ivs Intervals) Difference(iv Interval) Intervals {
    // ???
    return ivs
}

func main() {
    var ivs Intervals
    for _, input := range inputsFromSomewhere { // in reality, I don't have all these inputs at once, they come in one at a time
        iv := Interval{input.start, input.end}
        diffs := ivs.Difference(iv) // not yet implemented...
        // do something with diffs
        ivs = ivs.Merge(iv)
    }
}

我发现上面的 Intervals.Merge() 比 MergeUsingSort() 快 2 倍，所以我想知道是否还有一种简单的非排序方式来回答我的问题。

Answer 1

问答代码不完整，无法编译。没有基准。快速浏览一下代码，它可能效率低下。

interval.go 和 interval_test.go 的可用代码是从 https://github.com/VertebrateResequencing/wr/tree/develop/minfys 获得的。

让我们从为区间差示例编写基准开始。

package minfys

import (
    "fmt"
    "testing"
)

// Example
var (
    xA = Intervals{{10, 15}, {30, 35}, {20, 25}}
    xB = Interval{5, 32}
    xD = Intervals{{5, 9}, {16, 19}, {26, 29}}
    xR = Intervals{}
)

func BenchmarkExample(b *testing.B) {
    b.ReportAllocs()
    a := make(Intervals, len(xA))
    b.ResetTimer()
    for i := 0; i < b.N; i++ {
        copy(a, xA)
        xR = a.Difference(xB)
    }
    b.StopTimer()
    if fmt.Sprint(xD) != fmt.Sprint(xR) {
        b.Fatal(xD, xR)
    }
}

接下来，写一个Difference方法。

package minfys

func (a Intervals) Difference(b Interval) Intervals {
    // If A and B are sets, then the relative complement of A in B
    // is the set of elements in B but not in A.
    // The relative complement of A in B is denoted B ∖  A:
    //     B \ A = {x ∈ A | x ∉ B}
    //     B \ A = B ∩ A'
    //
    // For example. d = a\b,
    //     a: [{10, 15}, {30, 35}, {20, 25}]
    //     b: {5,32}
    //     d: [{5,9}, {16,19}, {26,29}]
    // The elements of set a are non-overlapping, non-adjacent,
    // and unsorted intervals.

    if len(a) <= 0 {
        return Intervals{b}
    }

    d := make(Intervals, 0, 3)
    for ; len(a) > 0; a = a[1:] {
        for i := 1; i < len(a); i++ {
            if a[i].Start < a[0].Start {
                a[i], a[0] = a[0], a[i]
            }
        }

        if b.Start < a[0].Start {
            if b.End < a[0].Start {
                d = append(d, b)
                break
            }
            d = append(d, Interval{b.Start, a[0].Start - 1})
            b.Start = a[0].Start
        }
        if b.End <= a[0].End {
            break
        }
        if b.Start <= a[0].End {
            b.Start = a[0].End + 1
        }
        if len(a) == 1 {
            if b.Start <= a[0].End {
                b.Start = a[0].End + 1
            }
            d = append(d, b)
            break
        }
    }
    return d
}

现在，对差异法进行基准测试。

BenchmarkExample-4     20000000     62.4 ns/op    48 B/op      1 allocs/op

sbs 写了一个Difference方法。

// Interval struct is used to describe something with a start and end. End must
// be greater than start.
type Interval struct {
    Start int64
    End   int64
}

// Overlaps returns true if this interval overlaps with the supplied one.
func (i *Interval) Overlaps(j Interval) bool {
    // https://nedbatchelder.com/blog/201310/range_overlap_in_two_compares.html
    return i.End >= j.Start && j.End >= i.Start
}

// Intervals type is a slice of Interval.
type Intervals []Interval

// Difference returns any portions of iv that do not overlap with any of our
// intervals. Assumes that all of our intervals have been Merge()d in.
func (ivs Intervals) Difference(iv Interval) (diffs Intervals) {
    diffs = append(diffs, iv)
    for _, prior := range ivs {
        for i := 0; i < len(diffs); {
            if left, right, overlapped := prior.Difference(diffs[i]); overlapped {
                if len(diffs) == 1 {
                    diffs = nil
                } else {
                    diffs = append(diffs[:i], diffs[i+1:]...)
                }

                if left != nil {
                    diffs = append(diffs, *left)
                }
                if right != nil {
                    diffs = append(diffs, *right)
                }
            } else {
                i++
            }
        }
        if len(diffs) == 0 {
            break
        }
    }

    return
}

基准sbs的Difference方法。

BenchmarkExample-4      5000000    365 ns/op     128 B/op      4 allocs/op

peterSO 的 Difference 方法明显更快。

old.txt (sbs) versus new.txt (peterSO):

benchmark              old ns/op     new ns/op     delta
BenchmarkExample-4     365           62.4          -82.90%

benchmark              old allocs     new allocs   delta
BenchmarkExample-4     4              1            -75.00%

benchmark              old bytes     new bytes     delta
BenchmarkExample-4     128           48            -62.50%

这只是一个开始。可能还有其他可以改进的地方。

interval_test.go 中有一些错误。 ShouldBeNil 用于指针； ShouldBeEmpty 用于收藏。 ShouldResemble 不处理集合相等（包含相同元素的两个集合是同一个集合）。更改 ShouldResemble 顺序以匹配实现相关的顺序。

$ go test
..........................................................................................................................x......................................................x................x
Failures:

  * interval_test.go 
  Line 247:
  Expected: nil
  Actual:   '[]'

  * interval_test.go 
  Line 375:
  Expected: 'minfys.Intervals{minfys.Interval{Start:5, End:6}, minfys.Interval{Start:31, End:32}, minfys.Interval{Start:11, End:14}, minfys.Interval{Start:19, End:19}}'
  Actual:   'minfys.Intervals{minfys.Interval{Start:5, End:6}, minfys.Interval{Start:11, End:14}, minfys.Interval{Start:19, End:19}, minfys.Interval{Start:31, End:32}}'
  (Should resemble)!

  * interval_test.go 
  Line 413:
  Expected: 'minfys.Intervals{minfys.Interval{Start:7, End:10}, minfys.Interval{Start:1, End:3}, minfys.Interval{Start:15, End:17}}'
  Actual:   'minfys.Intervals{minfys.Interval{Start:1, End:3}, minfys.Interval{Start:7, End:10}, minfys.Interval{Start:15, End:17}}'
  (Should resemble)!


195 total assertions

...
198 total assertions

--- FAIL: TestIntervals (0.04s)
FAIL

.

$ diff -a -u ../interval_test.go interval_test.go
--- ../interval_test.go 2017-04-29 20:23:29.365344008 -0400
+++ interval_test.go    2017-04-29 20:54:14.349344903 -0400
@@ -244,19 +244,19 @@
            So(len(ivs), ShouldEqual, 1)

            newIvs = ivs.Difference(twoSix)
-           So(newIvs, ShouldBeNil)
+           So(newIvs, ShouldBeEmpty)
            ivs = ivs.Merge(twoSix)
            So(len(ivs), ShouldEqual, 1)

            newIvs = ivs.Difference(oneThree)
-           So(newIvs, ShouldBeNil)
+           So(newIvs, ShouldBeEmpty)
            ivs = ivs.Merge(oneThree)
            So(len(ivs), ShouldEqual, 1)

            oneSeven := Interval{1, 7}

            newIvs = ivs.Difference(oneSix)
-           So(newIvs, ShouldBeNil)
+           So(newIvs, ShouldBeEmpty)
            ivs = ivs.Merge(oneSix)
            So(len(ivs), ShouldEqual, 1)

@@ -372,7 +372,7 @@

            fiveThirtyTwo := Interval{5, 32}
            newIvs = ivs.Difference(fiveThirtyTwo)
-           So(newIvs, ShouldResemble, Intervals{Interval{5, 6}, Interval{31, 32}, Interval{11, 14}, Interval{19, 19}})
+           So(newIvs, ShouldResemble, Intervals{Interval{5, 6}, Interval{11, 14}, Interval{19, 19}, Interval{31, 32}})
            ivs = ivs.Merge(fiveThirtyTwo)
            So(len(ivs), ShouldEqual, 3)

@@ -409,7 +409,7 @@

            ivs = ivs.Truncate(17)

-           expected := Intervals{sevenTen, oneThree, Interval{15, 17}}
+           expected := Intervals{oneThree, sevenTen, Interval{15, 17}}
            So(ivs, ShouldResemble, expected)
        })
    })

.

$ go test
.............................................................................................................................................................................................................
205 total assertions

...
208 total assertions

PASS
$ 

---

我 [@sbs] 确认它比我的解决方案更快。虽然如果你只是 测量使用 Difference() 所花费的挂墙时间（在 := 之前放一个 在 interval_test.go 中最后一次 Difference() 调用之前的 time.Now()， 和一个 time.Since(before) 在它之后并将这些持续时间相加 循环），它似乎几乎没有什么不同（在我的机器上 我的解决方案大约需要 31 毫秒，而你的解决方案大约需要 29 毫秒）。

根据要求，修改了interval_test.go 以测量挂墙时间：

$ diff -a -u ../interval_test.go walltime_test.go
--- ../interval_test.go 2017-04-29 20:23:29.365344008 -0400
+++ walltime_test.go    2017-04-30 13:39:29.000000000 -0400
@@ -24,6 +24,7 @@
    "math/rand"
    "testing"
    "time"
+   "fmt"
 )

 func TestIntervals(t *testing.T) {
@@ -459,16 +460,20 @@

        var ivs Intervals
        errors := 0
+       var diffTime time.Duration
        t := time.Now()
        for i, input := range inputs {
            iv := NewInterval(int64(input), int64(readSize))
+           before := time.Now()
            newIvs := ivs.Difference(iv)
+           diffTime += time.Since(before)
            if (len(newIvs) == 1) != exepectedNew[i] {
                errors++
            }
            ivs = ivs.Merge(iv)
        }
-       // fmt.Printf("\ntook %s\n", time.Since(t))
+       fmt.Printf("took %s\n", time.Since(t))
+       fmt.Printf("\n  Difference took %s\n", diffTime)
        So(errors, ShouldEqual, 0)
        So(len(ivs), ShouldEqual, 1)
        So(time.Since(t).Seconds(), ShouldBeLessThan, 1) // 42ms on my machine
$ 

interval_test.go 基准输入大小和频率是

size    frequency
0       1
1       94929
2       50072
3       4998

输出大小和频率是

size    frequency
0       50000
1       100000

为此分布调整 peterSo 的差分方法给出了

package minfys

func (a Intervals) Difference(b Interval) Intervals {
    // If A and B are sets, then the relative complement of A in B
    // is the set of elements in B but not in A.
    // The relative complement of A in B is denoted B ∖  A:
    //     B \ A = {x ∈ A | x ∉ B}
    //     B \ A = B ∩ A'
    //
    // For example. d = a\b,
    //     a: [{10, 15}, {30, 35}, {20, 25}]
    //     b: {5,32}
    //     d: [{5,9}, {16,19}, {26,29}]
    // The elements of set a are non-overlapping, non-adjacent,
    // and unsorted intervals.

    if len(a) <= 0 {
        return Intervals{b}
    }

    var d Intervals
    for ; len(a) > 0; a = a[1:] {
        for i := 1; i < len(a); i++ {
            if a[i].Start < a[0].Start {
                a[i], a[0] = a[0], a[i]
            }
        }

        if b.Start < a[0].Start {
            if b.End < a[0].Start {
                d = append(d, b)
                break
            }
            d = append(d, Interval{b.Start, a[0].Start - 1})
            b.Start = a[0].Start
        }
        if b.End <= a[0].End {
            break
        }
        if b.Start <= a[0].End {
            b.Start = a[0].End + 1
        }
        if len(a) == 1 {
            if b.Start <= a[0].End {
                b.Start = a[0].End + 1
            }
            d = append(d, b)
            break
        }
    }
    return d
}

运行 interval_test.go 基准测试 peterSO 和 sbs 的差异方法给出了

$ go test -v

  Merging many intervals is fast took 26.208614ms

  Difference took 10.706858ms

和

$ go test -v

  Merging many intervals is fast took 30.799216ms

  Difference took 14.414488ms

peterSo 的差值方法明显快于 sbs 的方法：10.706858 毫秒与 14.414488 毫秒或负 25.7%。

更新 peterSO 修改后的差值方法的早期示例基准测试结果

old.txt (sbs) versus new.txt (peterSO):

benchmark              old ns/op     new ns/op     delta
BenchmarkExample-4     365           221           -39.45%

benchmark              old allocs     new allocs   delta
BenchmarkExample-4     4              3            -25.00%

benchmark              old bytes     new bytes     delta
BenchmarkExample-4     128           112           -12.50%

Answer 2

为了回答我自己的问题，这是我的 Difference() 实现，它（在我的输入数据上）比例如更快。 JimB 的建议需要排序。

func (i *Interval) Overlaps(j Interval) bool {
    return i.End >= j.Start && j.End >= i.Start
}

func (i *Interval) Difference(j Interval) (left *Interval, right *Interval, overlapped bool) {
    if !i.Overlaps(j) {
        return
    }

    overlapped = true
    if j.Start < i.Start {
        left = &Interval{j.Start, i.Start - 1}
    }
    if j.End > i.End {
        right = &Interval{i.End + 1, j.End}
    }
    return
}

func (ivs Intervals) Difference(iv Interval) (diffs Intervals) {
    diffs = append(diffs, iv)
    for _, prior := range ivs {
        for i := 0; i < len(diffs); {
            if left, right, overlapped := prior.Difference(diffs[i]); overlapped {
                if len(diffs) == 1 {
                    diffs = nil
                } else {
                    diffs = append(diffs[:i], diffs[i+1:]...)
                }

                if left != nil {
                    diffs = append(diffs, *left)
                }
                if right != nil {
                    diffs = append(diffs, *right)
                }
            } else {
                i++
            }
        }
        if len(diffs) == 0 {
            break
        }
    }
    return
}

它适用于我尝试过的数据，但我有点担心我可能会错过一个得到错误答案的极端情况。