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1aa9ac14cc
This algorithm is the usual first step when generating diffs. This package is a bit of a strange home for it, but since it works with changes to cty.Value this feels more natural than any other place it could be.
105 lines
2.5 KiB
Go
105 lines
2.5 KiB
Go
package objchange
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import (
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"github.com/zclconf/go-cty/cty"
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)
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// LongestCommonSubsequence finds a sequence of values that are common to both
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// x and y, with the same relative ordering as in both collections. This result
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// is useful as a first step towards computing a diff showing added/removed
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// elements in a sequence.
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//
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// The approached used here is a "naive" one, assuming that both xs and ys will
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// generally be small in most reasonable Terraform configurations. For larger
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// lists the time/space usage may be sub-optimal.
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//
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// A pair of lists may have multiple longest common subsequences. In that
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// case, the one selected by this function is undefined.
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func LongestCommonSubsequence(xs, ys []cty.Value) []cty.Value {
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if len(xs) == 0 || len(ys) == 0 {
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return make([]cty.Value, 0)
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}
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c := make([]int, len(xs)*len(ys))
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eqs := make([]bool, len(xs)*len(ys))
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w := len(xs)
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for y := 0; y < len(ys); y++ {
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for x := 0; x < len(xs); x++ {
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eqV := xs[x].Equals(ys[y])
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eq := false
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if eqV.IsKnown() && eqV.True() {
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eq = true
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eqs[(w*y)+x] = true // equality tests can be expensive, so cache it
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}
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if eq {
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// Sequence gets one longer than for the cell at top left,
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// since we'd append a new item to the sequence here.
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if x == 0 || y == 0 {
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c[(w*y)+x] = 1
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} else {
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c[(w*y)+x] = c[(w*(y-1))+(x-1)] + 1
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}
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} else {
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// We follow the longest of the sequence above and the sequence
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// to the left of us in the matrix.
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l := 0
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u := 0
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if x > 0 {
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l = c[(w*y)+(x-1)]
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}
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if y > 0 {
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u = c[(w*(y-1))+x]
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}
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if l > u {
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c[(w*y)+x] = l
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} else {
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c[(w*y)+x] = u
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}
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}
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}
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}
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// The bottom right cell tells us how long our longest sequence will be
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seq := make([]cty.Value, c[len(c)-1])
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// Now we will walk back from the bottom right cell, finding again all
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// of the equal pairs to construct our sequence.
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x := len(xs) - 1
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y := len(ys) - 1
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i := len(seq) - 1
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for x > -1 && y > -1 {
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if eqs[(w*y)+x] {
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// Add the value to our result list and then walk diagonally
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// up and to the left.
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seq[i] = xs[x]
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x--
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y--
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i--
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} else {
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// Take the path with the greatest sequence length in the matrix.
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l := 0
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u := 0
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if x > 0 {
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l = c[(w*y)+(x-1)]
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}
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if y > 0 {
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u = c[(w*(y-1))+x]
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}
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if l > u {
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x--
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} else {
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y--
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}
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}
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}
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if i > -1 {
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// should never happen if the matrix was constructed properly
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panic("not enough elements in sequence")
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}
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return seq
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}
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