Some NP-Competes, NP-Hards and Reductions

Kavosh Havaledarnejad Icarus.2012@yahoo.com

Abstract:

It is not clear for human race that NP is equal to P or NP is not equal to P. But however we

can extend knowledge regarding NP problems. In this article we review some new NP-

Complete and NP-Hard problems and reductions among them. These give us deeper insight

to some problems like 3-SAT, Timetabling, Boolean Satisfiability and Maximum Clique.

Keywords:

Boolean Satisfiability, RSS, 3-RSS, Max 2-SAT, Timetabling, k-SAT, k-MHC, Maximum

Clique, Max True 2-SAT

1. Introduction

Every theorem can have only two conditions: "Theorem be true" or "Theorem be false".

is a theorem. It is not clear for human race that or but if we

consider that there is a pure consciousness then this pure consciousness certainly knows that

what is the answer: or . Thus we know that exactly one of these cases is

true. Researcher believes that there is anything wrong with this question and that the question

is decidable thus has a correct answer. But we cannot prove that: If we have a question that

has an answer then one can prove the correctness of this answer. Thus we cannot conclude

that the question is provable although we know it has exactly one answer. However we can

hope that one day a lucky scientist may prove the or prove the . Thus

continue to research is essential. We review in this paper some new NP-Hards ( that some of

them are also NP-Completes ) and some reductions between them. We will begin with

Boolean Satisfiability and showing that how it can map to Rules States Satisfiability ( RSS ).

Then we will show how RSS can transform to 3-RSS. Then we will show how we can

reformulate a 3-SAT instance to a 3-RSS instance and vise versa. We will explain a reduction

from Timetabling problem to 3-RSS. We describe that every k-SAT instance is immediately

an instance of k-Maximum Hyper Clique ( k-MHC ). Then we will show that we can obtain a

3-MHC instance from a Maximum Clique instance. We will explain Max True 2-SAT and a

reduction to this problem from Maximum Clique. Finally we propose a new NP-Hard

problem namely Max Var 2-SAT and show how Max 2-SAT problem can reduce to this

problem.

2. Boolean Satisfiability to RSS

Consider a Boolean Satisfiability problem in the form of conjunction of several Boolean

formulas. Let they be . Consider one of them like that defines a Boolean

relation on some variables. For these variables has a truth table. Every true row of this truth

table is a state of this formula . Thus we can consider rules corresponding Boolean

Formulas and in each rule: we can consider several states corresponding true rows of its truth

table. Consider 2 states and from two rules and . Every or denotes an assignment 1

and 0 to some Boolean variables of main problem. If there exist one variable like that one

of them select be 1 and one of them select be 0 then we say and have conflict and are

disconnected otherwise we say they are connected.

For the sake of convenience we detail an example. Consider we have the formula below as an

instance of Boolean Satisfiability.

( ) ( ) ( )

This Formula is conjunction of three Boolean Formulas. We label them respectively by

. Then we draw truth tables for them respectively Table. 1, Table. 2 and Tale. 3.

x y R

s 1 0 1

t 0 1 1

Table. 1

y z Q

s 1 0 1

t 0 1 1

Table. 2

x z P

s 1 1 1

t 0 0 1

Table. 3

We also labeled every true row of truth tables. Consider know state of rule and state of

rule . These two states have conflict and are disconnected because in of : is 0 and in

of : is 1. In this way we can draw the structure of RSS instance like Fig. 1.

Fig. 1

Fig. 1 shows a 2-RSS problem. 2-RSS problem is polynomial solvable but every -RSS

problem is a NP-Complete problem. If we solve the reduced RSS instance then we can easily

find the answer of main Boolean Satisfiability problem.

3. Reducing RSS to 3-RSS

Fig. 2 Expanding a 4 state rule to two 3 state rule

A k-RSS instance is a RSS instance with at most k states per each rule. We can reduce a k-RSS

instance to a 3-RSS instance where every rule has at most 3 states. Consider a rule with states. We

can divide these states to two sets and where they satisfy:

| | | | | | | |

Then we can replace old rule with 2 new rules that first consists of ( ) and a new extra state and

second consists of ( ) and a new extra state. Extra states have conflict together and states of ( ) and

states of ( ) have conflict together too ( Fig. 2) . Tues system must select only one of states of ( ) or

states of ( ) means mechanism works. We continue this division to access to the case that all rules

have exactly 3 states.

4. Reformulating a 3-SAT instance to a 3-RSS instance

Consider an instance of 3-SAT in form:

( ) ( ) ( ) ( )

⋀(⋁

)

If we extend this formula we have:

⋁ (⋀

)

{ }

( ) ( ) ( )

Every satisfying assignment on 3-SAT instances must select one literal from each clauses.

The formula 4.1 and 4.2 are the same and formula 4.2 will be true if at least one of

parenthesis be true. But a parenthesis is not true when it contains two literals that be a

variable and its negation like and ̅, In this case we say that literals have conflict. We wish

to find a parenthesis that be true. This is exactly 3-RSS problem. Consider rules

corresponding clauses and in each rule 3 states corresponding 3 literals of that clause. Two

4.1

4.2

states are disconnect iff corresponding literals be a variable and its negation. Now we have a

3-RSS problem.

5. Reformulating a 3-RSS instance to a 3-SAT instance

We can easily show a 3-RSS problem using Ternary Algebra ( see [1] ). In this case we have

several clauses defined on ternary variables. Each clause contains two literal and each

literal is negation of a color for example:

( ̅ ̅ ) ( ̅ ̅ ) ( ̅ ̅ )

For example the clause ( ̅ ̅ ) shows a conflict between red state of with green state

of . In general we can show a 3-RSS instance with this formula:

( ) ( ) ( ) ( )

⋀(⋁

)

If we extend this formula we have:

⋁ (⋀

)

{ }

( ) ( )

The formula 5.1 and 5.2 are the same and formula 5.2 will be true if at least one of

parenthesis be true. But a parenthesis is not true when it contains three literals that be a

variable like in the forms ̅ and ̅ and ̅ ( If only two of them exist then can be

the third color ), In this case we say that literals have conflict. We wish to find a parenthesis

that be true. This is exactly 3-SAT problem. Please note that every literal is of the form ̅

or ̅ or ̅ . Consider Boolean variable corresponding clauses of 3-RSS and in each

Boolean variable 2 states corresponding 2 literals of that clause. Three states have conflict iff

corresponding literals be a negation of three colors of a variable. Now we have a 2,3-CSP

problem that is immediately a 3-SAT problem.

6. Reducing a type of Timetabling to 3-RSS

Timetabling problem is problem of assigning a program based on limitations between

resources of an education center like a high school or a university. Here we propose a type of

timetabling problem that is a NP problem and reduce this type of timetabling to 3-RSS

problem. Aim is to produce a weekly program for the center and resources are: 1- lessons, 2-

Times, 3-Teechers and 4-Rooms. For example lesson can be "Computational Complexity" or

"Designing Algorithms" and a time can be "Friday 9:30-11:00". We consider six conflict

between these resources. Fig. 3 shows these conflicts as every line defines a conflicts for

5.1

5.2

example when a teacher cannot teach a special lesson it is a conflict between that teacher and

that lesson and a teacher may say that doesn’t teach in a special room or for example one

special teacher may not be available in some times. We save these conflicts in the tables of a

data base.

Fig. 3

We cross the set times with the set rooms having a new set time-room and withdraw the

combinations that time and room have conflict and for each element of this set we establish

one rule for teacher ( time-room-teacher ) that its states are teachers and on rule for lesson (

time-room-lesson ) that its states are lessons. Also in the rules time-room-teacher we omit the

combinations ( states ) that teacher have conflict with that time or that room and in a rule

time-room-lesson we omit the combinations ( states ) that lesson have conflict with that time

or that room. Now it remains to use conflict between teachers and lessons. For every teacher

and lesson that have conflict we let corresponding states have conflict. Also whereas a

teacher cannot teach in a same time in two rooms, in the rules time-room-teacher we consider

conflicts between the states that belong to same teacher from the rules that have the same

times. Now we have a RSS problem that solving this causes solving the main timetabling

problem using the method described in prior sessions we can reduce this problem to a 3-RSS

problem.

7. Maximum Hype Clique problem

Maximum Hyper Clique problem is a NP-Complete problem defined on Hype Graphs. In a

graph every connection defines exactly between two nodes where for example in a 3-Graph,

connections defines exactly between three nodes. We call such graphs: hyper-graphs. A k-

graph is a graph that connection defines exactly between nodes. We can show these

connections with sets of size of nodes.

Maximum Clique is problem of finding the maximum subset of nodes that whole of them be

connected where a Maximum Hyper Clique problem in a k-graph ( k-MHC ) is finding

maximum subset of nodes that whole of them be connected by connections in this graph. For

be more precise, consider a subset containing nodes and that the graph is a k-graph. This set

of nodes of size has ( ) subsets of size . If whole of them be in the list of connections of k-

graph then it is a hyper clique. The problem is to find such a subset that is maximum.

8. k-SAT to k-MHC

Here we introduce a reduction from k-SAT to k-MHC. We establish a k-graph as follows. For

every variable of k-SAT problem we consider two nodes in the k-graph: one corresponding

true state and one corresponding false state of that variable. A clause of form ( ) in

the k-SAT defines a conflict between ̅ and ̅ and … thus we let a subset of nodes of size

be in the list of connections iff satisfies two conditions:

1- Its corresponding states of variables be exactly between deferent variables. It means

that we have any variable that two states of it be in the set of corresponding nodes.

2- This subset of nodes doesn’t be corresponding with a conflict in the main problem.

Thus we have a k-MHC problem if we solve this problem its answer will be correspond to the

answer of main problem.

9. Maximum Clique to 3-MHC

Here we define a reduction from Maximum Clique to 3-MHC ( the method is the same for k-

MHC in general ). For doing this we establish a 3-graph that nodes are the same as main

Maximum Clique problem. Every connection in the new problem is a set of nodes of size 3.

Iff they be two by two connected in the main problem we let it be in the list of connections

but if at least one of these don’t be, then it is not in the list of connections.

10. Max True 2-SAT

A Max True 2-SAT problem is a NP-Hard optimization problem. Question is to find a

satisfying assignment for the 2-SAT problem that satisfies whole the clauses ( exactly like a

classical 2-SAT ) and that maximum number of variables be in true state. Also this problem

has a decision edition k True 2-SAT that is problem of finding a satisfying assignment that

exactly variables be in True state.

11. A Reduction from Maximum Clique to Max True 2-SAT

Maximum Clique and Max True 2-SAT are deferent NP-Hard problems. However we

introduce a reduction from Maximum Clique to Max True 2-SAT. For every node in the

Maximum Clique problem we consider a variable in Max True 2-SAT. If two nodes like

and are disconnect from each other then, in corresponding variables there exist a conflict

between true state of and true state of . We can show such a conflict with the clause

( ̅ ̅). Then if we solve this problem as a Max True 2-SAT we have maximum number of

variables that are true and they are the nodes that we select as a maximum clique in the main

problem.

12. Max Var 2-SAT

Here we introduce a new NP-Complete problem: Max Var 2-SAT ( Maximum Variable 2-

SAT ) that there is a reduction from Max-2-SAT to this problem. In this problem we assign

values ( True or False ) to a sub-set of variables ( Assigned Set ) so that whole clauses exactly

between these variables be satisfied. Aim is to finding maximum such a sub set that have

such a property.

13. Reduction from Max 2-SAT to Max Var 2-SAT

Consider a Max 2-SAT problem. Let every clause in Max 2-SAT be a variable in Max Var 2-

SAT. Whereas in Max 2-SAT clauses are binary in Max Var 2-SAT variables will be

Boolean. We let every state True or False be corresponding to a literal of binary clause. For

every two literal that are a variable and its negation in Max 2-SAT, we consider a conflict

between corresponding states in Max Var 2-SAT ( note that a clause in form is a conflict

between ̅and ̅ in 2-SAT problems in general ). Now we have a Max Var 2-SAT problem

and if we assign values to some variables that variables are exactly clauses in main Max 2-

SAT problem thus such an assignment correspond to a solution for Max 2-SAT problem.

14. Conclusion

In this article we introduced some new reductions and new NP-Hard problems. To dealing

question there exist two approaches. One of them is to find a polynomial algorithm

for existing problems. And one of them is to find a new problem that be polynomial solvable

with an algorithm. Researcher is working in both approaches and the present paper was result

of these efforts. Also some of these reductions give us more precise insight regarding NP-

Hard problems.

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