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.
;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 n generates a randomised Latin square of size n.
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:
- [[wp:Latin_square|Latin square]]
- [http://oeis.org/A002860 OEIS A002860]
=={{header|F_Sharp|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?
// 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()
[|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.
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
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.
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
}
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 [https://rosettacode.org/wiki/Latin_Squares_in_reduced_form#Go Latin Squares in Reduced Form] and [https://rosettacode.org/wiki/Permutations#non-recursive.2C_lexicographical_order Permutations] tasks.
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)
}
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
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);
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.
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)
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
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.
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
Hard Way
for 5x5 need some miliseconds
for 16X16 need 56 seconds
for 20X20 need 22 min (as for 9.8 version)
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$(ArrayToString)
Next
Print TimeCount
Refresh
Next
close #f
Flush ' empty stack again
End
Sub ResetColumns()
Local i
For i=1 To n:line(i)=(,):Next
End Sub
Sub newColumn()
Local i
For i=1 To n : Push i: Next
End Sub
Sub shuffle()
Local i
For i=1 To n div 2: Shift Random(2, n): Next
End Sub
Sub shuffleLocal(x)
If Stack.size<=x Then Exit Sub
Shift Random(x+1, Stack.size)
Shiftback x
End Sub
Sub PlaceColumn()
Local i, a, b, k
shuffle()
Do
data number ' rotate one position
k=0
For i=1 To n
a=line(i) ' get the pointer
Do
If a#Pos(Stackitem(i))=-1 Then k=0 :Exit Do
shuffleLocal(i)
k++
Until k>Stack.size-i
If k>0 Then Exit For
Next
Until k=0
For i=1 To n
a=line(i)
Append a, (Stackitem(i),)
Next
End Sub
}
Form 100,50
LatinSquare 5, 2, True
LatinSquare 16
A latin square of 5 by 5 4 5 3 1 2 5 4 2 3 1 2 1 5 4 3 1 3 4 2 5 3 2 1 5 4 A latin square of 5 by 5 4 3 5 1 2 2 4 3 5 1 1 2 4 3 5 5 1 2 4 3 3 5 1 2 4 A latin square of 16 by 16 12 14 5 16 1 2 7 15 9 11 10 8 13 3 6 4 3 13 16 12 7 4 1 11 5 6 15 2 8 14 10 9 13 2 8 3 4 12 5 9 14 7 16 10 6 1 15 11 8 3 13 9 2 10 16 1 15 14 5 4 11 7 12 6 4 12 2 7 5 3 6 10 1 9 11 16 14 8 13 15 16 8 3 4 14 6 13 7 11 10 9 15 1 12 2 5 15 4 14 1 16 8 2 13 6 12 7 9 10 11 5 3 11 16 12 10 15 9 4 5 7 1 8 6 3 13 14 2 10 15 4 5 12 16 3 6 8 13 1 11 7 2 9 14 9 11 15 8 3 1 14 12 13 4 6 5 2 16 7 10 7 10 11 13 9 14 15 4 3 5 2 12 16 6 1 8 6 7 10 2 8 13 9 16 12 15 14 3 5 4 11 1 5 6 1 14 13 11 8 2 10 3 12 7 15 9 4 16 2 5 6 15 11 7 12 14 4 8 3 1 9 10 16 13 1 9 7 11 6 15 10 8 2 16 13 14 4 5 3 12 14 1 9 6 10 5 11 3 16 2 4 13 12 15 8 7## Perl 6 ```perl6 sub latin-square { [[0],] }; sub random ( @ls, :$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; ``` ```txt 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 ``` ## Phix Brute force, begins to struggle above 42. ```Phix 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)