Linear Combination vs Common Divisor - PowerPoint Presentation

Slide1

Linear Combination vs Common Divisor

Greatest common divisor

d is a common divisor of a and b if d|a and d|b

gcd(a,b) = greatest common divisor of a and b

d is an integer linear combination of a and b if d=sa+tb for integers s,t.spc(a,b) = smallest positive integer linear combination of a and b

Smallest positive integer linear combination

Theorem: gcd(a,b) = spc(a,b)Slide2

Theorem: gcd(a,b) = spc(a,b)

Linear Combination vs Common Divisor

For example, the greatest common divisor of 52 and 44 is 4. And 4 is a linear combination of 52 and 44:

6 · 52 + (−7) · 44 = 4Furthermore, no linear combination of 52 and 44 is equal to a smaller positive integer.

To prove the theorem, we will prove:gcd(a,b) <= spc(a,b)

spc(a,b) <= gcd(a,b)

gcd(a,b) | spc(a,b)

spc(a,b) is a common divisor of a and bSlide3

3. If d | a and d | b, then d | sa + tb for all s and t.

GCD <= SPC

Proof of (3)d | a => a = dk1

d | b => b = dk2 sa + tb = sdk1 + tdk2 = d(sk1 + tk2) => d|(sa+tb)

Let d = gcd(a,b). By definition, d | a and d | b.

Let f = spc(a,b) = sa+tb

GCD | SPC

By (3), d | f. This implies d <= f. That is gcd(a,b) <= spc(a,b).Slide4

SPC <= GCD

We will prove that spc(a,b) is actually a common divisor of a and b.

First, show that spc(a,b) | a.

Suppose, by way of contradiction, that spc(a,b) does not divide a.Then, by the Division Theorem,

a = q x spc(a,b) + r and spc(a,b) > r > 0Let spc(a,b) = sa + tb.So r = a – q x spc(a,b) = a – q x (sa + tb) = (1-qs)a + qtb.Thus r is an integer linear combination of a and b, and spc(a,b) > r.This contradicts the definition of spc(a,b), and so r must be zero.

Similarly, spa(a,b) | b.

So, spc(a,b) is a common divisor of a and b, thus by definition spc(a,b) <= gcd(a,b).Slide5

Extended GCD Algorithm

How can we write gcd(a,b) as an integer linear combination?

This can be done by extending the Euclidean’s algorithm.

Example:

a = 259, b=70259 = 3·70 + 49 70 = 1·49 + 21 49 = 2·21 + 7 21 = 7·3 + 0 done, gcd = 7

49 = a – 3b

21 = 70 - 49

21 = b – (a-3b) = -a+4b

7 = 49 - 2·21

7 = (a-3b) – 2(-a+4b) = 3a – 11bSlide6

Example:

a = 899, b=493899 = 1·493 + 406 so 406 = a - b493 = 1·406 + 87 so 87 = 493 – 406 = b – (a-b) = -a + 2b406 = 4·87 + 58 so 58 = 406 - 4·87 = (a-b) – 4(-a+2b) = 5a - 9b87 = 1·58 + 29 so 29 = 87 – 1·58 = (-a+2b) - (5a-9b) = -6a + 11b58 = 2·29 + 0 done, gcd = 29

Extended GCD AlgorithmSlide7

This Lecture

Quotient remainder theorem Greatest common divisor & Euclidean algorithm Linear combination and GCD, extended Euclidean algorithm Prime factorization and other applicationsSlide8

Theorem: gcd(a,b) = spc(a,b)

Application of the Theorem

Why is this theorem useful?

we can now “write down” gcd(a,b) as some concrete equation, (i.e. gcd(a,b) = sa+tb for some integers s and t),

and this allows us to reason about gcd(a,b) much easier.(2) If we can find integers s and t so that sa+tb=c, then we can conclude that gcd(a,b) <= c. In particular, if c=1, then we can conclude that gcd(a,b)=1.Slide9

Prime Divisibility

pf

: say p does not divide a. so gcd(p,a)=1.So by the Theorem, there exist s and t such that sa + tp = 1 (sa)b + (tp)b = b

Lemma: p prime and p|a·b implies p|a or p|b.

p|ab

p|p

Cor

: If

p

is prime, and

p| a

1

·a

2

···a

m

then

p|a

i

for some

i

.

Theorem: gcd(a,b) = spc(a,b)

Hence p|bSlide10

Every integer,

n>1, has a unique factorization into primes:p0 ≤ p1 ≤ ··· ≤ pk p0 p1 ··· pk = n

Fundamental Theorem of Arithmetic

Example:

61394323221 = 3·3·3·7·11·11·37·37·37·53Slide11

Theorem: There is a unique factorization.

Unique Factorization

proof

: suppose, by contradiction, that there are numbers with two different factorization. By the well-ordering principle, we choose the smallest such n >1:n = p1·p2···pk = q1·q2···qm Since n is smallest, we must have that pi  qj

all i,j (Otherwise, we can obtain a smaller counterexample.) Since p1|n = q1·q2···qm, so by Cor., p1|qi for some i. Since both p1 = qi are prime numbers, we must have p1 = qi.

contradiction!Slide12

Lemma. If gcd(a,b)=1 and gcd(a,c)=1, then gcd(a,bc)=1.

Theorem: gcd(a,b) = spc(a,b)

Application of the Theorem

By the Theorem, there exist s,t,u,v such that

sa + tb = 1ua + vc = 1

Multiplying, we have (sa + tb)(ua + vc) = 1 saua + savc + tbua + tbvc = 1

(sau + svc + tbu)a + (tv)bc = 1

By the

Theorem

, since spc(a,bc)=1, we have gcd(a,bc)=1Slide13

Die Hard

Simon says:

On the fountain, there should be 2 jugs, do you see them? A 5-gallon and a 3-gallon. Fill one of the jugs with exactly 4 gallons of water and place it on the scale and the timer will stop. You must be precise; one ounce more or less will result in detonation. If you're still alive in 5 minutes, we'll speak.Slide14

3 Gallon Jug

5 Gallon Jug

Start with empty jugs: (0,0)

Fill the big jug: (0,5)

Die HardSlide15

3 Gallon Jug

5 Gallon Jug

Pour from big to little:

(3,2)

Die HardSlide16

3 Gallon Jug

5 Gallon Jug

Empty the little: (0,2)

Die HardSlide17

3 Gallon Jug

5 Gallon Jug

Pour from big to little: (2,0)

Die HardSlide18

3 Gallon Jug

5 Gallon Jug

Fill the big jug: (2,5)

Die HardSlide19

3 Gallon Jug

5 Gallon Jug

Pour from big to little:

(3,4)

Done!!

Die HardSlide20

3 Gallon Jug

5 Gallon Jug

What if you have a 9 gallon jug instead?

9 Gallon Jug

Can you do it? Can you prove it?

Die HardSlide21

3

Gallon Jug

9 Gallon Jug

Supplies:

Water

Die HardSlide22

Invariant Method

Invariant: the number of gallons in each jug is a multiple of 3.

i.e., 3|b and 3|l (3 divides b and 3 divides l)Corollary: it is impossible to have exactly 4 gallons in one jug.

Bruce Dies!Slide23

Generalized Die Hard

Can Bruce form 3 gallons using 21 and 26-gallon jugs?

This question is not so easy to answer without number theory.Slide24

Invariant in Die Hard Transition:Suppose that we have water jugs with capacities B and L. Then the amount of water in each jug is always an integer linear combination of B and L.

General Solution for Die Hard

Theorem: gcd(a,b) = spc(a,b)

Corollary: The amount of water in each jug is a multiple of gcd(a,b).

Corollary: Every linear combination of a and b is a multiple of gcd(a, b). Slide25

General Solution for Die Hard

Corollary: The amount of water in each jug is a multiple of gcd(a,b).

Given jug of 3 and jug of 9, is it possible to have exactly 4 gallons in one jug?

NO, because gcd(3,9)=3, and 4 is not a multiple of 3.

Given jug of 21 and jug of 26, is it possible to have exactly 3 gallons in one jug?

gcd(21,26)=1, and 3 is a multiple of 1,so this possibility has not been ruled out yet.

Theorem.

Given water jugs of capacity a and b,

it is possible to have exactly k gallons in one jug

if and only if k is a multiple of gcd(a,b).Slide26

Theorem. Given water jugs of capacity a and b, it is possible to have exactly k gallons in one jug if and only if k is a multiple of gcd(a,b).

General Solution for Die Hard

Given jug of 21 and jug of 26, is it possible to have exactly 3 gallons in one jug?

gcd(21,26) = 1

5x21 – 4x26 = 1 15x21 – 12x26 = 3

Repeat 15 times:1. Fill the 21-gallon jug.2. Pour all the water in the 21-gallon jug into the 26-gallon jug. Whenever the 26-gallon jug becomes full, empty it out.Slide27

15x21 –

12x26 = 3

Repeat 15 times:1. Fill the 21-gallon jug.2. Pour all the water in the 21-gallon jug into the 26-gallon jug. Whenever the 26-gallon jug becomes full, empty it out.

There must be exactly 3 gallons left after this process.Totally we have filled 15x21 gallons.We pour out some multiple t of 26 gallons.The 26 gallon jug can only hold somewhere between 0 and 26.So t must be equal to 12.And there are exactly 3 gallons left.

General Solution for Die HardSlide28

Repeat s times:

1. Fill the A-gallon jug.2. Pour all the water in the A-gallon jug into the B-gallon jug. Whenever the B-gallon jug becomes full, empty it out.

General Solution for Die Hard

Given two jugs with capacity A and B with A < B, the target is C.

If gcd(A,B) does not divide C, then it is impossible.Otherwise, compute C = sA +

tB.

The B-gallon jug will be emptied exactly

t

times.

After that, there will be exactly C gallons in the B-gallon jug.Slide29

Modular Arithmetic Slide30

This Lecture

Modular arithmetic is an arithmetic about remainders.

It is very useful in coding theory and cryptography. In this lecture we will focus on additions and multiplications,while in the next lecture we will talk about “divisions”.This lecture is short. We will talk about:

Basic rule of modular addition and modular multiplication Applications: Fast exponentiation and fast division testSlide31

Def

: a  b (mod n) iff n|(a - b) iff a mod n = b mod n.

Modular Arithmetic

e.g. 12  2 (mod 10) 107  207 (mod 10) 7  3 (mod 2)

7  -1 (mod 2) 13  -1 (mod 7) -15  10 (mod 5)

Be careful, a mod n means “the remainder when a is divided by n”.a  b (mod n) means “a and b have the same remainder when divided by n”.

12

mod 10 = 2

207 mod 10 = 7

7 mod 2 = 1

-1 mod 2 = 1

-1 mod 7 = 6

-15 mod 5 = 0

Fact

:

a

 a mod n (mod n)

as

a

and

a mod n

have the same remainder

mod n

Fact

: if

a

 b (mod n)

, then

a = b + nx

for some integer

x

.

Slide32

Lemma

: If a  c (mod n), and b  d (mod n) then a+b  c+d (mod n).

Modular Addition

When you try to understand a statement like this,

first think about the familiar cases, e.g. n=10 or n=2.When n=2, it says that if a and c have the same parity,and b and d have the same parity,then a+b and c+d have the same parity.When n=10, it says that if a and c have the same last digit,and b and d have the same last digit,then a+b and c+d have the same last digit.

And the lemma says that the same principle applied for all n.Slide33

Lemma

: If a  c (mod n), and b  d (mod n) then a+b  c+d (mod n).

Modular Addition

Example 1 13  1 (mod 3), 25  1 (mod 3)

=> 12 + 25 (mod 3)  1 + 1 (mod 3)  2 (mod 3)Example 2 87  2 (mod 17), 222  1 (mod 17) => 87 + 222 (mod 17)  2 + 1 (mod 17)  3 (mod 17)Example 3 101  2 (mod 11), 141  -2 (mod 11) => 101 + 141 (mod 11)  0 (mod 11)

In particular, when computing a+b mod n, we can first replace a by a mod n and b by b mod n, so that the computation is faster.Slide34

Lemma

: If a  c (mod n), and b  d (mod n) then a+b  c+d (mod n).

Modular Addition

a  c (mod n) => a = c + nx for some integer x

b  d (mod n) => b = d + ny for some integer yTo show a+b  c+d (mod n), it is equivalent to showing that n | (a+b-c-d).Consider a+b-c-d.a+b-c-d = (c+nx) + (d+ny) – c –d = nx + ny.It is clear that n | nx + ny.Therefore, n | a+b-c-d.We conclude that a+b  c+d (mod n).

Proof

Slide35

Lemma

: If a  c (mod n), and b  d (mod n) then ab  cd (mod n).

Modular Multiplication

Example 1 9876  6 (mod 10), 17642  2 (mod 10)

=> 9876 * 17642 (mod 10)  6 * 2 (mod 10)  2 (mod 10)Example 2 10987  1 (mod 2), 28663  1 (mod 2) => 10987 * 28663 (mod 2)  1 (mod 2)Example 3 1000  -1 (mod 7), 1000000  1 (mod 7) => 1000 * 1000000 (mod 7)  -1 * 1 (mod 7)  -1 (mod 7)

In particular, when computing ab mod n, we can first replace a by a mod n and b by b mod n, so that the computation is faster.Slide36

Lemma

: If a  c (mod n), and b  d (mod n) then ab  cd (mod n).

Modular Multiplication

a  c (mod n) => a = c + nx for some integer x

b  d (mod n) => b = d + ny for some integer yTo show ab  cd (mod n), it is equivalent to showing that n | (ab-cd).Consider ab-cd.ab-cd = (c+nx) (d+ny) – cd = cd + dnx + cny + n2xy – cd = n(dx + cy + nxy).It is clear that n | n(dx + cy + nxy). Therefore, n | ab-cd.We conclude that ab  cd (mod n).

Proof

Slide37

This Lecture

Basic rule of modular addition and modular multiplication Applications: Fast exponentiation and fast division testSlide38

Fast Exponentiation

144

4 mod 713 = 144 * 144 * 144 * 144 mod 713= 20736 * 144 * 144 mod 713= 59 * 144 * 144 mod 713= 8496 * 144 mod 713= 653 * 144 mod 713= 94032 mod 713= 629 mod 713

20736 * 20736 mod 713

= 59 * 59 mod 713= 3481 mod 713= 629 mod 713

Because 20736  59 (mod 713)

Because 653

 8496 (mod 713)

shortcutSlide39

Repeated Squaring

144

50 mod 713 = 14432 14416 1442 mod 713= 648·485·59 mod 713= 242

1442 mod 713 = 591444 mod 713 = 1442 ·144

2 mod 713= 59·59 mod 713= 629 1448 mod 713= 1444·1444 mod 713= 629·629 mod 713= 63914416 mod 713= 1448·1448 mod 713= 639·639 mod 713= 48514432 mod 713= 14416·14416 mod 713= 485·485 mod 713= 648

Note that 50 = 32 + 16 + 2Slide40

Fast Division Test

Using the basic rules for modular addition and modular multiplication,

we can derive some quick test to see if a big number is divisible by a small number.

Suppose we are given the decimal representation of a big number N.

To test if N is divisible by a small number n, of course we can do a division to check.But can we do faster?

If n = 2, we just need to check whether the last digit of N is even or not.If n = 10, we just need to check whether the last digit of N is 0 or not.If

n

= 5, we just need to check whether the last digit of

N

is either 5 or 0 or not.

What about when n=3? When n=7? When n=11?Slide41

Fast Division Test

A number written in decimal divisible by 9 if and only if

the sum of its digits is a multiple of 9?

Example 1. 9333234513171 is divisible by 9. 9+3+3+3+2+3+4+5+1+3+1+7+1 = 45 is divisible by 9.Example 2. 128573649683 is not divisible by 9. 1+2+8+5+7+3+6+4+9+6+8+3 = 62 is not divisible by 9.Slide42

Claim

. A number written in decimal is divisible by 9 if and only if the sum of its digits is a multiple of 9.

Hint: 10  1 (mod 9).

Let the decimal representation of N be dkdk-1dk-2…d1d0.This means that N = dk10k + dk-110k-1 + … + d110 + d0Note that di10i mod 9 = (di) (10

i mod 9) mod 9 = (di) (10 mod 9) (10 mod 9) … (10 mod 9) mod 9 = (di) (1 mod 9) (1 mod 9) … (1 mod 9) mod 9 = di mod 9

i terms

Fast Division Test

Rule of modular multiplicationSlide43

Let the decimal representation of n be d

kdk-1dk-2…d1d0.This means that N = dk10k + dk-110k-1 + … + d110 + d0Note that di10i mod 9 = di mod 9.Hence N mod 9 = (dk10k + dk-110k-1 + … + d1

10 + d0) mod 9 = (dk10k mod 9 + dk-110k-1 mod 9 + … + d110 mod 9 + d0 mod 9) mod 9 = (dk mod 9 + dk-1 mod 9 + … + d1 mod 9 + d0 mod 9) mod 9 = (dk + dk-1 + … + d1 + d0) mod 9

Hint: 10  1 (mod 9).

Fast Division Test

Claim. A number written in decimal is divisible by 9 if and only if the sum of its digits is a multiple of 9.

Rule of modular addition

By previous slideSlide44

The same procedure works to test whether N is divisible by n=3.

Fast Division Test

What about n=11?

Hint: 10

 -1 (mod 11).

Let the decimal representation of N be d92d91d90…d1d0

Then N is divisible by 11 if and only if

d

92

-d

91

+d

90

…-d

1

+d

0

is divisible by 11.

Why? Try to work it out

.

What about n=7?

Hint: 1000



-1 (mod 7).Slide45

Quick Summary

Need to know how to apply the basic rules effectively.

Understand the principle of fast division tests.Repeated squaring will be useful later.