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// Copyright 2011 The Go Authors.  All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.

package syntax

// Simplify returns a regexp equivalent to re but without counted repetitions
// and with various other simplifications, such as rewriting /(?:a+)+/ to /a+/.
// The resulting regexp will execute correctly but its string representation
// will not produce the same parse tree, because capturing parentheses
// may have been duplicated or removed.  For example, the simplified form
// for /(x){1,2}/ is /(x)(x)?/ but both parentheses capture as $1.
// The returned regexp may share structure with or be the original.
func (re *Regexp) Simplify() *Regexp {
        if re == nil {
                return nil
        }
        switch re.Op {
        case OpCapture, OpConcat, OpAlternate:
                // Simplify children, building new Regexp if children change.
                nre := re
                for i, sub := range re.Sub {
                        nsub := sub.Simplify()
                        if nre == re && nsub != sub {
                                // Start a copy.
                                nre = new(Regexp)
                                *nre = *re
                                nre.Rune = nil
                                nre.Sub = append(nre.Sub0[:0], re.Sub[:i]...)
                        }
                        if nre != re {
                                nre.Sub = append(nre.Sub, nsub)
                        }
                }
                return nre

        case OpStar, OpPlus, OpQuest:
                sub := re.Sub[0].Simplify()
                return simplify1(re.Op, re.Flags, sub, re)

        case OpRepeat:
                // Special special case: x{0} matches the empty string
                // and doesn't even need to consider x.
                if re.Min == 0 && re.Max == 0 {
                        return &Regexp{Op: OpEmptyMatch}
                }

                // The fun begins.
                sub := re.Sub[0].Simplify()

                // x{n,} means at least n matches of x.
                if re.Max == -1 {
                        // Special case: x{0,} is x*.
                        if re.Min == 0 {
                                return simplify1(OpStar, re.Flags, sub, nil)
                        }

                        // Special case: x{1,} is x+.
                        if re.Min == 1 {
                                return simplify1(OpPlus, re.Flags, sub, nil)
                        }

                        // General case: x{4,} is xxxx+.
                        nre := &Regexp{Op: OpConcat}
                        nre.Sub = nre.Sub0[:0]
                        for i := 0; i < re.Min-1; i++ {
                                nre.Sub = append(nre.Sub, sub)
                        }
                        nre.Sub = append(nre.Sub, simplify1(OpPlus, re.Flags, sub, nil))
                        return nre
                }

                // Special case x{0} handled above.

                // Special case: x{1} is just x.
                if re.Min == 1 && re.Max == 1 {
                        return sub
                }

                // General case: x{n,m} means n copies of x and m copies of x?
                // The machine will do less work if we nest the final m copies,
                // so that x{2,5} = xx(x(x(x)?)?)?

                // Build leading prefix: xx.
                var prefix *Regexp
                if re.Min > 0 {
                        prefix = &Regexp{Op: OpConcat}
                        prefix.Sub = prefix.Sub0[:0]
                        for i := 0; i < re.Min; i++ {
                                prefix.Sub = append(prefix.Sub, sub)
                        }
                }

                // Build and attach suffix: (x(x(x)?)?)?
                if re.Max > re.Min {
                        suffix := simplify1(OpQuest, re.Flags, sub, nil)
                        for i := re.Min + 1; i < re.Max; i++ {
                                nre2 := &Regexp{Op: OpConcat}
                                nre2.Sub = append(nre2.Sub0[:0], sub, suffix)
                                suffix = simplify1(OpQuest, re.Flags, nre2, nil)
                        }
                        if prefix == nil {
                                return suffix
                        }
                        prefix.Sub = append(prefix.Sub, suffix)
                }
                if prefix != nil {
                        return prefix
                }

                // Some degenerate case like min > max or min < max < 0.
                // Handle as impossible match.
                return &Regexp{Op: OpNoMatch}
        }

        return re
}

// simplify1 implements Simplify for the unary OpStar,
// OpPlus, and OpQuest operators.  It returns the simple regexp
// equivalent to
//
//      Regexp{Op: op, Flags: flags, Sub: {sub}}
//
// under the assumption that sub is already simple, and
// without first allocating that structure.  If the regexp
// to be returned turns out to be equivalent to re, simplify1
// returns re instead.
//
// simplify1 is factored out of Simplify because the implementation
// for other operators generates these unary expressions.
// Letting them call simplify1 makes sure the expressions they
// generate are simple.
func simplify1(op Op, flags Flags, sub, re *Regexp) *Regexp {
        // Special case: repeat the empty string as much as
        // you want, but it's still the empty string.
        if sub.Op == OpEmptyMatch {
                return sub
        }
        // The operators are idempotent if the flags match.
        if op == sub.Op && flags&NonGreedy == sub.Flags&NonGreedy {
                return sub
        }
        if re != nil && re.Op == op && re.Flags&NonGreedy == flags&NonGreedy && sub == re.Sub[0] {
                return re
        }

        re = &Regexp{Op: op, Flags: flags}
        re.Sub = append(re.Sub0[:0], sub)
        return re
}