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Randomized Algorithms

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Randomized Algorithms

Morteza ZadiMoghaddam

Amin Sayedi

- Las Vegas
- Monte Carlo

- Always gives the true answer.
- Running time is random.
- Running time is bounded.
- Quick sort is a Las Vegas algorithm.

- It may produce incorrect answer!
- We are able to bound its probability.
- By running it many times on independent random variables, we can make the failure probability arbitrarily small at the expense of running time.

- Suppose we want to find a number among n given numbers which is larger than or equal to the median.

Suppose A1 < … < An.

We want Ai, such that i ≥ n/2.

It’s obvious that the best deterministic algorithm needs O(n) time to produce the answer.

n may be very large!

Suppose n is 100,000,000,000 !

- Choose 100 of the numbers with equal probability.
- find the maximum among these numbers.
- Return the maximum.

- The running time of the given algorithm is O(1).
- The probability of Failure is 1/(2100).
- Consider that the algorithm may return a wrong answer but the probability is very smaller than the hardware failure or even an earthquake!

- Suppose the output is Yes or No.
- One sided error.
- Two sided error.

- Bounded polynomial time in the worst case.
- If the answer is Yes; Pr[ return Yes] > ½.
- If the answer is No; Pr[ return Yes] = 0.
- ½ is not actually important.

- Bounded polynomial time in worst case.
- If the answer is Yes; Pr[ return Yes] > ½.
- If the answer is No; Pr[ return Yes] < ½.
- Unfortunately the definition is weak because the distance to ½ is important but is not considered.

- There are n computers.
- Each computer has a packet.
- Each packet has a destination D(i).
- Packets can not follow the same edge simultaneously.
- An oblivious algorithm is required.

- For any deterministic oblivious algorithm on a network of N nodes each of outdegree d, there is an instance of permutation routing requiring (N/d) ½.

- Pick random intermediate destination.
- Packet i first travels to the intermediate destination and then to the final destination.
- With probability at least 1-(1/N), every packet reaches its destination in 14n of fewer steps in Qn.
- The expected number of steps is 15n.

- You have m clauses and n boolean variables.
- Each clause contains some of variables or some of complements.
- A clause is satisfied if at least one of it’s variables are satisfied.
- We want to set the variables such that the number of satisfied clauses is maximized.

- There are 3 variables A, B and C.
- M1 = (A) or (B)
- M2 = (A) or (not B) or (not C)
- M3 = (C)
- M4 = (B) or (not C)
- M5 = (not C)

- Set A = True
- Set B = True
- Set C = False
- Four of the clauses are satisfied.

- This problem is a famous problem which has no polynomial time algorithm yet. It’s NP-hard.

- For any set of m clauses, there is truth assignment for the variables that satisfies at least m/2 clauses.

- Let Zi =1 if the i-th clause is satisfied and 0 otherwise.
- Set the variables in a random way.
- The probability of a clause with k variables to be true is 1- (1/(2k)) >= ½.
- So E[Z1]+…+E[Zm] >= ½.
- Thus there exist at least one assignment such that Z1+…+Zm >= ½.

- This problem is NP-hard so we seek for approximation algorithms.
- We have an algorithm that produces an answer which is at least ½ of the best answer.
- If all clauses have at least 2 literals then we have an algorithm that produces an answer which is at least ¾ of the best one.

- We want to maximize Z1+…+Zm.
- We have some inequalities:
- ∑ yi (if Xi is in Zj and is uncomplemented)
- ∑ (1-yi) (if Xi is in Zj and is complemented).
- This inequality must be hold:
- ∑ yi + ∑ (1-yi) >= Zj
- This problem could be solved using integer linear programming.
- We have to use linear programming.

- Solve the problem using linear programming.
- You get a real number for each yi or zi.
- Assign Xi true with the probability yi.
- The expected number of clauses that are satisfied is (1- 1/e) of the best answer.

- Using both algorithms and choosing the better answer gives us an answer which is at least ¾ of the best answer!!! Which is better than ½ and 1- 1/e.

- Every clause has at most 2 literals.
- We want to check if all clauses can be satisfied.
- It has polynomial algorithm.
- Assign random values to the variables.
- If all clauses are satisfied we are finished.
- If there is an unsatisfied clause, the value of one of it’s literals is different from the best answer.
- Change the value of one of the variables in this clause. You may make a good change or bad one.

- You are walking on a path.
- If you are on 0 you go to 1.
- If you are on i you go to i+1 or i-1 with equal probability.
- The expected number of steps to reach the end of the path is O(n2).
- So the given algorithm is O(n3).

- You want to check if two vertices u and v are in the same connected component.
- Start a random walk from v.
- Have a random walk of length 2n3.
- If you haven’t visited u, the probability of u to be in this component is less than ½.
- By repeating this algorithm, you can make the probability of failure arbitrarily small.

- Running time of algorithm is O(n3).
- Required space is O(logn).

- You want to find the diameter of set of n points in the space.
- Suppose I(x) is the convex body formed by the intersection of n sphere centered at n points with radius x.
- F(p) is distance between p and the point in the set that is farthest from p.

- Consider I(x) when x=F(p).
- For any q in S, if q is in I(x) then F(q)<F(p).
- And if q is not in I(x) then F(p)<F(q).

- Pick a point p in s at random. Computer F(p). [O(n)]
- Set x=F(p). Compute I(x). [O(n logn)]
- Find the points outside I(x). Call this set T. [O(n logn)]
- If T is empty return x as the answer, else continue on T.

- The running time of algorithm above is O(n log n).
- In each step, all points that have smaller F(x) than the chosen point are removed.

- Let G(V,E) be an undirected, connected graph with V={1,…,n} and |E|=m.
- The adjacency matrix A is an n n 0-1 matrix with Aij=Aji=1 if and only if the edge (i,j) is present in E.
- We are going to compute matrix D which Dij equals the length of a shortest path from vertex i to vertex j.

- Z A2
- Compute matrix A’ such that A’ij=1 if and only if i ≠ j and (Aij=1 or Zij>0)
- If A’ij=1 for all i ≠ j then return D = 2A’-A.
- Recursively compute the APD matrix D’ for the graph G’ with adjacency matrix A’.
- S AD’
- Return matrix D with Dij=2D’ij if Sij≥D’ijZii , otherwise Dij=2D’ij-1.

- The APD algorithm computes the distance matrix for an n-vertex graph in time O(MM(n)log(n)) using integer matrix multiplication algorithm.
- Matrix multiplication algorithm running in time O(n2.376).

- Suppose A and B are nn boolean matrices and P=AB is their product under Boolean matrix multiplication.
- A witness for Pij is an index k {1,…,n} such that Aik=Akj=1. Observe that Pij=1 if and only if it has some witness k.

- W -AB
- 2. for t=0,…, log(n) do
- 2.1. r 2t
- 2.2. Repeat 3.77log(n) times
- 2.2.1. Choose random R {1,…,n} with |R|=r .
- 2.2.2. Compute AR and BR .
- 2.2.3. Z ARBR .
- 2.2.4. for all (i,j) do
- if Wij < 0 and Zij is witness then Wij Zij

- 3. for all (i,j) do
- if Wij < 0 then find witness Wij by brute force.

- The BPWM algorithm is a Las Vegas algorithm for the BPWM problem with expected running time O(MM(n)log2(n)).
- The probability that no witness is found for Pij before the end of Step 2 is at most
- (1-1/2e)3.77log(n) 1/n.

- A successor matrix S for an n-vertex graph G is an n n matrix such that Sij is the index of a neighbor of vertex i that lies on a shortest path from i to j.

- Compute the distance matrix D=APD(A).
- for s={0,1,2} do
- Compute 0-1 matrix Dkj(s)=1 if and only if Dkj+1 = s (mod 3)
- Compute the witness matrix W(s)=BPWM(A,D(s)).

- Compute successor matrix S for G.

- Algorithm APSP computes the successor matrix for an n-vertex graph G in expected time O(MM(n)log2(n)).

- H G
- While H has more than 2 vertices do
- Choose an edge (x,y) uniformly at random from the edges in H.
- F F {(x,y)}.
- H H / (x,y).

- (C,V/C) the sets of vertices corresponding to the two meta-vertices in H=G/F.

- n |v|
- if n 6 then compute min-cut of G by brute-force enumeration else
- t 1+n/2
- Using Algorithm Contract, perform two independent contraction sequences to obtain graphs H1 and H2 each with t vertices.
- Recursively compute min-cuts in each of H1 and H2.
- Return the smaller of the two min-cuts.

- Algorithm Fastcut succeeds in finding a min-cut with probability (1/log(n)).
- Algorithm Fastcut runs in O(n2log(n)) time and uses O(n2) space.

- Finding MST in a graph with n vertices and m edges has a Las Vegas algorithm which has the expected running time O(n+m).
- But We don’t have enough time to Explain it !!!

- Devise an algorithm for the all-pairs shortest paths problem that does not use matrix multiplication and runs in time O(n3-) for a positive constant .
- Devise an algorithm for computing the diameter of an unweighted graph that does not use matrix multiplication and runs in time O(n3-) for a positive constant .

- Devise a Las Vegas or a deterministic algorithm for min-cuts with running time close to O(n2).
- Is there a randomized algorithm for min-cuts with expected running time close to O(m)?

Have a nice randomized life