Part two of Levenshtein Distance

lev_part_two.go

package main
import "fmt"

// PART TWO: step through each string character and vector/cell of dynamic programming table to determine difference.
//// This handles the case where both characters are an exact match, and only the "no-change" condition is used.

func LevenshteinTwo(s1, s2 string) {
  m := len(s1)
	n := len(s2)
	width := n - 1
	fmt.Println("width =",width)
	vcell := make([]int, m * n)

	// cells for i of m(s1)
	for i := 1; i < m; i++ {
		vcell[i * width + 0] = i
		fmt.Println("i+=",i, "for len(str1)==",m, "at vector cell", vcell[i * width + 0]*width)
	}

	// cell for j of n(s2)
	for j := 1; j < n; j++ {
		vcell[0 * width + j] = j
		fmt.Println("j+=",j, "for len(str2)==",n, "at vector cell", vcell[0 * width + j])
	}
	fmt.Println("Vector Cell:\n", vcell)
	for j := 1; j < n; j++ {
		for i := 1; i < m; i++ {
			if s1[i] == s2[j] {
				//a fancy way of not changing cell by essentially 
				//vcell[i * width + j] = vcell[(i - 1) * width + (j - 1)]
				fmt.Println("\n\nChecking Index",(i * width + j), "with value =", vcell[i * width + j], )
				fmt.Println("doing math for ...:", "((",i, "- 1) =", i -1, "*", width, "+", "(", j, "-1) =", j - 1,")", "==", ((i -1)*width+(j-1)), "THIS IS THE INDEX WE OPERATE ON")
				fmt.Println("******* RESULT: Set index", (i*width+j), "To value=", vcell[(i - 1) * width + (j - 1)], "for comparison of string indexes", i, "," ,j)
				// find index and set result of maths to its value
				vcell[i * width + j] = vcell[(i - 1) * width + (j - 1)]
				fmt.Println("        indexes i/j are same char:", string(s1[i]), string(s2[j]))
				fmt.Println("Therefore make no changes", vcell)
			} else {
			}
		}
	}
}

func main() {
	LevenshteinTwo("string","string")
}

/*
*** OUTPUT
width = 5
i+= 1 for len(str1)== 6 at vector cell 5
i+= 2 for len(str1)== 6 at vector cell 10
i+= 3 for len(str1)== 6 at vector cell 15
i+= 4 for len(str1)== 6 at vector cell 20
i+= 5 for len(str1)== 6 at vector cell 25
j+= 1 for len(str2)== 6 at vector cell 1
j+= 2 for len(str2)== 6 at vector cell 2
j+= 3 for len(str2)== 6 at vector cell 3
j+= 4 for len(str2)== 6 at vector cell 4
j+= 5 for len(str2)== 6 at vector cell 5
Vector Cell:
 [0 1 2 3 4 5 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0]


Checking Index 6 with value = 0
doing math for ...: (( 1 - 1) = 0 * 5 + ( 1 -1) = 0 ) == 0 THIS IS THE INDEX WE OPERATE ON
******* RESULT: Set index 6 To value= 0 for comparison of string indexes 1 , 1
        indexes i/j are same char: t t
Therefore make no changes [0 1 2 3 4 5 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0]


Checking Index 12 with value = 0
doing math for ...: (( 2 - 1) = 1 * 5 + ( 2 -1) = 1 ) == 6 THIS IS THE INDEX WE OPERATE ON
******* RESULT: Set index 12 To value= 0 for comparison of string indexes 2 , 2
        indexes i/j are same char: r r
Therefore make no changes [0 1 2 3 4 5 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0]


Checking Index 18 with value = 0
doing math for ...: (( 3 - 1) = 2 * 5 + ( 3 -1) = 2 ) == 12 THIS IS THE INDEX WE OPERATE ON
******* RESULT: Set index 18 To value= 0 for comparison of string indexes 3 , 3
        indexes i/j are same char: i i
Therefore make no changes [0 1 2 3 4 5 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0]


Checking Index 24 with value = 0
doing math for ...: (( 4 - 1) = 3 * 5 + ( 4 -1) = 3 ) == 18 THIS IS THE INDEX WE OPERATE ON
******* RESULT: Set index 24 To value= 0 for comparison of string indexes 4 , 4
        indexes i/j are same char: n n
Therefore make no changes [0 1 2 3 4 5 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0]


Checking Index 30 with value = 0
doing math for ...: (( 5 - 1) = 4 * 5 + ( 5 -1) = 4 ) == 24 THIS IS THE INDEX WE OPERATE ON
******* RESULT: Set index 30 To value= 0 for comparison of string indexes 5 , 5
        indexes i/j are same char: g g
Therefore make no changes [0 1 2 3 4 5 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0]
*/