Random Latin squares
A Latin square of size n is an arrangement of n symbols in an n-by-n square in such a way that each row and column has each symbol appearing exactly once.
A randomised Latin square generates random configurations of the symbols for any given n.
You are encouraged to solve this task according to the task description, using any language you may know.
- Example n=4 randomised Latin square
0 2 3 1 2 1 0 3 3 0 1 2 1 3 2 0
- Task
- Create a function/routine/procedure/method/... that given
ngenerates a randomised Latin square of sizen. - Use the function to generate and show here, two randomly generated squares of size 5.
- Note
Strict Uniformity in the random generation is a hard problem and not a requirement of the task.
- Reference
11l
<lang 11l>F _transpose(matrix)
assert(matrix.len == matrix[0].len)
V r = [[0] * matrix.len] * matrix.len
L(i) 0 .< matrix.len
L(j) 0 .< matrix.len
r[i][j] = matrix[j][i]
R r
F _shuffle_transpose_shuffle(matrix)
V square = copy(matrix) random:shuffle(&square) V trans = _transpose(square) random:shuffle(&trans) R trans
F _rls(&symbols)
V n = symbols.len
I n == 1
R [symbols]
E
V sym = random:choice(symbols)
symbols.remove(sym)
V square = _rls(&symbols)
square.append(copy(square[0]))
L(i) 0 .< n
square[i].insert(i, sym)
R square
F rls(n)
V symbols = Array(0 .< n) V square = _rls(&symbols) R _shuffle_transpose_shuffle(square)
F _check_rows(square)
V set_row0 = Set(square[0]) R all(square.map(row -> row.len == Set(row).len & Set(row) == @set_row0))
F _check(square)
V transpose = _transpose(square) assert(_check_rows(square) & _check_rows(transpose), ‘Not a Latin square’)
L(i) [3, 3, 5, 5]
V square = rls(i)
print(square.map(row -> row.join(‘ ’)).join("\n"))
_check(square)
print()</lang>
- Output:
0 2 1 2 1 0 1 0 2 1 0 2 0 2 1 2 1 0 1 2 0 4 3 4 0 3 2 1 2 3 1 0 4 3 4 2 1 0 0 1 4 3 2 2 4 3 1 0 4 1 2 0 3 3 2 0 4 1 1 0 4 3 2 0 3 1 2 4
Action!
<lang Action!>DEFINE PTR="CARD" DEFINE DIMENSION="5"
TYPE Matrix=[
PTR data ;BYTE ARRAY BYTE dim]
PTR FUNC GetPtr(Matrix POINTER mat BYTE x,y) RETURN (mat.data+x+y*mat.dim)
PROC PrintMatrix(Matrix POINTER mat)
BYTE x,y BYTE POINTER d
d=GetPtr(mat,0,0)
FOR y=0 TO mat.dim-1
DO
FOR x=0 TO mat.dim-1
DO
PrintB(d^) Put(32)
d==+1
OD
PutE()
OD
RETURN
PROC KnuthShuffle(BYTE ARRAY tab BYTE size)
BYTE i,j,tmp
i=size-1 WHILE i>0 DO j=Rand(i+1) tmp=tab(i) tab(i)=tab(j) tab(j)=tmp i==-1 OD
RETURN
PROC LatinSquare(Matrix POINTER mat)
BYTE x,y,yy,shuffled BYTE POINTER ptr1,ptr2 BYTE ARRAY used(DIMENSION)
ptr1=GetPtr(mat,0,0)
FOR y=0 TO mat.dim-1
DO
FOR x=0 TO mat.dim-1
DO
ptr1^=x
ptr1==+1
OD
OD
;first row ptr1=GetPtr(mat,0,0) KnuthShuffle(ptr1,mat.dim)
;middle rows
FOR y=1 TO mat.dim-2
DO
shuffled=0
WHILE shuffled=0
DO
ptr1=GetPtr(mat,0,y)
KnuthShuffle(ptr1,mat.dim)
shuffled=1
yy=0
WHILE shuffled=1 AND yy<y
DO
x=0
WHILE shuffled=1 AND x<mat.dim
DO
ptr1=GetPtr(mat,x,yy)
ptr2=GetPtr(mat,x,y)
IF ptr1^=ptr2^ THEN
shuffled=0
FI
x==+1
OD
yy==+1
OD
OD
OD
;last row FOR x=0 TO mat.dim-1 DO Zero(used,mat.dim)
FOR y=0 TO mat.dim-2
DO
ptr1=GetPtr(mat,x,y)
yy=ptr1^ used(yy)=1
OD
FOR y=0 TO mat.dim-1
DO
IF used(y)=0 THEN
ptr1=GetPtr(mat,x,mat.dim-1)
ptr1^=y
EXIT
FI
OD
OD
RETURN
PROC Main()
BYTE ARRAY d(25) BYTE i Matrix mat
mat.data=d mat.dim=DIMENSION
FOR i=1 TO 2 DO LatinSquare(mat) PrintMatrix(mat) PutE() OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
3 1 2 4 0 1 4 0 2 3 4 0 3 1 2 2 3 1 0 4 0 2 4 3 1 2 1 3 4 0 3 0 4 2 1 4 3 1 0 2 1 2 0 3 4 0 4 2 1 3
Arturo
<lang rebol>latinSquare: function [n][
square: new []
variants: shuffle permutate 0..n-1
while -> n > size square [
row: sample variants
'square ++ @[row]
filter 'variants 'variant [
reject: false
loop.with:'i variant 'col [
if col = row\[i] ->
reject: true
]
reject
]
]
return square
]
loop 2 'x [
ls: latinSquare 5
loop ls 'row ->
print row
print "---------"
]</lang>
- Output:
2 4 0 1 3 3 0 2 4 1 4 3 1 0 2 1 2 4 3 0 0 1 3 2 4 --------- 3 2 1 4 0 2 1 4 0 3 4 3 0 2 1 1 0 2 3 4 0 4 3 1 2
C
<lang c>#include <stdbool.h>
- include <stdio.h>
- include <stdlib.h>
- include <string.h>
- include <time.h>
// low <= num < high int randInt(int low, int high) {
return (rand() % (high - low)) + low;
}
// shuffle an array of n elements void shuffle(int *const array, const int n) {
if (n > 1) {
int i;
for (i = 0; i < n - 1; i++) {
int j = randInt(i, n);
int t = array[i];
array[i] = array[j];
array[j] = t;
}
}
}
// print an n * n array void printSquare(const int *const latin, const int n) {
int i, j;
for (i = 0; i < n; i++) {
printf("[");
for (j = 0; j < n; j++) {
if (j > 0) {
printf(", ");
}
printf("%d", latin[i * n + j]);
}
printf("]\n");
}
printf("\n");
}
void latinSquare(const int n) {
int *latin, *used; int i, j, k;
if (n <= 0) {
printf("[]\n");
return;
}
// allocate
latin = (int *)malloc(n * n * sizeof(int));
if (!latin) {
printf("Failed to allocate memory.");
return;
}
// initialize
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
latin[i * n + j] = j;
}
}
// first row shuffle(latin, n);
// middle row(s)
for (i = 1; i < n - 1; i++) {
bool shuffled = false;
while (!shuffled) {
shuffle(&latin[i * n], n);
for (k = 0; k < i; k++) {
for (j = 0; j < n; j++) {
if (latin[k * n + j] == latin[i * n + j]) {
goto shuffling;
}
}
}
shuffled = true;
shuffling: {}
}
}
//last row
used = (int *)malloc(n * sizeof(int));
for (j = 0; j < n; j++) {
memset(used, 0, n * sizeof(int));
for (i = 0; i < n - 1; i++) {
used[latin[i * n + j]] = 1;
}
for (k = 0; k < n; k++) {
if (used[k] == 0) {
latin[(n - 1) * n + j] = k;
break;
}
}
}
free(used);
// print the result printSquare(latin, n); free(latin);
}
int main() {
// initialze the random number generator srand((unsigned int)time((time_t)0));
latinSquare(5); latinSquare(5); latinSquare(10);
return 0;
}</lang>
- Output:
[1, 4, 3, 0, 2] [3, 1, 0, 2, 4] [2, 3, 4, 1, 0] [4, 0, 2, 3, 1] [0, 2, 1, 4, 3] [0, 2, 4, 1, 3] [1, 3, 0, 4, 2] [4, 0, 3, 2, 1] [3, 1, 2, 0, 4] [2, 4, 1, 3, 0] [6, 8, 2, 1, 4, 7, 3, 9, 5, 0] [8, 7, 3, 2, 0, 1, 6, 5, 4, 9] [1, 3, 5, 6, 7, 9, 4, 0, 2, 8] [2, 5, 1, 9, 8, 0, 7, 4, 3, 6] [4, 9, 8, 5, 6, 2, 0, 1, 7, 3] [9, 6, 7, 8, 2, 4, 1, 3, 0, 5] [7, 2, 9, 0, 3, 6, 5, 8, 1, 4] [3, 1, 0, 4, 5, 8, 2, 6, 9, 7] [5, 0, 4, 7, 9, 3, 8, 2, 6, 1] [0, 4, 6, 3, 1, 5, 9, 7, 8, 2]
C++
<lang cpp>#include <algorithm>
- include <chrono>
- include <iostream>
- include <random>
- include <vector>
template <typename T> std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
auto it = v.cbegin(); auto end = v.cend();
os << '[';
if (it != end) {
os << *it;
it = std::next(it);
}
while (it != end) {
os << ", ";
os << *it;
it = std::next(it);
}
return os << ']';
}
void printSquare(const std::vector<std::vector<int>> &latin) {
for (auto &row : latin) {
std::cout << row << '\n';
}
std::cout << '\n';
}
void latinSquare(int n) {
if (n <= 0) {
std::cout << "[]\n";
return;
}
// obtain a time-based seed: unsigned seed = std::chrono::system_clock::now().time_since_epoch().count(); auto g = std::default_random_engine(seed);
std::vector<std::vector<int>> latin;
for (int i = 0; i < n; ++i) {
std::vector<int> inner;
for (int j = 0; j < n; ++j) {
inner.push_back(j);
}
latin.push_back(inner);
}
// first row
std::shuffle(latin[0].begin(), latin[0].end(), g);
// middle row(s)
for (int i = 1; i < n - 1; ++i) {
bool shuffled = false;
while (!shuffled) {
std::shuffle(latin[i].begin(), latin[i].end(), g);
for (int k = 0; k < i; ++k) {
for (int j = 0; j < n; ++j) {
if (latin[k][j] == latin[i][j]) {
goto shuffling;
}
}
}
shuffled = true;
shuffling: {}
}
}
// last row
for (int j = 0; j < n; ++j) {
std::vector<bool> used(n, false);
for (int i = 0; i < n - 1; ++i) {
used[latin[i][j]] = true;
}
for (int k = 0; k < n; ++k) {
if (!used[k]) {
latin[n - 1][j] = k;
break;
}
}
}
printSquare(latin);
}
int main() {
latinSquare(5); latinSquare(5); latinSquare(10); return 0;
}</lang>
- Output:
[4, 3, 1, 2, 0] [1, 0, 3, 4, 2] [3, 2, 0, 1, 4] [2, 1, 4, 0, 3] [0, 4, 2, 3, 1] [2, 4, 0, 3, 1] [0, 3, 4, 1, 2] [3, 0, 1, 2, 4] [1, 2, 3, 4, 0] [4, 1, 2, 0, 3] [9, 3, 5, 0, 8, 2, 7, 1, 6, 4] [0, 9, 4, 6, 1, 8, 5, 3, 7, 2] [4, 1, 9, 7, 3, 5, 2, 0, 8, 6] [2, 6, 8, 3, 4, 0, 9, 7, 1, 5] [6, 5, 3, 4, 7, 9, 1, 8, 2, 0] [1, 8, 6, 5, 2, 4, 0, 9, 3, 7] [7, 2, 0, 8, 9, 1, 4, 6, 5, 3] [3, 0, 7, 1, 5, 6, 8, 2, 4, 9] [8, 4, 2, 9, 6, 7, 3, 5, 0, 1] [5, 7, 1, 2, 0, 3, 6, 4, 9, 8]
C#
<lang csharp>using System; using System.Collections.Generic;
namespace RandomLatinSquares {
using Matrix = List<List<int>>;
// Taken from https://stackoverflow.com/a/1262619 static class Helper { private static readonly Random rng = new Random();
public static void Shuffle<T>(this IList<T> list) {
int n = list.Count;
while (n > 1) {
n--;
int k = rng.Next(n + 1);
T value = list[k];
list[k] = list[n];
list[n] = value;
}
}
}
class Program {
static void PrintSquare(Matrix latin) {
foreach (var row in latin) {
Console.Write('[');
var it = row.GetEnumerator();
if (it.MoveNext()) {
Console.Write(it.Current);
}
while (it.MoveNext()) {
Console.Write(", ");
Console.Write(it.Current);
}
Console.WriteLine(']');
}
Console.WriteLine();
}
static void LatinSquare(int n) {
if (n <= 0) {
Console.WriteLine("[]");
return;
}
var latin = new Matrix();
for (int i = 0; i < n; i++) {
List<int> temp = new List<int>();
for (int j = 0; j < n; j++) {
temp.Add(j);
}
latin.Add(temp);
}
// first row
latin[0].Shuffle();
// middle row(s)
for (int i = 1; i < n - 1; i++) {
bool shuffled = false;
while (!shuffled) {
latin[i].Shuffle();
for (int k = 0; k < i; k++) {
for (int j = 0; j < n; j++) {
if (latin[k][j] == latin[i][j]) {
goto shuffling;
}
}
}
shuffled = true;
shuffling: { }
}
}
// last row
for (int j = 0; j < n; j++) {
List<bool> used = new List<bool>();
for (int i = 0; i < n; i++) {
used.Add(false);
}
for (int i = 0; i < n-1; i++) {
used[latin[i][j]] = true;
}
for (int k = 0; k < n; k++) {
if (!used[k]) {
latin[n - 1][j] = k;
break;
}
}
}
PrintSquare(latin);
}
static void Main() {
LatinSquare(5);
LatinSquare(5);
LatinSquare(10); // for good measure
}
}
}</lang>
- Output:
[3, 0, 1, 4, 2] [4, 2, 3, 1, 0] [2, 1, 4, 0, 3] [0, 4, 2, 3, 1] [1, 3, 0, 2, 4] [4, 0, 2, 3, 1] [2, 1, 4, 0, 3] [0, 4, 3, 1, 2] [3, 2, 1, 4, 0] [1, 3, 0, 2, 4] [5, 9, 4, 2, 0, 3, 8, 6, 1, 7] [7, 1, 3, 9, 6, 2, 4, 0, 8, 5] [1, 8, 6, 3, 4, 0, 9, 5, 7, 2] [9, 0, 1, 4, 2, 7, 3, 8, 5, 6] [6, 7, 8, 0, 5, 1, 2, 3, 9, 4] [8, 2, 7, 5, 3, 4, 1, 9, 6, 0] [2, 4, 5, 6, 8, 9, 0, 7, 3, 1] [4, 5, 9, 8, 1, 6, 7, 2, 0, 3] [0, 3, 2, 7, 9, 5, 6, 1, 4, 8] [3, 6, 0, 1, 7, 8, 5, 4, 2, 9]
D
<lang d>import std.array; import std.random; import std.stdio;
alias Matrix = int[][];
void latinSquare(int n) {
if (n <= 0) {
writeln("[]");
return;
}
Matrix latin = uninitializedArray!Matrix(n, n);
foreach (row; latin) {
for (int i = 0; i < n; i++) {
row[i] = i;
}
}
// first row
latin[0].randomShuffle;
// middle row(s)
for (int i = 1; i < n - 1; i++) {
bool shuffled = false;
shuffling:
while (!shuffled) {
latin[i].randomShuffle;
for (int k = 0; k < i; k++) {
for (int j = 0; j < n; j++) {
if (latin[k][j] == latin[i][j]) {
continue shuffling;
}
}
}
shuffled = true;
}
}
// last row
for (int j = 0; j < n; j++) {
bool[] used = uninitializedArray!(bool[])(n);
used[] = false;
for (int i = 0; i < n - 1; i++) {
used[latin[i][j]] = true;
}
for (int k = 0; k < n; k++) {
if (!used[k]) {
latin[n - 1][j] = k;
break;
}
}
}
printSquare(latin);
}
void printSquare(Matrix latin) {
foreach (row; latin) {
writeln(row);
}
}
void main() {
latinSquare(5); writeln;
latinSquare(5); writeln;
latinSquare(10);
}</lang>
- Output:
[2, 4, 3, 1, 0] [1, 0, 2, 3, 4] [0, 1, 4, 2, 3] [3, 2, 0, 4, 1] [4, 3, 1, 0, 2] [4, 1, 0, 3, 2] [3, 4, 1, 2, 0] [0, 2, 3, 1, 4] [1, 0, 2, 4, 3] [2, 3, 4, 0, 1] [3, 5, 1, 9, 4, 2, 7, 0, 8, 6] [6, 2, 7, 8, 3, 9, 1, 4, 5, 0] [0, 4, 3, 6, 7, 8, 2, 5, 9, 1] [8, 3, 0, 5, 9, 7, 4, 1, 6, 2] [2, 8, 9, 0, 6, 1, 5, 3, 7, 4] [5, 6, 8, 1, 2, 0, 9, 7, 4, 3] [4, 1, 2, 3, 8, 5, 6, 9, 0, 7] [1, 9, 4, 7, 0, 6, 8, 2, 3, 5] [7, 0, 6, 2, 5, 4, 3, 8, 1, 9] [9, 7, 5, 4, 1, 3, 0, 6, 2, 8]
F#
This solution uses functions from Factorial_base_numbers_indexing_permutations_of_a_collection#F.23 and Latin_Squares_in_reduced_form#F.23. This solution generates completely random uniformly distributed Latin Squares from all possible Latin Squares of order 5. It takes 5 thousandths of a second can that really be called hard? <lang fsharp> // Generate 2 Random Latin Squares of order 5. Nigel Galloway: July 136th., 2019 let N=let N=System.Random() in (fun n->N.Next(n)) let rc()=let β=lN2p [|0;N 4;N 3;N 2|] [|0..4|] in Seq.item (N 56) (normLS 5) |> List.map(lN2p [|N 5;N 4;N 3;N 2|]) |> List.permute(fun n->β.[n]) |> List.iter(printfn "%A") rc(); printfn ""; rc() </lang>
- Output:
[|5; 3; 1; 4; 2|] [|1; 4; 5; 2; 3|] [|4; 1; 2; 3; 5|] [|2; 5; 3; 1; 4|] [|3; 2; 4; 5; 1|] [|4; 1; 2; 5; 3|] [|3; 5; 1; 2; 4|] [|2; 4; 5; 3; 1|] [|1; 2; 3; 4; 5|] [|5; 3; 4; 1; 2|]
I thought some statistics might be interesting so I generated 1 million Latin Squares of order 5. There are 161280 possible Latin Squares of which 3174 were not generated. The remainder were generated:
Times Generated Number of Latin Squares
1 1776
2 5669
3 11985
4 19128
5 24005
6 25333
7 22471
8 18267
9 12569
10 7924
11 4551
12 2452
13 1130
14 483
15 219
16 93
17 37
18 5
19 7
20 2
Factor
A brute force method for generating uniformly random Latin squares. Repeatedly select a random permutation of (0, 1,...n-1) and add it as the next row of the square. If at any point the rules for being a Latin square are violated, start the entire process over again from the beginning. <lang factor>USING: arrays combinators.extras fry io kernel math.matrices prettyprint random sequences sets ; IN: rosetta-code.random-latin-squares
- rand-permutation ( n -- seq ) <iota> >array randomize ;
- ls? ( n -- ? ) [ all-unique? ] column-map t [ and ] reduce ;
- (ls) ( n -- m ) dup '[ _ rand-permutation ] replicate ;
- ls ( n -- m ) dup (ls) dup ls? [ nip ] [ drop ls ] if ;
- random-latin-squares ( -- ) [ 5 ls simple-table. nl ] twice ;
MAIN: random-latin-squares</lang>
- Output:
0 4 3 2 1 3 0 2 1 4 4 2 1 3 0 2 1 4 0 3 1 3 0 4 2 4 0 1 3 2 0 2 4 1 3 1 3 0 2 4 2 4 3 0 1 3 1 2 4 0
Go
Restarting Row method
As the task is not asking for large squares to be generated and even n = 10 is virtually instant, we use a simple brute force approach here known as the 'Restarting Row' method (see Talk page). However, whilst easy to understand, this method does not produce uniformly random squares. <lang go>package main
import (
"fmt" "math/rand" "time"
)
type matrix [][]int
func shuffle(row []int, n int) {
rand.Shuffle(n, func(i, j int) {
row[i], row[j] = row[j], row[i]
})
}
func latinSquare(n int) {
if n <= 0 {
fmt.Println("[]\n")
return
}
latin := make(matrix, n)
for i := 0; i < n; i++ {
latin[i] = make([]int, n)
if i == n-1 {
break
}
for j := 0; j < n; j++ {
latin[i][j] = j
}
}
// first row
shuffle(latin[0], n)
// middle row(s)
for i := 1; i < n-1; i++ {
shuffled := false
shuffling:
for !shuffled {
shuffle(latin[i], n)
for k := 0; k < i; k++ {
for j := 0; j < n; j++ {
if latin[k][j] == latin[i][j] {
continue shuffling
}
}
}
shuffled = true
}
}
// last row
for j := 0; j < n; j++ {
used := make([]bool, n)
for i := 0; i < n-1; i++ {
used[latin[i][j]] = true
}
for k := 0; k < n; k++ {
if !used[k] {
latin[n-1][j] = k
break
}
}
}
printSquare(latin, n)
}
func printSquare(latin matrix, n int) {
for i := 0; i < n; i++ {
fmt.Println(latin[i])
}
fmt.Println()
}
func main() {
rand.Seed(time.Now().UnixNano()) latinSquare(5) latinSquare(5) latinSquare(10) // for good measure
}</lang>
- Output:
Sample run:
[3 2 1 0 4] [0 3 2 4 1] [4 1 0 3 2] [2 4 3 1 0] [1 0 4 2 3] [3 1 0 4 2] [1 0 2 3 4] [2 4 3 0 1] [4 3 1 2 0] [0 2 4 1 3] [9 2 8 4 6 1 7 5 0 3] [4 3 7 6 0 8 5 9 2 1] [2 1 9 7 3 4 6 0 5 8] [8 6 0 5 7 2 3 1 9 4] [5 0 6 8 1 3 9 2 4 7] [7 5 4 9 2 0 1 3 8 6] [3 9 2 1 5 6 8 4 7 0] [1 4 5 2 8 7 0 6 3 9] [6 8 3 0 4 9 2 7 1 5] [0 7 1 3 9 5 4 8 6 2]
Latin Squares in Reduced Form method
Unlike the "Restarting Row" method, this method does produce uniformly random Latin squares for n <= 6 (see Talk page) but is more involved and therefore slower. It reuses some (suitably adjusted) code from the Latin Squares in Reduced Form and Permutations tasks. <lang go>package main
import (
"fmt" "math/rand" "sort" "time"
)
type matrix [][]int
// generate derangements of first n numbers, with 'start' in first place. func dList(n, start int) (r matrix) {
start-- // use 0 basing
a := make([]int, n)
for i := range a {
a[i] = i
}
a[0], a[start] = start, a[0]
sort.Ints(a[1:])
first := a[1]
// recursive closure permutes a[1:]
var recurse func(last int)
recurse = func(last int) {
if last == first {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged.
for j, v := range a[1:] { // j starts from 0, not 1
if j+1 == v {
return // no, ignore it
}
}
// yes, save a copy
b := make([]int, n)
copy(b, a)
for i := range b {
b[i]++ // change back to 1 basing
}
r = append(r, b)
return
}
for i := last; i >= 1; i-- {
a[i], a[last] = a[last], a[i]
recurse(last - 1)
a[i], a[last] = a[last], a[i]
}
}
recurse(n - 1)
return
}
func reducedLatinSquares(n int) []matrix {
var rls []matrix
if n < 0 {
n = 0
}
rlatin := make(matrix, n)
for i := 0; i < n; i++ {
rlatin[i] = make([]int, n)
}
if n <= 1 {
return append(rls, rlatin)
}
// first row
for j := 0; j < n; j++ {
rlatin[0][j] = j + 1
}
// recursive closure to compute reduced latin squares
var recurse func(i int)
recurse = func(i int) {
rows := dList(n, i) // get derangements of first n numbers, with 'i' first.
outer:
for r := 0; r < len(rows); r++ {
copy(rlatin[i-1], rows[r])
for k := 0; k < i-1; k++ {
for j := 1; j < n; j++ {
if rlatin[k][j] == rlatin[i-1][j] {
if r < len(rows)-1 {
continue outer
} else if i > 2 {
return
}
}
}
}
if i < n {
recurse(i + 1)
} else {
rl := copyMatrix(rlatin)
rls = append(rls, rl)
}
}
return
}
// remaining rows recurse(2) return rls
}
func copyMatrix(m matrix) matrix {
le := len(m)
cpy := make(matrix, le)
for i := 0; i < le; i++ {
cpy[i] = make([]int, le)
copy(cpy[i], m[i])
}
return cpy
}
func printSquare(latin matrix, n int) {
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
fmt.Printf("%d ", latin[i][j]-1)
}
fmt.Println()
}
fmt.Println()
}
func factorial(n uint64) uint64 {
if n == 0 {
return 1
}
prod := uint64(1)
for i := uint64(2); i <= n; i++ {
prod *= i
}
return prod
}
// generate permutations of first n numbers, starting from 0. func pList(n int) matrix {
fact := factorial(uint64(n))
perms := make(matrix, fact)
a := make([]int, n)
for i := 0; i < n; i++ {
a[i] = i
}
t := make([]int, n)
copy(t, a)
perms[0] = t
n--
var i, j int
for c := uint64(1); c < fact; c++ {
i = n - 1
j = n
for a[i] > a[i+1] {
i--
}
for a[j] < a[i] {
j--
}
a[i], a[j] = a[j], a[i]
j = n
i++
for i < j {
a[i], a[j] = a[j], a[i]
i++
j--
}
t := make([]int, n+1)
copy(t, a)
perms[c] = t
}
return perms
}
func generateLatinSquares(n, tests, echo int) {
rls := reducedLatinSquares(n)
perms := pList(n)
perms2 := pList(n - 1)
for test := 0; test < tests; test++ {
rn := rand.Intn(len(rls))
rl := rls[rn] // select reduced random square at random
rn = rand.Intn(len(perms))
rp := perms[rn] // select a random permuation of 'rl's columns
// permute columns
t := make(matrix, n)
for i := 0; i < n; i++ {
t[i] = make([]int, n)
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
t[i][j] = rl[i][rp[j]]
}
}
rn = rand.Intn(len(perms2))
rp = perms2[rn] // select a random permutation of 't's rows 2 to n
// permute rows 2 to n
u := make(matrix, n)
for i := 0; i < n; i++ {
u[i] = make([]int, n)
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
if i == 0 {
u[i][j] = t[i][j]
} else {
u[i][j] = t[rp[i-1]+1][j]
}
}
}
if test < echo {
printSquare(u, n)
}
if n == 4 {
for i := 0; i < 4; i++ {
for j := 0; j < 4; j++ {
u[i][j]--
}
}
for i := 0; i < 4; i++ {
copy(a[4*i:], u[i])
}
for i := 0; i < 4; i++ {
if testSquares[i] == a {
counts[i]++
break
}
}
}
}
}
var testSquares = [4][16]int{
{0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 0, 1, 3, 2, 1, 0},
{0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 1, 0, 3, 2, 0, 1},
{0, 1, 2, 3, 1, 2, 3, 0, 2, 3, 0, 1, 3, 0, 1, 2},
{0, 1, 2, 3, 1, 3, 0, 2, 2, 0, 3, 1, 3, 2, 1, 0},
}
var (
counts [4]int a [16]int
)
func main() {
rand.Seed(time.Now().UnixNano())
fmt.Println("Two randomly generated latin squares of order 5 are:\n")
generateLatinSquares(5, 2, 2)
fmt.Println("Out of 1,000,000 randomly generated latin squares of order 4, ")
fmt.Println("of which there are 576 instances ( => expected 1736 per instance),")
fmt.Println("the following squares occurred the number of times shown:\n")
generateLatinSquares(4, 1e6, 0)
for i := 0; i < 4; i++ {
fmt.Println(testSquares[i][:], ":", counts[i])
}
fmt.Println("\nA randomly generated latin square of order 6 is:\n")
generateLatinSquares(6, 1, 1)
}</lang>
- Output:
Sample run:
Two randomly generated latin squares of order 5 are: 2 1 3 4 0 4 3 0 1 2 1 0 2 3 4 0 4 1 2 3 3 2 4 0 1 1 2 3 4 0 0 3 4 2 1 2 4 0 1 3 4 0 1 3 2 3 1 2 0 4 Out of 1,000,000 randomly generated latin squares of order 4, of which there are 576 instances ( => expected 1736 per instance), the following squares occurred the number of times shown: [0 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0] : 1737 [0 1 2 3 1 0 3 2 2 3 1 0 3 2 0 1] : 1736 [0 1 2 3 1 2 3 0 2 3 0 1 3 0 1 2] : 1726 [0 1 2 3 1 3 0 2 2 0 3 1 3 2 1 0] : 1799 A randomly generated latin square of order 6 is: 3 5 1 0 4 2 2 0 5 4 1 3 0 4 2 5 3 1 1 3 4 2 0 5 5 1 0 3 2 4 4 2 3 1 5 0
Haskell
Pure functional Haskell encourages programmer to separate randomness and deterministic business logic. So first we determine a function which returns a Latin square which is built according to given first row and first column.
<lang haskell>import Data.List (permutations, (\\))
latinSquare :: Eq a => [a] -> [a] -> a latinSquare [] [] = [] latinSquare c r
| head r /= head c = []
| otherwise = reverse <$> foldl addRow firstRow perms
where
-- permutations grouped by the first element
perms =
tail $
fmap
(fmap . (:) <*> (permutations . (r \\) . return))
c
firstRow = pure <$> r
addRow tbl rows =
head
[ zipWith (:) row tbl
| row <- rows,
and $ different (tail row) (tail tbl)
]
different = zipWith $ (not .) . elem
printTable :: Show a => a -> IO () printTable tbl =
putStrLn $
unlines $
unwords . map show <$> tbl</lang>
λ> printTable $ latinSquare [1,2,3,4,5] [1,3,2,5,4] 1 2 3 4 5 3 4 1 5 2 2 5 4 3 1 5 3 2 1 4
Now whatever random generator is used, the construction of a random Latin square may be done by feeding two appropriate random permutations to the deterministic algorithm.
<lang haskell>randomLatinSquare :: Eq a => [a] -> Random a randomLatinSquare set = do
r <- randomPermutation set c <- randomPermutation (tail r) return $ latinSquare r (head r:c)</lang>
For examples a naive linear congruent method in a State monad is used.
<lang haskell>import Control.Monad.State
type Random a = State Int a
random :: Integral a => a -> Random a random k = rescale <$> modify iter
where iter x = (x * a + c) `mod` m (a, c, m) = (1103515245, 12345, 2^31-1) rescale x = fromIntegral x `mod` k
randomPermutation :: Eq a => [a] -> Random [a] randomPermutation = go
where
go [] = pure []
go lst = do
x <- randomSample lst
(x :) <$> go (lst \\ [x])
randomSample :: [a] -> Random a randomSample lst = (lst !!) <$> random (length lst)</lang>
λ> printTable $ randomLatinSquare [0..4] `evalState` 42 3 2 0 1 4 0 1 4 3 2 1 4 3 2 0 2 0 1 4 3 4 3 2 0 1 λ> printTable $ randomLatinSquare [0..9] `evalState` 42 8 5 6 1 7 2 4 0 9 3 5 9 4 0 6 8 3 1 7 2 6 0 8 2 3 5 9 7 1 4 7 1 5 3 8 4 0 9 2 6 3 2 7 8 9 0 5 4 6 1 2 4 3 9 5 1 7 6 8 0 1 3 2 7 4 9 6 8 0 5 0 7 1 6 2 3 8 5 4 9 9 6 0 4 1 7 2 3 5 8 4 8 9 5 0 6 1 2 3 7
J
<lang j>rls=: 3 : 0
s=. ?~ y NB. "deal" y unique integers from 0 to y
for_ijk. i.<:y do.
NB. deal a new row. subtract it from all previous rows
NB. if you get a 0, some column has a matching integer, deal again
whilst. 0 = */ */ s -"1 r do.
r=. ?~ y
end.
s=. s ,,: r NB. "laminate" successful row to the square
end.
) </lang>
- Output:
rls 5 4 0 1 2 3 3 1 4 0 2 2 3 0 4 1 0 2 3 1 4 1 4 2 3 0 rls 5 0 4 2 1 3 4 1 3 2 0 1 0 4 3 2 2 3 1 0 4 3 2 0 4 1
Java
<lang java>import java.util.ArrayList; import java.util.Collections; import java.util.Iterator; import java.util.List; import java.util.Objects;
public class RandomLatinSquares {
private static void printSquare(List<List<Integer>> latin) {
for (List<Integer> row : latin) {
Iterator<Integer> it = row.iterator();
System.out.print("[");
if (it.hasNext()) {
Integer col = it.next();
System.out.print(col);
}
while (it.hasNext()) {
Integer col = it.next();
System.out.print(", ");
System.out.print(col);
}
System.out.println("]");
}
System.out.println();
}
private static void latinSquare(int n) {
if (n <= 0) {
System.out.println("[]");
return;
}
List<List<Integer>> latin = new ArrayList<>(n);
for (int i = 0; i < n; ++i) {
List<Integer> inner = new ArrayList<>(n);
for (int j = 0; j < n; ++j) {
inner.add(j);
}
latin.add(inner);
}
// first row
Collections.shuffle(latin.get(0));
// middle row(s)
for (int i = 1; i < n - 1; ++i) {
boolean shuffled = false;
shuffling:
while (!shuffled) {
Collections.shuffle(latin.get(i));
for (int k = 0; k < i; ++k) {
for (int j = 0; j < n; ++j) {
if (Objects.equals(latin.get(k).get(j), latin.get(i).get(j))) {
continue shuffling;
}
}
}
shuffled = true;
}
}
// last row
for (int j = 0; j < n; ++j) {
List<Boolean> used = new ArrayList<>(n);
for (int i = 0; i < n; ++i) {
used.add(false);
}
for (int i = 0; i < n - 1; ++i) {
used.set(latin.get(i).get(j), true);
}
for (int k = 0; k < n; ++k) {
if (!used.get(k)) {
latin.get(n - 1).set(j, k);
break;
}
}
}
printSquare(latin); }
public static void main(String[] args) {
latinSquare(5);
latinSquare(5);
latinSquare(10);
}
}</lang>
- Output:
[1, 3, 4, 0, 2] [4, 0, 2, 1, 3] [0, 2, 1, 3, 4] [2, 1, 3, 4, 0] [3, 4, 0, 2, 1] [4, 2, 1, 3, 0] [2, 1, 3, 0, 4] [0, 3, 2, 4, 1] [3, 4, 0, 1, 2] [1, 0, 4, 2, 3] [8, 7, 9, 0, 1, 2, 6, 5, 4, 3] [4, 6, 3, 8, 0, 5, 2, 9, 1, 7] [2, 1, 0, 4, 8, 9, 7, 3, 5, 6] [6, 4, 8, 1, 9, 7, 3, 0, 2, 5] [7, 9, 2, 6, 5, 3, 4, 8, 0, 1] [9, 5, 1, 3, 2, 6, 8, 4, 7, 0] [5, 0, 4, 9, 6, 8, 1, 7, 3, 2] [3, 8, 5, 2, 7, 1, 0, 6, 9, 4] [1, 3, 6, 7, 4, 0, 5, 2, 8, 9] [0, 2, 7, 5, 3, 4, 9, 1, 6, 8]
JavaScript
<lang javascript> class Latin {
constructor(size = 3) {
this.size = size;
this.mst = [...Array(this.size)].map((v, i) => i + 1);
this.square = Array(this.size).fill(0).map(() => Array(this.size).fill(0));
if (this.create(0, 0)) {
console.table(this.square);
}
}
create(c, r) {
const d = [...this.mst];
let s;
while (true) {
do {
s = d.splice(Math.floor(Math.random() * d.length), 1)[0];
if (!s) return false;
} while (this.check(s, c, r));
this.square[c][r] = s;
if (++c >= this.size) {
c = 0;
if (++r >= this.size) {
return true;
}
}
if (this.create(c, r)) return true;
if (--c < 0) {
c = this.size - 1;
if (--r < 0) {
return false;
}
}
}
}
check(d, c, r) {
for (let a = 0; a < this.size; a++) {
if (c - a > -1) {
if (this.square[c - a][r] === d)
return true;
}
if (r - a > -1) {
if (this.square[c][r - a] === d)
return true;
}
}
return false;
}
} new Latin(5); </lang>
- Output:
3 5 4 1 2 4 3 1 2 5 1 2 3 5 4 5 1 2 4 3 2 4 5 3 1 4 5 1 3 2 3 1 4 2 5 5 4 2 1 3 1 2 3 5 4 2 3 5 4 1
Julia
Using the Python algorithm as described in the discussion section. <lang julia>using Random
shufflerows(mat) = mat[shuffle(1:end), :] shufflecols(mat) = mat[:, shuffle(1:end)]
function addatdiagonal(mat)
n = size(mat)[1] + 1
newmat = similar(mat, size(mat) .+ 1)
for j in 1:n, i in 1:n
newmat[i, j] = (i == n && j < n) ? mat[1, j] : (i == j) ? n - 1 :
(i < j) ? mat[i, j - 1] : mat[i, j]
end
newmat
end
function makelatinsquare(N)
mat = [0 1; 1 0]
for i in 3:N
mat = addatdiagonal(mat)
end
shufflecols(shufflerows(mat))
end
function printlatinsquare(N)
mat = makelatinsquare(N)
for i in 1:N, j in 1:N
print(rpad(mat[i, j], 3), j == N ? "\n" : "")
end
end
printlatinsquare(5), println("\n"), printlatinsquare(5)
</lang>
- Output:
1 3 0 4 2 3 0 4 2 1 0 4 2 1 3 2 1 3 0 4 4 2 1 3 0 2 0 1 3 4 4 3 2 1 0 3 2 0 4 1 1 4 3 0 2 0 1 4 2 3
Kotlin
<lang scala>typealias matrix = MutableList<MutableList<Int>>
fun printSquare(latin: matrix) {
for (row in latin) {
println(row)
}
println()
}
fun latinSquare(n: Int) {
if (n <= 0) {
println("[]")
return
}
val latin = MutableList(n) { MutableList(n) { it } }
// first row
latin[0].shuffle()
// middle row(s)
for (i in 1 until n - 1) {
var shuffled = false
shuffling@
while (!shuffled) {
latin[i].shuffle()
for (k in 0 until i) {
for (j in 0 until n) {
if (latin[k][j] == latin[i][j]) {
continue@shuffling
}
}
}
shuffled = true
}
}
// last row
for (j in 0 until n) {
val used = MutableList(n) { false }
for (i in 0 until n - 1) {
used[latin[i][j]] = true
}
for (k in 0 until n) {
if (!used[k]) {
latin[n - 1][j] = k
break
}
}
}
printSquare(latin)
}
fun main() {
latinSquare(5) latinSquare(5) latinSquare(10) // for good measure
}</lang>
- Output:
[4, 1, 2, 3, 0] [1, 3, 0, 2, 4] [3, 2, 4, 0, 1] [0, 4, 3, 1, 2] [2, 0, 1, 4, 3] [2, 0, 3, 1, 4] [0, 4, 1, 3, 2] [1, 3, 2, 4, 0] [3, 2, 4, 0, 1] [4, 1, 0, 2, 3] [7, 8, 4, 6, 5, 2, 9, 3, 1, 0] [1, 5, 8, 2, 3, 0, 7, 9, 4, 6] [6, 9, 5, 8, 7, 1, 3, 4, 0, 2] [0, 6, 9, 4, 8, 3, 1, 2, 7, 5] [4, 1, 3, 0, 6, 5, 8, 7, 2, 9] [5, 0, 1, 7, 9, 4, 2, 6, 3, 8] [2, 3, 6, 9, 4, 7, 0, 8, 5, 1] [3, 7, 2, 5, 0, 9, 6, 1, 8, 4] [8, 2, 0, 3, 1, 6, 4, 5, 9, 7] [9, 4, 7, 1, 2, 8, 5, 0, 6, 3]
M2000 Interpreter
Easy Way
One row shuffled to be used as the destination row. One more shuffled and then n times rotated by one and stored to array
for 40x40 need 2~3 sec, including displaying to screen
We use the stack of values, a linked list, for pushing to top (Push) or to bottom (Data), and we can pop from top using Number or by using Read A to read A from stack. Also we can shift from a chosen position to top using Shift, or using shiftback to move an item from top to chosen position. So we shuffle items by shifting them.
<lang M2000 Interpreter> Module FastLatinSquare { n=5 For k=1 To 2 latin() Next n=40 latin() Sub latin() Local i,a, a(1 To n), b, k Profiler flush Print "latin square ";n;" by ";n For i=1 To n Push i Next i For i=1 To n div 2 Shiftback random(2, n) Next i a=[] Push ! stack(a) a=array(a) ' change a from stack to array For i=1 To n*10 Shiftback random(2, n) Next i For i=0 To n-1 Data number ' rotate by one the stack items b=[] ' move stack To b, leave empty stack a(a#val(i))=b Push ! stack(b) ' Push from a copy of b all items To stack Next i flush For k=1 To n div 2 z=random(2, n) For i=1 To n a=a(i) stack a { shift z } Next Next For i=1 To n a=a(i) a(i)=array(a) ' change To array from stack Next i For i=1 To n Print a(i) Next i Print TimeCount End Sub } FastLatinSquare</lang>
Hard Way
for 5x5 need some miliseconds
for 16X16 need 56 seconds
for 20X20 need 22 min (as for 9.8 version)
<lang M2000 Interpreter> Module LatinSquare (n, z=1, f$="latin.dat", NewFile As Boolean=False) { If Not Exist(f$) Or NewFile Then Open f$ For Wide Output As f Else Open f$ For Wide Append As f End If ArrayToString=Lambda -> { Shift 2 ' swap two top values in stack Push Letter$+Str$(Number) } Dim line(1 to n) flush ' erase current stack of value z=if(z<1->1, z) newColumn() For j=1 To z Profiler ResetColumns() For i=1 To n placeColumn() Next Print "A latin square of ";n;" by ";n For i=1 To n Print line(i) Print #f, line(i)#Fold$(ArrayToString) Next Print TimeCount Refresh Next close #f Flush ' empty stack again End Sub ResetColumns() Local i For i=1 To n:line(i)=(,):Next End Sub Sub newColumn() Local i For i=1 To n : Push i: Next End Sub Sub shuffle() Local i For i=1 To n div 2: Shift Random(2, n): Next End Sub Sub shuffleLocal(x) If Stack.size<=x Then Exit Sub Shift Random(x+1, Stack.size) Shiftback x End Sub Sub PlaceColumn() Local i, a, b, k shuffle() Do data number ' rotate one position k=0 For i=1 To n a=line(i) ' get the pointer Do If a#Pos(Stackitem(i))=-1 Then k=0 :Exit Do shuffleLocal(i) k++ Until k>Stack.size-i If k>0 Then Exit For Next Until k=0 For i=1 To n a=line(i) Append a, (Stackitem(i),) Next End Sub } Form 100,50 LatinSquare 5, 2, True LatinSquare 16
</lang>
- Output:
A latin square of 5 by 5 4 5 3 1 2 5 4 2 3 1 2 1 5 4 3 1 3 4 2 5 3 2 1 5 4 A latin square of 5 by 5 4 3 5 1 2 2 4 3 5 1 1 2 4 3 5 5 1 2 4 3 3 5 1 2 4 A latin square of 16 by 16 12 14 5 16 1 2 7 15 9 11 10 8 13 3 6 4 3 13 16 12 7 4 1 11 5 6 15 2 8 14 10 9 13 2 8 3 4 12 5 9 14 7 16 10 6 1 15 11 8 3 13 9 2 10 16 1 15 14 5 4 11 7 12 6 4 12 2 7 5 3 6 10 1 9 11 16 14 8 13 15 16 8 3 4 14 6 13 7 11 10 9 15 1 12 2 5 15 4 14 1 16 8 2 13 6 12 7 9 10 11 5 3 11 16 12 10 15 9 4 5 7 1 8 6 3 13 14 2 10 15 4 5 12 16 3 6 8 13 1 11 7 2 9 14 9 11 15 8 3 1 14 12 13 4 6 5 2 16 7 10 7 10 11 13 9 14 15 4 3 5 2 12 16 6 1 8 6 7 10 2 8 13 9 16 12 15 14 3 5 4 11 1 5 6 1 14 13 11 8 2 10 3 12 7 15 9 4 16 2 5 6 15 11 7 12 14 4 8 3 1 9 10 16 13 1 9 7 11 6 15 10 8 2 16 13 14 4 5 3 12 14 1 9 6 10 5 11 3 16 2 4 13 12 15 8 7
Mathematica/Wolfram Language
<lang Mathematica>Clear[RandomLatinSquare] RandomLatinSquare[n_] := Module[{out, ord},
out = Table[RotateLeft[Range[n], i], {i, n}];
out = RandomSample[out];
ord = RandomSample[Range[n]];
out = outAll, ord;
out
]
RandomLatinSquare[5] // Grid</lang>
- Output:
5 2 4 1 3 2 4 1 3 5 4 1 3 5 2 3 5 2 4 1 1 3 5 2 4
Nim
Not a straight translation of Kotlin version. There are many differences but the algorithm is the same.
Starting at n = 11, the execution time will be very variable as the program proceeds by trial and error. At least, the algorithm will be able to produce all the possible Latin squares but not in a uniform way.
<lang Nim>import random, sequtils, strutils
type LatinSquare = seq[seq[char]]
proc get[T](s: set[T]): T =
## Return the first element of a set. for n in s: return n
proc letterAt(n: Natural): char {.inline.} = chr(ord('A') - 1 + n)
proc latinSquare(n: Positive): LatinSquare =
result = newSeqWith(n, toSeq(letterAt(1)..letterAt(n))) result[0].shuffle()
for row in 1..(n - 2):
var ok = false
while not ok:
block shuffling:
result[row].shuffle()
for prev in 0..<row:
for col in 0..<n:
if result[row][col] == result[prev][col]:
break shuffling
ok = true
for col in 0..<n:
var s = {letterAt(1)..letterAt(n)}
for row in 0..(n - 2):
s.excl result[row][col]
result[^1][col] = s.get()
proc `$`(s: LatinSquare): string =
let n = s.len
for row in 0..<n:
result.add s[row].join(" ") & '\n'
randomize()
echo latinSquare(5)
echo latinSquare(5)
echo latinSquare(10)</lang>
- Output:
D A C E B A D B C E B E D A C E C A B D C B E D A E C D A B B A C E D C B A D E A D E B C D E B C A D I J B H F G E A C H E C G A I F D B J I J G A F E C B D H C D H J E A I G F B G B A F I H E C J D B F D E J C H I G A F C B H D G A J I E A H E I B D J F C G J A I C G B D H E F E G F D C J B A H I
Pascal
Jacobson-Matthews algorithm. Generates uniformly distributed random Latin squares (if used PRNG is good - Delphi/Pascal built-in PRNG is not).
Slightly modified translation of C code from https://brainwagon.org/2016/05/17/code-for-generating-a-random-latin-square/
Algorithm source: Jacobson, M. T.; Matthews, P. (1996). "Generating uniformly distributed random latin squares". Journal of Combinatorial Designs. 4 (6): 405–437.
<lang pascal>
{$APPTYPE CONSOLE}
const
Alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
Type
IncidenceCube = Array of Array Of Array of Integer;
Var Cube : IncidenceCube; DIM : Integer;
Procedure InitIncidenceCube(Var c:IncidenceCube; const Size:Integer);
var i, j, k : integer;
begin
DIM := Size;
SetLength(c,DIM,DIM,DIM);
for i := 0 to DIM-1 do
for j := 0 to DIM-1 do
for k := 0 to DIM-1 do c[i,j,k] := 0 ;
for i := 0 to DIM-1 do for j := 0 to DIM-1 do c[i,j,(i+j) mod DIM] := 1; end;
Procedure FreeIncidenceCube(Var c:IncidenceCube);
begin
Finalize(c);
end;
procedure PrintIncidenceCube(var c:IncidenceCube);
var i, j, k : integer;
begin
for i := 0 to DIM-1 do begin
for j := 0 to DIM-1 do begin
for k := 0 to DIM-1 do begin
if (c[i,j,k]=1) then begin
write(Alpha[k+1],' ');
break;
end;
end;
end;
Writeln;
end;
Writeln;
WriteLn; end;
procedure ShuffleIncidenceCube(var c:IncidenceCube);
var i, j, rx, ry, rz, ox, oy, oz : integer;
begin
for i := 0 to (DIM*DIM*DIM)-1 do begin
repeat
rx := Random(DIM);
ry := Random(DIM);
rz := Random(DIM);
until (c[rx,ry,rz]=0);
for j := 0 to DIM-1 do begin
if (c[j,ry,rz]=1) then ox := j;
if (c[rx,j,rz]=1) then oy := j;
if (c[rx,ry,j]=1) then oz := j;
end;
Inc(c[rx,ry,rz]);
Inc(c[rx,oy,oz]);
Inc(c[ox,ry,oz]);
Inc(c[ox,oy,rz]);
Dec(c[rx,ry,oz]);
Dec(c[rx,oy,rz]);
Dec(c[ox,ry,rz]);
Dec(c[ox,oy,oz]);
while (c[ox,oy,oz] < 0) do begin
rx := ox ;
ry := oy ;
rz := oz ;
if (random(2)=0) then begin
for j := 0 to DIM-1 do begin
if (c[j,ry,rz]=1) then ox := j;
end;
end else begin
for j := DIM-1 downto 0 do begin
if (c[j,ry,rz]=1) then ox := j;
end;
end;
if (random(2)=0) then begin
for j := 0 to DIM-1 do begin
if (c[rx,j,rz]=1) then oy := j;
end;
end else begin
for j := DIM-1 downto 0 do begin
if (c[rx,j,rz]=1) then oy := j;
end;
end;
if (random(2)=0) then begin
for j := 0 to DIM-1 do begin
if (c[rx,ry,j]=1) then oz := j;
end;
end else begin
for j := DIM-1 downto 0 do begin
if (c[rx,ry,j]=1) then oz := j;
end;
end;
Inc(c[rx,ry,rz]);
Inc(c[rx,oy,oz]);
Inc(c[ox,ry,oz]);
Inc(c[ox,oy,rz]);
Dec(c[rx,ry,oz]);
Dec(c[rx,oy,rz]);
Dec(c[ox,ry,rz]);
Dec(c[ox,oy,oz]);
end;
end;
end;
begin
Randomize; InitIncidenceCube(cube, 5); ShuffleIncidenceCube(cube); PrintIncidenceCube(cube); FreeIncidenceCube(Cube); InitIncidenceCube(cube, 5); ShuffleIncidenceCube(cube); PrintIncidenceCube(cube); FreeIncidenceCube(Cube); InitIncidenceCube(cube,10); ShuffleIncidenceCube(cube); PrintIncidenceCube(cube); FreeIncidenceCube(Cube); InitIncidenceCube(cube,26); ShuffleIncidenceCube(cube); PrintIncidenceCube(cube); FreeIncidenceCube(Cube);
end.
</lang>
- Output:
B A E D C D B C A E C D A E B A E B C D E C D B A A C D B E B E C D A D A B E C E D A C B C B E A D E F G C D A H B I J B J A H F D C E G I F I J A C E G H D B J A E D G F B I C H C E D I H G A J B F I D H E A B F C J G H G B J E C I D F A G C F B J I D A H E D H I F B J E G A C A B C G I H J F E D W D X Q V Z S A O T P K C Y M H J L F R U B I E N G E J R T D G P U C H F Y B Q W V I Z K L S X O N M A Y E B A W I T J U Z H F N G P L X M R K D Q C V S O H V F W Y S E P A N X M R O Q K B C L G J U T Z I D C Y E I G Q D X T S J L U M K B V P Z H N A F O R W L G J R O X F Q Y K C E W U V S A B D P H N Z I T M B M G D N F I R Z E L H Q K J U O T V C X Y A P W S G Z H S U L Q C K X Y V F I A O W J B M P R N T D E K O W L C T U I P V R A J N S E Z H X D M F G B Q Y D R A H X C K E W L S N V Z O P F Q Y T G M B J U I V Q Y O R D G B X U Z T H J E F K S C N I W M A P L U C M B A R Z F J O T G K X D N P I Q W L S V Y E H Z F N U T V M H R Q I B S P X D C A W E O L Y G K J J N D V M B X Z F C G O I S Y R L E P A W T K U H Q N T L Y S P J O B G D W E C Z I R F U X V H Q M A K A I C P J H B W Q D E S M R L Z G N T V Y K U F O X R K S N E W A V L M Q D G H T C Y U I F B O P X J Z O W I M F K R Y H B A Q X D U T N V G J Z P E S L C P B T X Q U N L S Y M I O W F J H K E Z A G D C V R X L U K I E H M N A B Z P V G W D Y S O R C J Q F T M H Z C K J Y T I F O P D A B X U W N S Q E R L G V Q S P G H Y O N V W U R A L C M T D J I E Z X K B F I A O F P M L K D J W U Z E N G Q X H Y T V S R C B T P K E B A V D G I N X L F R Y S O M Q C J H W Z U S U V Z L O C G M P K J Y T I Q E R A B F D W H X N F X Q J Z N W S E R V C T B H A M G O U K I L D Y P
Perl
<lang perl>use strict; use warnings; use feature 'say'; use List::Util 'shuffle';
sub random_ls {
my($n) = @_; my(@cols,@symbols,@ls_sym);
# build n-sized latin square
my @ls = [0,];
for my $i (1..$n-1) {
@{$ls[$i]} = @{$ls[0]};
splice(@{$ls[$_]}, $_, 0, $i) for 0..$i;
}
# shuffle rows and columns
@cols = shuffle 0..$n-1;
@ls = map [ @{$_}[@cols] ], @ls[shuffle 0..$n-1];
# replace numbers with symbols
@symbols = shuffle( ('A'..'Z')[0..$n-1] );
push @ls_sym, [@symbols[@$_]] for @ls;
@ls_sym
}
sub display {
my $str;
$str .= join(' ', @$_) . "\n" for @_;
$str
}
say display random_ls($_) for 2..5, 5;</lang>
- Output:
B A A B A B C B C A C A B A B D C B D C A C A B D D C A B C E A B D E A D C B B C E D A D B C A E A D B E C E D C B A C E B A D A B D E C B C A D E D A E C B
Phix
Brute force, begins to struggle above 42.
with javascript_semantics string aleph = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" function ls(integer n) if n>length(aleph) then ?9/0 end if -- too big... atom t1 = time()+1 sequence tn = tagset(n), -- {1..n} vcs = repeat(tn,n), -- valid for cols res = {} integer clashes = 0 while length(res)<n do sequence rn = {}, -- next row vr = tagset(n), -- valid for row (ie all) vc = deep_copy(vcs) -- copy (in case of clash) bool clash = false for c=1 to n do sequence v = {} for k=1 to n do -- collect all still valid options if vr[k] and vc[c][k] then v &= k end if end for if v={} then clash = true exit end if integer z = v[rand(length(v))] rn &= z vr[z] = 0 -- no longer valid vc[c][z] = 0 -- "" end for if not clash then res = append(res,rn) vcs = vc else clashes += 1 if time()>t1 then printf(1,"rows completed:%d/%d, clashes:%d\n", {length(res),n,clashes}) t1 = time()+1 end if end if end while for i=1 to n do string line = "" for j=1 to n do line &= aleph[res[i][j]] end for res[i] = line end for return res end function procedure latin_square(integer n) atom t0 = time() string res = join(ls(n),"\n"), e = elapsed(time()-t0) printf(1,"Latin square of order %d (%s):\n%s\n",{n,e,res}) end procedure latin_square(3) latin_square(5) latin_square(5) latin_square(10) if platform()!=JS then latin_square(42) end if
- Output:
Latin square of order 3 (0s): 231 123 312 Latin square of order 5 (0s): 15423 53142 24315 42531 31254 Latin square of order 5 (0s): 32514 21453 43125 15342 54231 Latin square of order 10 (0s): 3258A69417 9314275A86 586312479A 19A2753864 61294873A5 267194A538 473A618952 85973A6241 7A45831629 A486592173 rows completed:40/42, clashes:49854 rows completed:40/42, clashes:104051 Latin square of order 42 (2.1s): 5CMOgPTHbDBKGLU9d1aIREFNV8cQYeZ62AJXW3Sf74 VQEaBOIL2GefUYXbZWKP5cRDd3C4HS89MF7Jg6A1NT LIRc1AXPWKJH4bTNFC2VS935g6ZEDfaeOGdYQ8B7MU QPZ1LcN8O4I96dKRATfEWHaJUSM5e37GXBbDYFCV2g ATIZD2VY8gGRNHBO6P35QaEM1fS79dKFbCX4JWcUeL eE9Q7CJZ3VP254YMLHGOIBcdf1AbgXFKUR8WaTDNS6 7FfJdWGg4ZDC1UaVIcH9A5XLSb63MBTNPOQEKe2Y8R DMa53QWdB9bPSeEZJg6GK2A7YR1C4VLcIXTUFO8HfN G8d2eaARUPNEI6HCKQZSO4b1BM5g3L9VcTYFfXWJD7 9OHSGIRDEMf3YKb2cU1WVdNeTL8XA64gQPC57JZaBF 634YR52FINaeDQPKCXBAEZ8S7dgUG9OHJcfLMVTWb1 aVgXP7EO5f8JdFQ1RY9BHW42eTNGSUb3LMKZIC6AcD YBWLbFUAP1T8JSZ6N2gMeIQfHc3DC75d9EVOG4aKRX TXUBHe62JSCNA7WPfGLdYbOc9VRZ1IgD3aMQ5KEF48 dU1IS9D362V5cR8WTZCKfNPYA4OebQXJ7HBMLEFGga 3aDTUf5SCY6g8GOJbVIHZ1KWMNXBFPd7AQRc924LEe 1e2bJBHUDa9dTCRLYAWNXOMZFIK6QgfPE5S38G74Vc F786NJP1Gb4BLXVdUaAfcTgQC2eWKHYZ53D9ERISOM U1cRQGOeFBXWag6IPdVJ2Mf8L7YSNCHETK4bDZ539A 2DbgW6MfVAQY9acUX35ZNRBPIOJL81CSe4EHd7GTFK 8bNE2U7XTWHVBZJ5QMcePY9g6Ad1OaIf4DFKRSLC3G g5L8AbKN7JF6MD1SBEQU3fTXWHVdR2P4ZY9aCceIGO W2K9FHa5dT1OZ8C3gLScGVI4JB7YPDUbfe6RNAXMQE X6QdY49JZ7E1fB5THFD2MeGAPaI8VRSWNgUCcLbOK3 EASPTVLWRe7I35Mc24FX9gHO8KbJdN1aB6GfZQUDCY cSFM9ECbY65DW3GA17RTBQUVOg4K2ZNXadLPeIH8Jf O9GFf8QEXdAMVN47SRUD13ZB5PH26cWTCJaebYKgLI RcJUI319KFMXH2fea8dCL6WTG5EO74DYSNZAVgQBPb M4CHOZdIQUgGE9DB7KJaTLeF2WfN5A38V1c6XbPRYS IJBCad3cSQ27OEegM6XFb8LK4YDfTWGR1ZHNA59PUV SRXA4D8GeHUQ7c2aVbNgJP1E3FTILOMBKf5d69YZWC fNAeKMS7gR34bWIF5BEYDX69QCUTaGV28LO1HdJcZP PfOD6SeKaXcF2Id4W9bQCU73RZLMBYA5g81GTNVEHJ JZ5VMTcQfLRaP1SG8DOb4CYHNEFAXKeIW7g2BU36d9 4HVNEgbTLcKUCA7Y95eRFGJIaXBPZ8QMDW3SO1fd62 bgT7VKfBH5dLQJ3E4N86aSCRZUGFcM21Y9eIPDOXAW CW3fXRg6NOSTKV9HEJYL8D5GbQPcIFBUd2A74M1eaZ KGY3ZX4CA8OcFPNfeSM17JDbE92aW56LRVIgUHdQTB NYP4C1ZM9EWSROLDGI786KVaXeQHfJcAFU2T3Bgb5d HdeGcYB413LZgTFQDfP7UASCKJWVEbRO6IN82aM9X5 BK7W8LYacIZAeMgX3OT4dF26DG9RUEJCHbPVSfN51Q ZL6K5NFVMCYbXfA8Oe43g7dUcDa9JTEQGSWB1PR2IH
Picat
Using constraint modelling.
The used labeling (search strategy) is:
ff: select a variable to label using the first fail strategyrand_val: select a random (but valid) value.
Normally a non-random value strategy is preferred such as up, split, or updown, but the task requires random solutions.
<lang Picat>main =>
_ = random2(), % random seed N = 5, foreach(_ in 1..2) latin_square(N, X), pretty_print(X) end, % A larger random instance latin_square(62,X), pretty_print(X).
% Latin square latin_square(N, X) =>
X = new_array(N,N), X :: 1..N, foreach(I in 1..N) all_different([X[I,J] : J in 1..N]), all_different([X[J,I] : J in 1..N]) end, % rand_val for randomness solve($[ff,rand_val],X).
pretty_print(X) =>
N = X.len,
Alpha = "1234567890abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
foreach(I in 1..N)
foreach(J in 1..N)
if N > 20 then
printf("%w",Alpha[X[I,J]])
else
printf("%2w ",X[I,J])
end
end,
nl
end,
nl.</lang>
- Output:
5 1 3 4 2 1 2 4 3 5 4 5 2 1 3 2 3 1 5 4 3 4 5 2 1 5 2 3 4 1 3 4 5 1 2 2 5 1 3 4 4 1 2 5 3 1 3 4 2 5 9PRJhliyKjwkDVt60spfbz3aq8UmLMcSx1ed7nWIogENGr4uXHAvOQBTCYFZ52 CQEAorTdMWiDle2nXNg8YptFxBZwySGzURIPKubhHOJV49jmcqaLkfs06351v7 DEPvH2rfj7AVpyqx1ChMBQUGWsIoXzTedJFn3RL5YkZbK8Ni6cOmw90gauSlt4 wRSKyHhLiZeMztGcaETodVQprFjAvqN56Cmfxklg7UDYnOu1I49s8WPBJ032Xb SwCqxBOrFsjpYKb360dPzaMDIHgTRefQZNuUk4vV8mytE5GhWXLAno7219Jilc QcAaj62gZKf8TWPXY9Cv0yphwxDsF7SdkUoN4GtBEJnOMizI3blrume5RLqHV1 gFoxCYAT3U9nVGOEhu0m6WHdJ7wliQpks2atXLIfr4MeqbyzNB8PDj5RcSZv1K KVnYz1Lh98RGPvI5kQBrjlbqmNeJouMZ7FxOS0H3XCgwcpDTsay4Ut6EiAWf2d 5gFhdwmEWxHR4ziMsl1jZkOXU6JnpfrLoQYebKuctvGIAV29yTq8N3CDBP70aS RAq6edYJV40OuTctZGsWUXgPC5fiKb87manDIl2xyMB39kovpQr1hFENSzHjwL 1a4SA7XbOpr0IfKitRJke9sYGPxFQhuglj56Z8cLWNTyU2EwH3CnVDmvMBdqzo zt3LuMpFGYNwcEZUmjABDHRk9JXhri5I4lV7fvySs8xdW0TCO62qogabe1KPnQ I6Q4sfuxHw8EWX3jryDdoeSiplmBn9bVtvMTFz5KO72JLqk0G1cYAaUhZCgRPN HSiWqnUKre56ZJfFxOEQuCNIM10p3ws4TY2yLojz9DcmvaXlVht7bG8dPRkAgB flmnBts7CrL9eUzHMA6XguFVdj13N42TykhcaWoYZwvDpGKPJO5QE8RIxqiSb0 mvN5M4ReuItxGpnBOdLDTAlg3oc92rUwSb1QYsky0Ei8VhCJqZHKPXW6f7azFj AJwd3WCOx6PmEIjVog7TsN0MQykLztR2Y8BGqXU9SFlciDvHfr1a4pZKnbueh5 LdpikbWDNCSqAOQoBTycR29rheaUjY4PHMG3vtw6fIu5Zxls1g0EFJXn78VmKz pTzjrI52oMXWa0H4f6K1Owi7Lk38GUdNRgtmsFZJuQeBDnAEPCVcqv9SylbxYh XnYwNpyzbgua1A4CI3F9MvTElhLj05J8QdsVmcOHqiPWfR627UZBrSotKGxDke 2kHX4ujthGZb89vwpDPzxqcOBIKEsVo6AeJLN3QnUTa70FmyrfdMSYg15iRWCl uLGMfyksUS35CaRWqBHhtnIc7D4bAoVmXZ6j01ivJrQzPTFgKNExYd2Owelp98 MOh76z1k8tbYFcXeWvRxLomByU5dJjArnT4luI0PVG9HNgqpasDSCiwf3K2QEZ qUaEgXd47vQz0isyuFt6nLDmofhGe81cPpSwlNxT3RkrjCHYA5K29IbJVWOBZM iMeZtNKHwosU9npbDPGgC873SqWkOX6xzVyaTBAujdrE51IFYvRfQhlL024cmJ YW7fDeN8vcMAqbUpQHVlGxazi4SKPBZn9OCr5wF013dTJ6gjEokymLtX2IshuR l1vFmOP5Qy2TnBgYLKwCAGVHstbxIRe90Eqk8pfN6oU4ajrdMWShZcu7zJX3Di hYkRQKGiXFBCbu97y58w3djsnLHVSxvD2cgM6ePOpf40zZJqlEWta1IoArUTNm xbKrWQBl0AY3L2hzJt4naPECFus5TGqfV7NRgD8micwZ1SMekdIjpHv9X6oOyU yNrzODlnauKFsCoQ7U92fSdJcvTMbIthgmHpGPBiA05Lk4eR8VX61ZxqjEYwW3 TeLGwhFSRqV725rNUxuJ8MCA0Xza6WIEpibYDd4jvKmPlf1oQkBgHy3c9ZtnOs oXTOamVG6zUtg4Suv7Iih3WZEKR1fFl0rqjby9spwA8n2eckDLNJ5CHPdMBYQx ZmOliR6CfkTy5DJ83Io72c104MvWBKPjhArStqEFaHX9bwpQeYgzLsdVUnNuxG 6Gub5keN1dnvSYLqPcxtW0JoD98HlZziMIU2EA34TsKgryBOF7mwRVhjpQCXfa U0jeX8M1s5zuJZdICSvV7fYNtA9ycgEqaK3xnQDW4hboHBLrmRPOT2iFlwpkG6 J3fg1SnUcHCoyNTPzb5Y9OrKjQV786BsWDvqimaZdX0RIlwtu2MGexFAkhE4Lp v8soEP7At1WeBwC9VJnpc6LUKaM4qd3Y5ufZhOTGI2NQgH0xbzilXRjrFDmySk jZ5kpLJR4ByX3SuO2MUqibfn80NrtAFlEWw9chm7C1YvosPKdeGV6zDagxQITH BsymbGQvgnhcdHDZAaluVF6LzwiI9JOtCXkWrE1R2YjqSU3Mxp7Tf0KeoN854P 35687jzIYDxZKdVrHnOsSTX1NClvUkWMBfpuJ2qwe9FamEbALGQo04yihcPtRg nfJtv94jyL71HklsEerAQhPRXmq2ZNKow3z0MxCdbpVUYWia5SFDcuG8TgI6BO Vhb2FcEmk9GIMgNa4LiylRZ6uOCPdvwp8rAJjSKszWoft35BTxeH7qnQDX1U0Y ED2U8xtuIQO4rL51jkqFPYybVgAfHlCJNwKsRMpXBa76hzZS0iv3WecmGT9don W4Zy9gaw50mPUqYfRVNEX7vjbitDMnx3cBLKQHeolzS1CuOG2I6dJArk8FhspT G7xQSUgopacrN8w0bYM4kseW2dOXuCiAftEv95J1nPqKymh6ZljRBTLzHVDF3I NiWTIsSBPE4HfjALFzb01mKQaG2cY39UvxO8eJnC5yphXotDRuwZl7kMqdrg6V 7Ht1G5cWnJaBhl6dK2zNpIkufroewELb3yDAVCMUxZOFQXR4SmT0gPqYsvj8i9 0pXPcT9Mdlks6xWAGwaRKEo4vVYO5DyuqnigHrh2QS1jeJ7NtF3IzBfZmULb8C 8o9ulV3Qeb6fjM7giWXHNt4TZRyqDaYvF0P12UGkK5sxOcSLBAhCInzprmwJdE FjMVJoqYzRdgkPeKN1cGmi8v5pnux0XaIHThOfStDBCl7LWZ49sU2bQ3Ey6rAw OrVsKA0qLTg2tQkJlXZI5UneH3ENmpay1h9zBb78P6fSRvdciDuFxM4wYoGCjW kIc0TibaAfEKm1xG9Z2O4D58RSughsjBJPQozyVqLn6XdtU7vwYN3lpCWHeMrF Pq0CnEZVS2DJxm8hTfQ3yKAw1zBRgOkXe9ciU6daFltMuIsWojpbGNYH45vL7r axd9Yqw6DVJNRo0kc8eKI1hfgTpS72HFu4ZBPirAmLWs3MQXCnb5yEOlvjzGUt rKBcVCxplNFSOs1v5hmeEZq2TbPtky0HLo7XWaYDGjRiwAn3g8zud6JUQ4M9If tuUBLvDcmPIQw6aRnqYSFj2lAZr041gCisWEdVzbMx3p8N95hJoeKkTGOfy7HX bzgIZav0J3lhXFB2doSLr4wxkc6QEPmRD58CpT9MNuHGs7YfUynijK1WtOAVeq c9lpP0HXqOoLi3ESemjbJ5ByYWdZCT71GzRFAg6rhtI2xQa8nKUkvwV4usfNMD sB830FI92ipdv7MmgrfZwJGSOnQCaHDKb6l41jXERehATYVUzPxWt5NyLkcoqu dy1DU3fZTmqiorFlw4kaHBx5e2GYVLhWKSXIC7RQgbzu6P8njtJ9MOAsNp0Ecv e2INRZoPBX1jQhmD8i35vrut6E7zWcnGOL0HwYglkqACFdfV9M4psUSxbaTKJy 4CDH2J83Ehvl7RyTSpWUqgz9PYF61mQOjGd5oZNecVLkBKxbw0fXirMuItnasA CPU time 0.048 seconds.
Number of solutions
The number of solutions for a Latin square of size N is the OEIS sequence A002860: Number of Latin squares of order n; or labeled quasigroups. Here we check N = 1..6 which took 18min54s. <lang Picat>main =>
foreach(N in 1..6) Count = count_all(latin_square(N,_)), println(N=Count) end.</lang>
- Output:
1 = 1 2 = 2 3 = 12 4 = 576 5 = 161280 6 = 812851200 CPU time 1134.36 seconds.
Python
<lang python>from random import choice, shuffle from copy import deepcopy
def rls(n):
if n <= 0:
return []
else:
symbols = list(range(n))
square = _rls(symbols)
return _shuffle_transpose_shuffle(square)
def _shuffle_transpose_shuffle(matrix):
square = deepcopy(matrix) shuffle(square) trans = list(zip(*square)) shuffle(trans) return trans
def _rls(symbols):
n = len(symbols)
if n == 1:
return [symbols]
else:
sym = choice(symbols)
symbols.remove(sym)
square = _rls(symbols)
square.append(square[0].copy())
for i in range(n):
square[i].insert(i, sym)
return square
def _to_text(square):
if square:
width = max(len(str(sym)) for row in square for sym in row)
txt = '\n'.join(' '.join(f"{sym:>{width}}" for sym in row)
for row in square)
else:
txt =
return txt
def _check(square):
transpose = list(zip(*square))
assert _check_rows(square) and _check_rows(transpose), \
"Not a Latin square"
def _check_rows(square):
if not square:
return True
set_row0 = set(square[0])
return all(len(row) == len(set(row)) and set(row) == set_row0
for row in square)
if __name__ == '__main__':
for i in [3, 3, 5, 5, 12]:
square = rls(i)
print(_to_text(square))
_check(square)
print()</lang>
- Output:
2 1 0 0 2 1 1 0 2 1 0 2 0 2 1 2 1 0 1 0 3 2 4 3 4 2 0 1 4 2 1 3 0 2 1 0 4 3 0 3 4 1 2 2 1 0 4 3 0 4 3 2 1 3 2 1 0 4 4 3 2 1 0 1 0 4 3 2 6 2 4 8 11 9 3 1 7 0 5 10 1 11 5 2 8 6 0 9 4 10 7 3 2 7 10 5 4 8 9 11 0 6 3 1 8 5 0 4 7 11 1 2 3 9 10 6 11 4 3 7 5 2 6 8 10 1 0 9 10 1 8 6 9 0 7 3 11 4 2 5 7 0 1 3 10 5 8 4 6 2 9 11 9 8 7 11 2 1 10 6 5 3 4 0 3 9 2 1 6 10 4 0 8 5 11 7 5 3 6 10 0 4 11 7 9 8 1 2 4 10 9 0 3 7 2 5 1 11 6 8 0 6 11 9 1 3 5 10 2 7 8 4
Raku
(formerly Perl 6)
<lang perl6>sub latin-square { [[0],] };
sub random ( @ls, :$size = 5 ) {
# Build
for 1 ..^ $size -> $i {
@ls[$i] = @ls[0].clone;
@ls[$_].splice($_, 0, $i) for 0 .. $i;
}
# Shuffle @ls = @ls[^$size .pick(*)]; my @cols = ^$size .pick(*); @ls[$_] = @ls[$_][@cols] for ^@ls;
# Some random Latin glyphs
my @symbols = ('A' .. 'Z').pick($size);
@ls.deepmap: { $_ = @symbols[$_] };
}
sub display ( @array ) { $_.fmt("%2s ").put for |@array, }
- The Task
- Default size 5
display random latin-square;
- Specified size
display random :size($_), latin-square for 5, 3, 9;
- Or, if you'd prefer:
display random latin-square, :size($_) for 12, 2, 1;</lang>
- Sample output:
V Z M J U Z M U V J U J V M Z J V Z U M M U J Z V B H K U D H D U B K K U H D B U B D K H D K B H U I P Y P Y I Y I P Y J K E Z B I W H E Y B W K H J Z I B K Y H J E Z I W I H W J E Z B Y K J I Z Y W K H E B W E H Z B I Y K J H B E I Y W K J Z K Z J B I Y W H E Z W I K H J E B Y L Q E M A T Z C N Y R D Q R Y L N D C E M T A Z E Y M C D Q A N Z L T R M L C N R Y D Z A E Q T N M Z A Q E T D R C L Y T D Q Y C A M L E R Z N R A T Q M Z E Y L D N C D Z R T E N L Q Y A C M Y T L E Z R N M C Q D A A N D R L C Y T Q Z M E Z C A D Y M Q R T N E L C E N Z T L R A D M Y Q Y G G Y I
REXX
This REXX version produces a randomized Latin square similar to the Julia program.
The symbols could be any characters (except those that contain a blank), but the numbers from 0 ──► N-1 are used. <lang rexx>/*REXX program generates and displays a randomized Latin square. */ parse arg N seed . /*obtain the optional argument from CL.*/ if N== | N=="," then N= 5 /*Not specified? Then use the default.*/ if datatype(seed, 'W') then call random ,,seed /*Seed numeric? Then use it for seed.*/ w= length(N - 1) /*get the length of the largest number.*/ $= /*initialize $ string to null. */
do i=0 for N; $= $ right(i, w, '_') /*build a string of numbers (from zero)*/
end /*i*/ /* [↑] $ string is (so far) in order.*/
z= /*Z: will be the 1st row of the square*/
do N; ?= random(1,words($)) /*gen a random number from the $ string*/
z= z word($, ?); $= delword($, ?, 1) /*add the number to string; del from $.*/
end /*r*/
zz= z||z /*build a double-length string of Z. */
do j=1 for N /* [↓] display rows of random Latin sq*/
say translate(subword(zz, j, N), , '_') /*translate leading underbar to blank. */
end /*j*/ /*stick a fork in it, we're all done. */</lang>
- output for 1st run when using the default inputs:
4 1 3 0 2 1 3 0 2 4 3 0 2 4 1 0 2 4 1 3 2 4 1 3 0
- output for 2nd run when using the default inputs:
2 1 0 4 3 1 0 4 3 2 0 4 3 2 1 4 3 2 1 0 3 2 1 0 4
Ring
<lang ring> load "stdlib.ring" load "guilib.ring"
- ====================================================================================
size = 10
time1 = 0 bwidth = 0 bheight = 0
a2DSquare = newlist(size,size) a2DFinal = newlist(size,size)
aList = 1:size aList2 = RandomList(aList)
GenerateRows(aList2) ShuffleCols(a2DSquare, a2DFinal)
C_SPACING = 1 Button = newlist(size,size) LayoutButtonRow = list(size) C_ButtonOrangeStyle = 'border-radius:1x;color:black; background-color: orange'
- ====================================================================================
MyApp = New qApp {
StyleFusion()
win = new qWidget() {
workHeight = win.height() fontSize = 8 + (300/size)
wwidth = win.width() wheight = win.height() bwidth = wwidth/size bheight = wheight/size
setwindowtitle("Random Latin Squares") move(555,0) setfixedsize(1000,1000)
myfilter = new qallevents(win) myfilter.setResizeEvent("resizeBoard()") installeventfilter(myfilter)
LayoutButtonMain = new QVBoxLayout() { setSpacing(C_SPACING) setContentsmargins(50,50,50,50) }
LayoutButtonStart = new QHBoxLayout() { setSpacing(C_SPACING) setContentsmargins(0,0,0,0) }
btnStart = new qPushButton(win) { setFont(new qFont("Calibri",fontsize,2100,0)) resize(bwidth,bheight) settext(" Start ") setstylesheet(C_ButtonOrangeStyle) setclickevent("gameSolution()") }
sizeBtn = new qlabel(win) { setFont(new qFont("Calibri",fontsize,2100,0)) resize(bwidth,bheight) setStyleSheet("background-color:rgb(255,255,204)") setText(" Size: ") }
lineSize = new qLineEdit(win) { setFont(new qFont("Calibri",fontsize,2100,0)) resize(bwidth,bheight) setStyleSheet("background-color:rgb(255,255,204)") setAlignment( Qt_AlignHCenter) setAlignment( Qt_AlignVCenter) setreturnPressedEvent("newBoardSize()") setText(string(size)) }
btnExit = new qPushButton(win) { setFont(new qFont("Calibri",fontsize,2100,0)) resize(bwidth,bheight) settext(" Exit ") setstylesheet(C_ButtonOrangeStyle) setclickevent("pExit()") }
LayoutButtonStart.AddWidget(btnStart) LayoutButtonStart.AddWidget(sizeBtn) LayoutButtonStart.AddWidget(lineSize) LayoutButtonStart.AddWidget(btnExit)
LayoutButtonMain.AddLayout(LayoutButtonStart)
for Row = 1 to size
LayoutButtonRow[Row] = new QHBoxLayout() { setSpacing(C_SPACING) setContentsmargins(0,0,0,0) }
for Col = 1 to size Button[Row][Col] = new qlabel(win) { setFont(new qFont("Calibri",fontsize,2100,0)) resize(bwidth,bheight) } LayoutButtonRow[Row].AddWidget(Button[Row][Col]) next LayoutButtonMain.AddLayout(LayoutButtonRow[Row])
next
setLayout(LayoutButtonMain)
show()
}
exec() }
- ====================================================================================
func newBoardSize()
nrSize = number(lineSize.text())
if nrSize = 1 ? "Enter: Size > 1" return ok
for Row = 1 to size for Col = 1 to size Button[Row][Col].settext("") next next
newWindow(nrSize)
- ====================================================================================
func newWindow(newSize)
time1 = clock()
for Row = 1 to size for Col = 1 to size Button[Row][Col].delete() next next
size = newSize
bwidth = ceil((win.width() - 8) / size) bheight = ceil((win.height() - 32) / size)
fontSize = 8 + (300/size)
if size > 16 fontSize = 8 + (150/size) ok
if size < 8 fontSize = 30 + (150/size) ok
if size = 2 fontSize = 10 + (100/size) ok
btnStart.setFont(new qFont("Calibri",fontsize,2100,0)) sizeBtn.setFont(new qFont("Calibri",fontsize,2100,0)) lineSize.setFont(new qFont("Calibri",fontsize,2100,0)) btnExit.setFont(new qFont("Calibri",fontsize,2100,0))
LayoutButtonStart = new QHBoxLayout() { setSpacing(C_SPACING) setContentsmargins(0,0,0,0) }
Button = newlist(size,size) LayoutButtonRow = list(size)
for Row = 1 to size
LayoutButtonRow[Row] = new QHBoxLayout() { setSpacing(C_SPACING) setContentsmargins(0,0,0,0) }
for Col = 1 to size
Button[Row][Col] = new qlabel(win) { setFont(new qFont("Calibri",fontsize,2100,0)) resize(bwidth,bheight) } LayoutButtonRow[Row].AddWidget(Button[Row][Col])
next
LayoutButtonMain.AddLayout(LayoutButtonRow[Row])
next
win.setLayout(LayoutButtonMain)
return
- ====================================================================================
func resizeBoard
bwidth = ceil((win.width() - 8) / size) bheight = ceil((win.height() - 32) / size)
for Row = 1 to size for Col = 1 to size Button[Row][Col].resize(bwidth,bheight) next next
- ====================================================================================
Func pExit
MyApp.quit()
- ====================================================================================
func gameSolution()
a2DSquare = newlist(size,size) a2DFinal = newlist(size,size)
aList = 1:size aList2 = RandomList(aList)
GenerateRows(aList2) ShuffleCols(a2DSquare, a2DFinal)
for nRow = 1 to size for nCol = 1 to size Button[nRow][nCol].settext("-") next next
for nRow = 1 to size for nCol = 1 to size Button[nRow][nCol].resize(bwidth,bheight) Button[nRow][nCol].settext(string(a2DSquare[nRow][nCol])) next next
time2 = clock() time3 = (time2 - time1)/1000 ? "Elapsed time: " + time3 + " ms at size = " + size + nl
- ====================================================================================
// Scramble the numbers in the List // Uniq random picks, then shorten list by each pick
Func RandomList(aInput)
aOutput = []
while len(aInput) > 1
nIndex = random(len(aInput)-1)
nIndex++
aOutput + aInput[nIndex]
del(aInput,nIndex)
end
aOutput + aInput[1]
return aOutput
- ====================================================================================
// Generate Rows of data. Put them in the 2DArray
Func GenerateRows(aInput)
aOutput = [] size = len(aInput)
shift = 1
for k = 1 to size // Make 8 Rows of lists aOutput = []
for i = 1 to size // make a list pick = i + shift // shift every Row by +1 more if pick > size pick = pick - size ok
aOutput + aInput[pick] next
a2DSquare[k] = aOutput // Row of Output to a2DSquare
shift++ // shift next line by +1 more if shift > size shift = 1 ok next
return
- ====================================================================================
// Shift random Rows into a2DFinal, then random Cols
Func ShuffleCols(a2DSquare, a2DFinal)
aSuffle = 1:size
aSuffle2 = RandomList(aSuffle) // Pick random Col to insert in a2DFinal
for i = 1 to size // Row pick = aSuffle2[i]
for j = 1 to size // Col
a2DFinal[i][j] = a2DSquare[pick][j] // i-Row-Col j-Horz-Vert next next
a2DSquare = a2DFinal // Now do the verticals aSuffle = 1:size aSuffle2 = RandomList(aSuffle)
for i = 1 to size // Row pick = aSuffle2[i]
for j = 1 to size // Col
a2DFinal[j][i] = a2DSquare[j][pick] //Reverse i-j , i-Row-Col j-Horz-Vert next next
return
- ====================================================================================
</lang>
Ruby
This crude algorithm works fine up to a square size of 10; higher values take too much time and memory. It creates an array of all possible permutations, picks a random one as first row an weeds out all permutations which cannot appear in the remaining square. Repeat picking and weeding until there is a square. <lang ruby>N = 5
def generate_square
perms = (1..N).to_a.permutation(N).to_a.shuffle
square = []
N.times do
square << perms.pop
perms.reject!{|perm| perm.zip(square.last).any?{|el1, el2| el1 == el2} }
end
square
end
def print_square(square)
cell_size = N.digits.size + 1
strings = square.map!{|row| row.map!{|el| el.to_s.rjust(cell_size)}.join }
puts strings, "\n"
end
2.times{print_square( generate_square)} </lang>
- Output:
3 4 2 1 5 2 3 4 5 1 1 2 5 3 4 5 1 3 4 2 4 5 1 2 31 2 5 4 3 2 3 4 1 5 5 4 2 3 1 3 5 1 2 4 4 1 3 5 2
Wren
Restarting Row method
<lang ecmascript>import "random" for Random
var rand = Random.new()
var printSquare = Fn.new { |latin|
for (row in latin) System.print(row) System.print()
}
var latinSquare = Fn.new { |n|
if (n <= 0) {
System.print("[]\n")
return
}
var latin = List.filled(n, null)
for (i in 0...n) {
latin[i] = List.filled(n, 0)
if (i == n - 1) break
for (j in 0...n) latin[i][j] = j
}
// first row rand.shuffle(latin[0])
// middle row(s)
for (i in 1...n-1) {
var shuffled = false
while (!shuffled) {
rand.shuffle(latin[i])
var shuffling = false
for (k in 0...i) {
for (j in 0...n) {
if (latin[k][j] == latin[i][j]) {
shuffling = true
break
}
}
if (shuffling) break
}
if (!shuffling) shuffled = true
}
}
// last row
for (j in 0...n) {
var used = List.filled(n, false)
for (i in 0...n-1) used[latin[i][j]] = true
for (k in 0...n) {
if (!used[k]) {
latin[n-1][j] = k
break
}
}
}
printSquare.call(latin)
}
latinSquare.call(5) latinSquare.call(5) latinSquare.call(10) // for good measure</lang>
- Output:
Sample run:
[4, 1, 2, 0, 3] [3, 2, 0, 1, 4] [1, 0, 3, 4, 2] [0, 3, 4, 2, 1] [2, 4, 1, 3, 0] [1, 2, 0, 4, 3] [2, 4, 3, 0, 1] [4, 3, 1, 2, 0] [0, 1, 2, 3, 4] [3, 0, 4, 1, 2] [5, 3, 0, 6, 8, 2, 1, 7, 4, 9] [6, 5, 9, 8, 7, 1, 3, 4, 0, 2] [7, 6, 8, 4, 2, 0, 9, 1, 3, 5] [0, 1, 7, 2, 4, 6, 5, 8, 9, 3] [1, 0, 2, 3, 9, 5, 8, 6, 7, 4] [3, 7, 5, 0, 1, 9, 4, 2, 8, 6] [2, 4, 1, 5, 3, 8, 7, 9, 6, 0] [9, 2, 3, 7, 6, 4, 0, 5, 1, 8] [8, 9, 4, 1, 5, 3, 6, 0, 2, 7] [4, 8, 6, 9, 0, 7, 2, 3, 5, 1]
Latin Squares in Reduced Form method
<lang ecmascript>import "random" for Random import "/sort" for Sort import "/fmt" for Fmt import "/math" for Int
var rand = Random.new() var counts = List.filled(4, 0) var aa = List.filled(16, 0)
var testSquares = [
[0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 0, 1, 3, 2, 1, 0], [0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 1, 0, 3, 2, 0, 1], [0, 1, 2, 3, 1, 2, 3, 0, 2, 3, 0, 1, 3, 0, 1, 2], [0, 1, 2, 3, 1, 3, 0, 2, 2, 0, 3, 1, 3, 2, 1, 0]
]
// Checks whether two lists contain the same elements in the same order var areSame = Fn.new { |l1, l2|
if (l1.count != l2.count) return false
for (i in 0...l1.count) {
if (l1[i] != l2[i]) return false
}
return true
}
// generate derangements of first n numbers, with 'start' in first place. var dList = Fn.new { |n, start|
var r = []
start = start - 1 // use 0 basing
var a = List.filled(n, 0)
for (i in 0...n) a[i] = i
var t = a[0]
a[0] = start
a[start] = t
Sort.quick(a, 1, a.count-1, null)
var first = a[1]
var recurse // recursive closure permutes a[1..-1]
recurse = Fn.new { |last|
if (last == first) {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged.
var j = 0 // j starts from 0, not 1
for (v in a.skip(1)) {
if (j+1 == v) return r // no, ignore it
j = j + 1
}
// yes, save a copy
var b = a.toList
for (i in 0...b.count) b[i] = b[i] + 1 // change back to 1 basing
r.add(b)
return r
}
var i = last
while (i >= 1) {
a.swap(i, last)
recurse.call(last - 1)
a.swap(i, last)
i = i - 1
}
}
recurse.call(n - 1)
return r
}
var copyMatrix = Fn.new { |m|
var le = m.count var cpy = List.filled(le, null) for (i in 0...le) cpy[i] = m[i].toList return cpy
}
var reducedLatinSquares = Fn.new { |n|
var rls = []
if (n < 0) n = 0
var rlatin = List.filled(n, null)
for (i in 0...n) rlatin[i] = List.filled(n, 0)
if (n <= 1) {
rls.add(rlatin)
return rls
}
// first row for (j in 0...n) rlatin[0][j] = j + 1
// recursive closure to compute reduced latin squares
var recurse
recurse = Fn.new { |i|
var rows = dList.call(n, i) // get derangements of first n numbers, with 'i' first.
for (r in 0...rows.count) {
var outer = false
rlatin[i-1] = rows[r].toList
for (k in 0...i-1) {
for (j in 1...n) {
if (rlatin[k][j] == rlatin[i-1][j]) {
if (r < rows.count-1) {
outer = true
break
} else if (i > 2) {
return
}
}
}
if (outer) break
}
if (outer) continue
if (i < n) {
recurse.call(i + 1)
} else {
var rl = copyMatrix.call(rlatin)
rls.add(rl)
}
}
return
}
// remaining rows
recurse.call(2)
return rls
}
var printSquare = Fn.new { |latin, n|
for (i in 0...n) {
for (j in 0...n) Fmt.write("$d ", latin[i][j]-1)
System.print()
}
System.print()
}
// generate permutations of first n numbers, starting from 0. var pList = Fn.new { |n|
var fact = Int.factorial(n)
var perms = List.filled(fact, null)
var a = List.filled(n, 0)
for (i in 0...n) a[i] = i
var t = a.toList
perms[0] = t
n = n - 1
for (c in 1...fact) {
var i = n - 1
var j = n
while (a[i] > a[i+1]) i = i - 1
while (a[j] < a[i]) j = j - 1
a.swap(i, j)
j = n
i = i + 1
while (i < j) {
a.swap(i, j)
i = i + 1
j = j - 1
}
var t = a.toList
t.add(0)
perms[c] = t
}
return perms
}
var generateLatinSquares = Fn.new { |n, tests, echo|
var rls = reducedLatinSquares.call(n)
var perms = pList.call(n)
var perms2 = pList.call(n - 1)
for (test in 0...tests) {
var rn = rand.int(rls.count)
var rl = rls[rn] // select reduced random square at random
rn = rand.int(perms.count)
var rp = perms[rn] // select a random permuation of 'rl's columns
// permute columns
var t = List.filled(n, null)
for (i in 0...n) {
t[i] = List.filled(n, 0)
for (j in 0...n) t[i][j] = rl[i][rp[j]]
}
rn = rand.int(perms2.count)
rp = perms2[rn] // select a random permutation of 't's rows 2 to n
// permute rows 2 to n
var u = List.filled(n, null)
for (i in 0...n) {
u[i] = List.filled(n, 0)
for (j in 0...n) {
if (i == 0) {
u[i][j] = t[i][j]
} else {
u[i][j] = t[rp[i-1]+1][j]
}
}
}
if (test < echo) printSquare.call(u, n)
if (n == 4) {
for (i in 0..3) {
for (j in 0..3) u[i][j] = u[i][j] - 1
}
for (i in 0..3) {
for (j in 4*i...4*i+4) {
aa[j] = u[i][j - 4*i]
}
}
for (i in 0..3) {
if (areSame.call(testSquares[i], aa)) {
counts[i] = counts[i] + 1
break
}
}
}
}
}
System.print("Two randomly generated latin squares of order 5 are:\n") generateLatinSquares.call(5, 2, 2)
System.print("Out of 1,000,000 randomly generated latin squares of order 4, ") System.print("of which there are 576 instances ( => expected 1736 per instance),") System.print("the following squares occurred the number of times shown:\n") generateLatinSquares.call(4, 1e6, 0) for (i in 0..3) System.print("%(testSquares[i]) : %(counts[i])") System.print("\nA randomly generated latin square of order 6 is:\n") generateLatinSquares.call(6, 1, 1)</lang>
- Output:
Sample run:
Two randomly generated latin squares of order 5 are: 1 3 2 4 0 0 1 3 2 4 2 4 1 0 3 3 0 4 1 2 4 2 0 3 1 0 2 3 4 1 3 1 4 2 0 4 0 2 1 3 1 4 0 3 2 2 3 1 0 4 Out of 1,000,000 randomly generated latin squares of order 4, of which there are 576 instances ( => expected 1736 per instance), the following squares occurred the number of times shown: [0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 0, 1, 3, 2, 1, 0] : 1749 [0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 1, 0, 3, 2, 0, 1] : 1686 [0, 1, 2, 3, 1, 2, 3, 0, 2, 3, 0, 1, 3, 0, 1, 2] : 1702 [0, 1, 2, 3, 1, 3, 0, 2, 2, 0, 3, 1, 3, 2, 1, 0] : 1764 A randomly generated latin square of order 6 is: 4 5 2 0 1 3 1 4 5 3 0 2 3 0 4 5 2 1 2 1 3 4 5 0 5 2 0 1 3 4 0 3 1 2 4 5
zkl
<lang zkl>fcn randomLatinSquare(n,symbols=[1..]){ //--> list of lists
if(n<=0) return(T);
square,syms := List(), symbols.walker().walk(n);
do(n){ syms=syms.copy(); square.append(syms.append(syms.pop(0))) }
// shuffle rows, transpose & shuffle columns
T.zip(square.shuffle().xplode()).shuffle();
} fcn rls2String(square){ square.apply("concat"," ").concat("\n") }</lang> <lang zkl>foreach n in (T(1,2,5)){ randomLatinSquare(n) : rls2String(_).println("\n") } randomLatinSquare(5, ["A".."Z"]) : rls2String(_).println("\n"); randomLatinSquare(10,"!@#$%^&*()") : rls2String(_).println("\n");</lang>
- Output:
1 1 2 2 1 3 1 4 5 2 4 2 5 1 3 1 4 2 3 5 5 3 1 2 4 2 5 3 4 1 E D A B C D C E A B B A C D E A E B C D C B D E A & % # ! * @ ) $ ( ^ @ ) * ^ # & % ( $ ! ( & % # ) $ @ ^ ! * ! ( & % @ ^ $ * # ) % # ! ( ^ ) * @ & $ ^ $ @ ) & ! ( # * % # ! ( & $ * ^ ) % @ $ @ ) * % ( & ! ^ # ) * ^ $ ! % # & @ ( * ^ $ @ ( # ! % ) &