Number Theory Speaker: P.H. Wu Date: 21/04/11 1. Outline Introduction Residue number system Chinese...
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Transcript of Number Theory Speaker: P.H. Wu Date: 21/04/11 1. Outline Introduction Residue number system Chinese...
2
Outline
• Introduction• Residue number system• Chinese remainder theory• Euclid’s algorithm• Stern-Brocot tree• Continuants• Conclusion
3
Introduction
• gcd: greatest common divisor• lcm: least common multiple• primes: p• factorial: n!• module: mod• module congruence: a b(mod m)• Application: encryption , combination and
permutation
≡
4
Introduction
• Some primes1. Fermat prime
2. Mersenne prime
3.
22 1n
nF
2 1, p is prime.pnM
( ) : #
ln
x prime x
x
x
5
Residue number system
• Def: An integer x can be represented as a sequence of residue with respect to moduli that are prime to each other.
• Res(x) is unique if 0< x <m1m2…mr , or Res(x) = Res(x + km1m2…mr).
1 2 kRes( ) ( mod , mod ,..., mod ) for r jx x m x m x m m m
6
Residue Number systemX mod 12 X mod 3 X mod 4
0 0 0
1 1 1
2 2 2
3 0 3
4 1 0
5 2 1
6 0 2
7 1 3
8 2 0
9 0 1
10 1 2
11 2 3
Nice coding system!!!
Res( ) ( mod3, mod 4)
mod3
mod 4
Res( ) ( , )
Res(5) (2,1)
Res(10) (1,2)
Res(15) (3,3) (0,3)
Res(50) (2,2) (2,2)
x x x
a x
b x
x a b
7
Residue Number systemX mod 12 X mod 3 X mod 4
0 0 0
1 1 1
2 2 2
3 0 3
4 1 0
5 2 1
6 0 2
7 1 3
8 2 0
9 0 1
10 1 2
11 2 3
(1,0) 4
(0,1) 9
( , ) ( )mod12
(2,3) 2 (1,0) 3 (0,1)
2
(2,3)
3 35
35mod12 11
11 12
?
a
b
x y ax by
a b
x k
9
Euclid’s Algorithm
• 輾轉相除法192, 33
192 33 27
33 27 6
2
5
1
47 6
gcd( , ) gcd( , )
3
6 3 02
a q b r
a b b r
a b
5 192 133 1
165 27
4 27 6 2
24 6
3 0
10
Stern-Brocot tree
• Binary search tree1.
2. (n, m)=1
3.
4. 5/7LRRL
' '
' '
n n n n
m m m m
is unique.n
BSTm
n
m
'
'
n
m'
'
n n
m m
11
Stern-Brocot tree
• Example:
5 4
64 53 42 31
11 11 11 1120 9 9 7
11 11 2 25 3 1 1
2 2 2 11 64
1 11
R R R
R R L R
R R R L
R LR L
12
Stern-Brocot tree
• Binary search tree5.
n
m
'
'
n
m'
'
n n
m m
( ')
( ')
n n n
m m m
( ') '
( ') '
n n n
m m m
''
''
n nn n
m mm m
( ')
( ')
' ' 1 1
' ' 0 1
1 1
0 1
1 0
1 1
n n n
m m m
n n n n n
m m m m m
L
R
13
Stern-Brocot tree
• Example:
5 4
5 4
64 53 42 31
11 11 11 1120 9 9 7
11 11 2 25 3 1 1
2 2 2 11 64
1 11
0 1 1 0 1 1 1 0 1 1 29 35
1 0 1 1 0 1 1 1 0 1 5 6
R R R
R R L R
R R R L
R LR L
14
Continuants
• Any positive rational number can be represented as continuants.
0 i
1
2
34
1 a {0,1,2,...}
11
1...
r aa
aa
a
15
Continuants
• Continuant polynomial
• Example:
0
1 1 1
1 2 1 1 2 1 2 1 2 2
() 1;
( ) ;
( , ,..., ) ( , ,..., ) ( , ,..., );n n n n n n n
K
K x x
K x x x K x x x x K x x x
0
1 1 1
2 1 2 1 1 2 0 1 2
3 1 2 3 2 1 2 3 1 1 1 2 3 1 3
4 1 2 3 4 3 1 2 3 4 2 1 2
1 2 3 4 1 4 3 4 1 2
() 1;
( ) ;
( , ) ( ) () 1;
( , , ) ( , ) ( ) ;
( , , , ) ( , , ) ( , )
1;
K
K x x
K x x K x x K x x
K x x x K x x x K x x x x x x
K x x x x K x x x x K x x
x x x x x x x x x x
16
Continuants
• Continuant polynomial
• Example:
0
1 1 1
1 2 1 1 2 1 2 1 2 1
() 1;
( ) ;
( , ,..., ) ( , ,..., ) ( , ,..., );n n n n n n n
K
K x x
K x x x K x x x x K x x x
0
1 1 1
2 1 2 1 1 2 0 1 2
3 1 2 3 2 1 2 3 1 1 1 2 3 1 3
4 1 2 3 4 3 1 2 3 4 2 1 2
1 2 3
() 1 ;
( ) ;
( , ) ( ) () 1 ;
( , , ) ( , ) ( ) ;
( , , , ) ( , , ) ( , )
(1)
(1)
(2
)
(
5)
3)
(
K
K x x
K x x K x x K x x
K x x x K x x x K x x x x x x
K x x x x K x x x x K x x
x x x x
4 1 4 3 4 1 2 1
;
x x x x x x
17
Relation between Euclid’s Alg. and Continuants
1
1 0 2
2 1 3
3 2 4
4
3
(2)
(2) () (4,2)
(4,2
192, 33
192
) (2) (1,4,2)
(1,4,2) (4,2) (5,1,4,2)
(5,1,4
33 27
33 27 6
27 6 3
6 3 0
6 3 0
27 3 3 3
33 3 3 3
192 3 3 3
192 3
3
,2)
(1,4,2
5
1
4
2
4
)3 3
1
5
a b
K
K K K
K K K
K K K
K
K
1 2 2 1( , ,..., ) ( ,..., , )n n n nK x x x K x x x
18
Relation between Euclid’s Alg. and Continuants
192 27 1 15 5 5
633 33 33 / 27 127
1 1 1 5 5 5
1 1 11 1 1
3 127 / 6 4 46 6 / 3
1 5 5 1 4 2[ , , , ]
11
14
2
19
Relation between Stern-Brocot tree and Continuants
• Example:
• Path:
192 159 126 93
33 33 33 3360 27 27 21
33 33 6 615 9 3 3
6 6 6 3
3
0
R R R
R R L R
R L LR R
5 1 4 2 [ ,5 1, ]4,2R L R L
20
Relation between Stern-Brocot tree and Continuants
192 1 15 5
1 133 1 11 1
4 42 2 /1
1 5 [ , , , , ]
15 1 4 1 1
11
41
11
[ , ,5 1 4,1 1 5 1 4, ] [ , , , ]2
1 2 1 1 2( , ,..., 1) ( , ,..., ,1)n n n nK x x x K x x x
21
Other continuant property
• 1. mirror symmetry
• 2. more general recurrence
• 3. The number of terms is a Fibonacci number.
2 1 1 2( ,..., , ) ( , ,..., )n nK x x x K x x x
1 1
1 1
1 1 1 1 2
( ,..., , ,..., )
( ,..., ) ( ,..., )
+ ( ,..., ) ( ,..., )
m n m m m n
m m n m m n
m m n m m n
K x x x x
K x x K x x
K x x K x x
1(1,1,...,1)n nK F