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Ingy döt Net 2023-07-01 11:58:00 -04:00
parent 7387c8f97b
commit cb5bb5e222
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// A fairly literal translation of the example program on the referenced
// WP page. Well, it happened to be the example program the day I completed
// the task. It seems from the WP history that there has been some churn
// in the posted example program. The example program of the day was in
// Pascal and was credited to Niklaus Wirth, from his "Algorithms +
// Data Structures = Programs."
package main
import "fmt"
var (
i int
q bool
a [9]bool
b [17]bool
c [15]bool // offset by 7 relative to the Pascal version
x [9]int
)
func try(i int) {
for j := 1; ; j++ {
q = false
if a[j] && b[i+j] && c[i-j+7] {
x[i] = j
a[j] = false
b[i+j] = false
c[i-j+7] = false
if i < 8 {
try(i + 1)
if !q {
a[j] = true
b[i+j] = true
c[i-j+7] = true
}
} else {
q = true
}
}
if q || j == 8 {
break
}
}
}
func main() {
for i := 1; i <= 8; i++ {
a[i] = true
}
for i := 2; i <= 16; i++ {
b[i] = true
}
for i := 0; i <= 14; i++ {
c[i] = true
}
try(1)
if q {
for i := 1; i <= 8; i++ {
fmt.Println(i, x[i])
}
}
}

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/*
* N-Queens Problem
*
* For an NxN chess board, 'safely' place a chess queen in every column and row such that none can attack another.
* This solution is based Wirth Pascal solution, although a tad cleaner, thus easier to understand as it uses Go/C
* style indexing and naming, and also prints the Queen using a Unicode 'rune' (which other languages do not handle natively).
*
* N rows by N columns are number left to right top to bottom 0 - 7
*
* There are 2N-1 diagonals (showing an 8x8)
* the upper-right to lower-left are numbered row + col that is:
* 0 1 2 3 4 5 6 7
* 1 2 3 4 5 6 7 8
* 2 3 4 5 6 7 8 9
* 3 4 5 6 7 8 9 10
* 4 5 6 7 8 9 10 11
* 5 6 7 8 9 10 11 12
* 6 7 8 9 10 11 12 13
* 7 8 9 10 11 12 13 14
*
* the upper-left to lower-right are numbered N-1 + row - col
* 7 6 5 4 3 2 1 0
* 8 7 6 5 4 3 2 1
* 9 8 7 6 5 4 3 2
* 10 9 8 7 6 5 4 3
* 11 10 9 8 7 6 5 4
* 12 11 10 9 8 7 6 5
* 13 12 11 10 9 8 7 6
* 14 13 12 11 10 9 8 7
*/
package main
import "fmt"
const N = 8
const HAS_QUEEN = false
const EMPTY = true
const UNASSIGNED = -1
const white_queen = '\u2655'
var row_num[N]int // results, indexed by row will be the column where the queen lives (UNASSIGNED) is empty
var right_2_left_diag[(2*N-1)]bool // T if no queen in diag[idx]: row i, column col is diag i+col
var left_2_right_diag[(2*N-1)]bool // T is no queen in diag[idx], row i, column col is N-1 + i-col
func printresults() {
for col := 0; col < N; col++ {
if col != 0 {
fmt.Printf(" ");
}
fmt.Printf("%d,%d", col, row_num[col])
}
fmt.Printf("\n");
for row := 0; row < N; row++ {
for col := 0; col < N; col++ {
if col == row_num[row] {
fmt.Printf(" %c ", white_queen)
} else {
fmt.Printf(" . ")
}
}
fmt.Printf("\n")
}
}
/*
* save a queen on the board by saving where we think it should go, and marking the diagonals as occupied
*/
func savequeen(row int, col int) {
row_num[row] = col // save queen column for this row
right_2_left_diag[row+col] = HAS_QUEEN // mark forward diags as occupied
left_2_right_diag[row-col+(N-1)] = HAS_QUEEN // mark backward diags as occupied
}
/*
* backout a previously saved queen by clearing where we put it, and marking the diagonals as empty
*/
func clearqueen(row int, col int) {
row_num[row] = UNASSIGNED
right_2_left_diag[row+col] = EMPTY
left_2_right_diag[row-col+(N-1)] = EMPTY
}
/*
* for each column try the solutions
*/
func trycol(col int) bool {
// check each row to look for the first empty row that does not have a diagonal in use too
for row := 0; row < N; row++ {
if row_num[row] == UNASSIGNED && // has the row been used yet?
right_2_left_diag[row+col] == EMPTY && // check for the forward diags
left_2_right_diag[row-col+(N-1)] == EMPTY { // check for the backwards diags
savequeen(row, col) // this is a possible solution
// Tricky part here: going forward thru the col up to but not including the rightmost one
// if this fails, we are done, no need to search any more
if col < N-1 && !trycol(col+1) {
// ok this did not work - we need to try a different row, so undo the guess
clearqueen(row, col)
} else {
// we have a solution on this row/col, start popping the stack.
return true
}
}
}
return false // not a solution for this col, pop the stack, undo the last guess, and try the next one
}
func main() {
for i := 0; i < N ; i++ {
row_num[i] = UNASSIGNED
}
for i := 0; i < 2*N-1 ; i++ {
right_2_left_diag[i] = EMPTY
}
for i := 0; i < 2*N-1 ; i++ {
left_2_right_diag[i] = EMPTY
}
trycol(0)
printresults()
}

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package main
import (
"flag"
"fmt"
"log"
"os"
"time"
"rosettacode.org/dlx" // or where ever you put the dlx package
)
func main() {
log.SetPrefix("N-queens: ")
log.SetFlags(0)
profile := flag.Bool("profile", false, "show DLX profile")
flag.Parse()
for N := 2; N <= 18; N++ {
err := nqueens(N, N == 8, *profile)
if err != nil {
log.Fatal(err)
}
}
}
func nqueens(N int, printFirst, profile bool) error {
// Build a new DLX matrix with 2N primary columns and 4N-6 secondary
// columns: R0..R(N-1), F0..F(N-1), A1..A(2N-3), B1..B(2N-3).
// We also know the number of cells and solution rows required.
m := dlx.NewWithHint(2*N, 4*N-6, N*N*4-4, 8)
s := solution{
N: N,
renumFwd: make([]int, 0, 2*N),
renumBack: make([]int, 2*N),
printFirst: printFirst,
}
// column indexes
iR0 := 0
iF0 := iR0 + N
iA1 := iF0 + N
iB1 := iA1 + 2*N - 3
// Use "organ-pipe" ordering. E.g. for N=8:
// R4 F4 R3 F3 R5 F5 R2 F2 R6 F6 R1 F1 R7 F7 R0 F0
// This can reduce the number of link updates required by
// almost half for large N; see Knuth's paper for details.
mid := N / 2
for off := 0; off <= N-mid; off++ {
i := mid - off
if i >= 0 {
s.renumBack[iR0+i] = len(s.renumFwd)
s.renumBack[iF0+i] = len(s.renumFwd) + 1
s.renumFwd = append(s.renumFwd, iR0+i, iF0+i)
}
if i = mid + off; off != 0 && i < N {
s.renumBack[iR0+i] = len(s.renumFwd)
s.renumBack[iF0+i] = len(s.renumFwd) + 1
s.renumFwd = append(s.renumFwd, iR0+i, iF0+i)
}
}
// Add constraint rows.
// TODO: pre-eliminate symetrical possibilities.
cols := make([]int, 4)
for i := 0; i < N; i++ {
for j := 0; j < N; j++ {
cols[0] = iR0 + i // Ri, rank i
cols[1] = iF0 + j // Fj, file j
a := (i + j) // A(i+j), diagonals
b := (N - 1 - i + j) // B(N-1-i+j), reverse diagonals
cols = cols[:2]
// Do organ-pipe reordering for R and F.
for i, c := range cols {
cols[i] = s.renumBack[c]
}
// Only add diagonals with more than one space; that
// is we omit the corners: A0, A(2N-2), B0, and B(2N-2)
if 0 < a && a < 2*N-2 {
cols = append(cols, iA1+a-1)
}
if 0 < b && b < 2*N-2 {
cols = append(cols, iB1+b-1)
}
m.AddRow(cols)
}
}
// Search for solutions.
start := time.Now()
err := m.Search(s.found)
if err != nil {
return err
}
elapsed := time.Since(start)
fmt.Printf("%d×%d queens has %2d solutions, found in %v\n", N, N, s.count, elapsed)
if profile {
m.ProfileWrite(os.Stderr)
}
return nil
}
type solution struct {
N int
count int
renumFwd []int // for "organ-pipe" column ordering
renumBack []int
printFirst bool
}
func (s *solution) found(m *dlx.Matrix) error {
s.count++
if s.printFirst && s.count == 1 {
fmt.Printf("First %d×%d queens solution:\n", s.N, s.N)
for _, cols := range m.SolutionIDs(nil) {
var r, f int
for _, c := range cols {
// Undo organ-pipe reodering
if c < len(s.renumFwd) {
c = s.renumFwd[c]
}
if c < s.N {
r = c + 1
} else if c < 2*s.N {
f = c - s.N + 1
}
}
fmt.Printf(" R%d F%d\n", r, f)
}
}
return nil
}