Part 22
Combinatorial Designs
Printed version of the lecture Discrete Mathematics on 16. November 2009
Tommy R. Jensen, Department of Mathematics, KNU
22.1
Contents
1
Modular Arithmetic
1
2
Block Designs
5
3
Conclusion
7
22.2
1
Modular Arithmetic
Modular addition and multiplication
Definition
Let n be a positive integer. We write Zn = {0, 1, 2, . . . , n − 1}.
Division with remainder
If m is any integer, then there are uniquely determined numbers q and r such that
m = q · n + r and r ∈ Zn .
The number r is the remainder of m after division by n.
Definition
If a, b are two elements of Zn , we define a ⊕ b and a ⊗ b by:
a ⊕ b is the remainder of a + b after division by n, and a ⊗ b is the remainder of a · b after division by n.
If n = 7, then 3 ⊕ 4 = 0 and 3 ⊗ 4 = 5.
22.3
1
Inverses
Definition
Let n ≥ 2 and let a be an element of Zn .
An additive inverse of a is an element b of Zn such that
a ⊕ b = 0.
A multiplicative inverse of a is an element c of Zn such that
a ⊗ c = 1.
Additive inverses
If a is 0, then a ⊕ 0 = 0, so 0 is an additive inverse of a.
If a is not 0, then a ⊕ (n − a) = 0, so n − a is an additive inverse of a.
For each a in Zn there is only one additive inverse, we can write −a to denote the additive inverse of a.
22.4
Multiplicative inverses
Existence question
If a = 0, then a ⊗ b = 0 ⊗ b = 0 for all b in Zn . So 0 has no multiplicative inverse.
If n = 4, then 1 ⊗ 1 = 1 and 3 ⊗ 3 = 1, so 1 and 3 have themselves as multiplicative inverses.
But 2 has no multiplicative inverse in Z4 .
How can we know whether an element of Zn has a multiplicative inverse or not?
Definition
The greatest common divisor gcd(a, b) of two positive numbers a and b is defined as the largest number d
such that both a and b are multiples of d.
Examples: gcd(3, 6) = 3 and gcd(12, 20) = 4.
22.5
Theorem 10.1.2
Let n ≥ 2 and let a be an element of Zn .
Then a has a multiplicative inverse in Zn if and only if
gcd(n, a) = 1.
Proof of “only if”
Assume that a has a multiplicative inverse c; then a ⊗ c = 1.
Let q be the integer such that a · c = q · n + 1.
Let d = gcd(n, a).
Then a · c is a multiple of d, because a is a multiple of d.
Also q · n is a multiple of d, because n is a multiple of d.
It follows that 1 = a · c − q · n is a multiple of d.
This implies d = 1, since 1 is not a multiple of any integer larger than 1.
This shows gcd(n, a) = 1.
22.6
2
Proof of “if”
Assume that gcd(n, a) = 1.
We have to show: (∗) There exists an element c of Zn and an integer q such that
a · c = q · n + 1.
We apply induction on n to prove (∗).
If a = 1, then we let c = 1 and q = 0. Then a · c = 1 and q · n + 1 = 1.
This proves (∗) for n = 2, since Z2 = {0, 1}, so a = 1 is the only element with gcd(n, a) = 1.
22.7
Induction step
Now we assume n > 2 and the statement (∗) is true for all smaller numbers. We may also assume a ≥ 2.
We use Division with Remainder on n and a :
n = x · a + y, where 0 < x ≤ n/a and 0 < y < a.
Let d = gcd(a, y). Since a and y are multiples of d, and since n = x · a + y, it follows that n is a multiple
of d.
We know that 1 is the largest number for which both n and a are its multiples. This implies d = 1. So
gcd(a, y) = 1.
Since 2 ≤ a < n, the statement (∗) is true for a and y. Let c and q be integers with c ∈ Za = {0, 1, . . . , a −
1} such that
y · c = q · a + 1.
22.8
Induction step
Using division we found x and y such that
n = x · a + y, where 0 ≤ x < n/a and 0 < y < a,
and by the induction hypothesis we found c and q such that
y · c = q · a + 1, where 0 ≤ c < a.
We calculate 1
=
y·c−q·a
=
(n − x · a) · c − q · a
=
n · c − (x · c + q) · a
Now we have −(x · c + q) · a = −c · n + 1, but −(x · c + q) is not in Zn .
So let m be the largest number such that m · n − (x · c + q) < n.
Then c0 = m · n − (x · c + q) is in Zn , and c0 · a = (m · a − c) · n + 1.
It follows by induction that (∗) is true for all n ≥ 2.
22.9
3
Example
Find the multiplicative inverse of 11 in Z28 , if it exists.
We calculate:
28
=
2 · 11 + 6
11
=
1·6+5
6
=
1·5+1
5
=
5·1+0
And now backwards:
1
=
6−1·5
=
6 − 1 · (11 − 1 · 6) = −1 · 11 + 2 · 6
=
−1 · 11 + 2 · (28 − 2 · 11) = 2 · 28 − 5 · 11.
Then −5 · 11 = −2 · 28 + 1 implies (28 − 5) · 11 = (11 − 2) · 28 + 1 ⇔ 23 · 11 = 9 · 28 + 1.
Therefore the multiplicative inverse of 11 in Z28 is equal to 23.
22.10
Algebraic fields
Corollary 10.1.3
Every nonzero element of Zn has a multiplicative inverse if and only if n is a prime number.
Proof of Corollary 10.1.3
Since n is a prime number if and only if n is not divisible by any of the numbers 2, 3, . . . , n − 1, the statement
follows directly from Theorem 10.1.2.
Fields
The addition ⊕ and the multiplication ⊗ defined on the elements of Zn obey the usual rules for addition and
multiplication.
If n is a prime number p, since every nonzero element of Z p has a multiplicative inverse, we call
(Z p , ⊕, ⊗) a field.
22.11
Finite fields
Method to construct other fields
To construct a field with pk elements you will have to:
Find a polynomial P(x) = a0 + a1 x + a2 x2 + · · · + ak−1 xk−1 + xk of degree k, such that
• the coefficients of P are in Z p and addition and multiplication in P(x) are ⊕ and ⊗,
• there are no polynomials Q(x) and R(x) both having positive degree, such that P(x) = Q(x)R(x),
then add a new element i to Z p such that P(i) = 0.
In the new set Z p [i],
a2 i2 + · · · + ak−1 ik−1 ).
calculate as normal, but every time a term ik appears, replace it by −(a0 + a1 i +
22.12
4
2
Block Designs
A testing problem
Example: A company that manufactures seven different kinds of soju wants to export one kind of their
soju to Europe.
They do not know which kind of soju will be liked by european people, so first they have to test all
seven kinds.
The test can be designed in different ways, for example:
Seven european testpersons are chosen, and each tests one bottle of soju.
Tester \ Type
1
2
3
4
5
6
7
0
x
1
2
3
4
5
6
x
x
x
x
x
x
Problem: If all testpersons say “it was good”, then you do not know which is the best. You need
testpersons to make comparisons between different types.
22.13
Another possible design of the test:
Only one person will test all seven bottles together.
Tester \ Type
1
0
x
1
x
2
x
3
x
4
x
5
x
6
x
New problem: Probably a non-korean person will be confused after drinking seven bottles of soju. The
answer to the test could get unreliable.
22.14
A third possible design:
For each way of choosing two types to compare, choose one testperson to test both types together.
Since 72 = 21, there will be 21 testpersons.
Tester \ Type
1
2
3
4
..
.
21
0
x
x
x
x
1
x
2
3
4
5
6
x
x
x
x
x
..
.
Problem: Although this looks better, the large number of testpersons makes the test expensive.
22.15
5
Finally, if we can make each testperson compare three types of soju together, then the test may not
require as many testpersons, but it will still give good information about comparisons.
We should design the experiment so that any two different kinds of soju get compared by at least on
testperson.
The table for the test can be constructed like this:
Tester \ Type 0 1
1
x x
2
x
3
x
4
x
5
x
6
7
2
x
3
4
x
x
x
5
6
x
x
x
x
x
x
x
x
x
x
x
In set notation, we have selected seven 3-subsets of Z7 = {0, 1, 2, 3, 4, 5, 6} :
{0, 1, 2}, {0, 3, 4}, {0, 5, 6}, {1, 3, 5}, {1, 4, 6}, {2, 3, 6}, {2, 4, 5}.
Every 2-subset of Z7 is a subset of exactly one of them.
22.16
Block designs
Definition
Let V be a set of v elements, called varieties.
Let k be a number with 0 < k ≤ v.
Let B = {B1 , B2 , . . . , Bb } be a multiset of k-subsets of V , called blocks.
Then B is a balanced block design if there is a number λ such that every 2-subset of V is a subset of
precisely λ of the blocks.The number λ is called index of the design.
If B = {V,V, . . . ,V } (b = λ times), then B is called complete (not very interesting).
In any other case B is called balanced incomplete block design.
The expression ‘Balanced Incomplete Block Design’ is almost always shortened to BIBD.
22.17
Number of blocks with a given element
Lemma 10.2.1
Let B be a BIBD with parameters v, k, λ on a set V.
Then every element of V is an element of precisely r blocks of B, where
r=
λ (v − 1)
.
k−1
Proof of Lemma 10.2.1
Let x be any element of V.
Since V has v − 1 other elements, the number of 2-subsets of V containing x is equal to (v − 1).
Each 2-subset containing x appears in λ blocks. So there are λ (v − 1) 2-subsets containing x present in
the blocks.
The blocks of B are k-subsets of V, so if one of the blocks contains x, then it contains precisely (k − 1)
of the 2-subsets which contain x.
λ (v−1)
Hence the number of blocks containing x is equal to k−1 .
22.18
6
Some more facts about BIBD’s
Corollary 10.2.2
If B is a BIBD with parameters v, k, λ , r, then
bk = vr.
Proof of Corollary 10.2.2
We have b blocks each with k elements, so the sum of the sizes of the blocks is equal to bk.
We also have v elements in total, and each element belongs to r blocks.
So the sum of the sizes of the blocks is also equal to vr.
22.19
Corollary 10.2.3
If B is a BIBD with parameters v, k, λ , r, then
λ < r.
Proof of Corollary 10.2.3
The block design is not complete, so k < v.
Lemma 10.2.1 implies
r=λ·
v−1
> λ ·1
k−1
22.20
Example
Is there a BIBD with parameters v = 12, k = 4, and b = 16?
If such a BIBD exists, then Corollary 10.2.2 implies
r=
bk
16 · 4 64 16
=
=
= .
v
12
12
3
But the parameter r of a BIBD is a natural number.
So the answer is NO!
22.21
3
Conclusion
Conclusion
This ends the lecture!
7
22.22
Next time:
More BIBD’s
Steiner Triple Systems
22.23
8