# Example: Boolean expression Boolean Expressions The Satisfiability Problem

The Above Picture is Related Image of Another Journal

## Example: Boolean expression Boolean Expressions The Satisfiability Problem

Chaminade University of Honolulu, US has reference to this Academic Journal, The Satisfiability Problem Cook?s Theorem: An NP-Complete Problem Restricted SAT: CSAT, 3SAT Boolean Expressions Boolean, or propositional-logic expressions are built from variables in addition to constants using the operators AND, OR, in addition to NOT. Constants are true in addition to false, represented by 1 in addition to 0, respectively. We?ll use concatenation (juxtaposition) in consideration of AND, + in consideration of OR, – in consideration of NOT, unlike the text. Example: Boolean expression (x+y)(-x + -y) is true only when variables x in addition to y have opposite truth values. Note: parentheses can be used at will, in addition to are needed so that modify the precedence order NOT (highest), AND, OR.

Related University That Contributed for this Journal are Acknowledged in the above Image

The Satisfiability Problem (SAT ) Study of boolean functions generally is concerned alongside the set of truth assignments (assignments of 0 or 1 so that each of the variables) that make the function true. NP-completeness needs only a simpler question (SAT): does there exist a truth assignment making the function true? Example: SAT (x+y)(-x + -y) is satisfiable. There are, in fact, two satisfying truth assignments: x=0; y=1. x=1; y=0. x(-x) is not satisfiable. SAT as a Language/Problem An instance of SAT is a boolean function. Must be coded in a finite alphabet. Use special symbols (, ), +, – as themselves. Represent the i-th variable by symbol x followed by integer i in binary.

Example: Encoding in consideration of SAT (x+y)(-x + -y) would be encoded by the string (x1+x10)(-x1+-x10) SAT is in NP There is a multitape NTM that can decide if a Boolean formula of length n is satisfiable. The NTM takes O(n2) time along any path. Use nondeterminism so that guess a truth assignment on a second tape. Replace all variables by guessed truth values. Evaluate the formula in consideration of this assignment. Accept if true. Cook?s Theorem SAT is NP-complete. Really a stronger result: formulas may be in conjunctive normal form (CSAT) ? later. To prove, we must show how so that construct a polytime reduction from each language L in NP so that SAT. Start by assuming the most resticted possible form of NTM in consideration of L (next slide).

??????????? ???? ??????? ??????? ???? ? ???? ???? ? ???? & ?? ?????? ????? ??? ???? ??? ??? ???????????????????????????????????? ????? ??? ?????? ?????? ???????? ????? ???? ? ?? ????? ???? ????? ????? ???? ???? ???? ? ??-1 < ?????????? ???? ? ??-2 < ?????????? ??????? ?? ????????? ??????? ? ? !

Assumptions About NTM in consideration of L One tape only. Head never moves left of the initial position. States in addition to tape symbols are disjoint. Key Points: States can be named arbitrarily, in addition to the constructions many-tapes-to-one in addition to two-way-infinite-tape-to-one at most square the time. More About the NTM M in consideration of L Let p(n) be a polynomial time bound in consideration of M. Let w be an input of length n so that M. If M accepts w, it does so through a sequence I0?I1???Ip(n) of p(n)+1 ID?s. Assume trivial move from a final state. Each ID is of length at most p(n)+1, counting the state. From ID Sequences so that Boolean Functions The Boolean function that the transducer in consideration of L will construct from w will have (p(n)+1)2 ?variables.? Let variable Xij represent the j-th position of the i-th ID in the accepting sequence in consideration of w, if there is one. i in addition to j each range from 0 so that p(n).

Picture of Computation as an Array Initial ID X00 X01 ? X0p(n) X10 X11 ? X1p(n) I1 Ip(n) Xp(n)0 Xp(n)1 ? Xp(n)p(n) . . . . . . Intuition From M in addition to w we construct a boolean formula that forces the X?s so that represent one of the possible ID sequences of NTM M alongside input w, if it is so that be satisfiable. It is satisfiable iff some sequence leads so that acceptance. From ID?s so that Boolean Variables The Xij?s are not boolean variables; they are states in addition to tape symbols of M. However, we can represent the value of each Xij by a family of Boolean variables yijA, in consideration of each possible state or tape symbol A. yijA is true if in addition to only if Xij = A.

Points so that Remember The boolean function has components that depend on n. These must be of size polynomial in n. Other pieces depend only on M. No matter how many states/symbols m has, these are of constant size. Any logical formula about a set of variables whose size is independent of n can be written somehow. Designing the Function We want the Boolean function that describes the Xij?s so that be satisfiable if in addition to only if the NTM M accepts w. Four conditions: Unique: only one symbol per position. Starts right: initial ID is q0w. Moves right: each ID follows from the next by a move of M. Finishes right: M accepts. Unique Take the AND over all i, j, Y, in addition to Z of (-yijY+ -yijZ). That is, it is not possible in consideration of Xij so that be both symbols Y in addition to Z.

Starts Right The Boolean Function needs so that assert that the first ID is the correct one alongside w = a1?an as input. X00 = q0. X0i = ai in consideration of i = 1,?, n. X0i = B (blank) in consideration of i = n+1,?, p(n). Formula is the AND of y0iZ in consideration of all i, where Z is the symbol in position i. Finishes Right Somewhere, there must be an accepting state. Form the OR of Boolean variables yijq, where i in addition to j are arbitrary in addition to q is an accepting state. Note: differs from text. Running Time So Far Unique requires O(p2(n)) symbols be written. Parentheses, signs, propositional variables. Algorithm is easy, so it takes no more time than O(p2(n)). Starts Right takes O(p(n)) time. Finishes Right takes O(p2(n)) time.

Running Time ? (2) Caveat: Technically, the propositions that are output of the transducer must be coded in a fixed alphabet, e.g., x10011 rather than yijA. Thus, the time in addition to output length have an additional factor O(log n) because there are O(p2(n)) variables. But log factors do not affect polynomials Moves Right Xij = Xi-1,j whenever the state is none of Xi-1,j-1, Xi-1,j, or Xi-1,j+1. For each i in addition to j, construct a formula that says (in propositional variables) the OR of ?Xij = Xi-1,j? in addition to all yi-1,k,A where A is a state symbol (k = i-1, i, or i+1). Note: Xij = Xi-1,j is the OR of yijA.yi-1,jA in consideration of all symbols A. Constraining the Next Symbol ? A B C ? ? A q C ?

Moves Right ? (2) In the case where the state is nearby, we need so that write an expression that: Picks one of the possible moves of the NTM M. Enforces the condition that when Xi-1,j is the state, the values of Xi,j-1, Xi,j, in addition to Xi,j+1. are related so that Xi-1,j-1, Xi-1,j, in addition to Xi-1,j+1 in a way that reflects the move. Example: Moves Right Suppose ?(q, A) contains (p, B, L). Then one option in consideration of any i, j, in addition to C is: C q A p C B If ?(q, A) contains (p, B, R), then an option in consideration of any i, j, in addition to C is: C q A C B p Moves Right ? (3) For each possible move, express the constraints on the six X?s by a Boolean formula. For each i in addition to j, take the OR over all possible moves. Take the AND over all i in addition to j. Small point: in consideration of edges (e.g., state at 0), assume invisible symbols are blank.

Running Time We have so that generate O(p2(n)) Boolean formulas, but each is constructed from the moves of the NTM M, which is fixed in size, independent of the input w. Takes time O(p2(n)) in addition to generates an output of that length. Times log n, because variables must be coded in a fixed alphabet. Cook?s Theorem ? Finale In time O(p2(n) log n) the transducer produces a boolean formula, the AND of the four components: Unique, Starts, Finishes, in addition to Moves Right. If M accepts w, the ID sequence gives us a satisfying truth assignment. If satisfiable, the truth values tell us an accepting computation of M. Picture So Far We have one NP-complete problem: SAT. In the future, we shall do polytime reductions of SAT so that other problems, thereby showing them NP-complete. Why? If we polytime reduce SAT so that X, in addition to X is in P, then so is SAT, in addition to therefore so is all of NP.

CSAT so that 3SAT Running Time This reduction is surely polynomial. In fact it is linear in the length of the CSAT instance. Thus, we have polytime-reduced CSAT so that 3SAT. Since CSAT is NP-complete, so is 3SAT.

## Nugent, Ted Interim Program Director

Nugent, Ted is from United States and they belong to Interim Program Director and work for KWFM-AM in the AZ state United States got related to this Particular Article.

## Journal Ratings by Chaminade University of Honolulu

This Particular Journal got reviewed and rated by Moves Right ? (2) In the case where the state is nearby, we need so that write an expression that: Picks one of the possible moves of the NTM M. Enforces the condition that when Xi-1,j is the state, the values of Xi,j-1, Xi,j, in addition to Xi,j+1. are related so that Xi-1,j-1, Xi-1,j, in addition to Xi-1,j+1 in a way that reflects the move. Example: Moves Right Suppose ?(q, A) contains (p, B, L). Then one option in consideration of any i, j, in addition to C is: C q A p C B If ?(q, A) contains (p, B, R), then an option in consideration of any i, j, in addition to C is: C q A C B p Moves Right ? (3) For each possible move, express the constraints on the six X?s by a Boolean formula. For each i in addition to j, take the OR over all possible moves. Take the AND over all i in addition to j. Small point: in consideration of edges (e.g., state at 0), assume invisible symbols are blank. and short form of this particular Institution is US and gave this Journal an Excellent Rating.