Euclidean Algorithm, Divide and Conquer - Notes | POLI 10, Study notes of Political Science

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CSE 202 - More Math

CSE 202 - Algorithms

Euclidean Algorithm

Divide and Conquer

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Euclidean Algorithm

GCD stands for “greatest common divisor”.

E.g. GCD(10, 25) = 5.

If GCD(A,B)=1, we say A and B are relatively prime.

Note that GCD(N, 0) = N. (At least for N>0.)

Thm: If A + kB = C and B

∫

0, then GCD(A,B) = GCD(B,C).

Proof. Let g = GCD(A,B) and h = GCD(B,C).Since g divides A and B, it must divide A + kB = C.Since g is a common divisor of B and C, it must be

their

greatest common divisor, i.e. g

h.

Reversing the roles of A and C, we conclude that h

g.

Thus, g = h. QED

The Euclidean Algorithm finds GCD(A,B).Given A > B > 0, set C = A mod B (so C = A - kB for some k and C<B).

This reduces the problem of finding GCD(A,B) to the smaller problem,

finding GCD(B,C). Eventually, we’ll get to GCD(X,0) = X.

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Euclidean Algorithm

Example: Find GCD of 38 and 10.38 mod 10 = 810 mod 8 = 28 mod 2 = 0GCD(2,0) = 2

Can write “2” as linear combination of 10 and 38.

8 = 38 - 3x102 = 10 – 1x

= 10 – 1x(38–3x10)

2 = 4x10 – 1x

Lets you write 8 in terms of 38 and 10Lets you write 2 in terms of 10 and 8.

Change 8’s to 10’s

and 38’s

Voila!

Extended

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Euclidean Magic

Solve Diophantine Equations:

• E.g. “Write 6 as an integer combination of 38 and 10”
  • First write GCD(38,10) as a combination (2 =

4x10 – 1x38),

then multiply by 6/GCD = 3

(6 = 12x10 – 3x38).

Rational approximations:

E.g. 31416 = 3x10000 + 1416

10000 = 7x1416 + 88

Let’s pretend 88

1. We then have 1416

Thus, 31416

3x10000 + 10000/7 = 10000x(3+1/7)

Thus, 3.

We can get all “close” approximations this way.

Linear equations over integers

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Divide & Conquer

• Basic Idea
  • Break big problem into subproblems– Solve each subproblem
    • Directly if it’s very small• Otherwise recursively break it up, etc.
      • Combine these solutions to solve big problem

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Faster Multiplication ??

Standard method for multiplying long numbers:

(1000a+b)x(1000c+d) = 1,000,000 ac

• 1000 (ad + bc)

• bd

Clever trick:(1000a+b)x(1000c+d) = 1,000,000 ac

• 1000 ( (a+b)(c+d) – ac - bd)

• bd

One length-k multiply = 3 length-k/2 multiplies and a

bunch of additions and shifting.

On computer, might use 2

16

in place of 1000

Fine print: If “a+b”

has 17 bits, drop the top bit, but add extra c+d in the right

place. Handle overflow for c+d similarly.

4 multiplies: ac, ad, bc, bd

3 multiplies:

ac, (a+b)(c+d), bd

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Recursion Tree

c n

c n/

c n/

c n/

c n/

c n/

c n/

c n/

c n/

1 depth-0 node 3 depth-1 nodes

9 depth-2 nodes

...

3

lg n

depth-lg nnodes

...

...

c

c

c

...

c

c

c

...

...

c

c

c

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From the picture, we know ...

T(n) < cn ( 1 + 3(1/2) + 9(1/4) + ... + 3

lg n

(1/ 2

lg n

) )

< cn ( 1 + 3/2 + (3/2)

2

• ... + (3/2)

lg n

).

Lemma 1: For a

∫

1, a

k

• a

k-

• ... 1 = (a

k+

• 1. / (a-1)

Proof:

(a

k

• a

k-

• ... 1) (a-1) =

a

k+

• 1

It’s obvious! (Multiply it out – middle terms cancel.)

By lemma 1, T(n) < cn ( (3/2)

(lg n + 1)

-

1. / ((3/2)-1)

< cn ( (3/2)

lg n

(3/2) – 0) / (1/2)

= c n (3/2)

lg n

(3/2)/(1/2)

= 3 c n (3/2)

lg n

.

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Key points

• Divide & Conquer can magnify a small

advantage.

• Recursion tree gives a method of solving

recurrence equations.

• We’ll look at other methods next.
  • Lemmas 1 and 2 are useful.
    • Make sure you’re comfortable with the math.