1 Lecture 23 Decision problems about regular languages –Programs can be inputs to other programs...

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Lecture 23

• Decision problems about regular languages– Programs can be inputs to other programs

• FSA’s, NFA’s, regular expressions

– Basic problems are solvable• halting, accepting, and emptiness problems

– Solvability of other problems• many-one reductions to basic problems

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Problems with programs as inputs

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Decision problems with numbers as inputs

• Examples– Primality:

• Input: integer n

• Yes/No question: Is n prime?

– Zero:• Input: integer n

• Yes/No question: Is n = 0?

– Equal:• Input: integers m, n

• Yes/No question: Is m=n?

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Decision problems with programs as inputs

• Examples– Lines of code:

• Input: program P, integer n

• Yes/No question: Does P have less than n lines of code?

– Empty language:• Input: program P

• Yes/No question: Is L(P) = {}?

– Equal:• Input: programs P, Q

• Yes/No question: Is L(P) = L(Q)?

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Language representation

• Empty language problem:– Input: program P

– Yes/No question: Is L(P) = {}?

• How might we represent input program P?– As a string P over ASCII alphabet

• What is the language that corresponds to empty language decision problem?– The set of strings representing programs which are yes

instances to the empty language problem

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Programs

• In this unit, our programs are the following three types of objects– FSA’s– NFA’s– regular expressions

• Previously, they were C++ programs– Review those topics after mastering today’s

examples

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Basic Decision Problems(and algorithms for solving them)

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Divisibility Problem

• Input– integers m,n

• Question– Is m a divisor of n?

• Algorithm DIV for solving divisibility problem– Apply Euclid’s algorithm for finding the greatest common divisor

to m and n

– If greatest common divisor is m THEN yes ELSE no

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Halting Problem

• Input– FSA M

– Input string x to M

• Question– Does M halt on x?

• Algorithm for solving halting problem– Output yes

• Correct because an FSA always halts on all input strings

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Accepting Problem

• Input– FSA M

– Input string x to M

• Question– Is x in L(M)?

• Algorithm ACCEPT for solving accepting problem– Run M on x

– If halting configuration is accepting THEN yes ELSE no

• Note this algorithm actually has to do something

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Primality Problem

• Input– integer n

• Question– Is n a prime number?

• Algorithm for solving primality problem– for i = 2 to n-1 Do

• apply algorithm DIV to inputs i,n

• If answer for any i is yes THEN no ELSE yes

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Empty Language Problem

• Input– FSA M

• Question– Is L(M)={}?

• How would you solve this problem?

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Algorithms for solving empty language problem

• Algorithm 1– View FSA M as a directed graph (nodes, arcs)

– See if any accepting node is reachable from the start node

• Algorithm 2– Let n be the number of states in FSA M

– Run ACCEPT(M,x) for all input strings of length < n

– If any are accepted THEN no ELSE yes• Why is algorithm 2 correct?

– Same underlying reason for why algorithm 1 works.

– If any string is accepted by M, some string x must be accepted where M never is in the same state twice while processing x

– This string x will have length at most n-1

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Solving Other Problems(using many-one reductions)

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Complement Empty Problem

• Input– FSA M

• Question– Is (L(M))c={}?

• Use many-one reductions to solve this problem– We will show that the Complement Empty problem many-one

reduces to the Empty Language problem

– How do we do this?• Apply the construction which showed that LFSA is closed under set

complement

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Algorithm Description

• Convert input FSA M into an FSA M’ such that L(M’) = (L(M))c

– We do this by applying the algorithm which we used to show that LFSA is closed under complement

• Feed FSA M’ into algorithm which solves the empty language problem

• If that algorithm returns yes THEN yes ELSE no

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Reduction Illustrated

Algorithm forsolving empty

language problemFSA M Complement

Construction

FSA M’Yes/No

Algorithm for complement empty problem

The complement construction algorithm is the many-one reduction function.If M is a yes input instance of CE, then M’ is a yes input instance of EL.If M is a no input instance of CE, then M’ is a no input instance of EL.

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NFA Empty Problem

• Input– NFA M

• Question– Is L(M)={}?

• Use many-one reductions to solve this problem– We will show that the NFA Empty problem many-one reduces to

the Empty Language problem

– How do we do this?• Apply the construction which showed that any NFA can be converted

into an equivalent FSA

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Algorithm Description

• Convert input NFA M into an FSA M’ such that L(M’) = L(M)– We do this by applying the algorithms which we used

to show that LNFA is a subset of LFSA

• Feed FSA M’ into algorithm which solves the empty language problem

• If that algorithm returns yes THEN yes ELSE no

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Reduction Illustrated

Algorithm forsolving empty

language problemNFA M Subset

Construction

FSA M’Yes/No

Algorithm for NFA empty problem

The subset construction algorithm is the many-one reduction function.If M is a yes input instance of NE, then M’ is a yes input instance of EL.If M is a no input instance of NE, then M’ is a no input instance of EL.

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Equal Problem

• Input– FSA’s M1 and M2

• Question– Is L(M1) = L(M2)?

• Use many-one reductions to solve this problem– Try and reduce this problem to the empty language problem

– If L(M1) = L(M2), then what combination of L(M1) and L(M2) must be empty?

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Algorithm Description

• Convert input FSA’s M1 and M2 into an FSA M3 such that L(M3) = (L(M1) - L(M2)) union (L(M2) - L(M1)) – We do this by applying the algorithm which we used to

show that LFSA is closed under symmetric difference (similar to intersection)

• Feed FSA M3 into algorithm which solves the empty language problem

• If that algorithm returns yes THEN yes ELSE no

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Reduction Illustrated

Algorithm forsolving empty

language problem

FSA M1

FSA M2

2FSA -> 1FSAConstruction

FSA M3Yes/No

Algorithm for Equal problem

The 2FSA->1FSA construction algorithm is the many-one reduction function. If (M1,M2) is a yes input instance of EQ, then M3 is a yes input instance of EL. If (M1,M2) is a no input instance of EQ, then M3 is a no input instance of EL.

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Summary

• Decision problems with programs as inputs

• Basic problems– You need to develop algorithms from scratch

based on properties of FSA’s

• Solving new problems– You need to figure out how to combine the

various algorithms we have seen in this unit to solve the given problem