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Latin Squares in reduced form

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

Translation of: Python
F dList(n, =start)
start--
V a = Array(0 .< n)
a[start] = a
a = start
a.sort_range(1..)
V first = a
[[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()
R 1

V rlatin = [ * n] * n
L(j) 0 .< n
rlatin[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))
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#

Translation of: D
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;
a = start;
Array.Sort(a, 1, a.Length - 1);
var first = a;
// 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();
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("\n");
}
return 1;
}

matrix rlatin = new matrix();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
}
}
// first row
for (int j = 0; j < n; j++) {
rlatin[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);
}
}
}
}
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++

Translation of: C#
#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;
a = start;
std::sort(a.begin() + 1, a.end());
auto first = a;
// 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 << "\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[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;
}
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

Translation of: Go
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;
a = start;
sort(a[1..\$]);
auto first = a;
// 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("\n");
}
return 1;
}

matrix rlatin = uninitializedArray!matrix(n);
foreach (i; 0..n) {
rlatin[i] = uninitializedArray!(int[])(n);
}
// first row
foreach (j; 0..n) {
rlatin[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);
}
}
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

// 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.)|>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.

normLS 4 |> Seq.iter(fun n->List.iter(printfn "%A") n;printfn "");;

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|]

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)))

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.

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, a[start] = start, a
sort.Ints(a[1:])
first := a
// 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("\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[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)
}
}
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

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.

import Data.List (permutations, (\\))

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

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

putStrLn "Latin squares of order 4:"
mapM_ printTable \$ latinSquares [1..4]

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
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

J

Implementation:

redlat=: {{
perms=: (A.&i.~ !)~ y
sqs=. i.1 1,y
for_j.}.i.y do.
p=. (j={."1 perms)#perms
sel=.-.+./"1 p +./@:="1/"2 sqs
sqs=.(#~ 1-0*/ .="1{:"2),/sqs,"2 1 sel#"2 p
end.
}}

redlat 4
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
#@redlat every 1 2 3 4 5 6
1 1 1 4 56 9408
(#@redlat every 1 2 3 4 5 6)*(!1 2 3 4 5 6x)*(!0 1 2 3 4 5x)
1 2 12 576 161280 812851200

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<>();

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+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);
}
}
}
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;
}
}

}

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: jq

Preliminaries

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 ;

Latin Squares

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]);

# 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 ;

"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)"
) ;

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

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
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()

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

Translation of: D
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.also { a = a[start] }
a.subList(1, a.size).sort()

val first = a
// 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 }
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("\n")
}
return 1
}

val rlatin = MutableList(n) { MutableList(n) { it } }
// first row
for (j in 0 until n) {
rlatin[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))
}
}
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)

%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]));

displaying the four reduced Latin Squares of order 4

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"];

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

Translation of: Go, Python, D, Kotlin

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.

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 `\$`(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, a[start]
a.sort(2, n)
let first = a
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.
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 
return 1

var rlatin = newMatrix(n, n)
# Initialize first row.
for i in 1..n: rlatin[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:"

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}"
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.

#!/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 { \$_ > 1 ? \$_ * fact(\$_ - 1) : 1 }

sub perm
{
my \$s = shift;
length \$s <= 1 ? \$s :
map { my \$f = \$_; map "\$f\$_", perm( \$s =~ s/\$_//r ) } split //, \$s;
}
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 = {}     -- (not strictly necessary)
vrs = {}     --          """
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

Translation of: D
def dList(n, start):
start -= 1 # use 0 basing
a = range(n)
a[start] = a
a = start
a[1:] = sorted(a[1:])
first = a
# 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 
return 1

rlatin = [None] * n
for i in xrange(n):
rlatin[i] = [None] * n
# first row
for j in xrange(0, n):
rlatin[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)
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)

# 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  if \$n == 1;

my @LS;
my @l = 1 X+ ^\$n;
my %D = derangements(@l).classify(*.);

for [X] (^(!\$n/(\$n-1))) xx \$n-1 -> \$tuple {
my @d.push: @l;
@d.push: %D{2}[\$tuple];
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
}
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

Translation of: D
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, a[start] = a[start], a
a[1..] = a[1..].sort
first = a

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 "\n\n"
end
return 1
end

rlatin = Array.new(n) { Array.new(n, Float::NAN)}

# first row
for j in 0 .. n - 1
rlatin[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
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

Translation of: C#
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
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("")
Console.WriteLine()
End If
Return 1
End If

Dim rlatin As New Matrix
For i = 0 To n - 1
For j = 0 To n - 1
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
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

Translation of: Go
Library: Wren-sort
Library: Wren-math
Library: Wren-fmt
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 =  * n
for (i in 1...n) a[i] = i
a[start] = a
a = start
Sort.quick(a, 1, a.count - 1, false)
var first = a
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
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("\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[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)
}
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

Translation of: Go

This reuses the dList function from the Permutations/Derangements#zkl task, suitably adjusted for the present one.

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[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]!=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) }
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));
}
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