CSCI381 Fall 2004

1
CS Theory Review

• P/NP
• Complexity of some example algorithms

• CSCI381 Fall 2004
• GWU

2
Example Traveling Salesman
Given a set of points and the cost of going from
each to another, find a least cost path that
covers all points
n points Try all combinations T(n) n!
3
Definitions

• Polynomial time When the worst-case running time
  of an algorithm is polynomial in the size of the
  input
• Exponential time When the worst-case running
  time of an algorithm is exponential in the size
  of the input

4
Polynomial vs. Exponential
5
Non-deterministic Polynomial NP

• When you can check that a solution is correct in
  polynomial time
• Example Travelling Salesman, all polynomial-time
  algorithms, discrete log, roots of a polynomial

6
Computational Complexity

• An NP-complete problem is one which, if solved in
  polynomial time, can be used to solve all other
  NP problems in polynomial time.
• Example?

7
Computational Complexity

• any NP-complete problem is solved in
  polynomial-time,
• ? P NP
• i.e. (problems solvable in polynomial time)
  (problems for which solutions can be checked in
  polynomial time)

8
Aside Computational Complexity

• There are NP problems not known to have
  polynomial-time solutions which are also not
  known to be NP-complete i.e. they are difficult,
  but perhaps not among the most difficult

9
Computational Complexity

• Computational complexity of the following
  operations on x (k bit) and y (l bit), k ? l
• x y
• x y
• xy
• Floor(x/y) O(l(k-l))

10
Euclidean Algorithm

• gcd(m, n) / m gt n /
• (a, b) (m, n) / Initialize /
• while (b?0) (a, b) (b, a bq) /Where q
  ?a/b? /
• return(a)
• Complexity?

11
Computational Complexity mod n

• Computational complexity of the following
  operations mod n, where n is a k-bit integer
• x y
• x y
• xy

12
Efficient exponentiation(from Memon notes)

• Usual approach to computing xc mod n is
  inefficient when c is large.
• Example 551 involves 51 multiplications mod n
• Instead, represent c as bit string bk-1 b0 and
  use the following algorithm
• z 1
• For i k-1 downto 0 do
• z z2 mod n
• if bi 1 then z z x mod n

13
Example

• Calculate 751 mod 11 efficiently
• 51 110011 25 24 21 20
• 751 ((((72)2)2)2)2 ? (((72)2)2)2 ? 72 ? 71
• How many multiplications did you need?

14
751 mod 11
i bi z
5 1
4 1
3 0
2 0
1 1
0 1