simulating 2 fsa’s with 1 fsa
DESCRIPTION
Simulating 2 FSA’s with 1 FSA. Purpose. This presentation presents an example execution of the algorithm which takes as input two FSA’s and produces as output an FSA which “simulates” both input FSA’s on any input string Algorithm Specification Input: FSA’s M 1 and M 2 Output: FSA M 3. - PowerPoint PPT PresentationTRANSCRIPT
Simulating 2 FSA’s with 1 FSA
Purpose
• This presentation presents an example execution of the algorithm which takes as input two FSA’s and produces as output an FSA which “simulates” both input FSA’s on any input string
• Algorithm Specification
– Input: FSA’s M1 and M2
– Output: FSA M3
The two input FSA’s
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III
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M1 M2
Initializationa
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a bI,A
The initial state of the output FSA M3 is the ordered pair of the initial states of M1 and M2.
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Expand State I,Aa
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a bI,A I,C II,B
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Add New Statesa
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a,b
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a bI,A I,C II,BI,CII,B
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Expand State I,Ca
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a bI,A I,C II,BI,C I,B II,DII,B
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Add New Statesa
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a
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a,b
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a bI,A I,C II,BI,C I,B II,DII,BI,BII,D
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Expand State II,Ba
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,BII,D
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Add One New Statea
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b b
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a,b
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,BII,DIII,B
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Expand State I,Ba
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,DIII,B
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No New States Addeda
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,DIII,B
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Expand State II,Da
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B
I
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Add One New Statea
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b b
a
a
a,b
aa
b
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a,b
a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,BIII,D
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Expand State III,Ba
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D
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No New States Addeda
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b b
a
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a,b
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a,b
a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D
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Expand State III,Da
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D III,D I,D
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Add 1 New Statea
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b b
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a,b
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D III,D I,DI,D
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Expand State I,Da
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b b
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D III,D I,DI,D I,D II,D
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No More New Statesa
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a bI,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D III,D I,DI,D I,D II,D
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Determining Final Statesa b
I,A I,C II,BI,C I,B II,DII,B II,B III,BI,B I,B II,BII,D II,D III,DIII,B III,B I,BIII,D III,D I,DI,D I,D II,D
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Set IntersectionBoth input machines should accept in order for new machine to acceptThe one accepting state of the new machine M3 is I,D
Set UnionIf either input machine accepts, new machine should acceptThe six accepting states of the new machine M3 are I,A; I,C; I,B; II,D; III,D; I,D
Set Difference: M1 should accept but M2 should notThe three accepting states of the new machine M3 are I,A; I,C; I,B
Symmetric Difference: M1 should accept but M2 should not or M2 should accept but M1 should notThe five accepting states of the new machine M3 are I,A; I,C; I,B; II,D; III,D