Latin Squares in reduced form
You are encouraged to solve this task according to the task description, using any language you may know.
A Latin Square is in its reduced form if the first row and first column contain items in their natural order. The order n is the number of items. For any given n there is a set of reduced Latin Squares whose size increases rapidly with n. g is a number which identifies a unique element within the set of reduced Latin Squares of order n. The objective of this task is to construct the set of all Latin Squares of a given order and to provide a means which given suitable values for g any element within the set may be obtained.
For a reduced Latin Square the first row is always 1 to n. The second row is all Permutations/Derangements of 1 to n starting with 2. The third row is all Permutations/Derangements of 1 to n starting with 3 which do not clash (do not have the same item in any column) with row 2. The fourth row is all Permutations/Derangements of 1 to n starting with 4 which do not clash with rows 2 or 3. Likewise continuing to the nth row.
Demonstrate by:
- displaying the four reduced Latin Squares of order 4.
- for n = 1 to 6 (or more) produce the set of reduced Latin Squares; produce a table which shows the size of the set of reduced Latin Squares and compares this value times n! times (n-1)! with the values in OEIS A002860.
11l
<lang 11l>F dList(n, =start)
start-- V a = Array(0 .< n) a[start] = a[0] a[0] = start a.sort_range(1..) V first = a[1] Int r F recurse(Int last) -> N I (last == @first) L(v) @a[1..] I L.index + 1 == v R V b = @a.map(x -> x + 1) @r.append(b) R L(i) (last .< 0).step(-1) swap(&@a[i], &@a[last]) @recurse(last - 1) swap(&@a[i], &@a[last]) recurse(n - 1) R r
F printSquare(latin, n)
L(row) latin print(row) print()
F reducedLatinSquares(n, echo)
I n <= 0 I echo print(‘[]’) R 0 E I n == 1 I echo print([1]) R 1
V rlatin = [[0] * n] * n L(j) 0 .< n rlatin[0][j] = j + 1
V count = 0 F recurse(Int i) -> N V rows = dList(@n, i)
L(r) 0 .< rows.len @rlatin[i - 1] = rows[r] V justContinue = 0B V k = 0 L !justContinue & k < i - 1 L(j) 1 .< @n I @rlatin[k][j] == @rlatin[i - 1][j] I r < rows.len - 1 justContinue = 1B L.break I i > 2 R k++ I !justContinue I i < @n @recurse(i + 1) E @count++ I @echo printSquare(@rlatin, @n)
recurse(2) R count
print("The four reduced latin squares of order 4 are:\n") reducedLatinSquares(4, 1B)
print(‘The size of the set of reduced latin squares for the following orders’) print("and hence the total number of latin squares of these orders are:\n") L(n) 1..6
V size = reducedLatinSquares(n, 0B) V f = factorial(n - 1) f *= f * n * size print(‘Order #.: Size #<4 x #.! x #.! => Total #.’.format(n, size, n, n - 1, f))</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
namespace LatinSquares {
using matrix = List<List<int>>;
class Program { static void Swap<T>(ref T a, ref T b) { var t = a; a = b; b = t; }
static matrix DList(int n, int start) { start--; // use 0 basing var a = Enumerable.Range(0, n).ToArray(); a[start] = a[0]; a[0] = start; Array.Sort(a, 1, a.Length - 1); var first = a[1]; // recursive closure permutes a[1:] matrix r = new matrix(); void recurse(int last) { if (last == first) { // bottom of recursion. you get here once for each permutation. // test if permutation is deranged. for (int j = 1; j < a.Length; j++) { var v = a[j]; if (j == v) { return; //no, ignore it } } // yes, save a copy with 1 based indexing var b = a.Select(v => v + 1).ToArray(); r.Add(b.ToList()); return; } for (int i = last; i >= 1; i--) { Swap(ref a[i], ref a[last]); recurse(last - 1); Swap(ref a[i], ref a[last]); } } recurse(n - 1); return r; }
static ulong ReducedLatinSquares(int n, bool echo) { if (n <= 0) { if (echo) { Console.WriteLine("[]\n"); } return 0; } else if (n == 1) { if (echo) { Console.WriteLine("[1]\n"); } return 1; }
matrix rlatin = new matrix(); for (int i = 0; i < n; i++) { rlatin.Add(new List<int>()); for (int j = 0; j < n; j++) { rlatin[i].Add(0); } } // first row for (int j = 0; j < n; j++) { rlatin[0][j] = j + 1; }
ulong count = 0; void recurse(int i) { var rows = DList(n, i);
for (int r = 0; r < rows.Count; r++) { rlatin[i - 1] = rows[r]; for (int k = 0; k < i - 1; k++) { for (int j = 1; j < n; j++) { if (rlatin[k][j] == rlatin[i - 1][j]) { if (r < rows.Count - 1) { goto outer; } if (i > 2) { return; } } } } if (i < n) { recurse(i + 1); } else { count++; if (echo) { PrintSquare(rlatin, n); } } outer: { } } }
//remaing rows recurse(2); return count; }
static void PrintSquare(matrix latin, int n) { foreach (var row in latin) { var it = row.GetEnumerator(); Console.Write("["); if (it.MoveNext()) { Console.Write(it.Current); } while (it.MoveNext()) { Console.Write(", {0}", it.Current); } Console.WriteLine("]"); } Console.WriteLine(); }
static ulong Factorial(ulong n) { if (n <= 0) { return 1; } ulong prod = 1; for (ulong i = 2; i < n + 1; i++) { prod *= i; } return prod; }
static void Main() { Console.WriteLine("The four reduced latin squares of order 4 are:\n"); ReducedLatinSquares(4, true);
Console.WriteLine("The size of the set of reduced latin squares for the following orders"); Console.WriteLine("and hence the total number of latin squares of these orders are:\n"); for (int n = 1; n < 7; n++) { ulong nu = (ulong)n;
var size = ReducedLatinSquares(n, false); var f = Factorial(nu - 1); f *= f * nu * size; Console.WriteLine("Order {0}: Size {1} x {2}! x {3}! => Total {4}", n, size, n, n - 1, f); } } }
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
C++
<lang cpp>#include <algorithm>
- include <functional>
- include <iostream>
- include <numeric>
- include <vector>
typedef std::vector<std::vector<int>> matrix;
matrix dList(int n, int start) {
start--; // use 0 basing
std::vector<int> a(n); std::iota(a.begin(), a.end(), 0); a[start] = a[0]; a[0] = start; std::sort(a.begin() + 1, a.end()); auto first = a[1]; // recursive closure permutes a[1:] matrix r; std::function<void(int)> recurse; recurse = [&](int last) { if (last == first) { // bottom of recursion you get here once for each permutation. // test if permutation is deranged. for (size_t j = 1; j < a.size(); j++) { auto v = a[j]; if (j == v) { return; //no, ignore it } } // yes, save a copy with 1 based indexing std::vector<int> b; std::transform(a.cbegin(), a.cend(), std::back_inserter(b), [](int v) { return v + 1; }); r.push_back(b); return; } for (int i = last; i >= 1; i--) { std::swap(a[i], a[last]); recurse(last - 1); std::swap(a[i], a[last]); } }; recurse(n - 1); return r;
}
void printSquare(const matrix &latin, int n) {
for (auto &row : latin) { auto it = row.cbegin(); auto end = row.cend(); std::cout << '['; if (it != end) { std::cout << *it; it = std::next(it); } while (it != end) { std::cout << ", " << *it; it = std::next(it); } std::cout << "]\n"; } std::cout << '\n';
}
unsigned long reducedLatinSquares(int n, bool echo) {
if (n <= 0) { if (echo) { std::cout << "[]\n"; } return 0; } else if (n == 1) { if (echo) { std::cout << "[1]\n"; } return 1; }
matrix rlatin; for (int i = 0; i < n; i++) { rlatin.push_back({}); for (int j = 0; j < n; j++) { rlatin[i].push_back(j); } } // first row for (int j = 0; j < n; j++) { rlatin[0][j] = j + 1; }
unsigned long count = 0; std::function<void(int)> recurse; recurse = [&](int i) { auto rows = dList(n, i);
for (size_t r = 0; r < rows.size(); r++) { rlatin[i - 1] = rows[r]; for (int k = 0; k < i - 1; k++) { for (int j = 1; j < n; j++) { if (rlatin[k][j] == rlatin[i - 1][j]) { if (r < rows.size() - 1) { goto outer; } if (i > 2) { return; } } } } if (i < n) { recurse(i + 1); } else { count++; if (echo) { printSquare(rlatin, n); } } outer: {} } };
//remaining rows recurse(2); return count;
}
unsigned long factorial(unsigned long n) {
if (n <= 0) return 1; unsigned long prod = 1; for (unsigned long i = 2; i <= n; i++) { prod *= i; } return prod;
}
int main() {
std::cout << "The four reduced lating squares of order 4 are:\n"; reducedLatinSquares(4, true);
std::cout << "The size of the set of reduced latin squares for the following orders\n"; std::cout << "and hence the total number of latin squares of these orders are:\n\n"; for (int n = 1; n < 7; n++) { auto size = reducedLatinSquares(n, false); auto f = factorial(n - 1); f *= f * n * size; std::cout << "Order " << n << ": Size " << size << " x " << n << "! x " << (n - 1) << "! => Total " << f << '\n'; }
return 0;
}</lang>
- Output:
The four reduced lating squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
D
<lang d>import std.algorithm; import std.array; import std.range; import std.stdio;
alias matrix = int[][];
auto dList(int n, int start) {
start--; // use 0 basing auto a = iota(0, n).array; a[start] = a[0]; a[0] = start; sort(a[1..$]); auto first = a[1]; // recursive closure permutes a[1:] matrix r; void recurse(int last) { if (last == first) { // bottom of recursion. you get here once for each permutation. // test if permutation is deranged. foreach (j,v; a[1..$]) { if (j + 1 == v) { return; //no, ignore it } } // yes, save a copy with 1 based indexing auto b = a.map!"a+1".array; r ~= b; return; } for (int i = last; i >= 1; i--) { swap(a[i], a[last]); recurse(last -1); swap(a[i], a[last]); } } recurse(n - 1); return r;
}
ulong reducedLatinSquares(int n, bool echo) {
if (n <= 0) { if (echo) { writeln("[]\n"); } return 0; } else if (n == 1) { if (echo) { writeln("[1]\n"); } return 1; }
matrix rlatin = uninitializedArray!matrix(n); foreach (i; 0..n) { rlatin[i] = uninitializedArray!(int[])(n); } // first row foreach (j; 0..n) { rlatin[0][j] = j + 1; }
ulong count; void recurse(int i) { auto rows = dList(n, i);
outer: foreach (r; 0..rows.length) { rlatin[i-1] = rows[r].dup; foreach (k; 0..i-1) { foreach (j; 1..n) { if (rlatin[k][j] == rlatin[i - 1][j]) { if (r < rows.length - 1) { continue outer; } if (i > 2) { return; } } } } if (i < n) { recurse(i + 1); } else { count++; if (echo) { printSquare(rlatin, n); } } } }
// remaining rows recurse(2); return count;
}
void printSquare(matrix latin, int n) {
foreach (row; latin) { writeln(row); } writeln;
}
ulong factorial(ulong n) {
if (n == 0) { return 1; } ulong prod = 1; foreach (i; 2..n+1) { prod *= i; } return prod;
}
void main() {
writeln("The four reduced latin squares of order 4 are:\n"); reducedLatinSquares(4, true);
writeln("The size of the set of reduced latin squares for the following orders"); writeln("and hence the total number of latin squares of these orders are:\n"); foreach (n; 1..7) { auto size = reducedLatinSquares(n, false); auto f = factorial(n - 1); f *= f * n * size; writefln("Order %d: Size %-4d x %d! x %d! => Total %d", n, size, n, n - 1, f); }
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
F#
The Function
This task uses Permutations/Derangements#F.23 <lang fsharp> // Generate Latin Squares in reduced form. Nigel Galloway: July 10th., 2019 let normLS α=
let N=derange α|>List.ofSeq|>List.groupBy(fun n->n.[0])|>List.sortBy(fun(n,_)->n)|>List.map(fun(_,n)->n)|>Array.ofList let rec fG n g=match n with h::t->fG t (g|>List.filter(fun g->Array.forall2((<>)) h g )) |_->g let rec normLS n g=seq{for i in fG n N.[g] do if g=α-2 then yield [|1..α|]::(List.rev (i::n)) else yield! normLS (i::n) (g+1)} match α with 1->seq[[[|1|]]] |2-> seq[[[|1;2|];[|2;1|]]] |_->Seq.collect(fun n->normLS [n] 1) N.[0]
</lang>
The Task
<lang fsharp> normLS 4 |> Seq.iter(fun n->List.iter(printfn "%A") n;printfn "");; </lang>
- Output:
[|1; 2; 3; 4|] [|2; 3; 4; 1|] [|3; 4; 1; 2|] [|4; 1; 2; 3|] [|1; 2; 3; 4|] [|2; 1; 4; 3|] [|3; 4; 2; 1|] [|4; 3; 1; 2|] [|1; 2; 3; 4|] [|2; 1; 4; 3|] [|3; 4; 1; 2|] [|4; 3; 2; 1|] [|1; 2; 3; 4|] [|2; 4; 1; 3|] [|3; 1; 4; 2|] [|4; 3; 2; 1|]
<lang fsharp> let rec fact n g=if n<2 then g else fact (n-1) n*g [1..6] |> List.iter(fun n->let nLS=normLS n|>Seq.length in printfn "order=%d number of Reduced Latin Squares nLS=%d nLS*n!*(n-1)!=%d" n nLS (nLS*(fact n 1)*(fact (n-1) 1))) </lang>
- Output:
order=1 number of Reduced Latin Squares nLS=1 nLS*n!*(n-1)!=1 order=2 number of Reduced Latin Squares nLS=1 nLS*n!*(n-1)!=2 order=3 number of Reduced Latin Squares nLS=1 nLS*n!*(n-1)!=12 order=4 number of Reduced Latin Squares nLS=4 nLS*n!*(n-1)!=576 order=5 number of Reduced Latin Squares nLS=56 nLS*n!*(n-1)!=161280 order=6 number of Reduced Latin Squares nLS=9408 nLS*n!*(n-1)!=812851200
Go
This reuses the dList function from the Permutations/Derangements#Go task, suitably adjusted for the present one. <lang go>package main
import (
"fmt" "sort"
)
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 reducedLatinSquare(n int, echo bool) uint64 {
if n <= 0 { if echo { fmt.Println("[]\n") } return 0 } else if n == 1 { if echo { fmt.Println("[1]\n") } return 1 } rlatin := make(matrix, n) for i := 0; i < n; i++ { rlatin[i] = make([]int, n) } // first row for j := 0; j < n; j++ { rlatin[0][j] = j + 1 }
count := uint64(0) // recursive closure to compute reduced latin squares and count or print them 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 { count++ if echo { printSquare(rlatin, n) } } } return }
// remaining rows recurse(2) return count
}
func printSquare(latin matrix, n int) {
for i := 0; i < n; i++ { fmt.Println(latin[i]) } 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
}
func main() {
fmt.Println("The four reduced latin squares of order 4 are:\n") reducedLatinSquare(4, true)
fmt.Println("The size of the set of reduced latin squares for the following orders") fmt.Println("and hence the total number of latin squares of these orders are:\n") for n := uint64(1); n <= 6; n++ { size := reducedLatinSquare(int(n), false) f := factorial(n - 1) f *= f * n * size fmt.Printf("Order %d: Size %-4d x %d! x %d! => Total %d\n", n, size, n, n-1, f) }
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1 2 3 4] [2 1 4 3] [3 4 1 2] [4 3 2 1] [1 2 3 4] [2 1 4 3] [3 4 2 1] [4 3 1 2] [1 2 3 4] [2 4 1 3] [3 1 4 2] [4 3 2 1] [1 2 3 4] [2 3 4 1] [3 4 1 2] [4 1 2 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Haskell
The solution uses permutation generator given by Data.List package and List monad for generating all possible latin squares as a fold of permutation list.
<lang haskell>import Data.List (permutations, (\\)) import Control.Monad (foldM, forM_)
latinSquares :: Eq a => [a] -> [[[a]]] latinSquares [] = [] latinSquares set = map reverse <$> squares
where squares = foldM addRow firstRow perm perm = tail (groupedPermutations set) firstRow = pure <$> set addRow tbl rows = [ zipWith (:) row tbl | row <- rows , and $ different (tail row) (tail tbl) ] different = zipWith $ (not .) . elem
groupedPermutations :: Eq a => [a] -> [[[a]]] groupedPermutations lst = map (\x -> (x :) <$> permutations (lst \\ [x])) lst
printTable :: Show a => a -> IO () printTable tbl = putStrLn $ unlines $ unwords . map show <$> tbl </lang>
It is slightly optimized by grouping permutations by the first element according to a set order. Partitioning reduces the filtering procedure by factor of an initial set size.
Examples
λ> latinSquares "abc" [["abc","bca","cab"]] λ> mapM_ printTable $ take 3 $ latinSquares [1..9] 1 2 3 4 5 6 7 8 9 2 9 4 8 1 7 3 6 5 3 8 2 5 9 1 4 7 6 4 7 5 6 2 9 8 1 3 5 6 9 1 3 8 2 4 7 6 5 1 7 4 2 9 3 8 7 4 6 3 8 5 1 9 2 8 3 7 9 6 4 5 2 1 9 1 8 2 7 3 6 5 4 1 2 3 4 5 6 7 8 9 2 9 4 8 1 7 3 5 6 3 8 2 5 9 1 4 6 7 4 7 5 6 2 9 8 1 3 5 6 9 1 3 8 2 7 4 6 5 1 7 4 2 9 3 8 7 4 6 3 8 5 1 9 2 8 3 7 9 6 4 5 2 1 9 1 8 2 7 3 6 4 5 1 2 3 4 5 6 7 8 9 2 9 4 8 1 7 3 6 5 3 8 2 5 9 1 4 7 6 4 7 5 6 2 9 1 3 8 5 6 9 1 3 8 2 4 7 6 5 1 7 4 2 8 9 3 7 4 6 3 8 5 9 1 2 8 3 7 9 6 4 5 2 1 9 1 8 2 7 3 6 5 4
Tasks <lang haskell>task1 = do
putStrLn "Latin squares of order 4:" mapM_ printTable $ latinSquares [1..4]
task2 = do
putStrLn "Sizes of latin squares sets for different orders:" forM_ [1..6] $ \n -> let size = length $ latinSquares [1..n] total = fact n * fact (n-1) * size fact i = product [1..i] in printf "Order %v: %v*%v!*%v!=%v\n" n size n (n-1) total</lang>
λ> task1 >> task2 Latin squares of order 4: 1 2 3 4 4 1 2 3 3 4 1 2 2 3 4 1 1 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 1 2 3 4 2 1 4 3 4 3 1 2 3 4 2 1 1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 Sizes of latin squares sets for different orders: Order 1: 1*1!*0!=1 Order 2: 1*2!*1!=2 Order 3: 1*3!*2!=12 Order 4: 4*4!*3!=576 Order 5: 56*5!*4!=161280 Order 6: 9408*6!*5!=812851200
Java
<lang java> import java.math.BigInteger; import java.util.ArrayList; import java.util.Arrays; import java.util.List;
public class LatinSquaresInReducedForm {
public static void main(String[] args) { System.out.printf("Reduced latin squares of order 4:%n"); for ( LatinSquare square : getReducedLatinSquares(4) ) { System.out.printf("%s%n", square); } System.out.printf("Compute the number of latin squares from count of reduced latin squares:%n(Reduced Latin Square Count) * n! * (n-1)! = Latin Square Count%n"); for ( int n = 1 ; n <= 6 ; n++ ) { List<LatinSquare> list = getReducedLatinSquares(n); System.out.printf("Size = %d, %d * %d * %d = %,d%n", n, list.size(), fact(n), fact(n-1), list.size()*fact(n)*fact(n-1)); } } private static long fact(int n) { if ( n == 0 ) { return 1; } int prod = 1; for ( int i = 1 ; i <= n ; i++ ) { prod *= i; } return prod; } private static List<LatinSquare> getReducedLatinSquares(int n) { List<LatinSquare> squares = new ArrayList<>(); squares.add(new LatinSquare(n)); PermutationGenerator permGen = new PermutationGenerator(n); for ( int fillRow = 1 ; fillRow < n ; fillRow++ ) { List<LatinSquare> squaresNext = new ArrayList<>(); for ( LatinSquare square : squares ) { while ( permGen.hasMore() ) { int[] perm = permGen.getNext(); // If not the correct row - next permutation. if ( (perm[0]+1) != (fillRow+1) ) { continue; } // Check permutation against current square. boolean permOk = true; done: for ( int row = 0 ; row < fillRow ; row++ ) { for ( int col = 0 ; col < n ; col++ ) { if ( square.get(row, col) == (perm[col]+1) ) { permOk = false; break done; } } } if ( permOk ) { LatinSquare newSquare = new LatinSquare(square); for ( int col = 0 ; col < n ; col++ ) { newSquare.set(fillRow, col, perm[col]+1); } squaresNext.add(newSquare); } } permGen.reset(); } squares = squaresNext; } return squares; } @SuppressWarnings("unused") private static int[] display(int[] in) { int [] out = new int[in.length]; for ( int i = 0 ; i < in.length ; i++ ) { out[i] = in[i] + 1; } return out; } private static class LatinSquare { int[][] square; int size; public LatinSquare(int n) { square = new int[n][n]; size = n; for ( int col = 0 ; col < n ; col++ ) { set(0, col, col + 1); } } public LatinSquare(LatinSquare ls) { int n = ls.size; square = new int[n][n]; size = n; for ( int row = 0 ; row < n ; row++ ) { for ( int col = 0 ; col < n ; col++ ) { set(row, col, ls.get(row, col)); } } } public void set(int row, int col, int value) { square[row][col] = value; }
public int get(int row, int col) { return square[row][col]; }
@Override public String toString() { StringBuilder sb = new StringBuilder(); for ( int row = 0 ; row < size ; row++ ) { sb.append(Arrays.toString(square[row])); sb.append("\n"); } return sb.toString(); } }
private static class PermutationGenerator {
private int[] a; private BigInteger numLeft; private BigInteger total;
public PermutationGenerator (int n) { if (n < 1) { throw new IllegalArgumentException ("Min 1"); } a = new int[n]; total = getFactorial(n); reset(); }
private void reset () { for ( int i = 0 ; i < a.length ; i++ ) { a[i] = i; } numLeft = new BigInteger(total.toString()); }
public boolean hasMore() { return numLeft.compareTo(BigInteger.ZERO) == 1; }
private static BigInteger getFactorial (int n) { BigInteger fact = BigInteger.ONE; for ( int i = n ; i > 1 ; i-- ) { fact = fact.multiply(new BigInteger(Integer.toString(i))); } return fact; }
/*-------------------------------------------------------- * Generate next permutation (algorithm from Rosen p. 284) *-------------------------------------------------------- */ public int[] getNext() { if ( numLeft.equals(total) ) { numLeft = numLeft.subtract (BigInteger.ONE); return a; }
// Find largest index j with a[j] < a[j+1] int j = a.length - 2; while ( a[j] > a[j+1] ) { j--; }
// Find index k such that a[k] is smallest integer greater than a[j] to the right of a[j] int k = a.length - 1; while ( a[j] > a[k] ) { k--; }
// Interchange a[j] and a[k] int temp = a[k]; a[k] = a[j]; a[j] = temp;
// Put tail end of permutation after jth position in increasing order int r = a.length - 1; int s = j + 1; while (r > s) { int temp2 = a[s]; a[s] = a[r]; a[r] = temp2; r--; s++; }
numLeft = numLeft.subtract(BigInteger.ONE); return a; } }
} </lang>
- Output:
Reduced latin squares of order 4: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] Compute the number of latin squares from count of reduced latin squares: (Reduced Latin Square Count) * n! * (n-1)! = Latin Square Count Size = 1, 1 * 1 * 1 = 1 Size = 2, 1 * 2 * 1 = 2 Size = 3, 1 * 6 * 2 = 12 Size = 4, 4 * 24 * 6 = 576 Size = 5, 56 * 120 * 24 = 161,280 Size = 6, 9408 * 720 * 120 = 812,851,200
jq
Works with recent versions of gojq (e.g. f0faa22 (August 22, 2021))
Preliminaries <lang jq>def count(s): reduce s as $x (0; .+1);
def factorial: reduce range(2;.+1) as $i (1; . * $i);
def permutations:
if length == 0 then [] else range(0;length) as $i | [.[$i]] + (del(.[$i])|permutations) end ;
</lang> Latin Squares <lang jq>def clash($row2; $row1):
any(range(0;$row2|length); $row1[.] == $row2[.]);
- Input is a row; stream is a stream of rows
def clash(stream):
. as $row | any(stream; clash($row; .)) ;
- Emit a stream of latin squares of size .
def latin_squares:
. as $n
# Emit a stream of arrays of permutation of 1 .. $n inclusive, and beginning with $i | def permutations_beginning_with($i): [$i] + ([range(1; $i), range($i+1; $n + 1)] | permutations);
# input: an array of rows, $rows # output: a stream of all the permutations starting with $i # that are permissible relative to $rows def filter_permuted($i): . as $rows | permutations_beginning_with($i) | select( clash($rows[]) | not ) ;
# input: an array of the first few rows (at least one) of a latin square # output: a stream of possible immediate-successor rows def next_latin_square_row: filter_permuted(1 + .[-1][0]);
# recursion makes completing a latin square a snap def complete_latin_square: if length == $n then . else next_latin_square_row as $next | . + [$next] | complete_latin_square end;
range(1;$n+1) | complete_latin_square ;
</lang> The Task <lang jq>def task:
"The reduced latin squares of order 4 are:", (4 | latin_squares), "", (range(1; 7) | . as $i | count(latin_squares) as $c | ($c * factorial * ((.-1)|factorial)) as $total | "There are \($c) reduced latin squares of order \(.); \($c) * \(.)! * \(.-1)! is \($total)" ) ;
task</lang>
- Output:
Invocation: jq -nrc -f latin-squares.jq
The reduced latin squares of order 4 are: [[1,2,3,4],[2,1,4,3],[3,4,1,2],[4,3,2,1]] [[1,2,3,4],[2,1,4,3],[3,4,2,1],[4,3,1,2]] [[1,2,3,4],[2,3,4,1],[3,4,1,2],[4,1,2,3]] [[1,2,3,4],[2,4,1,3],[3,1,4,2],[4,3,2,1]] There are 1 reduced latin squares of order 1; 1 * 1! * 0! is 1 There are 1 reduced latin squares of order 2; 1 * 2! * 1! is 2 There are 1 reduced latin squares of order 3; 1 * 3! * 2! is 12 There are 4 reduced latin squares of order 4; 4 * 4! * 3! is 576 There are 56 reduced latin squares of order 5; 56 * 5! * 4! is 161280 There are 9408 reduced latin squares of order 6; 9408 * 6! * 5! is 812851200
Julia
<lang julia>using Combinatorics
clash(row2, row1::Vector{Int}) = any(i -> row1[i] == row2[i], 1:length(row2))
clash(row, rows::Vector{Vector{Int}}) = any(r -> clash(row, r), rows)
permute_onefixed(i, n) = map(vec -> vcat(i, vec), permutations(filter(x -> x != i, 1:n)))
filter_permuted(rows, i, n) = filter(v -> !clash(v, rows), permute_onefixed(i, n))
function makereducedlatinsquares(n)
matarray = [reshape(collect(1:n), 1, n)] for i in 2:n newmatarray = Vector{Matrix{Int}}() for mat in matarray r = size(mat)[1] + 1 newrows = filter_permuted(collect(row[:] for row in eachrow(mat)), r, n) newmat = zeros(Int, r, n) newmat[1:r-1, :] .= mat append!(newmatarray, [deepcopy(begin newmat[i, :] .= row; newmat end) for row in newrows]) end matarray = newmatarray end matarray, length(matarray)
end
function testlatinsquares()
squares, count = makereducedlatinsquares(4) println("The four reduced latin squares of order 4 are:") for sq in squares, (i, row) in enumerate(eachrow(sq)), j in 1:4 print(row[j], j == 4 ? (i == 4 ? "\n\n" : "\n") : " ") end for i in 1:6 squares, count = makereducedlatinsquares(i) println("Order $i: Size ", rpad(count, 5), "* $(i)! * $(i - 1)! = ", count * factorial(i) * factorial(i - 1)) end
end
testlatinsquares()
</lang>
- Output:
The four reduced latin squares of order 4 are: 1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 2 1 4 3 3 4 2 1 4 3 1 2 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 1 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 Order 1: Size 1 * 1! * 0! = 1 Order 2: Size 1 * 2! * 1! = 2 Order 3: Size 1 * 3! * 2! = 12 Order 4: Size 4 * 4! * 3! = 576 Order 5: Size 56 * 5! * 4! = 161280 Order 6: Size 9408 * 6! * 5! = 812851200
Kotlin
<lang scala>typealias Matrix = MutableList<MutableList<Int>>
fun dList(n: Int, sp: Int): Matrix {
val start = sp - 1 // use 0 basing
val a = generateSequence(0) { it + 1 }.take(n).toMutableList() a[start] = a[0].also { a[0] = a[start] } a.subList(1, a.size).sort()
val first = a[1] // recursive closure permutes a[1:] val r = mutableListOf<MutableList<Int>>() fun recurse(last: Int) { if (last == first) { // bottom of recursion. you get here once for each permutation. // test if permutation is deranged for (jv in a.subList(1, a.size).withIndex()) { if (jv.index + 1 == jv.value) { return // no, ignore it } } // yes, save a copy with 1 based indexing val b = a.map { it + 1 } r.add(b.toMutableList()) return } for (i in last.downTo(1)) { a[i] = a[last].also { a[last] = a[i] } recurse(last - 1) a[i] = a[last].also { a[last] = a[i] } } } recurse(n - 1) return r
}
fun reducedLatinSquares(n: Int, echo: Boolean): Long {
if (n <= 0) { if (echo) { println("[]\n") } return 0 } else if (n == 1) { if (echo) { println("[1]\n") } return 1 }
val rlatin = MutableList(n) { MutableList(n) { it } } // first row for (j in 0 until n) { rlatin[0][j] = j + 1 }
var count = 0L fun recurse(i: Int) { val rows = dList(n, i)
outer@ for (r in 0 until rows.size) { rlatin[i - 1] = rows[r].toMutableList() for (k in 0 until i - 1) { for (j in 1 until n) { if (rlatin[k][j] == rlatin[i - 1][j]) { if (r < rows.size - 1) { continue@outer } if (i > 2) { return } } } } if (i < n) { recurse(i + 1) } else { count++ if (echo) { printSquare(rlatin) } } } }
// remaining rows recurse(2) return count
}
fun printSquare(latin: Matrix) {
for (row in latin) { println(row) } println()
}
fun factorial(n: Long): Long {
if (n == 0L) { return 1 } var prod = 1L for (i in 2..n) { prod *= i } return prod
}
fun main() {
println("The four reduced latin squares of order 4 are:\n") reducedLatinSquares(4, true)
println("The size of the set of reduced latin squares for the following orders") println("and hence the total number of latin squares of these orders are:\n") for (n in 1 until 7) { val size = reducedLatinSquares(n, false) var f = factorial(n - 1.toLong()) f *= f * n * size println("Order $n: Size %-4d x $n! x ${n - 1}! => Total $f".format(size)) }
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
MiniZinc
The Model (lsRF.mnz)
<lang MiniZinc> %Latin Squares in Reduced Form. Nigel Galloway, September 5th., 2019 include "alldifferent.mzn"; int: N; array[1..N,1..N] of var 1..N: p; constraint forall(n in 1..N)(p[1,n]=n /\ p[n,1]=n); constraint forall(n in 1..N)(alldifferent([p[n,g]|g in 1..N])/\alldifferent([p[g,n]|g in 1..N])); </lang>
The Tasks
- displaying the four reduced Latin Squares of order 4
<lang MiniZinc> include "lsRF.mzn";
output [show_int(1,p[i,j])++ if j == 4 then if i != 4 then "\n" else "" endif else "" endif | i,j in 1..4 ] ++ ["\n"];
</lang> When the above is run using minizinc --all-solutions -DN=4 the following is produced:
- Output:
1234 2143 3421 4312 ---------- 1234 2143 3412 4321 ---------- 1234 2413 3142 4321 ---------- 1234 2341 3412 4123 ---------- ==========
- counting the solutions
minizinc.exe --all-solutions -DN=5 -s lsRF.mzn produces the following:
. . . p = array2d(1..5, 1..5, [1, 2, 3, 4, 5, 2, 3, 4, 5, 1, 3, 1, 5, 2, 4, 4, 5, 2, 1, 3, 5, 4, 1, 3, 2]); ---------- p = array2d(1..5, 1..5, [1, 2, 3, 4, 5, 2, 3, 5, 1, 4, 3, 5, 4, 2, 1, 4, 1, 2, 5, 3, 5, 4, 1, 3, 2]); ---------- p = array2d(1..5, 1..5, [1, 2, 3, 4, 5, 2, 3, 4, 5, 1, 3, 5, 2, 1, 4, 4, 1, 5, 2, 3, 5, 4, 1, 3, 2]); ---------- ========== %%%mzn-stat: initTime=0.057 %%%mzn-stat: solveTime=0.003 %%%mzn-stat: solutions=56 %%%mzn-stat: variables=43 %%%mzn-stat: propagators=8 %%%mzn-stat: propagations=960 %%%mzn-stat: nodes=111 %%%mzn-stat: failures=0 %%%mzn-stat: restarts=0 %%%mzn-stat: peakDepth=7 %%%mzn-stat-end %%%mzn-stat: nSolutions=56
and minizinc.exe --all-solutions -DN=6 -s lsRF.mzn produces the following:
. . . p = array2d(1..6, 1..6, [1, 2, 3, 4, 5, 6, 2, 4, 5, 6, 3, 1, 3, 1, 4, 2, 6, 5, 4, 6, 2, 5, 1, 3, 5, 3, 6, 1, 2, 4, 6, 5, 1, 3, 4, 2]); ---------- p = array2d(1..6, 1..6, [1, 2, 3, 4, 5, 6, 2, 1, 4, 6, 3, 5, 3, 4, 5, 2, 6, 1, 4, 6, 2, 5, 1, 3, 5, 3, 6, 1, 2, 4, 6, 5, 1, 3, 4, 2]); ---------- ========== %%%mzn-stat: initTime=0.003 %%%mzn-stat: solveTime=6.669 %%%mzn-stat: solutions=9408 %%%mzn-stat: variables=58 %%%mzn-stat: propagators=10 %%%mzn-stat: propagations=179635 %%%mzn-stat: nodes=19035 %%%mzn-stat: failures=110 %%%mzn-stat: restarts=0 %%%mzn-stat: peakDepth=17 %%%mzn-stat-end %%%mzn-stat: nSolutions=9408
The only way to complete the tasks requirement to produce a table is with another language. Ruby has the ability to run an external program, capture the output, and text handling ability to format it to this tasks requirements. Othe scripting languages are available.
Nim
We use the Go algorithm but have chosen to create two types, Row and Matrix, to simulate sequences starting at index 1. So, the indexes and tests are somewhat different.
<lang Nim>import algorithm, math, sequtils, strformat
type
# Row managed as a sequence of ints with base index 1. Row = object value: seq[int]
# Matrix managed as a sequence of rows with base index 1. Matrix = object value: seq[Row]
func newRow(n: Natural = 0): Row =
## Create a new row of length "n". Row(value: newSeq[int](n))
- Create a new matrix of length "n" containing rows of length "p".
func newMatrix(n, p: Natural = 0): Matrix = Matrix(value: newSeqWith(n, newRow(p)))
- Functions for rows.
func `[]`(r: var Row; i: int): var int = r.value[i - 1] func `[]=`(r: var Row; i, n: int) = r.value[i - 1] = n func sort(r: var Row; low, high: Positive) =
r.value.toOpenArray(low - 1, high - 1).sort()
func `$`(r: Row): string = ($r.value)[1..^1]
- Functions for matrices.
func `[]`(m: Matrix; i: int): Row = m.value[i - 1] func `[]`(m: var Matrix; i: int): var Row = m.value[i - 1] func `[]=`(m: var Matrix; i: int; r: Row) = m.value[i - 1] = r func high(m: Matrix): Natural = m.value.len func add(m: var Matrix; r: Row) = m.value.add r func `$`(m: Matrix): string =
for row in m.value: result.add $row & '\n'
func dList(n, start: Positive): Matrix =
## Generate derangements of first 'n' numbers, with 'start' in first place.
var a = Row(value: toSeq(1..n))
swap a[1], a[start] a.sort(2, n) let first = a[2] var r: Matrix
func recurse(last: int) = ## Recursive closure permutes a[2..^1]. if last == first: # Bottom of recursion. You get here once for each permutation. # Test if permutation is deranged. for i in 2..n: if a[i] == i: return # No: ignore it. r.add a return for i in countdown(last, 2): swap a[i], a[last] recurse(last - 1) swap a[i], a[last]
recurse(n) result = r
proc reducedLatinSquares(n: Positive; print: bool): int =
if n == 1: if print: echo [1] return 1
var rlatin = newMatrix(n, n) # Initialize first row. for i in 1..n: rlatin[1][i] = i
var count = 0
proc recurse(i: int) = let rows = dList(n, i) for r in 1..rows.high: block inner: rlatin[i] = rows[r] for k in 1..<i: for j in 2..n: if rlatin[k][j] == rlatin[i][j]: if r < rows.high: break inner if i > 2: return if i < n: recurse(i + 1) else: inc count if print: echo rlatin
# Remaining rows. recurse(2) result = count
when isMainModule:
echo "The four reduced latin squares of order 4 are:" discard reducedLatinSquares(4, true)
echo "The size of the set of reduced latin squares for the following orders" echo "and hence the total number of latin squares of these orders are:" for n in 1..6: let size = reducedLatinSquares(n, false) let f = fac(n - 1)^2 * n * size echo &"Order {n}: Size {size:<4} x {n}! x {n - 1}! => Total {f}"</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Perl
It takes a little under 2 minutes to find order 7. <lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Latin_Squares_in_reduced_form use warnings;
my $n = 0; my $count; our @perms;
while( ++$n <= 7 )
{ $count = 0; @perms = perm( my $start = join , 1 .. $n ); find( $start ); print "order $n size $count total @{[$count * fact($n) * fact($n-1)]}\n\n"; }
sub find
{ @_ >= $n and return $count += ($n != 4) || print join "\n", @_, "\n"; local @perms = grep 0 == ($_[-1] ^ $_) =~ tr/\0//, @perms; my $row = @_ + 1; find( @_, $_ ) for grep /^$row/, @perms; }
sub fact { $_[0] > 1 ? $_[0] * fact($_[0] - 1) : 1 }
sub perm
{ my $s = shift; length $s <= 1 ? $s : map { my $f = $_; map "$f$_", perm( $s =~ s/$_//r ) } split //, $s; }</lang>
- Output:
order 1 size 1 total 1 order 2 size 1 total 2 order 3 size 1 total 12 1234 2143 3412 4321 1234 2143 3421 4312 1234 2341 3412 4123 1234 2413 3142 4321 order 4 size 4 total 576 order 5 size 56 total 161280 order 6 size 9408 total 812851200 order 7 size 16942080 total 61479419904000
Phix
A Simple backtracking search.
aside: in phix here is no difference between res[r][c] and res[r,c]. I mixed them here, using whichever felt the more natural to me.
string aleph = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" function rfls(integer n, bool count_only=true) if n>length(aleph) then ?9/0 end if -- too big... if n=1 then return iff(count_only?1:{{1}}) end if sequence tn = tagset(n), -- {1..n} vcs = repeat(tn,n), -- valid for cols vrs = repeat(tn,n), -- valid for rows res = repeat(tn,n) -- (main workspace/one element of result) object result = iff(count_only?0:{}) vcs[1] = {} -- (not strictly necessary) vrs[1] = {} -- """ for i=2 to n do res[i] = i & repeat(0,n-1) vrs[i][i] = 0 vcs[i][i] = 0 end for integer r = 2, c = 2 while true do -- place with backtrack: -- if we successfully place [n,n] add to results and backtrack -- terminate when we fail to place or backtrack from [2,2] integer rrc = res[r,c] if rrc!=0 then -- backtrack (/undo) if vrs[r][rrc]!=0 then ?9/0 end if -- sanity check if vcs[c][rrc]!=0 then ?9/0 end if -- "" res[r,c] = 0 vrs[r][rrc] = rrc vcs[c][rrc] = rrc end if bool found = false for i=rrc+1 to n do if vrs[r][i] and vcs[c][i] then res[r,c] = i vrs[r][i] = 0 vcs[c][i] = 0 found = true exit end if end for if found then if r=n and c=n then if count_only then result += 1 else result = append(result,res) end if -- (here, backtracking == not advancing) elsif c=n then c = 2 r += 1 else c += 1 end if else -- backtrack if r=2 and c=2 then exit end if c -= 1 if c=1 then r -= 1 c = n end if end if end while return result end function procedure reduced_form_latin_squares(integer n) sequence res = rfls(n,false) for k=1 to length(res) do for i=1 to n do string line = "" for j=1 to n do line &= aleph[res[k][i][j]] end for res[k][i] = line end for res[k] = join(res[k],"\n") end for string r = join(res,"\n\n") printf(1,"There are %d reduced form latin squares of order %d:\n%s\n",{length(res),n,r}) end procedure reduced_form_latin_squares(4) puts(1,"\n") for n=1 to 6 do integer size = rfls(n) atom f = factorial(n)*factorial(n-1)*size printf(1,"Order %d: Size %-4d x %d! x %d! => Total %d\n", {n, size, n, n-1, f}) end for
- Output:
There are 4 reduced form latin squares of order 4: 1234 2143 3412 4321 1234 2143 3421 4312 1234 2341 3412 4123 1234 2413 3142 4321 Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Whle the above finishes near-instantly, if you push it to 7 and add an elapsed(), you'll get:
Order 7: Size 16942080 x 7! x 6! => Total 61479419904000 "2 minutes and 23s"
Python
<lang python>def dList(n, start):
start -= 1 # use 0 basing a = range(n) a[start] = a[0] a[0] = start a[1:] = sorted(a[1:]) first = a[1] # rescursive closure permutes a[1:] r = [] def recurse(last): if (last == first): # bottom of recursion. you get here once for each permutation. # test if permutation is deranged. # yes, save a copy with 1 based indexing for j,v in enumerate(a[1:]): if j + 1 == v: return # no, ignore it b = [x + 1 for x in a] r.append(b) return for i in xrange(last, 0, -1): a[i], a[last] = a[last], a[i] recurse(last - 1) a[i], a[last] = a[last], a[i] recurse(n - 1) return r
def printSquare(latin,n):
for row in latin: print row print
def reducedLatinSquares(n,echo):
if n <= 0: if echo: print [] return 0 elif n == 1: if echo: print [1] return 1
rlatin = [None] * n for i in xrange(n): rlatin[i] = [None] * n # first row for j in xrange(0, n): rlatin[0][j] = j + 1
class OuterScope: count = 0 def recurse(i): rows = dList(n, i)
for r in xrange(len(rows)): rlatin[i - 1] = rows[r] justContinue = False k = 0 while not justContinue and k < i - 1: for j in xrange(1, n): if rlatin[k][j] == rlatin[i - 1][j]: if r < len(rows) - 1: justContinue = True break if i > 2: return k += 1 if not justContinue: if i < n: recurse(i + 1) else: OuterScope.count += 1 if echo: printSquare(rlatin, n)
# remaining rows recurse(2) return OuterScope.count
def factorial(n):
if n == 0: return 1 prod = 1 for i in xrange(2, n + 1): prod *= i return prod
print "The four reduced latin squares of order 4 are:\n" reducedLatinSquares(4,True)
print "The size of the set of reduced latin squares for the following orders" print "and hence the total number of latin squares of these orders are:\n" for n in xrange(1, 7):
size = reducedLatinSquares(n, False) f = factorial(n - 1) f *= f * n * size print "Order %d: Size %-4d x %d! x %d! => Total %d" % (n, size, n, n - 1, f)</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Raku
(formerly Perl 6) <lang perl6># utilities: factorial, sub-factorial, derangements sub postfix:<!>($n) { (constant f = 1, |[\×] 1..*)[$n] } sub prefix:<!>($n) { (1, 0, 1, -> $a, $b { ($++ + 2) × ($b + $a) } ... *)[$n] } sub derangements(@l) { @l.permutations.grep(-> @p { none(@p Zeqv @l) }) }
sub LS-reduced (Int $n) {
return [1] if $n == 1;
my @LS; my @l = 1 X+ ^$n; my %D = derangements(@l).classify(*.[0]);
for [X] (^(!$n/($n-1))) xx $n-1 -> $tuple { my @d.push: @l; @d.push: %D{2}[$tuple[0]]; LOOP: for 3 .. $n -> $x { my @try = |%D{$x}[$tuple[$x-2]]; last LOOP if any @try »==« @d[$_] for 1..@d-1; @d.push: @try; } next unless @d == $n and [==] [Z+] @d; @LS.push: @d; } @LS
}
say .join("\n") ~ "\n" for LS-reduced(4); for 1..6 -> $n {
printf "Order $n: Size %-4d x $n! x {$n-1}! => Total %d\n", $_, $_ * $n! * ($n-1)! given LS-reduced($n).elems
}</lang>
- Output:
1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 2 1 4 3 3 4 2 1 4 3 1 2 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 1 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Ruby
<lang ruby>def printSquare(a)
for row in a print row, "\n" end print "\n"
end
def dList(n, start)
start = start - 1 # use 0 based indexing a = Array.new(n) {|i| i} a[0], a[start] = a[start], a[0] a[1..] = a[1..].sort first = a[1]
r = [] recurse = lambda {|last| if last == first then # bottom of recursion, reached once for each permutation # test if permutation is deranged a[1..].each_with_index {|v, j| if j + 1 == v then return # no, ignore it end } # yes, save a copy with 1 based indexing b = a.map { |i| i + 1 } r << b return end
i = last while i >= 1 do a[i], a[last] = a[last], a[i] recurse.call(last - 1) a[i], a[last] = a[last], a[i] i = i - 1 end }
recurse.call(n - 1) return r
end
def reducedLatinSquares(n, echo)
if n <= 0 then if echo then print "[]\n\n" end return 0 end if n == 1 then if echo then print "[1]\n\n" end return 1 end
rlatin = Array.new(n) { Array.new(n, Float::NAN)}
# first row for j in 0 .. n - 1 rlatin[0][j] = j + 1 end
count = 0 recurse = lambda {|i| rows = dList(n, i)
for r in 0 .. rows.length - 1 rlatin[i - 1] = rows[r].dup catch (:outer) do for k in 0 .. i - 2 for j in 1 .. n - 1 if rlatin[k][j] == rlatin[i - 1][j] then if r < rows.length - 1 then throw :outer end if i > 2 then return end end end end if i < n then recurse.call(i + 1) else count = count + 1 if echo then printSquare(rlatin) end end end end }
# remaining rows recurse.call(2) return count
end
def factorial(n)
if n == 0 then return 1 end prod = 1 for i in 2 .. n prod = prod * i end return prod
end
print "The four reduced latin squares of order 4 are:\n" reducedLatinSquares(4, true)
print "The size of the set of reduced latin squares for the following orders\n" print "and hence the total number of latin squares of these orders are:\n" for n in 1 .. 6
size = reducedLatinSquares(n, false) f = factorial(n - 1) f = f * f * n * size print "Order %d Size %-4d x %d! x %d! => Total %d\n" % [n, size, n, n - 1, f]
end</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1 Size 1 x 1! x 0! => Total 1 Order 2 Size 1 x 2! x 1! => Total 2 Order 3 Size 1 x 3! x 2! => Total 12 Order 4 Size 4 x 4! x 3! => Total 576 Order 5 Size 56 x 5! x 4! => Total 161280 Order 6 Size 9408 x 6! x 5! => Total 812851200
Visual Basic .NET
<lang vbnet>Option Strict On
Imports Matrix = System.Collections.Generic.List(Of System.Collections.Generic.List(Of Integer))
Module Module1
Sub Swap(Of T)(ByRef a As T, ByRef b As T) Dim u = a a = b b = u End Sub
Sub PrintSquare(latin As Matrix) For Each row In latin Dim it = row.GetEnumerator Console.Write("[") If it.MoveNext Then Console.Write(it.Current) End If While it.MoveNext Console.Write(", ") Console.Write(it.Current) End While Console.WriteLine("]") Next Console.WriteLine() End Sub
Function DList(n As Integer, start As Integer) As Matrix start -= 1 REM use 0 based indexes Dim a = Enumerable.Range(0, n).ToArray a(start) = a(0) a(0) = start Array.Sort(a, 1, a.Length - 1) Dim first = a(1) REM recursive closure permutes a[1:] Dim r As New Matrix
Dim Recurse As Action(Of Integer) = Sub(last As Integer) If last = first Then REM bottom of recursion. you get here once for each permutation REM test if permutation is deranged. For j = 1 To a.Length - 1 Dim v = a(j) If j = v Then Return REM no, ignore it End If Next REM yes, save a copy with 1 based indexing Dim b = a.Select(Function(v) v + 1).ToArray r.Add(b.ToList) Return End If For i = last To 1 Step -1 Swap(a(i), a(last)) Recurse(last - 1) Swap(a(i), a(last)) Next End Sub Recurse(n - 1) Return r End Function
Function ReducedLatinSquares(n As Integer, echo As Boolean) As ULong If n <= 0 Then If echo Then Console.WriteLine("[]") Console.WriteLine() End If Return 0 End If If n = 1 Then If echo Then Console.WriteLine("[1]") Console.WriteLine() End If Return 1 End If
Dim rlatin As New Matrix For i = 0 To n - 1 rlatin.Add(New List(Of Integer)) For j = 0 To n - 1 rlatin(i).Add(0) Next Next REM first row For j = 0 To n - 1 rlatin(0)(j) = j + 1 Next
Dim count As ULong = 0 Dim Recurse As Action(Of Integer) = Sub(i As Integer) Dim rows = DList(n, i)
For r = 0 To rows.Count - 1 rlatin(i - 1) = rows(r) For k = 0 To i - 2 For j = 1 To n - 1 If rlatin(k)(j) = rlatin(i - 1)(j) Then If r < rows.Count - 1 Then GoTo outer End If If i > 2 Then Return End If End If Next Next If i < n Then Recurse(i + 1) Else count += 1UL If echo Then PrintSquare(rlatin) End If End If
outer:
While False REM empty End While Next End Sub
REM remiain rows Recurse(2) Return count End Function
Function Factorial(n As ULong) As ULong If n <= 0 Then Return 1 End If Dim prod = 1UL For i = 2UL To n prod *= i Next Return prod End Function
Sub Main() Console.WriteLine("The four reduced latin squares of order 4 are:") Console.WriteLine() ReducedLatinSquares(4, True)
Console.WriteLine("The size of the set of reduced latin squares for the following orders") Console.WriteLine("and hence the total number of latin squares of these orders are:") Console.WriteLine() For n = 1 To 6 Dim nu As ULong = CULng(n)
Dim size = ReducedLatinSquares(n, False) Dim f = Factorial(nu - 1UL) f *= f * nu * size Console.WriteLine("Order {0}: Size {1} x {2}! x {3}! => Total {4}", n, size, n, n - 1, f) Next End Sub
End Module</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Wren
<lang ecmascript>import "/sort" for Sort import "/math" for Int import "/fmt" for Fmt
// 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 = [0] * n for (i in 1...n) a[i] = i a[start] = a[0] a[0] = start Sort.quick(a, 1, a.count - 1, false) 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 = 1 for (v in a.skip(1)) { if (j == v) return // 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 } var i = last while (i >= 1) { var t = a[i] a[i] = a[last] a[last] = t recurse.call(last-1) t = a[i] a[i] = a[last] a[last] = t i = i - 1 } } recurse.call(n-1) return r
}
var printSquare = Fn.new { |latin, n|
System.print(latin.join("\n")) System.print()
}
var reducedLatinSquare = Fn.new { |n, echo|
if (n <= 0) { if (echo) System.print("[]\n") return 0 } if (n == 1) { if (echo) System.print("[1]\n") return 1 } var rlatin = List.filled(n, null) for (i in 0...n) rlatin[i] = List.filled(n, 0) // first row for (j in 0...n) rlatin[0][j] = j + 1 var count = 0 var recurse // // recursive closure to compute reduced latin squares and count or print them 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 for (rr in 0...rows[r].count) rlatin[i-1][rr] = rows[r][rr] var k = 0 while (k < i-1) { var j = 1 while (j < n) { if (rlatin[k][j] == rlatin[i-1][j]) { if (r < rows.count - 1) { outer = true break } else if (i > 2) { return } } j = j + 1 } if (outer) break k = k + 1 } if (!outer) { if (i < n) { recurse.call(i + 1) } else { count = count + 1 if (echo) printSquare.call(rlatin, n) } } } }
// remaining rows recurse.call(2) return count
}
System.print("The four reduced latin squares of order 4 are:\n") reducedLatinSquare.call(4, true)
System.print("The size of the set of reduced latin squares for the following orders") System.print("and hence the total number of latin squares of these orders are:\n") for (n in 1..6) {
var size = reducedLatinSquare.call(n, false) var f = Int.factorial(n-1) f = f * f * n * size Fmt.print("Order $d: Size $-4d x $d! x $d! => Total $d", n, size, n, n-1, f)
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
zkl
This reuses the dList function from the Permutations/Derangements#zkl task, suitably adjusted for the present one. <lang zkl>fcn reducedLatinSquare(n,write=False){
if(n<=1) return(n); rlatin:=n.pump(List(), List.createLong(n,0).copy); // matrix of zeros foreach i in (n){ rlatin[0][i]=i+1 } // first row: (1,2,3..n)
count:=Ref(0); // recursive closure to compute reduced latin squares and count or print them rows,rsz := derangements(n), rows.len(); recurse:='wrap(i){ foreach r in (rsz){ // top if(rows[r][0]!=i) continue; // filter by first column, ignore all but i rlatin[i-1]=rows[r].copy();
foreach k,j in ([0..i-2],[1..n-1]){ // nested loop: foreach foreach if(rlatin[k][j] == rlatin[i-1][j]){ if(r < rsz-1) continue(3); // -->top if(i>2) return(); } } if(i<n) self.fcn(i + 1, vm.pasteArgs(1)); // 'wrap hides local data (ie count, rows, etc) else{ count.inc(); if(write) printSquare(rlatin,n); }
} }; recurse(2); // remaining rows return(count.value);
} fcn derangements(n,i){
enum:=[1..n].pump(List); Utils.Helpers.permuteW(enum).tweak('wrap(perm){ if(perm.zipWith('==,enum).sum(0)) Void.Skip else perm }).pump(List);
} fcn printSquare(matrix,n){
matrix.pump(Console.println,fcn(l){ l.concat(", ","[","]") }); println();
} fcn fact(n){ ([1..n]).reduce('*,1) }</lang> <lang zkl>println("The four reduced latin squares of order 4 are:"); reducedLatinSquare(4,True);
println("The size of the set of reduced latin squares for the following orders"); println("and hence the total number of latin squares of these orders are:"); foreach n in ([1..6]){
size,f,f := reducedLatinSquare(n), fact(n - 1), f*f*n*size;; println("Order %d: Size %-4d x %d! x %d! -> Total %,d".fmt(n,size,n,n-1,f));
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! -> Total 1 Order 2: Size 1 x 2! x 1! -> Total 2 Order 3: Size 1 x 3! x 2! -> Total 12 Order 4: Size 4 x 4! x 3! -> Total 576 Order 5: Size 56 x 5! x 4! -> Total 161,280 Order 6: Size 9408 x 6! x 5! -> Total 812,851,200