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115 lines
2.4 KiB
115 lines
2.4 KiB
8 years ago
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// Copyright (c) 2014 Couchbase, Inc.
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//
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// Licensed under the Apache License, Version 2.0 (the "License");
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// you may not use this file except in compliance with the License.
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// You may obtain a copy of the License at
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//
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// http://www.apache.org/licenses/LICENSE-2.0
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//
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// Unless required by applicable law or agreed to in writing, software
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// distributed under the License is distributed on an "AS IS" BASIS,
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// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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// See the License for the specific language governing permissions and
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// limitations under the License.
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package search
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import (
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"math"
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)
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func LevenshteinDistance(a, b string) int {
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la := len(a)
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lb := len(b)
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d := make([]int, la+1)
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var lastdiag, olddiag, temp int
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for i := 1; i <= la; i++ {
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d[i] = i
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}
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for i := 1; i <= lb; i++ {
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d[0] = i
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lastdiag = i - 1
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for j := 1; j <= la; j++ {
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olddiag = d[j]
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min := d[j] + 1
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if (d[j-1] + 1) < min {
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min = d[j-1] + 1
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}
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if a[j-1] == b[i-1] {
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temp = 0
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} else {
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temp = 1
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}
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if (lastdiag + temp) < min {
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min = lastdiag + temp
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}
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d[j] = min
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lastdiag = olddiag
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}
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}
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return d[la]
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}
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// LevenshteinDistanceMax same as LevenshteinDistance but
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// attempts to bail early once we know the distance
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// will be greater than max
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// in which case the first return val will be the max
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// and the second will be true, indicating max was exceeded
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func LevenshteinDistanceMax(a, b string, max int) (int, bool) {
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6 years ago
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v, wasMax, _ := LevenshteinDistanceMaxReuseSlice(a, b, max, nil)
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return v, wasMax
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}
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func LevenshteinDistanceMaxReuseSlice(a, b string, max int, d []int) (int, bool, []int) {
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8 years ago
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la := len(a)
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lb := len(b)
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ld := int(math.Abs(float64(la - lb)))
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if ld > max {
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6 years ago
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return max, true, d
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8 years ago
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}
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6 years ago
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if cap(d) < la+1 {
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d = make([]int, la+1)
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}
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d = d[:la+1]
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8 years ago
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var lastdiag, olddiag, temp int
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for i := 1; i <= la; i++ {
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d[i] = i
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}
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for i := 1; i <= lb; i++ {
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d[0] = i
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lastdiag = i - 1
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rowmin := max + 1
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for j := 1; j <= la; j++ {
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olddiag = d[j]
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min := d[j] + 1
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if (d[j-1] + 1) < min {
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min = d[j-1] + 1
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}
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if a[j-1] == b[i-1] {
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temp = 0
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} else {
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temp = 1
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}
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if (lastdiag + temp) < min {
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min = lastdiag + temp
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}
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if min < rowmin {
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rowmin = min
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}
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d[j] = min
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lastdiag = olddiag
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}
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// after each row if rowmin isn't less than max stop
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if rowmin > max {
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6 years ago
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return max, true, d
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8 years ago
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}
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}
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6 years ago
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return d[la], false, d
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8 years ago
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}
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