glob/syntax/ast/optimize.go

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2018-02-16 17:36:02 +03:00
package ast
import (
"reflect"
)
// Minimize tries to apply some heuristics to minimize number of nodes in given
// t
func Minimize(t *Node) *Node {
switch t.Kind {
case KindAnyOf:
return minimizeAnyOf(t)
default:
return nil
}
}
// minimizeAnyOf tries to find common children of given node of AnyOf pattern
// it searches for common children from left and from right
// if any common children are found then it returns new optimized ast t
// else it returns nil
func minimizeAnyOf(t *Node) *Node {
if !SameKind(t.Children, KindPattern) {
return nil
}
commonLeft, commonRight := CommonChildren(t.Children)
commonLeftCount, commonRightCount := len(commonLeft), len(commonRight)
if commonLeftCount == 0 && commonRightCount == 0 { // there are no common parts
return nil
}
var result []*Node
if commonLeftCount > 0 {
result = append(result, NewNode(KindPattern, nil, commonLeft...))
}
var anyOf []*Node
for _, child := range t.Children {
reuse := child.Children[commonLeftCount : len(child.Children)-commonRightCount]
var node *Node
if len(reuse) == 0 {
// this pattern is completely reduced by commonLeft and commonRight patterns
// so it become nothing
node = NewNode(KindNothing, nil)
} else {
node = NewNode(KindPattern, nil, reuse...)
}
anyOf = AppendUnique(anyOf, node)
}
switch {
case len(anyOf) == 1 && anyOf[0].Kind != KindNothing:
result = append(result, anyOf[0])
case len(anyOf) > 1:
result = append(result, NewNode(KindAnyOf, nil, anyOf...))
}
if commonRightCount > 0 {
result = append(result, NewNode(KindPattern, nil, commonRight...))
}
return NewNode(KindPattern, nil, result...)
}
func CommonChildren(nodes []*Node) (commonLeft, commonRight []*Node) {
if len(nodes) <= 1 {
return
}
// find node that has least number of children
idx := OneWithLeastChildren(nodes)
if idx == -1 {
return
}
tree := nodes[idx]
treeLength := len(tree.Children)
// allocate max able size for rightCommon slice
// to get ability insert elements in reverse order (from end to start)
// without sorting
commonRight = make([]*Node, treeLength)
lastRight := treeLength // will use this to get results as commonRight[lastRight:]
var (
breakLeft bool
breakRight bool
commonTotal int
)
for i, j := 0, treeLength-1; commonTotal < treeLength && j >= 0 && !(breakLeft && breakRight); i, j = i+1, j-1 {
treeLeft := tree.Children[i]
treeRight := tree.Children[j]
for k := 0; k < len(nodes) && !(breakLeft && breakRight); k++ {
// skip least children node
if k == idx {
continue
}
restLeft := nodes[k].Children[i]
restRight := nodes[k].Children[j+len(nodes[k].Children)-treeLength]
breakLeft = breakLeft || !treeLeft.Equal(restLeft)
// disable searching for right common parts, if left part is already overlapping
breakRight = breakRight || (!breakLeft && j <= i)
breakRight = breakRight || !treeRight.Equal(restRight)
}
if !breakLeft {
commonTotal++
commonLeft = append(commonLeft, treeLeft)
}
if !breakRight {
commonTotal++
lastRight = j
commonRight[j] = treeRight
}
}
commonRight = commonRight[lastRight:]
return
}
func AppendUnique(target []*Node, val *Node) []*Node {
for _, n := range target {
if reflect.DeepEqual(n, val) {
return target
}
}
return append(target, val)
}
func SameKind(nodes []*Node, kind Kind) bool {
for _, n := range nodes {
if n.Kind != kind {
return false
}
}
return true
}
func OneWithLeastChildren(nodes []*Node) int {
min := -1
idx := -1
for i, n := range nodes {
if idx == -1 || (len(n.Children) < min) {
min = len(n.Children)
idx = i
}
}
return idx
}
func Equal(a, b []*Node) bool {
if len(a) != len(b) {
return false
}
for i, av := range a {
if !av.Equal(b[i]) {
return false
}
}
return true
}