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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