Euclidean Algorithm Supplement Number Theory

8/19/2019 Euclidean Algorithm Supplement Number Theory

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Applied Number Theory

Professor Ellen Gethner

February 2, 2016

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Euclidean Algorithm; supplementary material for week 3

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Warmup: The Division Algorithm (not really an algorithm)

Division Algorithm. For any a, b ∈ Z+ with a ≥ b , ∃ uniqueintegers q and r such that

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a = bq + r with 0 ≤ r


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Warmup, continued.

A direct and easy-to-prove consequence of the DivisionAlgorithm is the following.

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Warmup, continued.


Theorem D. If a, b , c , d , r , s ∈ Z with d = 0 such that d |aand d |b then d |(ra + sb ).

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Warmup, continued.



It then follows that d |(a + b ), d |(a − b ) and d |ra (amongmany other things...).

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Warmup, continued.



It then follows that d |(a + b ), d |(a − b ) and d |ra (amongmany other things...).

In other words, if d divides both a and b , then d divides anyinteger linear combination of a and b . The above are just afew special cases.



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Warmup, greatest common divisor

Definition. Let a, b ∈ Z+. The greatest common divisor of aand b , written gcd (a, b ), is the largest positive integer d that

divides both a and b .

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That is, gcd (a, b ) = d .



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Example 1. gcd (4, 6) = 2, gcd (3, 7) = 1, gcd (100, 4) = 4,and gcd (100001, 0) = 100001, etc.

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One place (of many) where gcd is needed is in RSAencryption.

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One place (of many) where gcd is needed is in RSAencryption.

Thus the actual computation of gcd (a, b ) is necessary and

important.

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Example 2. (Illustration of the Euclidean Algorithm)

Problem. Find gcd (54, 21). That is, compute d = gcd (54, 21)where a = 54 and b = 21.

54 = 2 × 21 + 12 and note that

0 ≤ 12


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Example 2. (Illustration of the Euclidean Algorithm)

Problem. Find gcd (54, 21). That is, compute d = gcd (54, 21)where a = 54 and b = 21.

54 = 2 × 21 + 12 and note that

0 ≤ 12


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Euclidean algorithm example, continued

21 = 1 × 12 + 9 and note that 0 ≤ 9


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21 = 1 × 12 + 9 and note that 0 ≤ 9


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21 = 1 × 12 + 9 and note that 0 ≤ 9


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21 = 1 × 12 + 9 and note that 0 ≤ 9


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21 = 1 × 12 + 9 and note that 0 ≤ 9


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21 = 1 × 12 + 9 and note that 0 ≤ 9


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The Actual Euclidean Algorithm

Recall that a (mod b ) is the remainder upon dividing a by b . Thatis if a = qb + r with 0 ≤ r


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Euclidean Algorithm Written out in Stages

a = q 1b + r 1 0 ≤ r 1


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a = q 1b + r 1 0 ≤ r 1


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a = q 1b + r 1 0 ≤ r 1


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a = q 1b + r 1 0 ≤ r 1


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a = q 1b + r 1 0 ≤ r 1


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a = q 1b + r 1 0 ≤ r 1


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a = q 1b + r 1 0 ≤ r 1


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Correctness of the Euclidean Algorithm

Does the algorithm halt?


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g


If so, does it produce the desired output, namely gcd (a, b )?


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g


If so, does it produce the desired output, namely gcd (a, b )?

The following lemma will help answer the latter question.

Euclidean Algorithm: A helpful lemma

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g p

Lemma. Let a, b ∈ Z+ satisfy a = bq + r with 0 ≤ r


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g p



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Thus it remains to show that d 0|d .

Proof of lemma, continued. So far we have d |d 0

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Here goes. By the definition of gcd we have that d 0|b andthat d 0|(a−

a

b b )


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a

b b )

⇒ d |a (by Theorem D)


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a

b b )


⇒ d 0|gcd (a, b )


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a

b b )


⇒ d 0|gcd (a, b )

⇒ d 0|d .


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a

b b )


⇒ d 0|gcd (a, b )

⇒ d 0|d .

In total, we have d |d 0 and d 0|d


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a

b b )


⇒ d 0|gcd (a, b )

⇒ d 0|d .

In total, we have d |d 0 and d 0|d

⇒ d = d 0. That is, gcd (a, b ) = gcd (b , r ). QED

Why is the Euclidean Algorithm Correct?

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First, by the lemma, we have r k −1 = gcd (a, b ) because


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gcd (a, b ) = gcd (b , r 1) = gcd (r 1, r 2) = gcd (r 2, r 3) = · · · =gcd (r k −1, r k ) = gcd (r k −1, 0) = r k −1.




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Finally, and importantly, the algorithm halts becauser 1 > r 2 > r 3 > · · · is a strictly decreasing sequence of non-negative integers, which means that, eventually, r k = 0for some k ≥ 1.




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Finally, and importantly, the algorithm halts becauser 1 > r 2 > r 3 > · · · is a strictly decreasing sequence of non-negative integers, which means that, eventually, r k = 0for some k ≥ 1.

In particular, the last non-zero remainder, namely r k −1, isexactly gcd (a, b ).

Euclidean Algorithm and it’s relation to the FibonacciNumbers

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Numbers

Optional Material

Lemma 31.10 in Introduction to Algorithms by Cormen,Lieserson Rivest and Stein: number of recursive calls



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Lieserson, Rivest, and Stein: number of recursive calls

Lemma 31.10 in CLRS If a, b ≥ 1 are integers and

Euclid(a, b ) performs n ≥ 1 recursive calls, then a ≥ F n+2 andb ≥ F n+1, where F i is the i th Fibonacci number.




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Proof by induction on n.


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The statement P (n) is exactly the statement of the lemma.


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Base Case. n = 1. Since one recursion occurs we haveb > 0. That is, b ≥ 1 = F 2.

Lemma 31.10 in Introduction to Algorithms by Cormen,Lieserson, Rivest, and Stein: number of recursive calls

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By assumption a > b ⇒ a ≥ 2 = F 3.

Lemma 31.10 in Introduction to Algorithms by Cormen,Lieserson, Rivest, and Stein: number of recursive calls



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By assumption a > b ⇒ a ≥ 2 = F 3.

Also, since b > a (mod b ) (why?) in each recursive callEuclid(â, b̂ ) we have â > b̂ .

Proof of lemma 31.10, continued.

Induction Hypothesis Assume the lemma is true if k 1

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Induction Hypothesis. Assume the lemma is true if k − 1recursive calls have made (i.e., assume P (k − 1) is true).


Induction Hypothesis Assume the lemma is true if k 1

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Inductive Step. We must prove the lemma is true when k recursive calls have been made (i.e., verify P (k ) is true).


Induction Hypothesis Assume the lemma is true if k − 1

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To that end, we have k > 0 and b > 0.


Induction Hypothesis Assume the lemma is true if k − 1



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Induction Hypothesis. Assume the lemma is true if k 1recursive calls have made (i.e., assume P (k − 1) is true).



Assume Euclid(a, b ) makes k recursive calls.


Induction Hypothesis. Assume the lemma is true if k − 1

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Euclid(a, b ) first returns Euclid(b , a (mod b )), which makesk − 1 recursive calls.





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By the induction hypothesis, we have b ≥ F k +1 and a(mod b ) ≥ F k .



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duct o ypot es s. ssu e t e e a s t uerecursive calls have made (i.e., assume P (k − 1) is true).





By the induction hypothesis, we have b ≥ F k +1 and a(mod b ) ≥ F k .

It remains to show that a ≥ F k +2.

Proof of lemma 31.10, continued. Must show a ≥ F k +2.

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Note that b + a (mod b )


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= b + r


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= b + r

= b + (a−

a

b b )




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= b + r

= b + (a−

a

b b )

≤ a


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= b + r

= b + (a−

a

b b )

≤ a

because ab > 1.


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Thus we have that a ≥ b + a (mod b )


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≥ F k +1 + F k


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≥ F k +1 + F k

= F k +2, as desired.


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≥ F k +1 + F k


Conclusion. If Euclid(a, b ) makes n recursive calls thena ≥ F n+2 and b ≥ F n+1.


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≥ F k +1 + F k


Conclusion. If Euclid(a, b ) makes n recursive calls thena ≥ F n+2 and b ≥ F n+1.

QED

Direct Consequence of Lemma 31.10

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Theorem 31.11. If a > b ≥ 1 and b


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Recall that a = F n+2 and b = F n+1 .

Best possible theorem

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Then gcd (F n+1,F n) = gcd (F n,F n−1) = · · · = gcd (F 1, F 0) =gcd (1, 0) = 1.


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Thus n recursive calls will be made ⇒ the theorem is best

possible.


R ll h F d b F

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possible.

Finally, given that F n ≈ φn

5 , where φ = 1+

√ 5

2 , we see that


R ll h F d b F

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possible.


5 , where φ = 1+

√ 5

2 , we see that

Euclid(a, b ) makes C log b recursive calls.


R ll th t F d b F



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possible.


5 , where φ = 1+

√ 5

2 , we see that

Euclid(a, b ) makes C log b recursive calls.

Moral of the story: recursion is not always bad!

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Euclidean Algorithm Supplement Number Theory