[PDF] - a (mod m ) b is congruent to modulo a m b mod PDF Document

SLIDE 1

1

Number Theory and its Applications

Modular Exponentiation
Euclidean Algorithm for GCD
Solving Linear Congruences
Chinese Remainder Theorem and Application

to Arithmetic with large numbers

Covered in Sections 3.6 and 3.7

1 Based on Rosen and slides by K. Busch

“ is congruent to modulo ”

2

Modular Arithmetic Recap

Z b a  ,



Z m

) (mod m b a 

a b m

Examples:

) 12 (mod 13 1

) 6 (mod 5 11

m b m a mod mod 

) (mod m m k   ) (mod m m 

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2

3

) (mod m b a  km b a Z k     ,

b a m  | m b m a mod mod 

Equivalent statements

4

3 8 mod 3 

1 2 3 4 5 6 7

3 Length of line represents number

SLIDE 3

3

5

3 8 mod 11 

1 2 3 4 5 6 7

11 Length of line represents number

6

3 8 mod 19 

1 2 3 4 5 6 7

19 Length of line represents number

SLIDE 4

4

7 1 2 3 4 5 6 7

3

1 2 3 4 5 6 7

11

1 2 3 4 5 6 7

19

) 8 (mod 19 11 3  

All lines terminate in same number

8

)} (mod | { m b a b Sa  

“Congruence class” of modulo :

a m

There are congruence classes:

m

1 1

, , ,

 m

S S S 

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9

) (mod m b a 

) (mod m d c 

) (mod m d b c a   

) (mod m b a  sm b a   ) (mod m d c  tm d c   m t s b d c a ) (     

Proof sketch: Closure under addition:

10

) (mod m b a 

) (mod m d c 

) (mod m d b c a   

) (mod m b a  sm b a   ) (mod m d c  tm d c   ) ( bd ) )( ( stm ds bt m tm d sm b c a        

Proof sketch: Closure under multiplication:

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Closure under mod: m m a m a mod ) mod ( mod  2 5 mod 2 5 mod ) 5 mod 7 ( 2 ) 5 mod 7 (    (Follows from definition of mod)

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m m b m a m b a mod )) mod ( ) mod (( mod ) (    m m b m a m ab mod )) mod )( mod (( mod 

(Follows from previous slides) Example: Useful results for arithmetic with large numbers: 35 50 mod 5 7 50 mod )) 50 mod 55 )( 50 mod 57 (( 50 mod 55 57     

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13

Modular exponentiation Compute efficiently using small numbers

m bn mod

1 1 1 1 1 1

2 2 2 2 a a a a a a n

b b b b b

k k k k



   

 

   Binary expansion of n

m m b m b m b m b b b m b

a a a a a a n

k k k k

mod )) mod ( ) mod ( ) mod (( mod mod

1 1 1 1 1 1

2 2 2 2

    

   

 

14

Example:

36 645 mod 3644 

2 7 9

2 2 2 1010000100 644    

2 7 9 2 7 9

2 2 2 2 2 2 644

3 3 3 3 3  

 

) 645 mod ) 645 mod 3 )( 645 mod 3 )( 645 mod 3 (( 645 mod ) 3 3 3 ( 645 mod 3

2 7 9 2 7 9

2 2 2 2 2 2 644

 

SLIDE 8

8

15

 

 111 645 mod 81 81 645 mod )) 645 mod 3 )( 645 mod 3 (( 645 mod 3 645 mod 3 81 ) 645 mod 9 9 ( 645 mod )) 645 mod 3 )( 645 mod 3 (( 645 mod 3 645 mod 3 9 645 mod 9 645 mod 3

2 2 2 3 2

2 2 2 2 2 2 2 2 2 2 2

           

36 645 mod 471 111 ) 645 mod ) 471 ) 645 mod 3 ((( ) 645 mod 471 3 ( ) 645 mod ) 645 mod ) 81 396 (( 3 ( ) 645 mod ) 645 mod ) 81 ) 645 mod 3 ((( 3 ( ) 645 mod 81 3 3 ( ) 645 mod ) 645 mod 3 ( 3 3 ( ) 645 mod 3 3 3 ( 3

9 9 9 7 9 7 9 2 7 9 2 7 9

2 2 2 2 2 2 2 2 2 2 2 2 2 644

            

Compute the powers of 3 efficiently Use the powers of 3 to get result efficiently

16

Modular_Exponentiation( ) { for to { if ( ) } return }

m n b , ,

2 1 2 1

) ( a a a a n

n n



 

 1  x m b power mod   i 1  k 1 

i

a m power x x mod ) (   m power power power mod ) (   ) mod ( m b x

n

SLIDE 9

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17

Recall: Greatest Common Divisor

 ) , gcd( b a

largest integer such that and

d a d | b d |

Examples: Common divisors of 24, 36: 1, 2, 3, 4, 6, 12

12 ) 36 , 24 gcd(  1 ) 22 , 17 gcd( 

Common divisors of 17, 22: 1

| | | | ,    b a Z b a

18

Trivial cases:

1 ) 1 , gcd(  m m m  ) , gcd(  m

If then are relatively prime

1 ) , ( gcd  b a b a,

and have no common factors

a b

Example:

1 ) 22 , 21 gcd( 

21, 22 are relatively prime

SLIDE 10

10

How do we compute GCD efficiently?

(Finding prime factorization is slow)

19 20

Theorem: If then

) , gcd( ) , gcd( r b b a 

r q b a   

Proof:

a d | b d | ds a  dt b  ) ( tq s d r   dt b 

) , ( b a ) , ( r b

r d | b d |

Thus, and have the same set of common divisors End of proof b r  

SLIDE 11

11

21

1 1 1 1 2 2 3 3 2 2 1 1 2 2 1 1

             

     n n n n n n n n n

q r r r r r q r r r r r q r r r r r q r r 

r a 

1

r b 

n n n n n n

r r r r r r r r r r r r b a       

  

) , gcd( ) , gcd( ) , gcd( ) , gcd( ) , gcd( ) , gcd( ) , gcd(

1 1 2 3 2 2 1 1

  first zero result

1 0 / r

r

2 1 / r

r

1 2 /   n n

r r

n n

r r /

1 



divisions remainder

22

41 2 82 82 2 2 2 82 166 166 82 82 1 166 248 248 166 166 1 248 414 414 248 248 1 414 662

4 5 3 4 2 3 1 2

                           r r r r r r r r

662  a 414  b

2 ) , 2 gcd( ) 2 , 82 gcd( ) 82 , 166 gcd( ) 166 , 248 gcd( ) 248 , 414 gcd( ) 414 , 662 gcd(       result

SLIDE 12

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23

Euclidean Algorithm for GCD gcd( ) { while ( ) { } return }

b a, a x 

b y 

 y

y x r mod  y x 

r y  x

Useful Result regarding GCDs

24

if then there are such that



Z b a,

Z t s  ,

tb sa b a   ) , gcd(

Example:

14 1 6 ) 2 ( 2 ) 14 , 6 gcd(      

(i.e., gcd is a linear combination of a and b)

SLIDE 13

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25

The linear combination can be found by reversing the Euclidian algorithm steps

198 5 252 4 18 ) 198 , 252 gcd(     

18 2 36 18 36 1 54 36 54 3 198 54 198 1 252             198 5 252 4 198 1 ) 198 1 252 ( 4 198 1 54 4 ) 54 3 198 ( 1 54 36 1 54 18 ) 198 , 252 gcd(                       

26

Linear congruences

) (modm b x a  

We want to solve this equation for

) (mod ? m x 

x

SLIDE 14

14

27

Inverse of :

) (mod 1 m a a 

a

) (modm b x a   m a a mod 

) (modm b a x a a  

) (modm x x 

) (mod 1 m a a  ) (mod 1 m x x a a   

) (modm b a x 

28

If and are relatively prime then the inverse modulo exists Theorem:

a

m

a

m

Proof:

tm sa m a   1 ) , gcd( ) (mod 1 m sa  s a 

End of proof

(linear combo theorem) (Def. of mod) (Def. of inverse mod m)

SLIDE 15

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29

Example: solve equation

) 7 (mod 4 3  x

7 1 3 2 1 ) 7 , 3 gcd(      

Inverse of 3:

) (mod 1 3 2 m    7 mod 6 ) 7 (mod 8 ) 7 (mod 4 2       x

7 , 4 , 3    m b a

) (modm b a x 

2   a A Chinese Puzzle

(by Sun-Tzu, 300-500 AD)

I have some things whose number you don’t know. If divided by 3, the remainder is 2 If divided by 5, the remainder is 3 If divided by 7, the remainder is 2 How many things do I have?

30

SLIDE 16

16

Sun-Tzu’s Puzzle

31

) 3 (mod 2  x ) 5 (mod 3  x ) 7 (mod 2  x

What is x?

32

Chinese remainder theorem (CRT)

n

m m m , , ,

2 1



:pairwise relatively prime

) (mod

1 1

m a x 

) (mod

2 2

m a x  ) (mod

n n

m a x 



Has unique solution for modulo

x

n

m m m m 

2 1 



SLIDE 17

17

33

Unique solution modulo :

k k

m m M 

k

y

:inverse of modulo

k

M

k

m

n n n

y M a y M a y M a x     

2 2 2 1 1 1

n

m m m m 

2 1 

 where

34

Explanation:

k k

m m M 

k k k

m y M mod 1 

:inverse of modulo

k

M

k

m

n n n

y M a y M a y M a x     

2 2 2 1 1 1

) (mod

1 1

m Mk 



) (mod

1

m ) (mod

1

m

) (mod ) (mod

1 1 1 1 1 1

m a x m y M a x  

k

y

Similar for any

j

m

k = 1:

1 1 1

mod 1 m y M 

i.e., x satisfies 1st equation

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35

Example:

) 3 (mod 2  x ) 5 (mod 3  x ) 7 (mod 2  x

105 7 5 3     m

15 7 / 105 7 / 21 5 / 105 5 / 35 3 / 105 3 /

3 2 1

         m M m M m M 1 1 2

3 2 1

   y y y

) 105 (mod 23 ) 7 5 3 (mod 23 233 1 15 2 1 21 3 2 35 2

3 3 3 2 2 2 1 1 1

                 y M a y M a y M a x

36

An Application of CRT Perform arithmetic with large numbers using arithmetic modulo small numbers Example: Suppose your CPU can only perform fast arithmetic for positive integers < 100, but your input numbers are huge.

SLIDE 19

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An Application of CRT Idea: Convert your large numbers to small numbers < 100 using mod, perform modular arithmetic, convert back using CRT.

95 , 97 , 98 , 99

4 3 2 1

    m m m m 930 , 403 , 89 95 97 98 99      m

Choose relatively prime numbers < 100

) 89 , 9 , 8 , 33 ( 684 , 123 

89 95 mod 684 , 123 9 97 mod 684 , 123 8 98 mod 684 , 123 33 99 mod 684 , 123    