1. CSE 20: Discrete Mathematics for Computer Science Prof. Shachar Lovett

2. Today’s Topics:
GCD
Euclid’s algorithm
Extended Euclid’s algorithm

3. 1. GCD
The poor man version of prime factorization

4. GCD Greatest common divisor
Given two positive integers a,b, their GCD is the largest integer n such that n|a and n|b

5. GCD
What is GCD(20,30)? 5 10 20 30
Other

6. GCD
GCD is the same as finding simplified rational expressions, just like you did in school
GCD(20,30)=10 means that
20 30 = 2∗10 3∗10 = 2 3
and it can’t be simplified further

7. GCD
What is GCD( , )?

8. GCD How can we compute GCD(a,b)? Simple way:
Compute prime factorization of a,b
Take common primes and prime powers
Example: if a= and b=21735 then GCD(a,b)=21035
However, we believe that computing the prime factorization of large numbers is hard…
Euclid’s algorithm provides a much faster way

9. 2. Euclid’s algorithm
Fast GCD

10. Euclid’s algorithm Euclid(a,b): Input: 𝑎,𝑏∈𝑁 Output: 𝑔=gcd⁡(𝑎,𝑏)
While b>0:
x=a mod b
a=b
b=x
Return a

11. Euclid’s algorithm Euclid(a,b): Input: 𝑎,𝑏∈𝑁 Output: 𝑔=gcd⁡(𝑎,𝑏)
While b>0:
x=a mod b
a=b
b=x
Return a
(a,b)
(b,a mod b)

12. Euclid’s algorithm Euclid(a,b): Input: 𝑎,𝑏∈𝑁 Output: 𝑔=gcd⁡(𝑎,𝑏)
While b>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a

13. Euclid’s algorithm Euclid(a,b): Input: 𝑎,𝑏∈𝑁 Output: 𝑔=gcd⁡(𝑎,𝑏)
While b>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
Example run: a=20, b=30 a 20 30 10 b 30 20 10

14. Euclid’s algorithm The same basic questions Does it always terminate?
Does it return the correct answer?
How fast is it?

15. Euclid’s algorithm: termination
Euclid(a,b):
Input: 𝑎,𝑏∈𝑁
Output: 𝑔=gcd⁡(𝑎,𝑏)
While 𝑏>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
Loop invariant (after 1st iteration): 𝑎>𝑏

16. Euclid’s algorithm: termination
Euclid(a,b):
Input: 𝑎,𝑏∈𝑁
Output: 𝑔=gcd⁡(𝑎,𝑏)
While 𝑏>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
Loop invariant (after 1st iteration): 𝑎>𝑏
Loop beginning: (a,b)
Loop end: (b, a mod b)
By definition, 𝑎 𝑚𝑜𝑑 𝑏∈ 0,…,𝑏−1
and hence (a mod b) < b

17. Euclid’s algorithm: termination
Euclid(a,b):
Input: 𝑎,𝑏∈𝑁
Output: 𝑔=gcd⁡(𝑎,𝑏)
While 𝑏>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
Loop invariant (after 1st iteration): 𝑎>𝑏
The value of a keeps decreasing,
which proves termination

18. Euclid’s algorithm: correctness
Euclid(a,b):
Input: 𝑎,𝑏∈𝑁
Output: 𝑔=gcd⁡(𝑎,𝑏)
While 𝑏>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
g=gcd(a,b)

19. Euclid’s algorithm: correctness
Need to prove the following lemma
Lemma: ∀𝑎,𝑏≥1, gcd 𝑎,𝑏 =gcd⁡(𝑏,𝑎 𝑚𝑜𝑑 𝑏)

20. Euclid’s algorithm: correctness
Need to prove the following lemma
Lemma: ∀𝑎,𝑏≥1, gcd 𝑎,𝑏 =gcd⁡(𝑏,𝑎 𝑚𝑜𝑑 𝑏)
Proof: we will actually show
∀𝑚≥1, 𝑚 𝑎∧𝑚 𝑏 ⇔(𝑚 𝑏∧𝑚 𝑎 𝑚𝑜𝑑 𝑏 )
In particular, the largest such m is (by definition) the GCD, and so gcd 𝑎,𝑏 =gcd⁡(𝑏,𝑎 𝑚𝑜𝑑 𝑏)

21. Euclid’s algorithm: correctness
Lemma: ∀𝑎,𝑏≥1, gcd 𝑎,𝑏 = gcd 𝑏,𝑎 𝑚𝑜𝑑 𝑏
Proof (⇒): 𝑚 𝑎∧𝑚 𝑏 ⇒(𝑚 𝑏∧𝑚 𝑎 𝑚𝑜𝑑 𝑏 )
Let 𝑎 = 𝑏𝑝+𝑞, where 𝑝=𝑎 𝑑𝑖𝑣 𝑏, 𝑞 = 𝑎 𝑚𝑜𝑑 𝑏.
If 𝑚|𝑎,𝑚|𝑏 then 𝑎=𝑚𝑥, 𝑏=𝑚𝑦, where 𝑥,𝑦∈𝑍.
Then:
𝑎 𝑚𝑜𝑑 𝑏=𝑞=𝑎−𝑏𝑝=𝑚𝑥−𝑚𝑦⋅𝑝=𝑚 𝑥−𝑦𝑝
So, 𝑚|(𝑎 𝑚𝑜𝑑 𝑏).

22. Euclid’s algorithm: correctness
Lemma: ∀𝑎,𝑏≥1, gcd 𝑎,𝑏 = gcd 𝑏,𝑎 𝑚𝑜𝑑 𝑏
Proof (⇐): 𝑚 𝑎∧𝑚 𝑏 ⇐(𝑚 𝑏∧𝑚 𝑎 𝑚𝑜𝑑 𝑏 )
Let 𝑎 = 𝑏𝑝+𝑞, where 𝑝=𝑎 𝑑𝑖𝑣 𝑏, 𝑞 = 𝑎 𝑚𝑜𝑑 𝑏.
If 𝑚|𝑏,𝑚|𝑞 then b=𝑚𝑥, q=𝑚𝑦, where 𝑥,𝑦∈𝑍.
Then:
𝑎=𝑏𝑝+𝑞=𝑚𝑥⋅𝑝+𝑚𝑦=𝑚(𝑥𝑝+𝑦)
So, 𝑚|𝑎.

23. Euclid’s algorithm: correctness
Euclid(a,b):
Input: 𝑎,𝑏∈𝑁
Output: 𝑔=gcd⁡(𝑎,𝑏)
While 𝑏>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
g=gcd(a,b)
Proved!

24. Euclid’s algorithm: speed
Euclid(a,b):
Input: 𝑎,𝑏∈𝑁
Output: 𝑔=gcd⁡(𝑎,𝑏)
While 𝑏>0: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Return a
How many iterations?

25. Euclid’s algorithm: speed
Consider one iteration: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
We know that a,b become smaller
But by how much?

26. Euclid’s algorithm: speed
Consider one iteration: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
We know that a,b become smaller
But by how much?
Intuition:
Since 𝑎 𝑚𝑜𝑑 𝑏∈{0,..,𝑏−1}, on “average” 𝑎 𝑚𝑜𝑑 𝑏≈𝑏/2. Hence, value of b decreases by a factor of 2 at each iteration
(so log(b) iterations will be needed)
Can we justify this?

27. Euclid’s algorithm: speed
Consider one iteration: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Lemma: 𝑎+𝑏≥ 3 2 (𝑏+ 𝑎 𝑚𝑜𝑑 𝑏 )
Sum decreases by 3/2 at each iteration;
So, we need at most ~log(a+b) iterations
In fact, at most ~log(b) iterations, since after the first iteration, both values are at most b

28. Euclid’s algorithm: speed
Consider one iteration: 𝑎,𝑏 ←(𝑏, 𝑎 𝑚𝑜𝑑 𝑏)
Lemma: 𝑎+𝑏≥ 3 2 (𝑏+ 𝑎 𝑚𝑜𝑑 𝑏 )
Proof: Since 𝑎>𝑏>𝑎 𝑚𝑜𝑑 𝑏, and since b|𝑎−(𝑎 𝑚𝑜𝑑 𝑏), we must have 𝑎− 𝑎 𝑚𝑜𝑑 𝑏 ≥𝑏. So:
𝑎≥𝑏+(𝑎 𝑚𝑜𝑑 𝑏)
𝑏= 𝑏 2 + 𝑏 2 ≥ 1 2 𝑏+ 𝑎 𝑚𝑜𝑑 𝑏
⇒𝑎+𝑏≥ 3 2 (𝑏+ 𝑎 𝑚𝑜𝑑 𝑏 )

29. 3. Extended Euclid’s algorithm
Using algorithms to do math!

30. Extended Euclid’s algorithm
Theorem: ∀𝑎,𝑏∈𝑁 ∃𝑥,𝑦∈𝑍 𝑎𝑥+𝑏𝑦=gcd⁡(𝑎,𝑏)
(this is called Extended Euclid’s algorithm)
Example: a=3, b=5, gcd(a,b)=1
3*(-3)+5*2 = 1 (solution: x=-3, y=2)

31. Extended Euclid’s algorithm
Theorem: ∀𝑎,𝑏∈𝑁 ∃𝑥,𝑦∈𝑍 𝑎𝑥+𝑏𝑦=gcd⁡(𝑎,𝑏)
(this is called Extended Euclid’s algorithm)
Example: a=3, b=5, gcd(a,b)=1
3*(-3)+5*2 = 1 (solution: x=-3, y=2)
The proof will use our analysis of Euclid’s algorithm
So, even though the theorem has nothing to do with algorithms, the proof will use an algorithm!

32. Extended Euclid’s algorithm
Theorem: ∀𝑎≥𝑏≥1 ∃𝑥,𝑦∈𝑍 𝑎𝑥+𝑏𝑦=gcd⁡(𝑎,𝑏)
Proof (by strong induction on b):
Base case: b=1, so gcd(a,1)=a, can take x=1,y=1
Inductive case: We use the identity: gcd 𝑎,𝑏 =gcd⁡(𝑏,𝑎 𝑚𝑜𝑑 𝑏).
Since 𝑏>𝑎 𝑚𝑜𝑑 𝑏, by the inductive assumption
∃ 𝑥 ′ , 𝑦 ′ ∈𝑍, 𝑏 𝑥 ′ + 𝑎 𝑚𝑜𝑑 𝑏 𝑦 ′ =gcd b,a mod b = gcd 𝑎,𝑏
Let 𝑎 = 𝑏𝑝+𝑞, where 𝑝=𝑎 𝑑𝑖𝑣 𝑏, 𝑞 = 𝑎 𝑚𝑜𝑑 𝑏. Then:
𝑏𝑥′+ 𝑎−𝑏𝑝 𝑦′= gcd 𝑎,𝑏
Take 𝑥= 𝑦 ′ ,𝑦= 𝑥 ′ −𝑝𝑦′ so that 𝑎𝑥+𝑏𝑦= gcd 𝑎,𝑏 .