Latin Squares in reduced form

Latin Squares in reduced form is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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.

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