nvidia-container-toolkit/vendor/github.com/xrash/smetrics/ukkonen.go

package smetrics

import (
	"math"
)

// The Ukkonen algorithm for calculating the Levenshtein distance. The algorithm is described in http://www.cs.helsinki.fi/u/ukkonen/InfCont85.PDF, or in docs/InfCont85.PDF. It runs on O(t . min(m, n)) where t is the actual distance between strings a and b. It needs O(min(t, m, n)) space. This function might be preferred over WagnerFischer() for *very* similar strings. But test it out yourself.
// The first two parameters are the two strings to be compared. The last three parameters are the insertion cost, the deletion cost and the substitution cost. These are normally defined as 1, 1 and 2 respectively.
func Ukkonen(a, b string, icost, dcost, scost int) int {
	var lowerCost int

	if icost < dcost && icost < scost {
		lowerCost = icost
	} else if dcost < scost {
		lowerCost = dcost
	} else {
		lowerCost = scost
	}

	infinite := math.MaxInt32 / 2

	var r []int
	var k, kprime, p, t int
	var ins, del, sub int

	if len(a) > len(b) {
		t = (len(a) - len(b) + 1) * lowerCost
	} else {
		t = (len(b) - len(a) + 1) * lowerCost
	}

	for {
		if (t / lowerCost) < (len(b) - len(a)) {
			continue
		}

		// This is the right damn thing since the original Ukkonen
		// paper minimizes the expression result only, but the uncommented version
		// doesn't need to deal with floats so it's faster.
		// p = int(math.Floor(0.5*((float64(t)/float64(lowerCost)) - float64(len(b) - len(a)))))
		p = ((t / lowerCost) - (len(b) - len(a))) / 2

		k = -p
		kprime = k

		rowlength := (len(b) - len(a)) + (2 * p)

		r = make([]int, rowlength+2)

		for i := 0; i < rowlength+2; i++ {
			r[i] = infinite
		}

		for i := 0; i <= len(a); i++ {
			for j := 0; j <= rowlength; j++ {
				if i == j+k && i == 0 {
					r[j] = 0
				} else {
					if j-1 < 0 {
						ins = infinite
					} else {
						ins = r[j-1] + icost
					}

					del = r[j+1] + dcost
					sub = r[j] + scost

					if i-1 < 0 || i-1 >= len(a) || j+k-1 >= len(b) || j+k-1 < 0 {
						sub = infinite
					} else if a[i-1] == b[j+k-1] {
						sub = r[j]
					}

					if ins < del && ins < sub {
						r[j] = ins
					} else if del < sub {
						r[j] = del
					} else {
						r[j] = sub
					}
				}
			}
			k++
		}

		if r[(len(b)-len(a))+(2*p)+kprime] <= t {
			break
		} else {
			t *= 2
		}
	}

	return r[(len(b)-len(a))+(2*p)+kprime]
}