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Finding a decomposition of a graph T into isomorphic copies of a graph G is a classical problem in Combinatorics.

The G-decomposition of T is balanced if every vertex of T is in the same number of copies of G.

In such G-decompositions additional properties may be required on vertices, for example:

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Construction of combinatorial designs can be considered as a decomposition where the pairs of points (edges) meet certain requirements. For example,

let Kv denote copies of the complete graph Kv and we may ask to decompose it into copies of Kk for some k to get a design, in fact, to get a balanced incomplete block design, BIBD(v,b,r,k,), as it IS a decomposition of copies of Kv into b copies of Kk.

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For example, one can ask for partitioning copies of Kv into triangles, i.e., K3, which is the same as constructing BIBD(v,3,) or a triple system TS(v,3,).

Of course, this well known problem is completely solved among many other values of k and .

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The question we consider is whether one can decompose copies of monochromatic Kv into copies of Kk such that each copy of Kk contains at most one edge from each Kv.

This is called proper edge-coloring or proper coloring of the design, BIBD(v,k, ).

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Example

10 times 3 equals 30 edges, need 10 triangles

Three colored K5 ’s on five vertices {1,2,3,4,5}

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11 1

2 2 2

334 44

555

1

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We can form a Red X, Blue X, and a Green X and from the three copies of any block , K3, swap edges, say cyclically and obtained three colored blocks.

Suppose we wish to decompose 3 copies of Kv colored Red, Blue and Green into properly colored BIBD(v,3,3). Suppose a BIBD(v,3,1) exists, say X.

Then the solution is easy:

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Theorem: If a BIBD(v,k,1) exists, then for any index ≥ k(k-1)/2, there exists a properly colored BIBD(v,k.

In general,

This result is not enough to solve the problem completely!

(I.e. edges of blocks can be taken from monochromatic copies of Kv so that no two edges in a block have the same color.)

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Going back to the case when k=3, it may not be clear that the index =1 requires v to be congruent to 1 or 3 mod 6.but for index =3, v has to be odd,

The necessary conditions for the existence of a BIBD(v,b,r,k are

vr=bk and (v-1) = r(k-1).

So the previous theorem does not answer the question even for k=3.

The reason is that there are some relationships which the parameters of a design should satisfy for the design to exist:

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Our first aim is to prove that the necessary conditions are also sufficient for properly colored BIBDs with block size three,

(i.e., we want to decompose different colored copies of Kv into triples such that each edge is of different color.),

then we will establish a relationship of these colored designs with so called Path designs,

and in the end use this connection to construct certain c- Bhaskar Rao Design.

The new terms will be defined where needed.

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Agrawal’s theorem

A 1966 result by Agrawal is very useful for our work.

Theorem: In every binary equi-replicate design of constant block size k such that vr=bk and b = mv, the points can be arranged into a rectangular array with columns as blocks so that every point occurs in a row m times. If r = mk+t where 0 < t ≤ k-1, then the points can be arranged into blocks so that every point occurs m or m+1 times in a row.

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For example, the theorem can be applied to the blocks of a BIBD and points can be arranged in a k by b matrix so that each point appears in each row the same number of times (or almost same number of times).

We will apply this theorem on an equireplicate design to be constructed from a BIBD.

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Theorem: Suppose there exists BIBD(v,b,r,k,), say X, with = mk(k-1)/2. Then the blocks of X can be colored with colors so that no two edges in any block have the same color.

Use each block , say,{1,2,3, …, k} of X

to construct a block {(1,2),(1,3), …,(2,3),(2,4), …,(k-1,k)} of Y.

Proof:Construct a new design Y on v(v-1)/2 pairs of the points of X as follows.

Note Y is an equi-replicate design with replication number .

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Therefore we can apply Agrawal’s result and order the blocks of Y such that each pair (a,b) occurs in the array the same number of times (m times) in each row.

Let the colors be indexed by c(i,j) where j = 1,2, …,m and i=1,2,. . . , k(k-1)/2.

Color the jth occurrence of the edge (a,b) in the ith row by c(i,j).

Note the block size k(k-1)/2 of Y divides the replication number for Y which is the index of X.

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Example:

BIBD(5,3,3), X: 1 1 1 1 1 1 2 2 2 3 2 2 2 3 3 4 3 3 4 4 3 4 5 4 5 5 4 5 5 5

12 12 12 13 13 14 23 23 24 34 13 14 15 14 15 15 24 25 25 35 23 24 25 34 35 45 34 35 45 45

Y=

Apply Agrawal’s theorem (here m=1)

and color!

12 24 15 13 35 14 23 25 45 34 13 12 25 14 15 45 34 23 24 35 23 14 12 34 13 15 24 35 25 45

12 24 15 13 35 14 23 25 45 34 13 12 25 14 15 45 34 23 24 35 23 14 12 34 13 15 24 35 25 45

c(1,1) c(2,1) c(3,1)

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As an aside, we can use just k(k-1)/2 colors to color the blocks of X so that no two edges in a block has the same color.

Use distinct colors for each row of the array of Y, it means use same color c(i,j) for j= 1,2…, m.

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Now some results specific to block size 3.

In view of the previous theorem, we need to consider the cases where the index is not divisible by k(k-1)/2 = 3.

Given that index has to be greater than 3, we need to the case consider of =4 first.

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A BIBD is simple if no two blocks have exactly the same points.It is known that simple BIBD(v,3, ) exists if the necessary conditions are satisfied, in particular it is known that simple BIBD(v,3,2) exists for all v greater than 3 for which a BIBD(v,3,2) exists.That is for all v>3 and v=0,1 (mod 3), a simple BIBD(v,3,2) exist.

Theorem: If a simple BIBD(v,3,2) exists, then a properly edge colored BIBD(v,3,4) exists.

Let us see how the proof works with an example:

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Consider simple BIBD(4,3,2), say X: the blocks are {1,2,3},{1,2,4}, {1,3,4}, {2,3,4}.

1 1 1 2 2 2 3 3 3 4 4 4

X =

Y=(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (2,4) (1,3) (2,3)

Note that we have reordered the pairs so that the first and second rows have all distinct pairs.This is possible because the blocks of ordered pairs have such two systems of distinct representatives. The proof is based on counting and using the principle of inclusion and exclusion.

We will write the blocks as columns and create Y, a design on unordered pairs, as well.

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1 1 1 2 2 2 3 3 3 4 4 4

X =

Y=(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (2,4) (1,3) (2,3)

Change the third row of Y and create a fourth row as follows: If Row 3 contains same pair twice,retain one copy and put the second copy directly below into Row 4.(for example , here pair (2,3))Retain those pairs in Row 3 which are not in Row 1,(For example (1,3))If the pair is in Row 1, put it in Row 4. (For example (2,4))

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (1,3) (2,4) (2,3)

M=

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1 1 1 2 2 2 3 3 3 4 4 4

X =

Y=(1,2) (1,4) (3,4) (2,4)(1,3) (1,2) (1,4) (3,4)(2,3) (2,4) (1,3) (2,3)

(1,2) (1,4) (3,4) (2,4)(1,3) (1,2) (1,4) (3,4)(2,3) (1,3)

(2,4) (2,3)

M=

Now we double the array , using the same rows of M but different order:Use Row 3 of M as the first row, then use Row 4, Row 1, and last Row 2.

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (1,3) . (2,4) (2,3)

(2,3) (1,3)(2,4) (2,3)

(1,2) (1,4) (3,4) (2,4)(1,3) (1,2) (1,4) (3,4)

M*=

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1 1 1 2 2 2 3 3 3 4 4 4

X =

Y=(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (2,4) (1,3) (2,3)

Note these two copies give the required colored BIBD(4,3,4).

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (1,3)

(2,4) (2,3)

M=

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (1,3) (2,4) (2,3)

(2,3) (1,3) (2,4) (2,3)

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4)

M*=

Now color the first row of M* by, say, red,

second row by,say, blue, etc.

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4) (2,3) (1,3) (2,4) (2,3)

(2,3) (1,3) (2,4) (2,3)

(1,2) (1,4) (3,4) (2,4) (1,3) (1,2) (1,4) (3,4)

M*=

r b g p

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Corollary: There exists a BIBD(v,3,8) with a proper coloring.

Proof: A simple BIBD(v,3,2) exists for every v for which a BIBD(v,3,8) exists. Use two disjoint sets of four colors in the colorings obtained in the theorem.

This theorem and corollary together with the results for general block size k are enough to prove:

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Theorem: The necessary conditions are sufficient for properly colored BIBD(v,3, ).

Proof: Recall,the necessary (and sufficient) conditions for the existence of a BIBD(v,3 are (i) if = 1,5 (mod 6) then v = 1,3 (mod 6), (ii) if = 2,4 (mod 6) then v = 0,1 (mod 3), (iii) if = 3(mod 6) then v = 1 (mod 2), (iv) if = 0 (mod 6) then v ≥ 3.

For proper coloring additional and obvious necessary condition is that index be greater than 2.

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so the first theorem applies as BIBD(v,3,1) exists, hence we can get a properly colored BIBD(v,3, ).

but in such a situation the index is a multiple of 3= k(k-1)/2 and so proper coloring exists.

Here we can use a properly colored BIBD(v,3,4) or a BIBD(v,3,8) along with a properly colored BIBD(v,3,6t) to get the required colored designs.

For = 2,4 (mod 6), we know v = 0,1 (mod 3).

For = 3(mod 6), we have v = 1 (mod 2), and = 0 (mod 6) any v ≥ 3 is allowed,

For = 1,5 (mod 6), v = 1,3 (mod 6),

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A Path design P(v,k is a collection of ordered subsets, blocks, of a set V of size v where a block <1,2,3,…k> is said to contain unordered pairs (1,2), (2,3), …, (k-1,k), such that each unordered pair of distinct elements from V occurs in blocks.

M.L. Yu posed an open question as per Heinrich in 1991 on a paper on Path Decompositions.For k=3, the question translates to: Is it possible to decompose the edges of two copies of Kv into paths of length 2 such that the edges have different colors?

We can answer this special case:

A path design is balanced if every point appears in the same number of blocks. (Handcuffed designs)

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Theorem: The necessary conditions are sufficient for the existence of a Path(v,3,2) with a proper coloring.

If a block contains a green edge {a,b} and points {a,b,c}, then order the block as <a,c,b>.

Proof: Suppose the blocks of a BIBD(v,3,3) are properly colored with red, blue and green.

One can check that these ordered blocks give a bi-colored Path(v,3,2).

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For index 3, our proof that a proper coloring exist for BIBD(v,3,3) does not depend on which design we have used. (Unlike the case when the index is 4 or 8)Therefore the result for Path design can be expressed as

Theorem: For every BIBD(v,3,3) there is a balanced Path(v,3,2) which can be tightly embedded into it.

(In other words the blocks of every BIBD(v,3,3) can be ordered to give a Path design.)

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Now replace blocks <2,3,4> and <4,1,3> covering pairs (2,3), (3,4), (4,1) and (1,3) with blocks <2,3,1> and <3,4,1> covering the same pairs and the new Path design is not embedded in a BIBD.

The converse is false:

Path design Path(5,3,2) with blocks<1,2,3>, <2,3,4>, <3,4,5>, <4,5,1>, <5,1,2>, <1,3,5>, <4,1,3>, <2,4,1>, <4,2,5>, <2,5,3>gives a BIBD(5,3,3) when we consider these blocks as unordered. (Embedded design)

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Bhaskar Rao Designs

A c-BRD(v,k, ) is a {0,1,-1}-matrix with v rows and b columns such that

1. Every pair of distinct rows has standard inner product equal to c, and

2. If the -1’s in the matrix are changed to +1’s, the matrix becomes the incidence matrix of a BIBD(v,k, ).

The incidence matrix of a BIBD is a {0,1} matrix whose rows are indexed by the points of V and whose columns are indexed by the blocks of the BIBD. If point ai belongs to block bj, then entry M(i,j)=1; otherwise M(i,j)=0.

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The process of changing certain 1’s in M to -1 in order to create a c-BRD is called signing.

The next theorem connects Path designs, Colored designs and BRDs.

Example:

Incidence matrix of a BIBD(4,3,2) Corresponding 0-BRD

1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 1

1 1 1 0 1 -1 0 1 1 0 -1 -1 0 1 -1 1

1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 1

1 1 1 0 1 -1 0 1 1 0 -1 -1 0 1 -1 1

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Theorem: The incidence matrix of every BIBD(v,3,3) can be signed so as to create a (-1)-BRD(v,3,3).

Proof: Let X denotes any BIBD(v,3,3).

Let us order the points in each block of X to get Path design Path(v,3,2), say Y.

As each block <a,b,c> of Y corresponds to a block {a,b,c} of X under the tight embedding,change the entry in the incidence matrix M of X corresponding to the point b and this block by -1.

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As the index for the Path design is two, the points a and c must occur as neighbors in two Path design blocks where either a or c will be the center points and therefore the corresponding entries in M will have the minus sign. So the inner product of rows a and c will be -1.

Recall the signing rule: A block <a,b,c> of Y corresponds to a block {a,b,c} of X under the tight embedding,change the entry of the incidence matrix M for X corresponding to the center point b and this block by -1.

To check that inner product of any two rows is -1, calculate the inner product of rows corresponding to two arbitrary points a and c.

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An Edge-Coloring Problem for Designs

Joint work with

These are some results we have on

Spencer P. Hurd