Random Latin squares: Difference between revisions
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[|1; 2; 3; 4; 5|] |
[|1; 2; 3; 4; 5|] |
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[|5; 3; 4; 1; 2|] |
[|5; 3; 4; 1; 2|] |
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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: |
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Times Generated Number of Latin Squares |
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1 1776 |
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2 5669 |
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3 11985 |
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4 19128 |
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5 24005 |
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6 25333 |
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7 22471 |
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8 18267 |
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9 12569 |
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10 7924 |
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11 4551 |
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12 2452 |
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13 1130 |
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14 483 |
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15 219 |
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16 93 |
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17 37 |
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18 5 |
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19 7 |
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20 2 |
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Revision as of 21:42, 16 July 2019
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 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
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. Unlike other solutions on this page, even those claiming to do so! see Talk:Random_Latin_Squares#.22restarting_row.22_method. 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
The initial approach is simple: generate a random permutation, one row at a time. If a row conflicts with any of the rows above it, generate a new random permutation for that row. The upside is that this is easy to understand and generates uniformly-random Latin squares. The downside is that it is an exponential time algorithm.
If larger sizes are desired, non-uniform approaches may be used. The Koscielny product is one such approach that "multiplies" two Latin squares together to produce a larger one. The algorithm is described here. The two initial order-5 squares are multiplied using this method to produce an order-25 square.
<lang factor>USING: arrays io kernel locals math math.matrices prettyprint random sequences sets vectors ; IN: rosetta-code.random-latin-squares
- add-row ( matrix -- matrix' )
dup last clone suffix! [ dup [ all-unique? ] column-map [ f = ] any? ] [ [ length 1 - ] [ [ [ randomize ] change-nth ] keep ] bi ] while ;
! Small optimization: there is only 1 choice for the last row.
- add-last-row ( matrix -- matrix' )
dup dim second <iota> over [ diff first ] with column-map suffix! ;
- last-row? ( matrix -- ? ) dim first2 = ;
- latin ( n -- matrix )
[ <iota> >array randomize 1vector ] [ 1 - [ dup last-row? [ add-last-row ] [ add-row ] if ] times ] bi ;
- koscielny-product ( ls1 ls2 -- prod )
ls1 ls2 [ dim first ] bi@ :> ( n1 n2 ) n1 n2 [ sq ] bi@ zero-matrix :> prod n1 n2 * <iota> [ :> i n1 n2 * <iota> [ :> j n2 i n2 /i j n2 /i ls1 nth nth * i n2 mod j n2 mod ls2 nth nth + { i j } prod set-index ] each ] each prod ;
- random-latin-squares ( -- )
"Random Latin squares via \"restarting row\" method:" print 5 5 [ latin dup simple-table. nl ] bi@ "Koscielny product of previous squares:" print koscielny-product simple-table. ;
MAIN: random-latin-squares</lang>
- Output:
Random Latin squares via "restarting row" method: 1 0 2 3 4 3 2 4 1 0 2 4 1 0 3 4 3 0 2 1 0 1 3 4 2 2 4 1 3 0 0 1 3 4 2 3 0 2 1 4 4 3 0 2 1 1 2 4 0 3 Koscielny product of previous squares: 7 5 8 9 6 17 15 18 19 16 12 10 13 14 11 22 20 23 24 21 2 0 3 4 1 9 6 5 8 7 19 16 15 18 17 14 11 10 13 12 24 21 20 23 22 4 1 0 3 2 6 8 7 5 9 16 18 17 15 19 11 13 12 10 14 21 23 22 20 24 1 3 2 0 4 8 9 6 7 5 18 19 16 17 15 13 14 11 12 10 23 24 21 22 20 3 4 1 2 0 5 7 9 6 8 15 17 19 16 18 10 12 14 11 13 20 22 24 21 23 0 2 4 1 3 2 0 3 4 1 12 10 13 14 11 22 20 23 24 21 17 15 18 19 16 7 5 8 9 6 4 1 0 3 2 14 11 10 13 12 24 21 20 23 22 19 16 15 18 17 9 6 5 8 7 1 3 2 0 4 11 13 12 10 14 21 23 22 20 24 16 18 17 15 19 6 8 7 5 9 3 4 1 2 0 13 14 11 12 10 23 24 21 22 20 18 19 16 17 15 8 9 6 7 5 0 2 4 1 3 10 12 14 11 13 20 22 24 21 23 15 17 19 16 18 5 7 9 6 8 12 10 13 14 11 22 20 23 24 21 7 5 8 9 6 2 0 3 4 1 17 15 18 19 16 14 11 10 13 12 24 21 20 23 22 9 6 5 8 7 4 1 0 3 2 19 16 15 18 17 11 13 12 10 14 21 23 22 20 24 6 8 7 5 9 1 3 2 0 4 16 18 17 15 19 13 14 11 12 10 23 24 21 22 20 8 9 6 7 5 3 4 1 2 0 18 19 16 17 15 10 12 14 11 13 20 22 24 21 23 5 7 9 6 8 0 2 4 1 3 15 17 19 16 18 17 15 18 19 16 7 5 8 9 6 2 0 3 4 1 12 10 13 14 11 22 20 23 24 21 19 16 15 18 17 9 6 5 8 7 4 1 0 3 2 14 11 10 13 12 24 21 20 23 22 16 18 17 15 19 6 8 7 5 9 1 3 2 0 4 11 13 12 10 14 21 23 22 20 24 18 19 16 17 15 8 9 6 7 5 3 4 1 2 0 13 14 11 12 10 23 24 21 22 20 15 17 19 16 18 5 7 9 6 8 0 2 4 1 3 10 12 14 11 13 20 22 24 21 23 22 20 23 24 21 2 0 3 4 1 17 15 18 19 16 7 5 8 9 6 12 10 13 14 11 24 21 20 23 22 4 1 0 3 2 19 16 15 18 17 9 6 5 8 7 14 11 10 13 12 21 23 22 20 24 1 3 2 0 4 16 18 17 15 19 6 8 7 5 9 11 13 12 10 14 23 24 21 22 20 3 4 1 2 0 18 19 16 17 15 8 9 6 7 5 13 14 11 12 10 20 22 24 21 23 0 2 4 1 3 15 17 19 16 18 5 7 9 6 8 10 12 14 11 13
Go
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 which has the merit of being completely random. <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:
[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]
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
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$(ArrayToString) Next Print TimeCount Refresh Next close #f Flush ' empty stack again End Sub ResetColumns() Local i For i=1 To n:line(i)=(,):Next End Sub Sub newColumn() Local i For i=1 To n : Push i: Next End Sub Sub shuffle() Local i For i=1 To n div 2: Shift Random(2, n): Next End Sub Sub shuffleLocal(x) If Stack.size<=x Then Exit Sub Shift Random(x+1, Stack.size) Shiftback x End Sub Sub PlaceColumn() Local i, a, b, k shuffle() Do data number ' rotate one position k=0 For i=1 To n a=line(i) ' get the pointer Do If a#Pos(Stackitem(i))=-1 Then k=0 :Exit Do shuffleLocal(i) k++ Until k>Stack.size-i If k>0 Then Exit For Next Until k=0 For i=1 To n a=line(i) Append a, (Stackitem(i),) Next End Sub } Form 100,50 LatinSquare 5, 2, True LatinSquare 16
</lang>
- Output:
A latin square of 5 by 5 4 5 3 1 2 5 4 2 3 1 2 1 5 4 3 1 3 4 2 5 3 2 1 5 4 A latin square of 5 by 5 4 3 5 1 2 2 4 3 5 1 1 2 4 3 5 5 1 2 4 3 3 5 1 2 4 A latin square of 16 by 16 12 14 5 16 1 2 7 15 9 11 10 8 13 3 6 4 3 13 16 12 7 4 1 11 5 6 15 2 8 14 10 9 13 2 8 3 4 12 5 9 14 7 16 10 6 1 15 11 8 3 13 9 2 10 16 1 15 14 5 4 11 7 12 6 4 12 2 7 5 3 6 10 1 9 11 16 14 8 13 15 16 8 3 4 14 6 13 7 11 10 9 15 1 12 2 5 15 4 14 1 16 8 2 13 6 12 7 9 10 11 5 3 11 16 12 10 15 9 4 5 7 1 8 6 3 13 14 2 10 15 4 5 12 16 3 6 8 13 1 11 7 2 9 14 9 11 15 8 3 1 14 12 13 4 6 5 2 16 7 10 7 10 11 13 9 14 15 4 3 5 2 12 16 6 1 8 6 7 10 2 8 13 9 16 12 15 14 3 5 4 11 1 5 6 1 14 13 11 8 2 10 3 12 7 15 9 4 16 2 5 6 15 11 7 12 14 4 8 3 1 9 10 16 13 1 9 7 11 6 15 10 8 2 16 13 14 4 5 3 12 14 1 9 6 10 5 11 3 16 2 4 13 12 15 8 7
Perl 6
<lang perl6>sub latin-square { [[0],] };
sub random ( @ls, :$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
Phix
Brute force, begins to struggle above 42. <lang 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)<n do sequence rn = {}, -- next row vr = tn, -- valid for row (ie all) vc = 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) latin_square(42)</lang>
- 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
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
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 $.*/ end /*r*/
zz= z||z /*build a double-length string of Z. */
do j=1 for N /* [↓] display rows of random Latin sq*/ say translate(subword(zz, j, N), , '_') /*translate leading underbar to blank. */ end /*j*/ /*stick a fork in it, we're all done. */</lang>
- output for the 1st run when using the default inputs:
4 1 3 0 2 1 3 0 2 4 3 0 2 4 1 0 2 4 1 3 2 4 1 3 0
- output for the 2nd run when using the default inputs:
2 1 0 4 3 1 0 4 3 2 0 4 3 2 1 4 3 2 1 0 3 2 1 0 4
Ruby
This crude algorithm works fine up to a square size of 10; higher values take too much time and memory. It creates an array of all possible permutations, picks a random one as first row an weeds out all permutations which cannot appear in the remaining square. Repeat picking and weeding until there is a square. <lang ruby>N = 5
def generate_square
perms = (1..N).to_a.permutation(N).to_a.shuffle square = [] N.times do square << perms.pop perms.reject!{|perm| perm.zip(square.last).any?{|el1, el2| el1 == el2} } end square
end
def print_square(square)
cell_size = N.digits.size + 1 strings = square.map!{|row| row.map!{|el| el.to_s.rjust(cell_size)}.join } puts strings, "\n"
end
2.times{print_square( generate_square)} </lang>
- Output:
3 4 2 1 5 2 3 4 5 1 1 2 5 3 4 5 1 3 4 2 4 5 1 2 31 2 5 4 3 2 3 4 1 5 5 4 2 3 1 3 5 1 2 4 4 1 3 5 2
zkl
<lang zkl>fcn randomLatinSquare(n,symbols=[1..]){ //--> list of lists
if(n<=0) return(T); square,syms := List(), symbols.walker().walk(n); do(n){ syms=syms.copy(); square.append(syms.append(syms.pop(0))) } // shuffle rows, transpose & shuffle columns T.zip(square.shuffle().xplode()).shuffle();
} fcn rls2String(square){ square.apply("concat"," ").concat("\n") }</lang> <lang zkl>foreach n in (T(1,2,5)){ randomLatinSquare(n) : rls2String(_).println("\n") } randomLatinSquare(5, ["A".."Z"]) : rls2String(_).println("\n"); randomLatinSquare(10,"!@#$%^&*()") : 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 & % # ! * @ ) $ ( ^ @ ) * ^ # & % ( $ ! ( & % # ) $ @ ^ ! * ! ( & % @ ^ $ * # ) % # ! ( ^ ) * @ & $ ^ $ @ ) & ! ( # * % # ! ( & $ * ^ ) % @ $ @ ) * % ( & ! ^ # ) * ^ $ ! % # & @ ( * ^ $ @ ( # ! % ) &