STS

Document Sample

```					                                                                                         1

Eduardo Mariscal
Professor Kundgen
Math 540
Due: December 9, 2004
Steiner Triple System

A Steiner Triple System, also know as STS, is a (v,3,1)-design with order v where V is the
set of varieties of size v, k = 3 is the number of element in each set, and λ= 1 is the
number of blocks a pair of distinct elements x,yV appears in STS. If a STS exist then
r = l(v-1)/(k-1), if we substitute l=1 and k=3 we get r = 1(v-1)/(3-1)=(v-1)/2. In order to
find the b lets use the formula b = vr/k=v((v-1)/2)/3)=v(v-1)/6. So we get that every
element xÎV appears in r = (v-1)/2 blocks and we have b = v(v-1)/ 6 blocks in a STS(v).

Example 1
We will show how you can construct the following designs later, but these are examples
of a STS(v):

a) (7,3,1)-design: {124}, {235}, {346}, {457}, {561}, {672}, and {731}. (7,3,1)-design
has every element appear in r = (7-1)/2=3 blocks and has b=7(7-1)/6=7 blocks.

b) (9,3,1)-design: {123}, {456}, {789}, {147}, {258}, {369}, {159}, {357}, {348},
{267}, {249}, and {168}. (9,3,1)-design has every element appear in
r = (9-1)/2=4 blocks and has b=9(9-1)/6=12 blocks.

c) (13,3,1)-design: {1,2,5}, {2,3,6}, {3,4,7}, {4,5,8}, {5,6,9}, {6,7,10}, {7,8,11},
{8,9,12}, {9,10,13}, {10,11,1}, {11,12,2}, {12,13,3}, {13,1,4}, {1,3,9}, {2,4,10},
{3,5,11}, {4,6,12}, {5,7,13}, {6,8,1}, {7,9,2}, {8,10,3}, {9,11,4}, {10,12,5},
{11,13,6}{12,1,7}, and {13,2,8}. (13,3,1)-design has every element appear in
r = (13-1)/2=6 blocks and has b=13(13-1)/6=26.

Since we cannot have a numbers of fractional blocks and x cannot appear in a fractional
number of blocks so we must have that b and r are integers, so then v must be odd, let v =
2r +1 then b=(1/3)r (2r+1), so it follows that r  0 or 1 (mod 3). Now we get that r = 3m
or r = 3m +1 by the definition of congruency, then we get that v = 6m +1 or v=6m+ 3. In
order to construct a STS we need that v  1 or 3 (mod 6), so either v =6m+1 or v = 6m +
3. See example 1a) for STS(7) 7=6(1)+1 for m=1 and for example 1b) for STS(9),
9=6(1)+3 for m=1.

Next let’s recall the concept of a difference set. This will helps us understand the
constructions of STS(v).

Definition 1
A (v,k,λ) difference set (mod v) or a cyclic (v,k,λ) difference set is a set D={d1,d2,…,dk}
of distinct elements of Zv such that each non-zero dZv can be expressed in the form d=
di-dj in precisely λ ways.
2

Example 2
{124} is a cyclic (7,3,1) difference set in Z7 , since 1-2=-16(mod 7),1-4=-34(mod 7),
2-1=11(mod 7), 2-4=-25(mod 7), 4-1=33(mod 7), and 4-2=22(mod 7).

Let’s generalize Definition 1.

Definition 2
Let D1,…,Dm be sets of size k in Zv such that the differences arising from the Di give
each non-zero element of Zv exactly λ times. Then D1,…,Dm are said to form a cyclic
(v,k,λ) difference system.

Example 3
{1,2,5} and {1,3,9} form a (13,3,1) difference system in Z13, since
1-2=-112(mod 7), 1-5=-49(mod 7), 2-1=11(mod 7), 2-5=-310(mod 7),
5-1=44(mod 7), 5-2=33(mod 7), 1-3=-211(mod 7), 1-9=-85(mod 7),
3-1=22(mod 7), 3-9=-67(mod 7), 9-1=88(mod 7), and 9-3=66(mod 7).

We can think of a cyclic difference system as more than one cyclic different set.

Observe
If D1,…,Dm is a cyclic difference system, then {Di + j: for 1£ i £m and 0£ j £v-1} is a (v,
k, l)-design.

Proof: Is similar to the proof for difference sets.

Example 4
a) Remember example 2, {124} is a cyclic difference set in Z7, such that m=1, {124} will
generate the (7,3,1)-design. We can generate a block by adding 0
(mod 7) to the block and another block by adding 1 (mod 7), and so on until we add 6
(mod 7). We will get this {1+0,2+0,4+0}={1,2,4}={124}, {1+1,2+1,4+1} ={2,3,5}=
{235}, {1+2,2+2,4+2}={3,4,6}={346}, and so on, we will get next {457}, {561}, {672},
example 1a).

b) Remember example 3, {1,2,5} and {1,3,9} is a cyclic difference system in Z13, such
that m = 2, {125} and {139} will generate the (13,3,1)-design. We can generate two
blocks by adding 0 (mod 13) to the cyclic blocks and another two more block by adding 1
(mod 13) to the cyclic blocks, and so on until we add 12 (mod 13) which is 0 to both
blocks. We will get that cyclic {1,2,5} and {1,3,9} different system will generate 13
blocks each, and they are
{1+0,2+0,5+0}={1,2,5}       similarly {1+0,3+0,9+0}={1,3,9}
{1+1,2+1,5+1}={2,3,6}                 {1+1,3+1,9+1}={2,4,10}
{1+2,2+2,5+2}={3,4,7}                 {1+2,3+2,9+2}={3,5,11}
{1+3,2+3,5+3}={4,5,8}                 {1+3,3+3,9+3}={4,6,12}
{1+4,2+4,5+4}={5,6,9}                 {1+4,3+4,9+4}= {5,7,13}
{1+5,2+5,5+5}={6,7,10}                {1+5,3+5,9+5}= {6,8,1}
3

{1+6,2+6,5+6}={7,8,11}                 {1+6,3+6,9+6}= {7,9,2}
{1+7,2+7,5+7}={8,9,12}                 {1+7,3+7,9+7}= {8,10,3}
{1+8,2+8,5+8}={9,10,13}                {1+8,3+8,9+8}= {9,11,4}
{1+9,2+9,5+9}={10,11,1}                {1+9,3+9,9+9}= {10,12,5}
{1+10,2+10,5+10}={11,12,2}             {1+10,3+10,9+10}= {11,13,6}
{1+11,2+11,5+11}={12,13,3}             {1+11,3+11,9+11}={12,1,7}
{1+12,2+12,5+12}={13,1,4}              {1+12,3+12,9+12}= {13,2,8}
How can we construct STS(v) in general? Well if you have a cyclic difference system
you can construct the designs. I just show how you can construct a design but I still need
to show how to find a cyclic difference system. But first let’s give a definition.

Definition 3
A design obtained in this fashion from a difference system is called a cyclic design or in
our case since we are studying STS(v) we can call them a cyclic STS(v).

Remark: Blocks are partitioned into “cycles.”

For which values of v can there be a cyclic STS(v)? Let m be the number of blocks in the
difference system requires completing the difference set. The m blocks will generate all
the b blocks by definition when you add (0,1,2,…,v-1) to the m blocks, so mv=b. Since
b=v(v-1)/6 then mv=v(v-1)/6, thus the number of blocks in the difference system is
m=(v-1)/6 blocks, so the only way we can construct a cyclic STS(v) is if v=6m+1. This
means that no cyclic difference system can be found for a cyclic STS(v) for v=6m+3
according to our definition.

Example 5
a) STS(7)=STS(6(1)+1) with m=1 cyclic difference set {124}.

b) STS(13)=STS(6(2)+1) with m=2 cyclic difference system {125} and {139}.

We showed in lecture 17 that a STS(v) exist for v=6m+3. So if we can show a cyclic
STS(6m+1) exists for all m³1 then we showed
a) Cyclic STS(v) exists for all vº1(mod 6), and
b) STS (v) exists for all vº1 or 3(mod 6).

We will study the existence of cyclic STS(6m+1). Our final goal is to prove that a cyclic
STS(6m+1) exist for all m1. In order to prove this lets start with a basic theorem.

Theorem 1
Suppose that {1,…,3m} can be partitioned into m triples {a,b,c} such that a+b=c or
a+b+c0 (mod 6m+1). Then the m triples {0,a,a+b}, also know as base blocks, form a
(6m+1,3,1) difference system and so lead to the construction of a cyclic STS(6m+1).

Proof: In order to complete a difference system we need that difference of the elements
in the blocks is equal to all possible non-zero elements in Z6m+1. A set of the form
4

{0,a,a+b} has a differences 0-a=-a, 0-(a+b)=-(a+b), a-0=a, a-(a+b)=-b, (a+b)-0=(a+b),
and (a+b)-a=b, so we get differences a, b, and (a+b). If a+b=c, then ±(a+b)=±c. If
a+b+cº 0 (mod 6m+1) then a+b=-c, so ±(a+b)º c(mod 6m+1). Either way we get the

+|
differences ±a, ±b, and ±c. Altogether the differences are just the non-zero elements
1,…,3m we will get Z6m+1, since we had a partition of {1, 2, …, 3m}.

Example 6
Let’s see how to find a cyclic difference system, using theorem 1.
a) For (7,3,1)-design we get that m=1 since v=6(1)+1 then we have a the set
{1,2,…,3m}={1,2,3}, since c=3=1+2=a+b so then {0,a,a+b}={0,1,3} is a cyclic
difference set, if we add 1 to the set, we get {0+1,1+1,4+1}={1,2,4}={124} where this
block is equal to the one in example 4a. Thus by showing how to find the cyclic
difference set we have shown how to construct a cyclic STS(7).

b) For (13,3,1)-design we get that m=2 since v=6(2)+1 then we must have 2 sets in the
cyclic difference system. We see that {1,2,…,3m}={1,2,3,4,5,6}. Here is where we start
to play it with since what we do is trial and error. We have that at least 1 and 2 are not in
the same set since we are going to be force to have 3 be the next element and we cannot
have that because the next block will be force to have {456} and 4+5≠6 and 4+5+6≠13,
so we can start with {1, , } and {2, , }, lets guess b=3 for the first set so we get
a+b=1+3=4=c and we are left with 2, 5, and 6 so 2+5≠6 but 2+6+5=13º0(mod 13). Now
{1,3,4}correspond to the difference set {0,a,a+b}={0,1,4}, similarly {2,5,6} corresponds
to the difference set {0,a,a+b}={0,2,7}. If we add 1 to both blocks we get that
{0+1,1+1,4+1}={1,2,5} and {0+1,2+1,7+1}={1,3,8}where this block do not equal to the
one in example 4b. But if a=2 and b=6 then this corresponds to difference set
{0,a,a+b}={0,2,8}. If we add 1 to both blocks we get that {0+1,2+1,8+1}={1,3,9}where
this block is equal to the one in example 4b. Thus by showing how to find the cyclic
difference system we have shown how to construct a cyclic STS(13).

Definition 4
A Skolem triple system of order m is a partition of {1,…,3m} into m triples {i, ai, i+ai},
1im. By theorem 1 this yields a cyclic STS(6m+1)

Example 7
a) When m=4, the triples {1,5,6}, {2,8,10}, {4,7,11}, and {3,9,12} form a partition of
{1,…,12}. As we can see all the elements {1,…,12} are partition into triples, every
element appears in a block once, and the sum of 1+5=6, 2+8=10, 4+7=11, and 3+9=12 so
we have a Skolem triple system of order 4 and hence yield a cyclic STS(25). The base
blocks for the designs are {0,1,6}, {0,2,10}, {0,4,11}, and {0,3,12}

b) When m=5, the triples {1,14,15}, {2,7,9}, {3,10,13}, {4,8,12}, and {5,6,11} form a
partition of {1,…,15}. Similarly as we can see all the elements {1,…,15} are partition
into triples, every element appears in a block once, and the sum of 1+14=15, 2+7=9,
3+10=13, 4+8=12, and5+6=11 so we have a Skolem triple system of order 5 and hence
yield a cyclic STS(31). The base blocks for the designs are {0,4,15}, {0,2,9}, {0,3,13},
{0,4,12}, and {0,5,11}.
5

Theorem 2
If mº 0 or 1 (mod 4) then a Skolem triples system of order m exists.

Proof: If m = 4k then we claim that the triples
A: 1, 12k-1, 12k
B: 2r+1, 10k-r-1, 10k+r             1 £ r £ k-1
C: 2k+2r-1, 5k-r, 7k+r-1             1 £ r £ k-1
D: 4k-1, 5k, 9k-1
E: 2r, 6k-r, 6k+r                    1 £ r £ k-1
F: 2k, 8k-1, 10k-1
G: 2k+2r, 9k-1-r, 11k+r-1            1 £ r £ k-1
H: 4k, 6k, 10k
form a Skolem triple system. Note that, if k=1, we omit those triples defined for 1£r£k-1;
the remaining triples are {1,11,12}, {3,5,8}, {2,7,9}, and {4,6,10} which form a Skolem
triple system of order 4 different from the one in example 8a. The base blocks for the
designs are {0,1,12}, {0,3,8}, {0,2,9}, and {0,4,10}.

Now we must verify that this is a Skolem triple system. Let first add all the triples and
we get
A: 1+(12k-1)=12k+1-1=12k
B: (2r+1)+(10k-r-1)=10k+2r-r-1+1=10k+r
C: (2k+2r-1)+(5k-r)=2k+5k+2r-r-1=7k+r-1
D: (4k-1)+5k=4k+5k-1=9k-1
E: 2r+(6k-r)=6k+2r-r=6k+r
F: 2k+(8k-1)=10k-1
G: (2k+2r)+(9k-1-r)=2k+9k+2r-r-1=11k+r-1
H: 4k+6k=10k

Now lets count the number of triples to make sure it is m=4k. We get a triple each from
A, D, F, and H. So far we have 4. We get k-1 triples each from B, C, E, and G. If we
add them together we get 4 + 4(k-1)=4 + 4k -4=4k=m triples.

Now lets verify that all the element in the {1,…,3m} are in the triples at least once
The letters will tell you from what triple the element or elements belong

1,                                                   A
2, 4, 6,…,2k-2                                       E
3,5,7,…,2k-1                                         B
2k,                                                  F
2k+1, 2k+3, 2k+5,…,4k-3                              C
2k+2, 2k+4, 2k+6,…,4k-2                              G
4k-1                                                 D
4k                                                   H
5k-1, 5k-2, 5k-3,…4k+2,4k+1                          C
5k                                                   D
6

6k-1, 6k-2, 6k-3,…,5k+2,5k+1                        E
6k                                                  H
6k+1, 6k+2, 6k+3,…,7k-1                             E
7k, 7k+1, 7k+2,…,8k-2                               C
8k-1                                                F
9k-2, 9k-3, 9k-4,…,8k+1,8k                          G
9k-1                                                D
10k-2,10k-3, 10k-4,…,9k+1,9k, 9k-1,9k-2             B
10k-1                                               F
10k                                                  H
10k+1, 10k+2, 10k+3,…,11k-1                          B
11k, 11k+1, 11k+2,…,12k-2                            G
12k-1                                               A
12k                                                  A

So we can see that all the element from {1,2,…,3m}={1,2,….12k} are in a triple at least
once. But since we have m triples each number appears exactly once and we have a
partition.

If m = 4k+1 then the triples
1, 12k+2, 12k+3
2r+1, 10k-r, 10k+r+1             1 £ r £ k-1
2k+2r-1, 5k-r+1, 7k+r            1 £ r £ k-1        NOTE: 4k+1=m triples, sum (check),
4k-1, 5k+1, 9k                                           partition is similarly as for m=4k.
4k+1, 8k, 12k+1
2r, 6k+1-r, 6k+1+r               1 £ r £ k-1
2k, 10k, 12k
2k+2r, 9k-r, 11k+r               1 £ r £ k-1
4k, 6k+1, 10k+1
form a Skolem triple system. Note that, if k=1, we omit those triples defined for 1£r£k-1;
the remaining triples are {1,14,15}, {3,6,9}, {5,8,13}, {2,10,12}, and {4,7,11} which
form a Skolem triple system of order 5 different from the one in example 8b. The base
blocks for the designs are {0,1,15}, {0,3,9}, {0,5,13}, {0,2,12}, and {0,4,11}.

Now let’s defined an O’Keefe triple system.

Definition 5
An O’Keefe triple system of order m is a partition of {1,…,3m-1;3m+1} into m triples
{i,ai,i+ai}, for 1 £ i £ m.

Observe
If we have an O’Keefe triple system of order m then we have a block {i, ai, ai +i } in
which 3m+1=ai+i. Replacing 3m+1 by 3m we get {i, ai, 3m }. Now we have partition of
{1,…,3m} and i+ai=ai+i except for special block where i + ai + 3m = (3m+1)+3mº0 (mod
6m+1) so by Theorem 1 we have a cyclic STS(6m+1).
7

Example 9
a) When m=2, the triples {134} and {257} form an O’Keefe system of order 2 and a
partition of {1,…,5;7} and hence yield a cyclic STS(13). As we can see all the elements
{1,…,5;7} are partition into triples, every element appears in a block once, and the sum
of 1+3=4, and 2+5=7. The base blocks for the designs are {0,1,4}, {0,2,7}.

b) When m=3, the triples {1,4,5}, {2,6,8}, and {3,7,10} form an O’Keefe system of order
3 and a partition of {1,…,8;10} and hence yield a cyclic STS(19). As we can see all the
elements {1,…,8;10} are partition into triples, every element appears in a block once, and
the sum of 1+4=5, 2+6=8, 3+7=10. The base blocks for the designs are {0,1,5}, {0,2,8},
and {0,3,10}.

c) When m=6, the triples {1,16,17}, {2,8,10}, {3,12,15}, {4,7,11}, {5,9,14}, and
{6,13,19} form an O’Keefe system of order 6 and a partition of {1,…,17;19} and hence
yield a cyclic STS(37). As we can see all the elements {1,…,17;19} are partition into
triples, every element appears in a block once, as you can see {i,ai,i+ai}, for 1 £ i £ m.
The base blocks for the designs are {0,1,17}, {0,2,10}, {0,3,15}, {0,4,11}, {0,5,14}, and
{0,6,19}.

d) When m=7, the triples {1,9,10}, {2,17,19}, {3,11,14}, {4,16,20}, {5,8,13}, {6,12,18},
and {7,15,22} form an O’Keefe system of order 7 and a partition of {1,…,20;22} and
hence yield a cyclic STS(43). As we can see all the elements {1,…,20;22} are partition
into triples, every element appears in a block once, as you can see {i,ai,i+ai}, for 1 £ i £ m.
The base blocks for the designs are {0,1,10}, {0,2,19}, {0,3,14}, {0,4,20}, {0,5,13},
{0,6,18}, and {0,7,22}

Theorem 3
If mº2 or 3 (mod4) then an O’Keefe triple system of order m exists.

Proof: We do not have to worry for m<10 since have shown in the examples that these
exist. If m=4k+2 with k³2 the triples
1, 11k+6, 11k+7
2r+1, 10k+4-r, 10k+r+5        1£r£k
2k+3, 10k+4, 12k+7                    NOTE:12k+7 is the biggest element for m=4k+2
2k+3+2r, 9k+4-r, 11k+r+7      1 £ r £ k-2
4k+1, 6k+4, 10k+5
2r, 6k+4-r, 6k+4+r             1 £ r £ 2k+1

form an O’Keefe system with k£2.

If m=4k+3 the triples
1, 9k+7, 9k+8
2r+1, 10k+8-r, 10k+9+r          1 £ r £ k-1
2k+1, 10k+9, 12k+10                     NOTE: 12k+10 is the biggest element for m=4k+3
2k+1+2r, 9k+7-r, 11k+8+r        1£r£k
4k+3, 6k+5, 10k+8
8

2r, 6k+5-r, 6k+5+r              1 £ r £ 2k+1
It can be shown that this works for all m³1:

The previous proofs can be combined to find the cyclic difference system for all
STS(6m+1).

Theorem 4
A cyclic STS(6m+1) exists for all m³1.

Proof: If mº0 or 1 (mod 4), then theorem 2 yields the triples of a Skolem system of order
m. If mº2 or 3 (mod 4), then theorem 3 yields the triples of a O’Keefe system of order m.

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 11 posted: 11/27/2011 language: English pages: 8