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RANDOMIZED ROUNDING
RAJEEV MOTWANI AND
PRABHAKAR RAGHAVAN
PRESENTER : PRASANNA GANESAN
Extracted from
Randomized Algorithms ,
By Rajeev Motwani and Prabhakar Raghavan
Randomized Rounding
- This is a probabilistic method to convert
a solution of a relaxed linear problem into
an approximate solution to the original
problem
- The basic idea is to interpret the fractional
solutions provided by the linear program
as probabilities for rounding them.
- This is used when a linear program is
NP-hard.
A language L is NP-hard if, for all L’  NP , there is
a polynomial reduction L’ to L.
It actually means that if you apply a polynomial
function over any of the language L  NP then it
results in another L’  NP.
For such problems we need to relax some of the
conditions in linear programming which demands the
application of randomized rounding over the solution
sets.
We shall see a simple example of this in detail.
WIRING PROBLEM
Assume that this is a grid of logic gates, which are
connected by four wires, and the gates are numbered
from 1 to n. The wires are assumed to run over the
boundaries.
Our problem is to connect the set of nets connected
together.
A net may be connect 2 or more gates.
As a simple example we consider a simple net which
connects just 2 gates.
1
3
2
4
2
1
3
4
Xi0 - When a net goes from horizontal to vertical
direction and
Xi1 - When a net goes from vertical and then to
Horizontal direction
Where i denotes the net or the wire in our example.
For example ,
X10 = 0 and X11 = 1 from the above fig.
Tb0 = { i | net i passes thro b if Xi0 = 1 }
Tb1 = { i | net i passes thro b if Xi1 = 1 }
They represent a set of the total no of Xi0 and Xi1
respectively.
ws(b) = No of wires that pass thro b in a solution S
ws = maxb ws(b) [the max no of wires thro any
boundary
Our integer problem can be expressed as,
Minimize w which is the total no of wires used.
Where Xi0 , Xi1  {0,1} (  nets i )
Subject to
Xi0 + Xi1 = 1 (  nets i )
 Xi0 +  Xi1  w (  boundaries b )
Let W0 be the value of w in the above equations
To solve this,
We do linear program relaxation, which is
by replacing
Xi0 , Xi1  [0,1] (  nets i ).
i.e, they assume real values instead of
integer values.
By doing so we can obtain a solution for the
linear program, since the original program is NPhard.


Let X i,0 , X i,1  [0,1] (  nets i ) be the solutions

and let W the value of the objective function.

We note here that
W
< w0 because of our assumption

that W takes real values between 0 and 1.
Now we have to round the solutions,
For each i ,

set X i0 to 1 and X i1 to 0 with probability X i,0
Otherwise set X i0 to 0 and X i1 to 1
Thus for each i,

Pr [ X i0 = 1] = X i0

Pr [ X i1 = 1] = X i1
Note : The value closer to 0(or 1) is likely set to be
0(or 1).
Theorem :
Let  be a real number such that 0 <  < 1. Then
with probability 1-  , the global wiring S produced
by randomized rounding satisfies
Ws 

W
(1+  (


W ,  /2n))
 W0 (1+  ( W0,  /2n))
Proof:
To prove this lets have a brief introduction on
chernoff bounds and poisson trials.
Poisson Trials
Unlike fixed probability in bernoullis trials,we allow every Xi to
have a different probability, Pr[Xi = 1] = pi,and
Pr[Xi = 0] =(1-pi), then these event are called Poisson trials. A
Poisson trial by itself is really just a Bernoulli trial. But when you
have a lot of them together with different probabilities, they
are called Poisson trials. But it is very important that the Xi must
still be independent.
Chernoff bounds are a kind of tail bound. Like Markoff and
Chebyshev, they bound the total amount of probability of some
random variable Y that is in the “tail”, i.e. far from the mean.
Let X1,X2,…..Xn be independent poisson trials with
Pr[XI=1]=pi, then if X is the sum of Xi, and if  is E[x], for any
  (0,1]:
Pr[X> (1+  )  ] < [e  /((1+  )(1+  ))]  -------(1)
Here  is deviation given by  =  (  ,  ) suffices to keep
Pr[X> (1+  )  ] below  irrespective of the values of n and pis.

Now lets prove the theorem.
Consider a boundary b,
Since the solutions satisfy the constraints, we can say
that


 X i0 +  X i1 

W
The no of wires passing through b is
Ws(b) =  X i0 +  X i1
Since they r independent poisson trials



E[Ws(b)] =  E[ X i0] +  E[ X i1] =  X i0 +  X i1  W
By the result from Chernoff bound given in
equation (1),
Pr[Ws(b) 

(1+  (

W

W ,  /2n))
]   /2n
Thus for any particular boundary the probability is
atmost  /2n.

Since W < w0
Ws 

W
(1+  (


W ,  /2n))
 W0 (1+  ( W0,  /2n))
In a n  n array there are fewer than 2n boundaries.
Therefore by summing up all the failure probability
we get the upper bound
Some of the applications of Randomized roundings-VLSI wiring Problem
-Max-Sat problems
-Primality Testing
-Routing