CSCI 3160
Design and Analysis of Algorithms
Tutorial 4
Chengyu Lin
Outline
• More examples about 2-SAT
• More randomized algorithms
– Verifying Matrix Multiplication
– String Equality Test
• Design Patterns
2-SAT
• Given (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
• (x1, x2, x3, x4, x5) = (T, T, T, T, T) is a satisfying assignment.
• Suppose you do not know about this solution
• You do not even know if there exists a solution for this
formula
• How to decide if there is one using randomness?
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
• Start with a random assignment,
– say (x1, x2, x3, x4, x5) = (F, T, F, F, T)
Number of xis that agrees with the solution (i.e. number of i such that xi = T)
0
1
2
3
4
5
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
current
x1
x2
x3
x4
x5
F
T
F
F
T
• Find an unsatisfied clause: (x4∨x3)
• flip one of the value of x3 and x4 randomly
– If we flip x3, then we jump from 2 to 3
– If we flip x4, then we jump from 2 to 3
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
current
x1
x2
x3
x4
x5
F
T
T
F
T
• Find an unsatisfied clause: (x1∨¬x2)
• flip one of the value of x1 and x2 randomly
– If we flip x1, then we jump from 3 to 4
– If we flip x2, then we jump from 3 to 2
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
current
x1
x2
x3
x4
x5
F
F
T
F
T
• Find an unsatisfied clause: (x2∨¬x3)
• flip one of the value of x2 and x3 randomly
– If we flip x2, then we jump from 2 to 3
– If we flip x3, then we jump from 2 to 1
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
current
x1
x2
x3
x4
x5
F
T
T
F
T
• Find an unsatisfied clause: (x1∨¬x2)
• flip one of the value of x1 and x2 randomly
– If we flip x1, then we jump from 3 to 4
– If we flip x2, then we jump from 3 to 2
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
current
x1
x2
x3
x4
x5
T
T
T
F
T
• Find an unsatisfied clause: (x4∨¬x5)
• flip one of the value of x4 and x5 randomly
– If we flip x4, then we jump from 4 to 5
– If we flip x5, then we jump from 4 to 3
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
current
x1
x2
x3
x4
x5
T
T
T
T
T
• Find an unsatisfied clause: none
• We have a satisfying assignment! =)
Analysis
• To do the analysis, assume we have a satisfying
assignment in mind, call it solution
– Of course, we do not know if a satisfying assignment exists
when we run the algorithm
– we do this for the analysis only
• We compare the current assignment with the
solution
• And record the number of variables that are assigned
to the same T/F value in both (current and solution)
assignments
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
x1
x2
x3
x4
x5
Solution
T
T
T
T
T
current
F
T
F
F
T
• Find an unsatisfied clause: (x4∨x3)
• flip one of the value of x3 and x4 randomly
– If we flip x3, then we jump from 2 to 3
– If we flip x4, then we jump from 2 to 3
Number of xis that agrees with the solution (i.e. xi = T)
with prob. 1
0
1
2
3
4
5
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
x1
x2
x3
x4
x5
Solution
T
T
T
T
T
current
F
T
T
F
T
• Find an unsatisfied clause: (x1∨¬x2)
• flip one of the value of x1 and x2 randomly
– If we flip x1, then we jump from 3 to 4
– If we flip x2, then we jump from 3 to 2
Number of xis that agrees with the solution (i.e. xi = T)
with prob. 1/2
0
1
2
3
with prob. 1/2
4
5
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
x1
x2
x3
x4
x5
Solution
T
T
T
T
T
current
F
F
T
F
T
• Find an unsatisfied clause: (x2∨¬x3)
• flip one of the value of x2 and x3 randomly
– If we flip x2, then we jump from 2 to 3
– If we flip x3, then we jump from 2 to 1
Number of xis that agrees with the solution (i.e. xi = T)
with prob. 1/2
0
1
with prob. 1/2
2
3
4
5
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
x1
x2
x3
x4
x5
Solution
T
T
T
T
T
current
F
T
T
F
T
• Find an unsatisfied clause: (x1∨¬x2)
• flip one of the value of x1 and x2 randomly
– If we flip x1, then we jump from 3 to 4
– If we flip x2, then we jump from 3 to 2
Number of xis that agrees with the solution (i.e. xi = T)
with prob. 1/2
0
1
2
3
with prob. 1/2
4
5
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
x1
x2
x3
x4
x5
Solution
T
T
T
T
T
current
T
T
T
F
T
• Find an unsatisfied clause: (x4∨¬x5)
• flip one of the value of x4 and x5 randomly
– If we flip x4, then we jump from 4 to 5
– If we flip x5, then we jump from 4 to 3
Number of xis that agrees with the solution (i.e. xi = T)
with prob. 1/2
0
1
2
3
with prob. 1/2
4
5
2-SAT
• (x1∨¬x2)∧(x2∨¬x3)∧(x4∨x3)∧(x5∨x1)∧(x4∨¬x5)
clauses
x1
x2
x3
x4
x5
Solution
T
T
T
T
T
current
T
T
T
T
T
• Find an unsatisfied clause: none
• We have a satisfying assignment! =)
Number of xis that agrees with the solution (i.e. xi = T)
0
1
2
3
4
5
2-SAT
with prob. 1
0
1
2
3
with prob. 1/2
0
1
2
with prob. 1/2
0
1
1
5
with prob. 1/2
3
4
5
4
5
with prob. 1/2
2
3
with prob. 1/2
0
4
2
with prob. 1/2
3
4
with prob. 1/2
5
with prob. 1/2
0
1
2
3
4
5
0
1
2
3
4
5
• We always move to the right with probability at least 1/2
2-SAT
• We always move forward with probably at least 1/2
– a satisfying assignment must satisfy every clause
e.g. (x1∨¬x2)
x1
x2
good
T
F
good
T
T
good
F
F
bad
F
T
• When we find an unsatisfied clause, it must be bad
• The values of the literals in a satisfying assignment must
be one of the good ones
• Hence, can always flip one to move to the right
2-SAT
• We always move forward with probably at least 1/2
• When we reach n (i.e. 5 in the example), we have a
satisfying assignment.
– Sufficient condition, not necessary.
– There may be other satisfying assignments
– But that will only improve the running time
• This is random walk on line
• and so we expect O(n2) jumps to hit n (refer to lecture
notes)
2-SAT
Randomized algorithm:
•Repeat O(n2) times,
– if we hit 5 in the middle of these O(n2) flips, return there is
an assignment
– otherwise, return there is NO satisfying assignment
Verifying Matrix Multiplication
• Given 3 n x n integer matrices A, B and C, verify if AB
=C
• Naïve algorithm
– Compute AB, check if it is equal to C
– Runs in time O(n3), dominated by computing AB
• How to do faster with high probability?
Verifying Matrix Multiplication
• Given 3 n x n integer matrices A, B and C, verify if AB
=C
• Consider AB – C
• Check if AB – C is a zero matrix
• Lecture: how to check if two polynomials are
identical?
– Evaluate the polynomials at different points
Verifying Matrix Multiplication
• Given 3 n x n integer matrices A, B and C, verify if AB
=C
• Now, multiply AB – C by different random vectors
• To avoid computing AB, we compute A(Bx) – Cx
We have
• Working over modulo 2,
• If AB ≠ C, then Pr[(AB - C)x ≠ 0] ≥ 1/2
String equality
• Alice has an n-bit string (a1, a2, …, an)
• Bob has an n-bit string (b1, b2, …, bn)
• Alice can communicate with Bob
• How to send as few bits as possible so that they
know the two strings are equal?
String equality
• Alice has an n-bit string (a1, a2, …, an)
• Bob has an n-bit string (b1, b2, …, bn)
• Represent
– (a1, a2, …, an) by a1 + a2x + … + anxn-1
– (b1, b2, …, bn) by b1 + b2x + … + bnxn-1
• Polynomial Identity Testing!
• This idea can be extended to pattern matching
Design Patterns
• Las Vegas algorithm
– Always produces the correct answer
– Usually only expected running time is guaranteed
– Example: randomized quick sort
Design Patterns
• Monte Carlo algorithm
– Running time is deterministic
– Small probability of error
– Usually repeats a simple procedure to get a large
success probability
– Example: min-cut
End
• Questions?