Latin Squares in reduced form: Difference between revisions
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Order 5: Size 56 x 5! x 4! => Total 161280 |
Order 5: Size 56 x 5! x 4! => Total 161280 |
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Order 6: Size 9408 x 6! x 5! => Total 812851200</pre> |
Order 6: Size 9408 x 6! x 5! => Total 812851200</pre> |
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=={{header|Wren}}== |
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{{trans|Go}} |
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{{libheader|Wren-sort}} |
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{{libheader|Wren-math}} |
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{{libheader|Wren-fmt}} |
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<lang ecmascript>import "/sort" for Sort |
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import "/math" for Int |
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import "/fmt" for Fmt |
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// generate derangements of first n numbers, with 'start' in first place. |
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var dList = Fn.new { |n, start| |
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var r = [] |
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start = start - 1 // use 0 basing |
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var a = [0] * n |
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for (i in 1...n) a[i] = i |
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a[start] = a[0] |
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a[0] = start |
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Sort.quick(a, 1, a.count - 1, false) |
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var first = a[1] |
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var recurse // recursive closure permutes a[1..-1] |
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recurse = Fn.new { |last| |
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if (last == first) { |
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// bottom of recursion. you get here once for each permutation. |
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// test if permutation is deranged. |
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var j = 1 |
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for (v in a.skip(1)) { |
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if (j == v) return // no, ignore it |
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j = j + 1 |
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} |
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// yes, save a copy |
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var b = a.toList |
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for (i in 0...b.count) b[i] = b[i] + 1 // change back to 1 basing |
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r.add(b) |
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return |
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} |
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var i = last |
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while (i >= 1) { |
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var t = a[i] |
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a[i] = a[last] |
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a[last] = t |
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System.write("") // guard against VM recursion bug |
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recurse.call(last-1) |
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t = a[i] |
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a[i] = a[last] |
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a[last] = t |
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i = i - 1 |
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} |
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} |
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recurse.call(n-1) |
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return r |
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} |
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var printSquare = Fn.new { |latin, n| |
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System.print(latin.join("\n")) |
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System.print() |
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} |
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var reducedLatinSquare = Fn.new { |n, echo| |
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if (n <= 0) { |
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if (echo) System.print("[]\n") |
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return 0 |
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} |
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if (n == 1) { |
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if (echo) System.print("[1]\n") |
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return 1 |
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} |
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var rlatin = List.filled(n, null) |
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for (i in 0...n) rlatin[i] = List.filled(n, 0) |
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// first row |
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for (j in 0...n) rlatin[0][j] = j + 1 |
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var count = 0 |
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var recurse // // recursive closure to compute reduced latin squares and count or print them |
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recurse = Fn.new { |i| |
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var rows = dList.call(n, i) // get derangements of first n numbers, with 'i' first. |
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for (r in 0...rows.count) { |
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var outer = false |
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for (rr in 0...rows[r].count) rlatin[i-1][rr] = rows[r][rr] |
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var k = 0 |
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while (k < i-1) { |
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var j = 1 |
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while (j < n) { |
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if (rlatin[k][j] == rlatin[i-1][j]) { |
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if (r < rows.count - 1) { |
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outer = true |
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break |
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} else if (i > 2) { |
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return |
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} |
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} |
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j = j + 1 |
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} |
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if (outer) break |
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k = k + 1 |
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} |
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if (!outer) { |
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if (i < n) { |
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System.write("") |
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recurse.call(i + 1) |
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} else { |
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count = count + 1 |
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if (echo) printSquare.call(rlatin, n) |
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} |
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} |
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} |
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} |
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// remaining rows |
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recurse.call(2) |
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return count |
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} |
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System.print("The four reduced latin squares of order 4 are:\n") |
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reducedLatinSquare.call(4, true) |
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System.print("The size of the set of reduced latin squares for the following orders") |
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System.print("and hence the total number of latin squares of these orders are:\n") |
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for (n in 1..6) { |
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var size = reducedLatinSquare.call(n, false) |
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var f = Int.factorial(n-1) |
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f = f * f * n * size |
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Fmt.print("Order $d: Size $-4d x $d! x $d! => Total $d", n, size, n, n-1, f) |
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}</lang> |
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{{out}} |
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<pre> |
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The four reduced latin squares of order 4 are: |
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[1, 2, 3, 4] |
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[2, 1, 4, 3] |
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[3, 4, 1, 2] |
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[4, 3, 2, 1] |
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[1, 2, 3, 4] |
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[2, 1, 4, 3] |
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[3, 4, 2, 1] |
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[4, 3, 1, 2] |
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[1, 2, 3, 4] |
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[2, 4, 1, 3] |
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[3, 1, 4, 2] |
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[4, 3, 2, 1] |
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[1, 2, 3, 4] |
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[2, 3, 4, 1] |
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[3, 4, 1, 2] |
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[4, 1, 2, 3] |
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The size of the set of reduced latin squares for the following orders |
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and hence the total number of latin squares of these orders are: |
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Order 1: Size 1 x 1! x 0! => Total 1 |
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Order 2: Size 1 x 2! x 1! => Total 2 |
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Order 3: Size 1 x 3! x 2! => Total 12 |
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Order 4: Size 4 x 4! x 3! => Total 576 |
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Order 5: Size 56 x 5! x 4! => Total 161280 |
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Order 6: Size 9408 x 6! x 5! => Total 812851200 |
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</pre> |
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=={{header|zkl}}== |
=={{header|zkl}}== |
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Revision as of 10:54, 18 September 2020
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.
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
namespace LatinSquares {
using matrix = List<List<int>>;
class Program {
static void Swap<T>(ref T a, ref T b) {
var t = a;
a = b;
b = t;
}
static matrix DList(int n, int start) {
start--; // use 0 basing
var a = Enumerable.Range(0, n).ToArray();
a[start] = a[0];
a[0] = start;
Array.Sort(a, 1, a.Length - 1);
var first = a[1];
// recursive closure permutes a[1:]
matrix r = new matrix();
void recurse(int last) {
if (last == first) {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged.
for (int j = 1; j < a.Length; j++) {
var v = a[j];
if (j == v) {
return; //no, ignore it
}
}
// yes, save a copy with 1 based indexing
var b = a.Select(v => v + 1).ToArray();
r.Add(b.ToList());
return;
}
for (int i = last; i >= 1; i--) {
Swap(ref a[i], ref a[last]);
recurse(last - 1);
Swap(ref a[i], ref a[last]);
}
}
recurse(n - 1);
return r;
}
static ulong ReducedLatinSquares(int n, bool echo) {
if (n <= 0) {
if (echo) {
Console.WriteLine("[]\n");
}
return 0;
} else if (n == 1) {
if (echo) {
Console.WriteLine("[1]\n");
}
return 1;
}
matrix rlatin = new matrix();
for (int i = 0; i < n; i++) {
rlatin.Add(new List<int>());
for (int j = 0; j < n; j++) {
rlatin[i].Add(0);
}
}
// first row
for (int j = 0; j < n; j++) {
rlatin[0][j] = j + 1;
}
ulong count = 0;
void recurse(int i) {
var rows = DList(n, i);
for (int r = 0; r < rows.Count; r++) {
rlatin[i - 1] = rows[r];
for (int k = 0; k < i - 1; k++) {
for (int j = 1; j < n; j++) {
if (rlatin[k][j] == rlatin[i - 1][j]) {
if (r < rows.Count - 1) {
goto outer;
}
if (i > 2) {
return;
}
}
}
}
if (i < n) {
recurse(i + 1);
} else {
count++;
if (echo) {
PrintSquare(rlatin, n);
}
}
outer: { }
}
}
//remaing rows
recurse(2);
return count;
}
static void PrintSquare(matrix latin, int n) {
foreach (var row in latin) {
var it = row.GetEnumerator();
Console.Write("[");
if (it.MoveNext()) {
Console.Write(it.Current);
}
while (it.MoveNext()) {
Console.Write(", {0}", it.Current);
}
Console.WriteLine("]");
}
Console.WriteLine();
}
static ulong Factorial(ulong n) {
if (n <= 0) {
return 1;
}
ulong prod = 1;
for (ulong i = 2; i < n + 1; i++) {
prod *= i;
}
return prod;
}
static void Main() {
Console.WriteLine("The four reduced latin squares of order 4 are:\n");
ReducedLatinSquares(4, true);
Console.WriteLine("The size of the set of reduced latin squares for the following orders");
Console.WriteLine("and hence the total number of latin squares of these orders are:\n");
for (int n = 1; n < 7; n++) {
ulong nu = (ulong)n;
var size = ReducedLatinSquares(n, false);
var f = Factorial(nu - 1);
f *= f * nu * size;
Console.WriteLine("Order {0}: Size {1} x {2}! x {3}! => Total {4}", n, size, n, n - 1, f);
}
}
}
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
C++
<lang cpp>#include <algorithm>
- include <functional>
- include <iostream>
- include <numeric>
- include <vector>
typedef std::vector<std::vector<int>> matrix;
matrix dList(int n, int start) {
start--; // use 0 basing
std::vector<int> a(n);
std::iota(a.begin(), a.end(), 0);
a[start] = a[0];
a[0] = start;
std::sort(a.begin() + 1, a.end());
auto first = a[1];
// recursive closure permutes a[1:]
matrix r;
std::function<void(int)> recurse;
recurse = [&](int last) {
if (last == first) {
// bottom of recursion you get here once for each permutation.
// test if permutation is deranged.
for (size_t j = 1; j < a.size(); j++) {
auto v = a[j];
if (j == v) {
return; //no, ignore it
}
}
// yes, save a copy with 1 based indexing
std::vector<int> b;
std::transform(a.cbegin(), a.cend(), std::back_inserter(b), [](int v) { return v + 1; });
r.push_back(b);
return;
}
for (int i = last; i >= 1; i--) {
std::swap(a[i], a[last]);
recurse(last - 1);
std::swap(a[i], a[last]);
}
};
recurse(n - 1);
return r;
}
void printSquare(const matrix &latin, int n) {
for (auto &row : latin) {
auto it = row.cbegin();
auto end = row.cend();
std::cout << '[';
if (it != end) {
std::cout << *it;
it = std::next(it);
}
while (it != end) {
std::cout << ", " << *it;
it = std::next(it);
}
std::cout << "]\n";
}
std::cout << '\n';
}
unsigned long reducedLatinSquares(int n, bool echo) {
if (n <= 0) {
if (echo) {
std::cout << "[]\n";
}
return 0;
} else if (n == 1) {
if (echo) {
std::cout << "[1]\n";
}
return 1;
}
matrix rlatin;
for (int i = 0; i < n; i++) {
rlatin.push_back({});
for (int j = 0; j < n; j++) {
rlatin[i].push_back(j);
}
}
// first row
for (int j = 0; j < n; j++) {
rlatin[0][j] = j + 1;
}
unsigned long count = 0;
std::function<void(int)> recurse;
recurse = [&](int i) {
auto rows = dList(n, i);
for (size_t r = 0; r < rows.size(); r++) {
rlatin[i - 1] = rows[r];
for (int k = 0; k < i - 1; k++) {
for (int j = 1; j < n; j++) {
if (rlatin[k][j] == rlatin[i - 1][j]) {
if (r < rows.size() - 1) {
goto outer;
}
if (i > 2) {
return;
}
}
}
}
if (i < n) {
recurse(i + 1);
} else {
count++;
if (echo) {
printSquare(rlatin, n);
}
}
outer: {}
}
};
//remaining rows recurse(2); return count;
}
unsigned long factorial(unsigned long n) {
if (n <= 0) return 1;
unsigned long prod = 1;
for (unsigned long i = 2; i <= n; i++) {
prod *= i;
}
return prod;
}
int main() {
std::cout << "The four reduced lating squares of order 4 are:\n"; reducedLatinSquares(4, true);
std::cout << "The size of the set of reduced latin squares for the following orders\n";
std::cout << "and hence the total number of latin squares of these orders are:\n\n";
for (int n = 1; n < 7; n++) {
auto size = reducedLatinSquares(n, false);
auto f = factorial(n - 1);
f *= f * n * size;
std::cout << "Order " << n << ": Size " << size << " x " << n << "! x " << (n - 1) << "! => Total " << f << '\n';
}
return 0;
}</lang>
- Output:
The four reduced lating squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
D
<lang d>import std.algorithm; import std.array; import std.range; import std.stdio;
alias matrix = int[][];
auto dList(int n, int start) {
start--; // use 0 basing
auto a = iota(0, n).array;
a[start] = a[0];
a[0] = start;
sort(a[1..$]);
auto first = a[1];
// recursive closure permutes a[1:]
matrix r;
void recurse(int last) {
if (last == first) {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged.
foreach (j,v; a[1..$]) {
if (j + 1 == v) {
return; //no, ignore it
}
}
// yes, save a copy with 1 based indexing
auto b = a.map!"a+1".array;
r ~= b;
return;
}
for (int i = last; i >= 1; i--) {
swap(a[i], a[last]);
recurse(last -1);
swap(a[i], a[last]);
}
}
recurse(n - 1);
return r;
}
ulong reducedLatinSquares(int n, bool echo) {
if (n <= 0) {
if (echo) {
writeln("[]\n");
}
return 0;
} else if (n == 1) {
if (echo) {
writeln("[1]\n");
}
return 1;
}
matrix rlatin = uninitializedArray!matrix(n);
foreach (i; 0..n) {
rlatin[i] = uninitializedArray!(int[])(n);
}
// first row
foreach (j; 0..n) {
rlatin[0][j] = j + 1;
}
ulong count;
void recurse(int i) {
auto rows = dList(n, i);
outer:
foreach (r; 0..rows.length) {
rlatin[i-1] = rows[r].dup;
foreach (k; 0..i-1) {
foreach (j; 1..n) {
if (rlatin[k][j] == rlatin[i - 1][j]) {
if (r < rows.length - 1) {
continue outer;
}
if (i > 2) {
return;
}
}
}
}
if (i < n) {
recurse(i + 1);
} else {
count++;
if (echo) {
printSquare(rlatin, n);
}
}
}
}
// remaining rows recurse(2); return count;
}
void printSquare(matrix latin, int n) {
foreach (row; latin) {
writeln(row);
}
writeln;
}
ulong factorial(ulong n) {
if (n == 0) {
return 1;
}
ulong prod = 1;
foreach (i; 2..n+1) {
prod *= i;
}
return prod;
}
void main() {
writeln("The four reduced latin squares of order 4 are:\n");
reducedLatinSquares(4, true);
writeln("The size of the set of reduced latin squares for the following orders");
writeln("and hence the total number of latin squares of these orders are:\n");
foreach (n; 1..7) {
auto size = reducedLatinSquares(n, false);
auto f = factorial(n - 1);
f *= f * n * size;
writefln("Order %d: Size %-4d x %d! x %d! => Total %d", n, size, n, n - 1, f);
}
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
F#
The Function
This task uses Permutations/Derangements#F.23 <lang fsharp> // Generate Latin Squares in reduced form. Nigel Galloway: July 10th., 2019 let normLS α=
let N=derange α|>List.ofSeq|>List.groupBy(fun n->n.[0])|>List.sortBy(fun(n,_)->n)|>List.map(fun(_,n)->n)|>Array.ofList
let rec fG n g=match n with h::t->fG t (g|>List.filter(fun g->Array.forall2((<>)) h g )) |_->g
let rec normLS n g=seq{for i in fG n N.[g] do if g=α-2 then yield [|1..α|]::(List.rev (i::n)) else yield! normLS (i::n) (g+1)}
match α with 1->seq[[[|1|]]] |2-> seq[[[|1;2|];[|2;1|]]] |_->Seq.collect(fun n->normLS [n] 1) N.[0]
</lang>
The Task
<lang fsharp> normLS 4 |> Seq.iter(fun n->List.iter(printfn "%A") n;printfn "");; </lang>
- Output:
[|1; 2; 3; 4|] [|2; 3; 4; 1|] [|3; 4; 1; 2|] [|4; 1; 2; 3|] [|1; 2; 3; 4|] [|2; 1; 4; 3|] [|3; 4; 2; 1|] [|4; 3; 1; 2|] [|1; 2; 3; 4|] [|2; 1; 4; 3|] [|3; 4; 1; 2|] [|4; 3; 2; 1|] [|1; 2; 3; 4|] [|2; 4; 1; 3|] [|3; 1; 4; 2|] [|4; 3; 2; 1|]
<lang fsharp> let rec fact n g=if n<2 then g else fact (n-1) n*g [1..6] |> List.iter(fun n->let nLS=normLS n|>Seq.length in printfn "order=%d number of Reduced Latin Squares nLS=%d nLS*n!*(n-1)!=%d" n nLS (nLS*(fact n 1)*(fact (n-1) 1))) </lang>
- Output:
order=1 number of Reduced Latin Squares nLS=1 nLS*n!*(n-1)!=1 order=2 number of Reduced Latin Squares nLS=1 nLS*n!*(n-1)!=2 order=3 number of Reduced Latin Squares nLS=1 nLS*n!*(n-1)!=12 order=4 number of Reduced Latin Squares nLS=4 nLS*n!*(n-1)!=576 order=5 number of Reduced Latin Squares nLS=56 nLS*n!*(n-1)!=161280 order=6 number of Reduced Latin Squares nLS=9408 nLS*n!*(n-1)!=812851200
Go
This reuses the dList function from the Permutations/Derangements#Go task, suitably adjusted for the present one. <lang go>package main
import (
"fmt" "sort"
)
type matrix [][]int
// generate derangements of first n numbers, with 'start' in first place. func dList(n, start int) (r matrix) {
start-- // use 0 basing
a := make([]int, n)
for i := range a {
a[i] = i
}
a[0], a[start] = start, a[0]
sort.Ints(a[1:])
first := a[1]
// recursive closure permutes a[1:]
var recurse func(last int)
recurse = func(last int) {
if last == first {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged.
for j, v := range a[1:] { // j starts from 0, not 1
if j+1 == v {
return // no, ignore it
}
}
// yes, save a copy
b := make([]int, n)
copy(b, a)
for i := range b {
b[i]++ // change back to 1 basing
}
r = append(r, b)
return
}
for i := last; i >= 1; i-- {
a[i], a[last] = a[last], a[i]
recurse(last - 1)
a[i], a[last] = a[last], a[i]
}
}
recurse(n - 1)
return
}
func reducedLatinSquare(n int, echo bool) uint64 {
if n <= 0 {
if echo {
fmt.Println("[]\n")
}
return 0
} else if n == 1 {
if echo {
fmt.Println("[1]\n")
}
return 1
}
rlatin := make(matrix, n)
for i := 0; i < n; i++ {
rlatin[i] = make([]int, n)
}
// first row
for j := 0; j < n; j++ {
rlatin[0][j] = j + 1
}
count := uint64(0)
// recursive closure to compute reduced latin squares and count or print them
var recurse func(i int)
recurse = func(i int) {
rows := dList(n, i) // get derangements of first n numbers, with 'i' first.
outer:
for r := 0; r < len(rows); r++ {
copy(rlatin[i-1], rows[r])
for k := 0; k < i-1; k++ {
for j := 1; j < n; j++ {
if rlatin[k][j] == rlatin[i-1][j] {
if r < len(rows)-1 {
continue outer
} else if i > 2 {
return
}
}
}
}
if i < n {
recurse(i + 1)
} else {
count++
if echo {
printSquare(rlatin, n)
}
}
}
return
}
// remaining rows recurse(2) return count
}
func printSquare(latin matrix, n int) {
for i := 0; i < n; i++ {
fmt.Println(latin[i])
}
fmt.Println()
}
func factorial(n uint64) uint64 {
if n == 0 {
return 1
}
prod := uint64(1)
for i := uint64(2); i <= n; i++ {
prod *= i
}
return prod
}
func main() {
fmt.Println("The four reduced latin squares of order 4 are:\n")
reducedLatinSquare(4, true)
fmt.Println("The size of the set of reduced latin squares for the following orders")
fmt.Println("and hence the total number of latin squares of these orders are:\n")
for n := uint64(1); n <= 6; n++ {
size := reducedLatinSquare(int(n), false)
f := factorial(n - 1)
f *= f * n * size
fmt.Printf("Order %d: Size %-4d x %d! x %d! => Total %d\n", n, size, n, n-1, f)
}
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1 2 3 4] [2 1 4 3] [3 4 1 2] [4 3 2 1] [1 2 3 4] [2 1 4 3] [3 4 2 1] [4 3 1 2] [1 2 3 4] [2 4 1 3] [3 1 4 2] [4 3 2 1] [1 2 3 4] [2 3 4 1] [3 4 1 2] [4 1 2 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Java
<lang java> import java.math.BigInteger; import java.util.ArrayList; import java.util.Arrays; import java.util.List;
public class LatinSquaresInReducedForm {
public static void main(String[] args) {
System.out.printf("Reduced latin squares of order 4:%n");
for ( LatinSquare square : getReducedLatinSquares(4) ) {
System.out.printf("%s%n", square);
}
System.out.printf("Compute the number of latin squares from count of reduced latin squares:%n(Reduced Latin Square Count) * n! * (n-1)! = Latin Square Count%n");
for ( int n = 1 ; n <= 6 ; n++ ) {
List<LatinSquare> list = getReducedLatinSquares(n);
System.out.printf("Size = %d, %d * %d * %d = %,d%n", n, list.size(), fact(n), fact(n-1), list.size()*fact(n)*fact(n-1));
}
}
private static long fact(int n) {
if ( n == 0 ) {
return 1;
}
int prod = 1;
for ( int i = 1 ; i <= n ; i++ ) {
prod *= i;
}
return prod;
}
private static List<LatinSquare> getReducedLatinSquares(int n) {
List<LatinSquare> squares = new ArrayList<>();
squares.add(new LatinSquare(n));
PermutationGenerator permGen = new PermutationGenerator(n);
for ( int fillRow = 1 ; fillRow < n ; fillRow++ ) {
List<LatinSquare> squaresNext = new ArrayList<>();
for ( LatinSquare square : squares ) {
while ( permGen.hasMore() ) {
int[] perm = permGen.getNext();
// If not the correct row - next permutation.
if ( (perm[0]+1) != (fillRow+1) ) {
continue;
}
// Check permutation against current square.
boolean permOk = true;
done:
for ( int row = 0 ; row < fillRow ; row++ ) {
for ( int col = 0 ; col < n ; col++ ) {
if ( square.get(row, col) == (perm[col]+1) ) {
permOk = false;
break done;
}
}
}
if ( permOk ) {
LatinSquare newSquare = new LatinSquare(square);
for ( int col = 0 ; col < n ; col++ ) {
newSquare.set(fillRow, col, perm[col]+1);
}
squaresNext.add(newSquare);
}
}
permGen.reset();
}
squares = squaresNext;
}
return squares;
}
@SuppressWarnings("unused")
private static int[] display(int[] in) {
int [] out = new int[in.length];
for ( int i = 0 ; i < in.length ; i++ ) {
out[i] = in[i] + 1;
}
return out;
}
private static class LatinSquare {
int[][] square;
int size;
public LatinSquare(int n) {
square = new int[n][n];
size = n;
for ( int col = 0 ; col < n ; col++ ) {
set(0, col, col + 1);
}
}
public LatinSquare(LatinSquare ls) {
int n = ls.size;
square = new int[n][n];
size = n;
for ( int row = 0 ; row < n ; row++ ) {
for ( int col = 0 ; col < n ; col++ ) {
set(row, col, ls.get(row, col));
}
}
}
public void set(int row, int col, int value) {
square[row][col] = value;
}
public int get(int row, int col) {
return square[row][col];
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder();
for ( int row = 0 ; row < size ; row++ ) {
sb.append(Arrays.toString(square[row]));
sb.append("\n");
}
return sb.toString();
}
}
private static class PermutationGenerator {
private int[] a;
private BigInteger numLeft;
private BigInteger total;
public PermutationGenerator (int n) {
if (n < 1) {
throw new IllegalArgumentException ("Min 1");
}
a = new int[n];
total = getFactorial(n);
reset();
}
private void reset () {
for ( int i = 0 ; i < a.length ; i++ ) {
a[i] = i;
}
numLeft = new BigInteger(total.toString());
}
public boolean hasMore() {
return numLeft.compareTo(BigInteger.ZERO) == 1;
}
private static BigInteger getFactorial (int n) {
BigInteger fact = BigInteger.ONE;
for ( int i = n ; i > 1 ; i-- ) {
fact = fact.multiply(new BigInteger(Integer.toString(i)));
}
return fact;
}
/*--------------------------------------------------------
* Generate next permutation (algorithm from Rosen p. 284)
*--------------------------------------------------------
*/
public int[] getNext() {
if ( numLeft.equals(total) ) {
numLeft = numLeft.subtract (BigInteger.ONE);
return a;
}
// Find largest index j with a[j] < a[j+1]
int j = a.length - 2;
while ( a[j] > a[j+1] ) {
j--;
}
// Find index k such that a[k] is smallest integer greater than a[j] to the right of a[j]
int k = a.length - 1;
while ( a[j] > a[k] ) {
k--;
}
// Interchange a[j] and a[k]
int temp = a[k];
a[k] = a[j];
a[j] = temp;
// Put tail end of permutation after jth position in increasing order
int r = a.length - 1;
int s = j + 1;
while (r > s) {
int temp2 = a[s];
a[s] = a[r];
a[r] = temp2;
r--;
s++;
}
numLeft = numLeft.subtract(BigInteger.ONE);
return a;
}
}
} </lang>
- Output:
Reduced latin squares of order 4: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] Compute the number of latin squares from count of reduced latin squares: (Reduced Latin Square Count) * n! * (n-1)! = Latin Square Count Size = 1, 1 * 1 * 1 = 1 Size = 2, 1 * 2 * 1 = 2 Size = 3, 1 * 6 * 2 = 12 Size = 4, 4 * 24 * 6 = 576 Size = 5, 56 * 120 * 24 = 161,280 Size = 6, 9408 * 720 * 120 = 812,851,200
Julia
<lang julia>using Combinatorics
clash(row2, row1::Vector{Int}) = any(i -> row1[i] == row2[i], 1:length(row2))
clash(row, rows::Vector{Vector{Int}}) = any(r -> clash(row, r), rows)
permute_onefixed(i, n) = map(vec -> vcat(i, vec), permutations(filter(x -> x != i, 1:n)))
filter_permuted(rows, i, n) = filter(v -> !clash(v, rows), permute_onefixed(i, n))
function makereducedlatinsquares(n)
matarray = [reshape(collect(1:n), 1, n)]
for i in 2:n
newmatarray = Vector{Matrix{Int}}()
for mat in matarray
r = size(mat)[1] + 1
newrows = filter_permuted(collect(row[:] for row in eachrow(mat)), r, n)
newmat = zeros(Int, r, n)
newmat[1:r-1, :] .= mat
append!(newmatarray,
[deepcopy(begin newmat[i, :] .= row; newmat end) for row in newrows])
end
matarray = newmatarray
end
matarray, length(matarray)
end
function testlatinsquares()
squares, count = makereducedlatinsquares(4)
println("The four reduced latin squares of order 4 are:")
for sq in squares, (i, row) in enumerate(eachrow(sq)), j in 1:4
print(row[j], j == 4 ? (i == 4 ? "\n\n" : "\n") : " ")
end
for i in 1:6
squares, count = makereducedlatinsquares(i)
println("Order $i: Size ", rpad(count, 5), "* $(i)! * $(i - 1)! = ",
count * factorial(i) * factorial(i - 1))
end
end
testlatinsquares()
</lang>
- Output:
The four reduced latin squares of order 4 are: 1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 2 1 4 3 3 4 2 1 4 3 1 2 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 1 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 Order 1: Size 1 * 1! * 0! = 1 Order 2: Size 1 * 2! * 1! = 2 Order 3: Size 1 * 3! * 2! = 12 Order 4: Size 4 * 4! * 3! = 576 Order 5: Size 56 * 5! * 4! = 161280 Order 6: Size 9408 * 6! * 5! = 812851200
Kotlin
<lang scala>typealias Matrix = MutableList<MutableList<Int>>
fun dList(n: Int, sp: Int): Matrix {
val start = sp - 1 // use 0 basing
val a = generateSequence(0) { it + 1 }.take(n).toMutableList()
a[start] = a[0].also { a[0] = a[start] }
a.subList(1, a.size).sort()
val first = a[1]
// recursive closure permutes a[1:]
val r = mutableListOf<MutableList<Int>>()
fun recurse(last: Int) {
if (last == first) {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged
for (jv in a.subList(1, a.size).withIndex()) {
if (jv.index + 1 == jv.value) {
return // no, ignore it
}
}
// yes, save a copy with 1 based indexing
val b = a.map { it + 1 }
r.add(b.toMutableList())
return
}
for (i in last.downTo(1)) {
a[i] = a[last].also { a[last] = a[i] }
recurse(last - 1)
a[i] = a[last].also { a[last] = a[i] }
}
}
recurse(n - 1)
return r
}
fun reducedLatinSquares(n: Int, echo: Boolean): Long {
if (n <= 0) {
if (echo) {
println("[]\n")
}
return 0
} else if (n == 1) {
if (echo) {
println("[1]\n")
}
return 1
}
val rlatin = MutableList(n) { MutableList(n) { it } }
// first row
for (j in 0 until n) {
rlatin[0][j] = j + 1
}
var count = 0L
fun recurse(i: Int) {
val rows = dList(n, i)
outer@
for (r in 0 until rows.size) {
rlatin[i - 1] = rows[r].toMutableList()
for (k in 0 until i - 1) {
for (j in 1 until n) {
if (rlatin[k][j] == rlatin[i - 1][j]) {
if (r < rows.size - 1) {
continue@outer
}
if (i > 2) {
return
}
}
}
}
if (i < n) {
recurse(i + 1)
} else {
count++
if (echo) {
printSquare(rlatin)
}
}
}
}
// remaining rows recurse(2) return count
}
fun printSquare(latin: Matrix) {
for (row in latin) {
println(row)
}
println()
}
fun factorial(n: Long): Long {
if (n == 0L) {
return 1
}
var prod = 1L
for (i in 2..n) {
prod *= i
}
return prod
}
fun main() {
println("The four reduced latin squares of order 4 are:\n")
reducedLatinSquares(4, true)
println("The size of the set of reduced latin squares for the following orders")
println("and hence the total number of latin squares of these orders are:\n")
for (n in 1 until 7) {
val size = reducedLatinSquares(n, false)
var f = factorial(n - 1.toLong())
f *= f * n * size
println("Order $n: Size %-4d x $n! x ${n - 1}! => Total $f".format(size))
}
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
MiniZinc
The Model (lsRF.mnz)
<lang MiniZinc> %Latin Squares in Reduced Form. Nigel Galloway, September 5th., 2019 include "alldifferent.mzn"; int: N; array[1..N,1..N] of var 1..N: p; constraint forall(n in 1..N)(p[1,n]=n /\ p[n,1]=n); constraint forall(n in 1..N)(alldifferent([p[n,g]|g in 1..N])/\alldifferent([p[g,n]|g in 1..N])); </lang>
The Tasks
- displaying the four reduced Latin Squares of order 4
<lang MiniZinc> include "lsRF.mzn";
output [show_int(1,p[i,j])++
if j == 4 then
if i != 4 then "\n"
else "" endif
else "" endif
| i,j in 1..4 ] ++ ["\n"];
</lang> When the above is run using minizinc --all-solutions -DN=4 the following is produced:
- Output:
1234 2143 3421 4312 ---------- 1234 2143 3412 4321 ---------- 1234 2413 3142 4321 ---------- 1234 2341 3412 4123 ---------- ==========
- counting the solutions
minizinc.exe --all-solutions -DN=5 -s lsRF.mzn produces the following:
. . . p = array2d(1..5, 1..5, [1, 2, 3, 4, 5, 2, 3, 4, 5, 1, 3, 1, 5, 2, 4, 4, 5, 2, 1, 3, 5, 4, 1, 3, 2]); ---------- p = array2d(1..5, 1..5, [1, 2, 3, 4, 5, 2, 3, 5, 1, 4, 3, 5, 4, 2, 1, 4, 1, 2, 5, 3, 5, 4, 1, 3, 2]); ---------- p = array2d(1..5, 1..5, [1, 2, 3, 4, 5, 2, 3, 4, 5, 1, 3, 5, 2, 1, 4, 4, 1, 5, 2, 3, 5, 4, 1, 3, 2]); ---------- ========== %%%mzn-stat: initTime=0.057 %%%mzn-stat: solveTime=0.003 %%%mzn-stat: solutions=56 %%%mzn-stat: variables=43 %%%mzn-stat: propagators=8 %%%mzn-stat: propagations=960 %%%mzn-stat: nodes=111 %%%mzn-stat: failures=0 %%%mzn-stat: restarts=0 %%%mzn-stat: peakDepth=7 %%%mzn-stat-end %%%mzn-stat: nSolutions=56
and minizinc.exe --all-solutions -DN=6 -s lsRF.mzn produces the following:
. . . p = array2d(1..6, 1..6, [1, 2, 3, 4, 5, 6, 2, 4, 5, 6, 3, 1, 3, 1, 4, 2, 6, 5, 4, 6, 2, 5, 1, 3, 5, 3, 6, 1, 2, 4, 6, 5, 1, 3, 4, 2]); ---------- p = array2d(1..6, 1..6, [1, 2, 3, 4, 5, 6, 2, 1, 4, 6, 3, 5, 3, 4, 5, 2, 6, 1, 4, 6, 2, 5, 1, 3, 5, 3, 6, 1, 2, 4, 6, 5, 1, 3, 4, 2]); ---------- ========== %%%mzn-stat: initTime=0.003 %%%mzn-stat: solveTime=6.669 %%%mzn-stat: solutions=9408 %%%mzn-stat: variables=58 %%%mzn-stat: propagators=10 %%%mzn-stat: propagations=179635 %%%mzn-stat: nodes=19035 %%%mzn-stat: failures=110 %%%mzn-stat: restarts=0 %%%mzn-stat: peakDepth=17 %%%mzn-stat-end %%%mzn-stat: nSolutions=9408
The only way to complete the tasks requirement to produce a table is with another language. Ruby has the ability to run an external program, capture the output, and text handling ability to format it to this tasks requirements. Othe scripting languages are available.
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.
<lang Phix>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:Template: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),
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</lang>
- 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
Python
<lang python>def dList(n, start):
start -= 1 # use 0 basing
a = range(n)
a[start] = a[0]
a[0] = start
a[1:] = sorted(a[1:])
first = a[1]
# rescursive closure permutes a[1:]
r = []
def recurse(last):
if (last == first):
# bottom of recursion. you get here once for each permutation.
# test if permutation is deranged.
# yes, save a copy with 1 based indexing
for j,v in enumerate(a[1:]):
if j + 1 == v:
return # no, ignore it
b = [x + 1 for x in a]
r.append(b)
return
for i in xrange(last, 0, -1):
a[i], a[last] = a[last], a[i]
recurse(last - 1)
a[i], a[last] = a[last], a[i]
recurse(n - 1)
return r
def printSquare(latin,n):
for row in latin:
print row
print
def reducedLatinSquares(n,echo):
if n <= 0:
if echo:
print []
return 0
elif n == 1:
if echo:
print [1]
return 1
rlatin = [None] * n
for i in xrange(n):
rlatin[i] = [None] * n
# first row
for j in xrange(0, n):
rlatin[0][j] = j + 1
class OuterScope:
count = 0
def recurse(i):
rows = dList(n, i)
for r in xrange(len(rows)):
rlatin[i - 1] = rows[r]
justContinue = False
k = 0
while not justContinue and k < i - 1:
for j in xrange(1, n):
if rlatin[k][j] == rlatin[i - 1][j]:
if r < len(rows) - 1:
justContinue = True
break
if i > 2:
return
k += 1
if not justContinue:
if i < n:
recurse(i + 1)
else:
OuterScope.count += 1
if echo:
printSquare(rlatin, n)
# remaining rows recurse(2) return OuterScope.count
def factorial(n):
if n == 0:
return 1
prod = 1
for i in xrange(2, n + 1):
prod *= i
return prod
print "The four reduced latin squares of order 4 are:\n" reducedLatinSquares(4,True)
print "The size of the set of reduced latin squares for the following orders" print "and hence the total number of latin squares of these orders are:\n" for n in xrange(1, 7):
size = reducedLatinSquares(n, False) f = factorial(n - 1) f *= f * n * size print "Order %d: Size %-4d x %d! x %d! => Total %d" % (n, size, n, n - 1, f)</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Raku
(formerly Perl 6) <lang perl6># utilities: factorial, sub-factorial, derangements sub postfix:<!>($n) { (constant f = 1, |[\×] 1..*)[$n] } sub prefix:<!>($n) { (1, 0, 1, -> $a, $b { ($++ + 2) × ($b + $a) } ... *)[$n] } sub derangements(@l) { @l.permutations.grep(-> @p { none(@p Zeqv @l) }) }
sub LS-reduced (Int $n) {
return [1] if $n == 1;
my @LS; my @l = 1 X+ ^$n; my %D = derangements(@l).classify(*.[0]);
for [X] (^(!$n/($n-1))) xx $n-1 -> $tuple {
my @d.push: @l;
@d.push: %D{2}[$tuple[0]];
LOOP:
for 3 .. $n -> $x {
my @try = |%D{$x}[$tuple[$x-2]];
last LOOP if any @try »==« @d[$_] for 1..@d-1;
@d.push: @try;
}
next unless @d == $n and [==] [Z+] @d;
@LS.push: @d;
}
@LS
}
say .join("\n") ~ "\n" for LS-reduced(4); for 1..6 -> $n {
printf "Order $n: Size %-4d x $n! x {$n-1}! => Total %d\n", $_, $_ * $n! * ($n-1)! given LS-reduced($n).elems
}</lang>
- Output:
1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 2 1 4 3 3 4 2 1 4 3 1 2 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 1 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
Wren
<lang ecmascript>import "/sort" for Sort import "/math" for Int import "/fmt" for Fmt
// generate derangements of first n numbers, with 'start' in first place. var dList = Fn.new { |n, start|
var r = []
start = start - 1 // use 0 basing
var a = [0] * n
for (i in 1...n) a[i] = i
a[start] = a[0]
a[0] = start
Sort.quick(a, 1, a.count - 1, false)
var first = a[1]
var recurse // recursive closure permutes a[1..-1]
recurse = Fn.new { |last|
if (last == first) {
// bottom of recursion. you get here once for each permutation.
// test if permutation is deranged.
var j = 1
for (v in a.skip(1)) {
if (j == v) return // no, ignore it
j = j + 1
}
// yes, save a copy
var b = a.toList
for (i in 0...b.count) b[i] = b[i] + 1 // change back to 1 basing
r.add(b)
return
}
var i = last
while (i >= 1) {
var t = a[i]
a[i] = a[last]
a[last] = t
System.write("") // guard against VM recursion bug
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) {
System.write("")
recurse.call(i + 1)
} else {
count = count + 1
if (echo) printSquare.call(rlatin, n)
}
}
}
}
// remaining rows recurse.call(2) return count
}
System.print("The four reduced latin squares of order 4 are:\n") reducedLatinSquare.call(4, true)
System.print("The size of the set of reduced latin squares for the following orders") System.print("and hence the total number of latin squares of these orders are:\n") for (n in 1..6) {
var size = reducedLatinSquare.call(n, false)
var f = Int.factorial(n-1)
f = f * f * n * size
Fmt.print("Order $d: Size $-4d x $d! x $d! => Total $d", n, size, n, n-1, f)
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! => Total 1 Order 2: Size 1 x 2! x 1! => Total 2 Order 3: Size 1 x 3! x 2! => Total 12 Order 4: Size 4 x 4! x 3! => Total 576 Order 5: Size 56 x 5! x 4! => Total 161280 Order 6: Size 9408 x 6! x 5! => Total 812851200
zkl
This reuses the dList function from the Permutations/Derangements#zkl task, suitably adjusted for the present one. <lang zkl>fcn reducedLatinSquare(n,write=False){
if(n<=1) return(n);
rlatin:=n.pump(List(), List.createLong(n,0).copy); // matrix of zeros
foreach i in (n){ rlatin[0][i]=i+1 } // first row: (1,2,3..n)
count:=Ref(0);
// recursive closure to compute reduced latin squares and count or print them
rows,rsz := derangements(n), rows.len();
recurse:='wrap(i){
foreach r in (rsz){ // top
if(rows[r][0]!=i) continue; // filter by first column, ignore all but i
rlatin[i-1]=rows[r].copy();
foreach k,j in ([0..i-2],[1..n-1]){ // nested loop: foreach foreach if(rlatin[k][j] == rlatin[i-1][j]){ if(r < rsz-1) continue(3); // -->top if(i>2) return(); } } if(i<n) self.fcn(i + 1, vm.pasteArgs(1)); // 'wrap hides local data (ie count, rows, etc) else{ count.inc(); if(write) printSquare(rlatin,n); }
} }; recurse(2); // remaining rows return(count.value);
} fcn derangements(n,i){
enum:=[1..n].pump(List);
Utils.Helpers.permuteW(enum).tweak('wrap(perm){
if(perm.zipWith('==,enum).sum(0)) Void.Skip
else perm
}).pump(List);
} fcn printSquare(matrix,n){
matrix.pump(Console.println,fcn(l){ l.concat(", ","[","]") });
println();
} fcn fact(n){ ([1..n]).reduce('*,1) }</lang> <lang zkl>println("The four reduced latin squares of order 4 are:"); reducedLatinSquare(4,True);
println("The size of the set of reduced latin squares for the following orders"); println("and hence the total number of latin squares of these orders are:"); foreach n in ([1..6]){
size,f,f := reducedLatinSquare(n), fact(n - 1), f*f*n*size;;
println("Order %d: Size %-4d x %d! x %d! -> Total %,d".fmt(n,size,n,n-1,f));
}</lang>
- Output:
The four reduced latin squares of order 4 are: [1, 2, 3, 4] [2, 3, 4, 1] [3, 4, 1, 2] [4, 1, 2, 3] [1, 2, 3, 4] [2, 4, 1, 3] [3, 1, 4, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 1, 2] [4, 3, 2, 1] [1, 2, 3, 4] [2, 1, 4, 3] [3, 4, 2, 1] [4, 3, 1, 2] The size of the set of reduced latin squares for the following orders and hence the total number of latin squares of these orders are: Order 1: Size 1 x 1! x 0! -> Total 1 Order 2: Size 1 x 2! x 1! -> Total 2 Order 3: Size 1 x 3! x 2! -> Total 12 Order 4: Size 4 x 4! x 3! -> Total 576 Order 5: Size 56 x 5! x 4! -> Total 161,280 Order 6: Size 9408 x 6! x 5! -> Total 812,851,200