组合数学课件Chap10
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A 48 48 48 18 18 6 6 0 B 126 78 30 30 12 12 6 6
10
Theorem 10.1.2
Let n be an integer with n ≥ 2 and let a be a non-zero integer in Zn = {0, 1, 2, …, n-1}. Then a has a multiplicative inverse in Zn iff the greatest common divisor (GCD) of a and n is 1. If a has a multiplicative inverse, then it is unique.
16
Example
• Construction of a field with 4 elements. • We start with Z2 and the polynomial x2+x+1 with coefficients in Z2 . The polynomial has no root in Z2 and cannot be factored in any nontrivial way. We adjoin a root i of this polynomial to Z2, getting i2+i+1 = 0, hence, i2 = i-1=i+1. • The four elements of the field is {a+bi: a, b in Z2} = {0, 1, i, 1+i}. With addition table and multiplication table given in the next page.
Chapter 10 Combinatorial designs
1
Summary
• • • • • Modular arithmetic Block designs Steiner triple systems Latin squares Assignments
2
Modular arithmetic
8
Algorithm to compute the GCD
GCD (a, b): Set A = a and B = b. While A × B ≠ 0, do if A ≥ B, then replace A by A – B. else, replace B by B – A. Set GCD = B. Question: How to compute GCD(a1, a2, a2 ,…, an)?
19
Exercise
• Start with Z2 and show that x3 + x +1 cannot be factored in a non-trivial way, and then use this polynomial to construct a field with 23 = 8 elements. • Continue to do the following computations: 1) (1+i) + (1+i+i2) 2) i2 × (1+i+i2) 3) (1+i)-1.
4
• •
ቤተ መጻሕፍቲ ባይዱ
Example
The simplest case is n = 2. we have Z2 = {0, 1}, and addition and multiplication mod 2 are given in the following tables:
0
1
0
1
0
0 0
1 0
0
1
1
0
3
Some notations
• • • Z: the set of integers {… -2, -1 , 0, 1, 2, …} Zn: the set of non-negative integers {0, 1, 2, …, n-1} which are less than n. If m is an integer, then there exist unique integers q (the quotient) and r (the remainder) such that m = q × n + r, 0 ≤ r ≤ n-1. Addition mod n a b : is the (unique) remainder when the ordinary sum a +b is divided by n Multiplication mod n a b : is the unique remainder when the ordinary product a × b is divided by n.
9
Example
• Compute GCD(48, 126). • By applying the algorithm, the results are displayed as shown in the table. We conclude that the GCD of 48 and 126 is the terminal value d = 6 of B.
15
Field
• The pk elements of the finite field is {a +bi+ci2+…+dik-1: a, b, c, and d in Zp}, where i is a root of the polynomial. • ik can be calculated by the polynomial.
11
Corollary 10.1.3
Let n be a prime number. Then each non-zero integer in Zn has a multiplicative inverse. If GCD (a, b) = 1, then we call a and b relatively prime.
20
Block designs
21
An Example
suppose 7 varieties of products need to be tested by 7 consumers. Each consumer is asked to compare a certain 3 of the varieties. The test is to have a property that each pair of the 7 varieties is compared by exactly one person. Can such a testing experiment be designed?
i2 = 1 + i;
1+i 0
1+i 1
i × (1 +i) = i + (1 + i) = 1+0×i = 1 ;
18
(1+i) ×(1+i) = i2 +(1+1) ×i + 1 = (1+i) + 1 = i
Another example
• Construction of a field of 33 = 27 elements. • We start with Z3 and the polynomial x3+2x+1 with coefficients in Z3 . The polynomial has no root in Z3 and cannot be factored in any nontrivial way. We adjoin a root i of this polynomial to Z3, getting i3+2i+1 = 0, hence, i3 = -2i-1=i+2. • The field with 27 elements is {a+bi+ci2: a, b and c in Z3}. The addition and multiplicative arithmetic satisfy the usual laws of mod 3 arithmetic.
A 30 8 8 B 11 11 3 30=2×11+8 11=1×8+3 8=2×3+2
2
2 0
3
1 1
3=1×2+1
2=2×1+0 d=1
13
Exercise
• Find the multiplicative inverse of 16 in Z45.
14
Fields
• For each prime p and each integer k ≥ 2, there exists a polynomial of degree k with coefficients in Zp which does not have a non-trivial factorization. • Example: for Z2, x3 + x + 1 =0 has no non-trivial factorization in Z2. • The polynomial has a root denoted as i, i.e., i3 + i + 1 = 0.
a b 1
7
Example
• In the integers modulo 10, the additive inverses are as follows: -0 = 0, -1 = 9, -2 = 8, -3 = 7, -4 = 6 -5= 5 , -6 = 4, -7 = 3, -8 = 2, -9 = 1 • The multiplication inverses are as follows: 1-1 = 1, 3-1 = 7, 7-1 = 3, 9-1 = 9, none of 0, 2, 4, 5, 6, and 8 has a multiplication inverse in Z10.
1
0
1
5
Exercises
Construct the addition and multiplication tables for the integers mod 3 and integers modulo 4.
6
Inverse
An additive inverse (denote as – a) of an integer a in Zn is an integer b such that a b 0. A multiplication inverse (denote as a-1) of an integer a in Zn is an integer b in Zn such that
12
Calculate the inverse
Determine if 11 has a multiplicative inverse in Z30 and, if so, calculate the multiplicative inverse. By applying the algorithm, GCD(11, 30) = 1 and the results are displayed in the table. By theorem 10.1.2, 11 has a multiplicative inverse in Z30. We use the equations in the table: 1= 3-1×2=3-1×(8-2×3) = 3×3 – 1×8 =3×(11-1×8) – 1 × 8 = 3×11 – 4×8 = 11×11 - 4×30 , hence, 11-1 = 11 in Z30.
17
Addition table and multiplication table of F4
+ 0 1 i 0 0 1 i 1 1 0 i i 1+i 1+i 1 0 × 0 0 1 i 0 0 0 1 0 1 i i 0 i 1+i 0 1+i i
1+i i 1
1+i 0
1+i 1
1+i 1+i i
22
Analysis for the example
We label the different varities 0, 1, 2, 3, 4, 5, 6. There are c(7, 2) = 21 pairs of the 7 varieties. Each tester gets 3 varieties and thus makes c(3, 2) = 3 comparisons. Since each pair is to be compared exactly once, the number of testers must be 21/3=7. Thus in this case, the design is possible.
10
Theorem 10.1.2
Let n be an integer with n ≥ 2 and let a be a non-zero integer in Zn = {0, 1, 2, …, n-1}. Then a has a multiplicative inverse in Zn iff the greatest common divisor (GCD) of a and n is 1. If a has a multiplicative inverse, then it is unique.
16
Example
• Construction of a field with 4 elements. • We start with Z2 and the polynomial x2+x+1 with coefficients in Z2 . The polynomial has no root in Z2 and cannot be factored in any nontrivial way. We adjoin a root i of this polynomial to Z2, getting i2+i+1 = 0, hence, i2 = i-1=i+1. • The four elements of the field is {a+bi: a, b in Z2} = {0, 1, i, 1+i}. With addition table and multiplication table given in the next page.
Chapter 10 Combinatorial designs
1
Summary
• • • • • Modular arithmetic Block designs Steiner triple systems Latin squares Assignments
2
Modular arithmetic
8
Algorithm to compute the GCD
GCD (a, b): Set A = a and B = b. While A × B ≠ 0, do if A ≥ B, then replace A by A – B. else, replace B by B – A. Set GCD = B. Question: How to compute GCD(a1, a2, a2 ,…, an)?
19
Exercise
• Start with Z2 and show that x3 + x +1 cannot be factored in a non-trivial way, and then use this polynomial to construct a field with 23 = 8 elements. • Continue to do the following computations: 1) (1+i) + (1+i+i2) 2) i2 × (1+i+i2) 3) (1+i)-1.
4
• •
ቤተ መጻሕፍቲ ባይዱ
Example
The simplest case is n = 2. we have Z2 = {0, 1}, and addition and multiplication mod 2 are given in the following tables:
0
1
0
1
0
0 0
1 0
0
1
1
0
3
Some notations
• • • Z: the set of integers {… -2, -1 , 0, 1, 2, …} Zn: the set of non-negative integers {0, 1, 2, …, n-1} which are less than n. If m is an integer, then there exist unique integers q (the quotient) and r (the remainder) such that m = q × n + r, 0 ≤ r ≤ n-1. Addition mod n a b : is the (unique) remainder when the ordinary sum a +b is divided by n Multiplication mod n a b : is the unique remainder when the ordinary product a × b is divided by n.
9
Example
• Compute GCD(48, 126). • By applying the algorithm, the results are displayed as shown in the table. We conclude that the GCD of 48 and 126 is the terminal value d = 6 of B.
15
Field
• The pk elements of the finite field is {a +bi+ci2+…+dik-1: a, b, c, and d in Zp}, where i is a root of the polynomial. • ik can be calculated by the polynomial.
11
Corollary 10.1.3
Let n be a prime number. Then each non-zero integer in Zn has a multiplicative inverse. If GCD (a, b) = 1, then we call a and b relatively prime.
20
Block designs
21
An Example
suppose 7 varieties of products need to be tested by 7 consumers. Each consumer is asked to compare a certain 3 of the varieties. The test is to have a property that each pair of the 7 varieties is compared by exactly one person. Can such a testing experiment be designed?
i2 = 1 + i;
1+i 0
1+i 1
i × (1 +i) = i + (1 + i) = 1+0×i = 1 ;
18
(1+i) ×(1+i) = i2 +(1+1) ×i + 1 = (1+i) + 1 = i
Another example
• Construction of a field of 33 = 27 elements. • We start with Z3 and the polynomial x3+2x+1 with coefficients in Z3 . The polynomial has no root in Z3 and cannot be factored in any nontrivial way. We adjoin a root i of this polynomial to Z3, getting i3+2i+1 = 0, hence, i3 = -2i-1=i+2. • The field with 27 elements is {a+bi+ci2: a, b and c in Z3}. The addition and multiplicative arithmetic satisfy the usual laws of mod 3 arithmetic.
A 30 8 8 B 11 11 3 30=2×11+8 11=1×8+3 8=2×3+2
2
2 0
3
1 1
3=1×2+1
2=2×1+0 d=1
13
Exercise
• Find the multiplicative inverse of 16 in Z45.
14
Fields
• For each prime p and each integer k ≥ 2, there exists a polynomial of degree k with coefficients in Zp which does not have a non-trivial factorization. • Example: for Z2, x3 + x + 1 =0 has no non-trivial factorization in Z2. • The polynomial has a root denoted as i, i.e., i3 + i + 1 = 0.
a b 1
7
Example
• In the integers modulo 10, the additive inverses are as follows: -0 = 0, -1 = 9, -2 = 8, -3 = 7, -4 = 6 -5= 5 , -6 = 4, -7 = 3, -8 = 2, -9 = 1 • The multiplication inverses are as follows: 1-1 = 1, 3-1 = 7, 7-1 = 3, 9-1 = 9, none of 0, 2, 4, 5, 6, and 8 has a multiplication inverse in Z10.
1
0
1
5
Exercises
Construct the addition and multiplication tables for the integers mod 3 and integers modulo 4.
6
Inverse
An additive inverse (denote as – a) of an integer a in Zn is an integer b such that a b 0. A multiplication inverse (denote as a-1) of an integer a in Zn is an integer b in Zn such that
12
Calculate the inverse
Determine if 11 has a multiplicative inverse in Z30 and, if so, calculate the multiplicative inverse. By applying the algorithm, GCD(11, 30) = 1 and the results are displayed in the table. By theorem 10.1.2, 11 has a multiplicative inverse in Z30. We use the equations in the table: 1= 3-1×2=3-1×(8-2×3) = 3×3 – 1×8 =3×(11-1×8) – 1 × 8 = 3×11 – 4×8 = 11×11 - 4×30 , hence, 11-1 = 11 in Z30.
17
Addition table and multiplication table of F4
+ 0 1 i 0 0 1 i 1 1 0 i i 1+i 1+i 1 0 × 0 0 1 i 0 0 0 1 0 1 i i 0 i 1+i 0 1+i i
1+i i 1
1+i 0
1+i 1
1+i 1+i i
22
Analysis for the example
We label the different varities 0, 1, 2, 3, 4, 5, 6. There are c(7, 2) = 21 pairs of the 7 varieties. Each tester gets 3 varieties and thus makes c(3, 2) = 3 comparisons. Since each pair is to be compared exactly once, the number of testers must be 21/3=7. Thus in this case, the design is possible.