DEALING WITH NP-C PROBLEMS

Randomized Algorithm

NP-Complete Problem

A problem that, right now, we need exhaustive search

Example: SAT TSP Vertex Cover Etc.

Travelling Salesman Problem (TSP) Find a sequence

of city to visit Minimize the cost

of travel

Example

Route CostA B C DA B D CA C B DA C D BA D B CA D C B

Example

Route CostA B C D 20 + 30 + 12 + 35A B D C 20 + 34 + 12 + 42A C B D 42 + 30 + 34 + 35A C D B 42 + 12 + 34 + 20A D B C 35 + 34 + 30 + 42A D C B 35 + 12 + 30 + 20

General Search Algorithm

while has_unexplored_solution() { Solution sol = gen_another_solution() int cost = evaluate(sol) if (cost < min_cost) { min_cost = cost; min_sol = sol; }}

Time = number of solutions * evaluation

TSP Solution Space

N cities (N-1)! / 2

assume symmetric graph

What to do if N is too large?

Relaxation

In practice, we don’t really need the “best” solution But we need it now

Something close to now (1 min, 1 hour, etc.. But not 100 years)

Just something being “near” the best solution

One who try to optimize everything is bounded to be unhappy

Approximation Algorithm

Bounded solution

If solving for “optimum” solution requires exponential time What polynomial time could give us? What can we say about the solution from

polynomial time?

Approximation Ratio

The “best” solution for IOur algorithm

Problem Instance

Approximation ratio, upper bound of our sub-optimal

Approximation Algorithm

Algorithm that run fast (polynomial time) that give good approximation ratio

Reasonable choice when dealing with NP complete problem

Clustering

Input: Set of points X Integer k

Output: Partition of points into k set Such that the diameter of each set is

minimized

Metric Property

A function d(x,y) such that1. d(x,y) >= 02. d(x,y) = 0 if and only if x = y3. d(x,y) = d(y,x)4. d(x,y) < d(x,z) + d(z,y)

TriangularInequality

Example

Approximated Version

Guarantee

Ratio = 2 i.e., the resulting diameter is not

more than twice of original

Why ?

Let p be the point in X that is farthest from μ1, μ 2, μ 3,…, μ k

Let r be the distance from p to it’s closest center

Then… by triangular inequality, every cluster has diameter at most 2r

So what?

How r relate to the optimal solution? There are k + 1 points

μ1, μ 2, μ 3,…, μ k, p Such that they are all at least r from the

others So, any partition into k set must have

some set that contains at least two of tem

That set must has diameter at least r

Approximated Euclidian TSP TSP such that distance between two

cities is a Metric What is closely relate to TSP and can

be easily compute?

MST and TSP

Given an answer for TSP It is a cycle Remove one edge from the answer The result is path that is also a

spanning tree (not minimal) Let p be that path

From MST to TSP

Given an MST Do DFS, the result of visiting is

simply a cycle that visit every vertex Also visit some vertex multiple times

From MST to TSP

Length of that path is at most twice of the best TSP path

Fix the path into TSP Simply skip the vertex that is about

to re-visit and move to the next vertex in the list

Fixing the PathBy triangular inequality, the new path is shorter

Approximated 01-Knapsack Input

A number W, the capacity of the sack n pairs of weight and price ((w1,p1),(w2,p2),…,(wn,pn))

wi = weight of the ith items pi = price of the ith item

Output A subset S of {1,2,3,…,n} such that

is maximum

Si

ip

Si

i Ww

Guarantee

Pick any ε > 0 Result value is at least (1 – ε) of the

maximum value

Approximated 01-Knapsack

Knapsack can be solved using dynamic programming

Using O(nW) W = sum of weight We can derive similar algorithm

using O(nV) where V = sum of values

O(nV) knapsack

Let K(v) be the “minimal weight” when sum of selected value is v

If the ith item is in the best solution K(v) = K(v – pi) + wi

But, we don’t really know that the ith

item is in the optimal solution So, we try everything K(v) = min1≤i ≤ n(K(v – pi) + wi)

Approximation

Since it is O(nV) Can we reduce V?

To improve running time

Value scaling

Scale the price by a constant Resulting price is at most n/ε Thus, the running time is (n3/ ε)

The optimal solution

Let S be the selected subset of the optimal

Let K* be the maximum value Find the result of the rescaled input

Approximation Ratio

Rewrite in terms of K*

Let Ŝ be the set of selected rescaled item

Search with bounded resource

RANDOM SEARCH

General Search Algorithm

while time_not_exceed() { Solution sol = random_a_solution() int cost = evaluate(sol) if (cost < min_cost) { min_cost = cost; min_sol = sol; }}

Time is bounded Best solution is not guaranteed

Does it work?

If we have “enough” time…

Eventually, we will hit the “right” answer

It’s ok to be greed

Hill Climbing

Can we improve Random Search? Anything better than randomly

generate a new answer?

0-1 Knapsack Problem

Pick a combination of items

Solution Space 00000 (0) 00001 (1) 00010 (2) 00011 (3) 00100 (4) . . 11111 (31) 32

solutions in total

Evaluation Function

Solution

eval

(1) (2) (3) (4)(5) (6) …

Neighbor of Solution

In solution space 00100 (4) is close to

00011 (3) and 00101 (5)

Hill Climbing Search Generate only the neighbor solution Move to the best neighbor solution

Solution sol = random_a_solution()while time_not_exceed() { Solution nb[] = gen_all_neighbors(sol) Solution nb_sol; int nb_min; For all x in nb { int cost = evaluate(x) if (cost < nb_min) { nb_min = cost; nb_sol = x; } } if (nb_min < min_cost) { min_cost = nb_min; sol = nb_sol; }}

If cost does not improve, we might stop

Several definition of neighbor

0

2

4

6 0

2

4

6

-1

-0.5

0

0.5

1

0

2

4

6

Best Problem for Hill Climbing Unimodal

Bad problem for hill climbing Multimodal

Local Minima

Hill climbing is a local search (solver can define their own “local”)

It could stuck at local minima Need something to fix it

If stuck Randomly generate another solution and

start from that solution

O’ mother nature, I worship thou

Simulated Annealing

Annealing

A material (metal) is heated and then slowly cooled down

Rearrangement of atom Because, if not heated, atom is stuck at

irregular position from inertia That’s like the “Local Minima”

Simulated Annealing

Just like hill climbing But solution is allowed to move to

lower position With some probability

With inverse proportional to the elapsed time

Help escape from local minima

Simulated AnnealingSolution sol = random_a_solution()while time_not_exceed() { Solution nb = gen_one_neighbors(sol) int cost = evaluate(nb); if (cost < min_cost) { sol = nb; min_cost = cost; } else { if (random() < chance_at_time(t)) { sol = nb; min_cost = cost; } }}

Fitness Function

Randomized for Decision Problem

Decision Problem

We look for “the solution” Not the best Answer is either yes/no

SAT Problem

Find assignment to x,y,z that make this expression true

evaluate(sol) is either “true” or “false”

Fitness of SAT

Use other function for hill climbing A function that is maximal when

evaluate(sol) is true Should return something near maximum

when sol is near the “correct” answer

Example for SAT

Fitness = number of clause that is true

GA? SA? Random? Hill Climbing? Lot’s more….

What to use?

No Free Lunch Theorem

algo A outperform algo B on average For example Hill Climbing >>>

Random? Here’s come the “No Free Lunch”

theorem No such A,B exist

Definition: Black Box algorithm Oracle-based model

Does not ask same question twice

X1

X2

F(X1)F(X2)

Definition: algorithm

Sample Dm = all sample of size m D = all sample = D1 U D2 U … Optimization Algorithm is a mapping

From D to a : d D { x | x dx }

Given Algorithm a m number of oracle call Cost function f

Performance = P(dym | f , m , a )

Performance measure in dym : Φ(dy

m ) e.g., minimization

Definition: performance

Desiredoutput Proble

mat hand

Resource

Algorithm

The NFL theorem

For any two algorithms a1, a2

The chance of getting any particular dym

P(dym |f,m,a)

Is equal when sum over all possible function

Why?

Why?

= finite search space = finite cost value space Opt. problem f : =

= all possible function Size = ||||

Large, but finite

Implication of NFL

NFL says There are no universally good algo if A do better than B on some problem

B must do better than A on other problem E.g. If A beat random search on some

set… A must performs worse than random

for the rest of problems

Implication of NFL

If someone write a paper saying “We design A for one particular Problem

F” “We compare it with B on test problems

f1,f2,f3” “Result show that A is better than B” “My algo should be better than B on all

F” It might be wrong…

Another view of NFL

Better indicator for performance NFL is proven when P(f) is flat

Flat P(f) is sufficient... But not necessary In real life, P(f) is not flat However

Some knowledge must put into A structure of problem does not justify the

choice

Conclusion

Randomized algorithm works well in practice

But you need to at least put some knowledge it in

Given a problem, if you really has no idea, try them first…