# Orthogonal arrays (build recursive constructions)¶

This module implements several constructions of Orthogonal Arrays. As their input can be complex, they all have a counterpart in the orthogonal_arrays_find_recursive module that automatically computes it.

All these constructions are automatically queried when the orthogonal_array() function is called.

 construction_3_3() Return an $$OA(k,nm+i)$$. construction_3_4() Return a $$OA(k,nm+rs)$$. construction_3_5() Return an $$OA(k,nm+r+s+t)$$. construction_3_6() Return a $$OA(k,nm+i)$$. construction_q_x() Return an $$OA(k,(q-1)*(q-x)+x+2)$$ using the $$q-x$$ construction. OA_and_oval() Return a $$OA(q+1,q)$$ whose blocks contains $$\leq 2$$ zeroes in the last $$q$$ columns. thwart_lemma_3_5() Returns an $$OA(k,nm+a+b+c+d)$$. thwart_lemma_4_1() Returns an $$OA(k,nm+4(n-2))$$. three_factor_product() Returns an $$OA(k+1,n_1n_2n_3)$$. brouwer_separable_design() Returns a $$OA(k,t(q^2+q+1)+x)$$ using Brouwer’s result on separable designs.

## Functions¶

sage.combinat.designs.orthogonal_arrays_build_recursive.OA_and_oval(q)

Return a $$OA(q+1,q)$$ whose blocks contains $$\leq 2$$ zeroes in the last $$q$$ columns.

This $$OA$$ is build from a projective plane of order $$q$$, in which there exists an oval $$O$$ of size $$q+1$$ (i.e. a set of $$q+1$$ points no three of which are [colinear/contained in a common set of the projective plane]).

Removing an element $$x\in O$$ and all sets that contain it, we obtain a $$TD(q+1,q)$$ in which $$O$$ intersects all columns except one. As $$O$$ is an oval, no block of the $$TD$$ intersects it more than twice.

INPUT:

• q – a prime power

Note

This function is called by construction_3_6(), an implementation of Construction 3.6 from [AC07].

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import OA_and_oval
sage: _ = OA_and_oval

sage.combinat.designs.orthogonal_arrays_build_recursive.brouwer_separable_design(k, t, q, x, check=False, verbose=False, explain_construction=False)

Returns a $$OA(k,t(q^2+q+1)+x)$$ using Brouwer’s result on separable designs.

This method is an implementation of Brouwer’s construction presented in [Brouwer80]. It consists in a systematic application of the usual transformation from PBD to OA, applied to a specific PBD.

Baer subplanes

When $$q$$ is a prime power, the projective plane $$PG(2,q^2)$$ can be partitionned into subplanes $$PG(2,q)$$ (called Baer subplanes), giving $$PG(2,q^2)=B_1\cup \dots\cup B_{q^2-q+1}$$. As a result, every line of the $$PG(2,q^2)$$ intersects one of the subplane on $$q+1$$ points and all others on $$1$$ point.

The $$OA$$ are built by considering $$B_1\cup\dots\cup B_t$$, for a total of $$t(q^2+q+1)$$ points (to which $$x$$ new points are then added). The blocks of this subdesign belong to two categories:

• The blocks of size $$t$$: they come from the lines which intersect a $$B_i$$ on $$q+1$$ points for some $$i>t$$. The blocks of size $$t$$ can be partitionned into $$q^2-q+t-1$$ parallel classes according to their associated subplane $$B_i$$ with $$i>t$$.
• The blocks of size $$q+t$$: those blocks form a symmetric design, as every point is incident with $$q+t$$ of them.

Constructions

In the following, we write $$N=t(q^2+q+1)+x$$. The code is also heavily commented, and will clear any doubt.

• i) $$x=0$$: in that case we build a resolvable $$OA(k-1,N)$$ that will then be completed into an $$OA(k,N)$$.

• Sets of size $$t$$)

We take the product of each parallel class with the parallel classes of a resolvable $$OA(k-1,t)-t.OA(k-1,t)$$, yielding new parallel classes.

• Sets of size $$q+t$$)

A $$N \times (q+t)$$ array is built whose rows are the sets of size $$q+t$$ such that every value appears once per column. For each block of a $$OA(k-1,q+t)-(q+t).OA(k-1,t)$$, the product with the rows of the matrix yields a parallel class.

1. $$x=q+t$$
• Sets of size $$t$$)

Each set of size $$t$$ gives a $$OA(k,t)-t.OA(k,1)$$, except if there is only one parallel class in which case a $$OA(k,t)$$ is sufficient.

• Sets of size $$q+t$$)

A $$(N-x) \times (q+t)$$ array $$M$$ is built whose $$N-x$$ rows are the sets of size $$q+t$$ such that every value appears once per column. For each of the new $$x=q+t$$ points $$p_1,\dots,p_{q+t}$$ we build a matrix $$M_i$$ obtained from $$M$$ by adding a column equal to $$(p_i,p_i,p_i\dots )$$. We add to the OA the product of all rows of the $$M_i$$ with the block of the $$x=q+t$$ parallel classes of a resolvable $$OA(k,t+q+1)-(t+q+1).OA(k,1)$$.

• Set of size $$x$$) An $$OA(k,x)$$

1. $$x = q^2-q+1-t$$
• Sets of size $$t$$)

All blocks of the $$i$$-th parallel class are extended with the $$i$$-th new point. The blocks are then replaced by a $$OA(k,t+1)-(t+1).OA(k,1)$$ or, if there is only one parallel class (i.e. $$x=1$$) by a $$OA(k,t+1)-OA(k,1)$$.

• Set of size $$q+t$$)

They are replaced by $$OA(k,q+t)-(q+t).OA(k,1)$$.

• Set of size $$x$$) An $$OA(k,x)$$

1. $$x = q^2+1$$
• Sets of size $$t$$)

All blocks of the $$i$$-th parallel class are extended with the $$i$$-th new point (the other $$x-q-t$$ new points are not touched at this step). The blocks are then replaced by a $$OA(k,t+1)-(t+1).OA(k,1)$$ or, if there is only one parallel class (i.e. $$x=1$$) by a $$OA(k,t+1)-OA(k,1)$$.

• Sets of size $$q+t$$) Same as for ii)

• Set of size $$x$$) An $$OA(k,x)$$

1. $$0<x<q^2-q+1-t$$
• Sets of size $$t$$)

The blocks of the first $$x$$ parallel class are extended with the $$x$$ new points, and replaced with $$OA(k.t+1)-(t+1).OA(k,1)$$ or, if $$x=1$$, by $$OA(k.t+1)-.OA(k,1)$$

The blocks of the other parallel classes are replaced by $$OA(k,t)-t.OA(k,t)$$ or, if there is only one class left, by $$OA(k,t)-OA(k,t)$$

• Sets of size $$q+t$$)

They are replaced with $$OA(k,q+t)-(q+t).OA(k,1)$$.

• Set of size $$x$$) An $$OA(k,x)$$

1. $$t+q<x<q^2+1$$
• Sets of size $$t$$) Same as in v) with an $$x$$ equal to $$x-q+t$$.
• Sets of size $$t$$) Same as in vii)
• Set of size $$x$$) An $$OA(k,x)$$

INPUT:

• k,t,q,x (integers)
• check – (boolean) Whether to check that output is correct before returning it. Set to False by default.
• verbose (boolean) – whether to print some information on the construction and parameters being used.
• explain_construction (boolean) – return a string describing the construction.

REFERENCES:

 [Brouwer80] A Series of Separable Designs with Application to Pairwise Orthogonal Latin Squares, Andries E. Brouwer, Vol. 1, n. 1, pp. 39-41, European Journal of Combinatorics, 1980 http://www.sciencedirect.com/science/article/pii/S0195669880800199

EXAMPLES:

Test all possible cases:

sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import brouwer_separable_design
sage: k,q,t=4,4,3; _=brouwer_separable_design(k,q,t,0,verbose=True)
Case i) with k=4,q=3,t=4,x=0
sage: k,q,t=3,3,3; _=brouwer_separable_design(k,t,q,t+q,verbose=True,check=True)
Case ii) with k=3,q=3,t=3,x=6,e3=1
sage: k,q,t=3,3,6; _=brouwer_separable_design(k,t,q,t+q,verbose=True,check=True)
Case ii) with k=3,q=3,t=6,x=9,e3=0
sage: k,q,t=3,3,6; _=brouwer_separable_design(k,t,q,q**2-q+1-t,verbose=True,check=True)
Case iii) with k=3,q=3,t=6,x=1,e2=0
sage: k,q,t=3,4,6; _=brouwer_separable_design(k,t,q,q**2-q+1-t,verbose=True,check=True)
Case iii) with k=3,q=4,t=6,x=7,e2=1
sage: k,q,t=3,4,6; _=brouwer_separable_design(k,t,q,q**2+1,verbose=True,check=True)
Case iv) with k=3,q=4,t=6,x=17,e4=1
sage: k,q,t=3,2,2; _=brouwer_separable_design(k,t,q,q**2+1,verbose=True,check=True)
Case iv) with k=3,q=2,t=2,x=5,e4=0
sage: k,q,t=3,4,7; _=brouwer_separable_design(k,t,q,3,verbose=True,check=True)
Case v) with k=3,q=4,t=7,x=3,e1=1,e2=1
sage: k,q,t=3,4,7; _=brouwer_separable_design(k,t,q,1,verbose=True,check=True)
Case v) with k=3,q=4,t=7,x=1,e1=1,e2=0
sage: k,q,t=3,4,7; _=brouwer_separable_design(k,t,q,q**2-q-t,verbose=True,check=True)
Case v) with k=3,q=4,t=7,x=5,e1=0,e2=1
sage: k,q,t=5,4,7; _=brouwer_separable_design(k,t,q,t+q+3,verbose=True,check=True)
Case vi) with k=5,q=4,t=7,x=14,e3=1,e4=1
sage: k,q,t=5,4,8; _=brouwer_separable_design(k,t,q,t+q+1,verbose=True,check=True)
Case vi) with k=5,q=4,t=8,x=13,e3=1,e4=0
sage: k,q,t=5,4,8; _=brouwer_separable_design(k,t,q,q**2,verbose=True,check=True)
Case vi) with k=5,q=4,t=8,x=16,e3=0,e4=1

sage: print(designs.orthogonal_arrays.explain_construction(10,189))
Brouwer's separable design construction with t=9,q=4,x=0 from:
Andries E. Brouwer,
A series of separable designs with application to pairwise orthogonal Latin squares
Vol. 1, n. 1, pp. 39-41,
European Journal of Combinatorics, 1980

sage.combinat.designs.orthogonal_arrays_build_recursive.construction_3_3(k, n, m, i, explain_construction=False)

Return an $$OA(k,nm+i)$$.

This is Wilson’s construction with $$i$$ truncated columns of size 1 and such that a block $$B_0$$ of the incomplete OA intersects all truncated columns. As a consequence, all other blocks intersect only $$0$$ or $$1$$ of the last $$i$$ columns. This allow to consider the block $$B_0$$ only up to its first $$k$$ coordinates and then use a $$OA(k,i)$$ instead of a $$OA(k,m+i) - i.OA(k,1)$$.

This is construction 3.3 from [AC07].

INPUT:

• k,n,m,i (integers) such that the following designs are available : $$OA(k,n)$$, $$OA(k,m)$$, $$OA(k,m+1)$$, $$OA(k,r)$$.
• explain_construction (boolean) – return a string describing the construction.

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_find_recursive import find_construction_3_3
sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import construction_3_3
sage: from sage.combinat.designs.orthogonal_arrays import is_orthogonal_array
sage: k=11;n=177
sage: is_orthogonal_array(construction_3_3(*find_construction_3_3(k,n)),k,n,2)
True

sage: print(designs.orthogonal_arrays.explain_construction(9,91))
Construction 3.3 with n=11,m=8,i=3 from:
Julian R. Abel, Nicholas Cavenagh
Concerning eight mutually orthogonal latin squares,
Vol. 15, n.3, pp. 255-261,
Journal of Combinatorial Designs, 2007

sage.combinat.designs.orthogonal_arrays_build_recursive.construction_3_4(k, n, m, r, s, explain_construction=False)

Return a $$OA(k,nm+rs)$$.

This is Wilson’s construction applied to a truncated $$OA(k+r+1,n)$$ with $$r$$ columns of size $$1$$ and one column of size $$s$$.

The unique elements of the $$r$$ truncated columns are picked so that a block $$B_0$$ contains them all.

• If there exists an $$OA(k,m+r+1)$$ the column of size $$s$$ is truncated in order to intersect $$B_0$$.
• Otherwise, if there exists an $$OA(k,m+r)$$, the last column must not intersect $$B_0$$

This is construction 3.4 from [AC07].

INPUT:

• k,n,m,r,s (integers) – we assume that $$s<n$$ and $$1\leq r,s$$

The following designs must be available: $$OA(k,n)$$, $$OA(k,m)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$, $$OA(k,s)$$. Additionally, it requires either a $$OA(k,m+r)$$ or a $$OA(k,m+r+1)$$.

• explain_construction (boolean) – return a string describing the construction.

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_find_recursive import find_construction_3_4
sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import construction_3_4
sage: from sage.combinat.designs.orthogonal_arrays import is_orthogonal_array
sage: k=8;n=196
sage: is_orthogonal_array(construction_3_4(*find_construction_3_4(k,n)),k,n,2)
True

sage: print(designs.orthogonal_arrays.explain_construction(8,164))
Construction 3.4 with n=23,m=7,r=2,s=1 from:
Julian R. Abel, Nicholas Cavenagh
Concerning eight mutually orthogonal latin squares,
Vol. 15, n.3, pp. 255-261,
Journal of Combinatorial Designs, 2007

sage.combinat.designs.orthogonal_arrays_build_recursive.construction_3_5(k, n, m, r, s, t, explain_construction=False)

Return an $$OA(k,nm+r+s+t)$$.

This is exactly Wilson’s construction with three truncated groups except we make sure that all blocks have size $$>k$$, so we don’t need a $$OA(k,m+0)$$ but only $$OA(k,m+1)$$, $$OA(k,m+2)$$ ,OA(k,m+3).

This is construction 3.5 from [AC07].

INPUT:

• k,n,m (integers)
• r,s,t (integers) – sizes of the three truncated groups, such that $$r\leq s$$ and $$(q-r-1)(q-s) \geq (q-s-1)*(q-r)$$.
• explain_construction (boolean) – return a string describing the construction.

The following designs must be available : $$OA(k,n)$$, $$OA(k,r)$$, $$OA(k,s)$$, $$OA(k,t)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$, $$OA(k,m+3)$$.

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_find_recursive import find_construction_3_5
sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import construction_3_5
sage: from sage.combinat.designs.orthogonal_arrays import is_orthogonal_array
sage: k=8;n=111
sage: is_orthogonal_array(construction_3_5(*find_construction_3_5(k,n)),k,n,2)
True

sage: print(designs.orthogonal_arrays.explain_construction(8,90))
Construction 3.5 with n=11,m=6,r=8,s=8,t=8 from:
Julian R. Abel, Nicholas Cavenagh
Concerning eight mutually orthogonal latin squares,
Vol. 15, n.3, pp. 255-261,
Journal of Combinatorial Designs, 2007

sage.combinat.designs.orthogonal_arrays_build_recursive.construction_3_6(k, n, m, i, explain_construction=False)

Return a $$OA(k,nm+i)$$

This is Wilson’s construction with $$r$$ columns of order $$1$$, in which each block intersects at most two truncated columns. Such a design exists when $$n$$ is a prime power and is returned by OA_and_oval().

INPUT:

• k,n,m,i (integers) – $$n$$ must be a prime power. The following designs must be available: $$OA(k+r,q)$$, $$OA(k,m)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$.
• explain_construction (boolean) – return a string describing the construction.

This is construction 3.6 from [AC07].

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_find_recursive import find_construction_3_6
sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import construction_3_6
sage: from sage.combinat.designs.orthogonal_arrays import is_orthogonal_array
sage: k=8;n=95
sage: is_orthogonal_array(construction_3_6(*find_construction_3_6(k,n)),k,n,2)
True

sage: print(designs.orthogonal_arrays.explain_construction(10,756))
Construction 3.6 with n=16,m=47,i=4 from:
Julian R. Abel, Nicholas Cavenagh
Concerning eight mutually orthogonal latin squares,
Vol. 15, n.3, pp. 255-261,
Journal of Combinatorial Designs, 2007

sage.combinat.designs.orthogonal_arrays_build_recursive.construction_q_x(k, q, x, check=True, explain_construction=False)

Return an $$OA(k,(q-1)*(q-x)+x+2)$$ using the $$q-x$$ construction.

Let $$v=(q-1)*(q-x)+x+2$$. If there exists a projective plane of order $$q$$ (e.g. when $$q$$ is a prime power) and $$0<x<q$$ then there exists a $$(v-1,\{q-x-1,q-x+1\})$$-GDD of type $$(q-1)^{q-x}(x+1)^1$$ (see [Greig99] or Theorem 2.50, section IV.2.3 of [DesignHandbook]). By adding to the ground set one point contained in all groups of the GDD, one obtains a $$(v,\{q-x-1,q-x+1,q,x+2\})$$-PBD with exactly one set of size $$x+2$$.

Thus, assuming that we have the following:

• $$OA(k,q-x-1)-(q-x-1).OA(k,1)$$
• $$OA(k,q-x+1)-(q-x+1).OA(k,1)$$
• $$OA(k,q)-q.OA(k,1)$$
• $$OA(k,x+2)$$

Then we can build from the PBD an $$OA(k,v)$$.

Construction of the PBD (shared by Julian R. Abel):

Start with a resolvable $$(q^2,q,1)$$-BIBD and put the points into a $$q\times q$$ array so that rows form a parallel class and columns form another.

Now delete:

• All $$x(q-1)$$ points from the first $$x$$ columns and not in the first row
• All $$q-x$$ points in the last $$q-x$$ columns AND the first row.

Then add a point $$p_1$$ to the blocks that are rows. Add a second point $$p_2$$ to the $$q-x$$ blocks that are columns of size $$q-1$$, plus the first row of size $$x+1$$.

INPUT:

• k,q,x – integers such that $$0<x<q$$ and such that Sage can build:

• A projective plane of order $$q$$
• $$OA(k,q-x-1)-(q-x-1).OA(k,1)$$
• $$OA(k,q-x+1)-(q-x+1).OA(k,1)$$
• $$OA(k,q)-q.OA(k,1)$$
• $$OA(k,x+2)$$
• check – (boolean) Whether to check that output is correct before returning it. As this is expected to be useless (but we are cautious guys), you may want to disable it whenever you want speed. Set to True by default.

• explain_construction (boolean) – return a string describing the construction.

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import construction_q_x
sage: _ = construction_q_x(9,16,6)

sage: print(designs.orthogonal_arrays.explain_construction(9,158))
(q-x)-construction with q=16,x=6 from:
Malcolm Greig,
Designs from projective planes and PBD bases,
vol. 7, num. 5, pp. 341--374,
Journal of Combinatorial Designs, 1999


REFERENCES:

 [Greig99] Designs from projective planes and PBD bases Malcolm Greig Journal of Combinatorial Designs vol. 7, num. 5, pp. 341–374 1999
sage.combinat.designs.orthogonal_arrays_build_recursive.three_factor_product(k, n1, n2, n3, check=False, explain_construction=False)

Returns an $$OA(k+1,n_1n_2n_3)$$

The three factor product construction from [DukesLing14] does the following:

If $$n_1\leq n_2\leq n_3$$ are such that there exists an $$OA(k,n_1)$$, $$OA(k+1,n_2)$$ and $$OA(k+1,n_3)$$, then there exists a $$OA(k+1,n_1n_2n_3)$$.

It works with a modified product of orthogonal arrays ([Rees93], [Rees00]) which keeps track of parallel classes in the $$OA$$ (the definition is given for transversal designs).

A subset of blocks in an $$TD(k,n)$$ is called a $$c$$-parallel class if every point is covered exactly $$c$$ times. A 1-parallel class is a parallel class.

The modified product:

If there exists an $$OA(k,n_1)$$, and if there exists an $$OA(k,n_2)$$ whose blocks are partitionned into $$s$$ $$n_1$$-parallel classes and $$n_2-sn_1$$ parallel classes, then there exists an $$OA(k,n_1n_2)$$ whose blocks can be partitionned into $$sn_1^2$$ parallel classes and $$(n_1n_2-sn_1^2)/n_1=n_2-sn_1$$ $$n_1$$-parallel classes.

Proof:

• The product of the blocks of a parallel class with an $$OA(k,n_1)$$ yields an $$n_1$$-parallel class of an $$OA(k,n_1n_2)$$.

• The product of the blocks of a $$n_1$$-parallel class of $$OA(k,n_2)$$ with an $$OA(k,n_1)$$ can be done in such a way that it yields $$n_1n_2$$ parallel classes of $$OA(k,n_1n_2)$$. Those classes cover exactly the pairs that would have been covered with the usual product.

This can be achieved by simple cyclic permutations. Let us build the product of the $$n_1$$-parallel class $$\mathcal P\subseteq OA(k,n_2)$$ with $$OA(k,n_1)$$: when computing the product of $$P\in\mathcal P$$ with $$B^1\in OA(k,n_1)$$ the $$i$$-th coordinate should not be $$(B^1_i,P_i)$$ but $$(B^1_i+r,P_i)$$ (the sum is mod $$n_1$$) where $$r$$ is the number of blocks of $$\mathcal P$$ we have already processed whose $$i$$-th coordinate is equal to $$P_i$$ (note that $$r< n_1$$ as $$\mathcal P$$ is $$n_1$$-parallel).

With these tools, one can obtain the designs promised by the three factors construction applied to $$k,n_1,n_2,n_3$$ (thanks to Julian R. Abel’s help):

1. Let $$s$$ be the largest integer $$\leq n_3/n_1$$. Apply the product construction to $$OA(k,n_1)$$ and a resolvable $$OA(k,n_3)$$ whose blocks are partitionned into $$s$$ $$n_1$$-parallel classes and $$n_3-sn_1$$ parallel classes. It results in a $$OA(k,n_1n_3)$$ partitionned into $$sn_1^2$$ parallel classes plus $$(n_1n_3-sn_1^2)/n_1=n_3-sn_1$$ $$n_1$$-parallel classes.
2. Add $$n_3-n_1$$ parallel classes to every $$n_1$$-parallel class to turn them into $$n_3$$-parallel classes. Apply the product construction to this partitionned $$OA(k,n_1n_3)$$ with a resolvable $$OA(k,n_2)$$.
3. As $$OA(k,n_2)$$ is resolvable, the $$n_2$$-parallel classes of $$OA(k,n_1n_2n_3)$$ are actually the union of $$n_2$$ parallel classes, thus the $$OA(k,n_1n_2n_3)$$ is resolvable and can be turned into an $$OA(k+1,n_1n_2n_3)$$

INPUT:

• k,n1,n2,n3 (integers)
• check – (boolean) Whether to check that everything is going smoothly while the design is being built. It is disabled by default, as the constructor of orthogonal arrays checks the final design anyway.
• explain_construction (boolean) – return a string describing the construction.

EXAMPLES:

sage: from sage.combinat.designs.designs_pyx import is_orthogonal_array
sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import three_factor_product

sage: OA = three_factor_product(4,4,4,4)
sage: is_orthogonal_array(OA,5,64)
True

sage: OA = three_factor_product(4,3,4,5)
sage: is_orthogonal_array(OA,5,60)
True

sage: OA = three_factor_product(5,4,5,7)
sage: is_orthogonal_array(OA,6,140)
True

sage: OA = three_factor_product(9,8,9,9) # long time
sage: is_orthogonal_array(OA,10,8*9*9)   # long time
True

sage: print(designs.orthogonal_arrays.explain_construction(10,648))
Three-factor product with n=8.9.9 from:
Peter J. Dukes, Alan C.H. Ling,
A three-factor product construction for mutually orthogonal latin squares,
https://arxiv.org/abs/1401.1466


REFERENCE:

 [DukesLing14] A three-factor product construction for mutually orthogonal latin squares, Peter J. Dukes, Alan C.H. Ling, arXiv 1401.1466
 [Rees00] Truncated Transversal Designs: A New Lower Bound on the Number of Idempotent MOLS of Side, Rolf S. Rees, Journal of Combinatorial Theory, Series A 90.2 (2000): 257-266.
 [Rees93] Two new direct product-type constructions for resolvable group-divisible designs, Rolf S. Rees, Journal of Combinatorial Designs 1.1 (1993): 15-26.
sage.combinat.designs.orthogonal_arrays_build_recursive.thwart_lemma_3_5(k, n, m, a, b, c, d=0, complement=False, explain_construction=False)

Returns an $$OA(k,nm+a+b+c+d)$$

(When d=0)

According to [Thwarts] when $$n$$ is a prime power and $$a+b+c\leq n+1$$, one can build an $$OA(k+3,n)$$ with three truncated columns of sizes $$a,b,c$$ in such a way that all blocks have size $$\leq k+2$$.

(in order to build a $$OA(k,nm+a+b+c)$$ the following designs must also exist: $$OA(k,a)$$, $$OA(k,b)$$, $$OA(k,c)$$, $$OA(k,m+0)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$)

Considering the complement of each truncated column, it is also possible to build an $$OA(k+3,n)$$ with three truncated columns of sizes $$a,b,c$$ in such a way that all blocks have size $$>k$$ whenever $$(n-a)+(n-b)+(n-c)\leq n+1$$.

(in order to build a $$OA(k,nm+a+b+c)$$ the following designs must also exist: $$OA(k,a)$$, $$OA(k,b)$$, $$OA(k,c)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$, $$OA(k,m+3)$$)

Here is the proof of Lemma 3.5 from [Thwarts] enriched with explanations from Julian R. Abel:

For any prime power $$n$$ one can build $$k-1$$ MOLS by associating to every nonzero $$x\in \mathbb F_n$$ the latin square:

$M_x(i,j) = i+x*j \text{ where }i,j\in \mathbb F_n$

In particular $$M_1(i,j)=i+j$$, whose $$n$$ columns and lines are indexed by the elements of $$\mathbb F_n$$. If we order the elements of $$\mathbb F_n$$ as $$0,1,...,n-1,x+0,...,x+n-1,x^2+0,...$$ and reorder the columns and lines of $$M_1$$ accordingly, the top-left $$a\times b$$ squares contains at most $$a+b-1$$ distinct symbols.

(When $$d\neq 0$$ )

If there exists an $$OA(k+3,n)$$ with three truncated columns of sizes $$a,b,c$$ in such a way that all blocks have size $$\leq k+2$$, by truncating arbitrarily another column to size $$d$$ one obtains an $$OA$$ with 4 truncated columns whose blocks miss at least one value. Thus, following the proof again one can build an $$OA(k+4)$$ with four truncated columns of sizes $$a,b,c,d$$ with blocks of size $$\leq k+3$$.

(in order to build a $$OA(k,nm+a+b+c+d)$$ the following designs must also exist: $$OA(k,a)$$, $$OA(k,b)$$, $$OA(k,c)$$, $$OA(k,d)$$, $$OA(k,m+0)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$, $$OA(k,m+3)$$)

As before, this also shows that one can build an $$OA(k+4,n)$$ with four truncated columns of sizes $$a,b,c,d$$ in such a way that all blocks have size $$>k$$ whenever $$(n-a)+(n-b)+(n-c)\leq n+1$$

(in order to build a $$OA(k,nm+a+b+c+d)$$ the following designs must also exist: $$OA(k,n-a)$$, $$OA(k,n-b)$$, $$OA(k,n-c)$$, $$OA(k,d)$$, $$OA(k,m+1)$$, $$OA(k,m+2)$$, $$OA(k,m+3)$$, $$OA(k,m+4)$$)

INPUT:

• k,n,m,a,b,c,d – integers which must satisfy the constraints above. In particular, $$a+b+c\leq n+1$$ must hold. By default, $$d=0$$.
• complement (boolean) – whether to complement the sets, i.e. follow the $$n-a,n-b,n-c$$ variant described above.
• explain_construction (boolean) – return a string describing the construction.

EXAMPLES:

sage: from sage.combinat.designs.orthogonal_arrays_build_recursive import thwart_lemma_3_5
sage: from sage.combinat.designs.designs_pyx import is_orthogonal_array
sage: OA = thwart_lemma_3_5(6,23,7,5,7,8)
sage: is_orthogonal_array(OA,6,23*7+5+7+8,2)
True

sage: print(designs.orthogonal_arrays.explain_construction(10,408))
Lemma 4.1 with n=13,m=28 from:
Charles J.Colbourn, Jeffrey H. Dinitz, Mieczyslaw Wojtas,
Thwarts in transversal designs,
Designs, Codes and Cryptography 5, no. 3 (1995): 189-197.


With sets of parameters from [Thwarts]:

sage: l = [
....:    [11, 27, 78, 16, 17, 25, 0],
....:    [12, 19, 208, 11, 13, 16, 0],
....:    [12, 19, 208, 13, 13, 16, 0],
....:    [10, 13, 78, 9, 9, 13, 1],
....:    [10, 13, 79, 9, 9, 13, 1]]
sage: for k,n,m,a,b,c,d in l:                                       # not tested -- too long
....:     OA = thwart_lemma_3_5(k,n,m,a,b,c,d,complement=True)      # not tested -- too long
....:     assert is_orthogonal_array(OA,k,n*m+a+b+c+d,verbose=True) # not tested -- too long

sage: print(designs.orthogonal_arrays.explain_construction(10,1046))
Lemma 3.5 with n=13,m=79,a=9,b=1,c=0,d=9 from:
Charles J.Colbourn, Jeffrey H. Dinitz, Mieczyslaw Wojtas,
Thwarts in transversal designs,
Designs, Codes and Cryptography 5, no. 3 (1995): 189-197.


REFERENCE:

 [Thwarts] (1, 2, 3, 4) Thwarts in transversal designs Charles J.Colbourn, Jeffrey H. Dinitz, Mieczyslaw Wojtas. Designs, Codes and Cryptography 5, no. 3 (1995): 189-197.
sage.combinat.designs.orthogonal_arrays_build_recursive.thwart_lemma_4_1(k, n, m, explain_construction=False)

Returns an $$OA(k,nm+4(n-2))$$.

Implements Lemma 4.1 from [Thwarts].

If $$n\equiv 0,1\pmod{3}$$ is a prime power, then there exists a truncated $$OA(n+1,n)$$ whose last four columns have size $$n-2$$ and intersect every block on $$1,3$$ or $$4$$ values. Consequently, if there exists an $$OA(k,m+1)$$, $$OA(k,m+3)$$, $$OA(k,m+4)$$ and a $$OA(k,n-2)$$ then there exists an $$OA(k,nm+4(n-2)$$

Proof: form the transversal design by removing one point of the $$AG(2,3)$$ (Affine Geometry) contained in the Desarguesian Projective Plane $$PG(2,n)$$.

The affine geometry on 9 points contained in the projective geometry $$PG(2,n)$$ is given explicitly in [OS64] (Thanks to Julian R. Abel for finding the reference!).

INPUT:

• k,n,m (integers)
• explain_construction (boolean) – return a string describing the construction.

EXAMPLES:

sage: print(designs.orthogonal_arrays.explain_construction(10,408))
Lemma 4.1 with n=13,m=28 from:
Charles J.Colbourn, Jeffrey H. Dinitz, Mieczyslaw Wojtas,
Thwarts in transversal designs,
Designs, Codes and Cryptography 5, no. 3 (1995): 189-197.


REFERENCES:

 [OS64] Finite projective planes with affine subplanes, T. G. Ostrom and F. A. Sherk. Canad. Math. Bull vol7 num.4 (1964)