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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[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))`
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[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);            }        }    }}`
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[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;}`
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[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);    }}`
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., 2019let 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] `

` 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[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)    }}`
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, (\\))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 `

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```

`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`
```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```

## 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 40 1 2 31 0 3 22 3 0 13 2 1 0 0 1 2 31 0 3 22 3 1 03 2 0 1 0 1 2 31 2 3 02 3 0 13 0 1 2 0 1 2 31 3 0 22 0 3 13 2 1 0   #@redlat every 1 2 3 4 5 61 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<>();         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;        }    } } `
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 rowsdef 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 ; `

`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`
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] + 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))     endend 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[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))    }}`
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., 2019include "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] = nfunc 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] = rfunc high(m: Matrix): Natural = m.value.lenfunc add(m: var Matrix; r: Row) = m.value.add rfunc `\$`(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}"`
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_formuse 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;  }`
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

Translation of: D
`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)`
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, derangementssub  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}`
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[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 rend 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 countend def factorial(n)    if n == 0 then        return 1    end    prod = 1    for i in 2 .. n        prod = prod * i    end    return prodend 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                                                    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 Ifouter:                                                    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 Sortimport "/math" for Intimport "/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)}`
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[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) }`
`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
```