# Go语言实现的最简单数独解法

soduku.go

package main
import (
"fmt"
)
type node []int
var sudokuMay [9][9]node
var Sudoku = [9][9]int{
{0, 0, 0, 0, 0, 0, 8, 0, 0},
{0, 8, 2, 4, 0, 0, 0, 0, 0},
{1, 9, 0, 0, 6, 3, 0, 0, 0},
{0, 5, 0, 0, 8, 0, 7, 0, 0},
{6, 7, 8, 2, 0, 9, 1, 4, 3},
{0, 0, 3, 0, 4, 0, 0, 8, 0},
{0, 0, 0, 6, 2, 0, 0, 9, 4},
{0, 0, 0, 0, 0, 5, 6, 1, 0},
{0, 0, 0, 6, 0, 0, 0, 0, 0}}
func main() {
n := inited(Sudoku)
SudokuSure, _ := sure(sudokuMay)
for n > 0 {
n = Subinit(SudokuSure)
// Output(sudokuMay)
// fmt.Println(n)
SudokuSure, _ = sure(sudokuMay)
}
Output(sudokuMay)
fmt.Println(isEnable(sudokuMay))
// test()
}
func isEnable(tn [9][9]node) bool {
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if len(tn[i][j]) == 0 {
return false
}
}
}
return true
}
func sure(may [9][9]node) (sure [9][9]int, n int) {
n = 0
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if len(may[i][j]) == 1 {
sure[i][j] = may[i][j][0]
n++
} else {
sure[i][j] = 0
}
}
}
return
}
func test() {
i, j := 1, 3
fmt.Println(Sudoku[i][j])
for k := ((i / 3) * 3); k < ((i/3)*3)+3; k++ {
for l := ((j / 3) * 3); l < ((j/3)*3)+3; l++ {
fmt.Print(Sudoku[k][l])
}
fmt.Println(" ")
}
}
func inited(Sud [9][9]int) (changeCount int) {
tmp := 0
changeCount = 0
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if Sud[i][j] != 0 {
sudokuMay[i][j] = append(sudokuMay[i][j], Sud[i][j])
} else {
for k := 0; k < 9; k++ {
sudokuMay[i][j] = append(sudokuMay[i][j], k+1)
}
sudokuMay[i][j], tmp = excludeMay(i, j, sudokuMay[i][j], Sud)
changeCount += tmp
}
}
}
return
}
func Subinit(Sud [9][9]int) (changeCount int) {
tmp := 0
changeCount = 0
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if Sud[i][j] != 0 {
sudokuMay[i][j][0] = Sud[i][j]
} else {
sudokuMay[i][j], tmp = excludeMay(i, j, sudokuMay[i][j], Sud)
changeCount += tmp
}
}
}
return
}
func excludeMay(ti, tj int, t node, S [9][9]int) (rmay node, changeCount int) {
changeCount = 0
var tmpChangeCount int
for i := 0; i < 9; i++ {
if S[i][tj] != 0 {
t, tmpChangeCount = exclude(t, S[i][tj])
changeCount += tmpChangeCount
}
if S[ti][i] != 0 {
t, tmpChangeCount = exclude(t, S[ti][i])
changeCount += tmpChangeCount
}
}
for k := ((ti / 3) * 3); k < ((ti/3)*3)+3; k++ {
for l := ((tj / 3) * 3); l < ((tj/3)*3)+3; l++ {
if S[k][l] != 0 {
t, tmpChangeCount = exclude(t, S[k][l])
changeCount += tmpChangeCount
}
}
}
rmay = t
return
}
func excludeFirstOne(smay node, n int) (rmay node, changeCount int) {
changeCount = 0
rmay = smay
for i := 0; i < len(smay); i++ {
if smay[i] == n {
changeCount++
rmay = append(smay[:i], smay[i+1:]...)
return
}
if i == len(smay)-1 {
return
}
}
return
}
func exclude(smay node, n int) (tmp node, changeCount int) {
var nc int
changeCount = 0
tmp, nc = excludeFirstOne(smay, n)
for nc > 0 {
tmp, nc = excludeFirstOne(tmp, n)
changeCount++
}
return
}
func Output(sudoku [9][9]node) {
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
fmt.Print(sudokuMay[i][j])
}
fmt.Println("")
}
}

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