LING 388: Language and Computers

Sandiway Fong

Lecture 12: 10/5

Today’s Topics

• Tasks – Practice how to write a FSA in Prolog– Dealing with string transitions

• Homework #4– Due next Thursday– Only two questions

• Derivational morphology and FSA• Converting NDFSA to (deterministic) FSA

Prolog FSA: One predicate per state

• from previous lectures• Prolog FSA implementation

– regular expression: ba+!

– query• ?- s([b,a,a,’!’]).

– code– s([b|L]) :- x(L).

s x

z

b

!

ya

a

>

– y([a|L]) :- y(L).

– y([‘!’|L]) :- z(L).– z([]).

– x([a|L]) :- y(L).

Exercise 1

Derivational Morphology• Write a Prolog FSA that accepts all words

ending in the verbal suffix -ize• use:

– atom_chars(Word,List)

– from the previous homework

• examples:– formalize– summarize– *formalizes

Exercise 1

• Given code:ends_ize(Word) :-

atom_chars(Word,List),

s(List).

s = start state

• Write s/1 such that?- ends_ize(formalize).

Yes

?- ends_ize(summarize).Yes

?- ends_ize(formalizes).No

Exercise 1

• Step 1: Draw the FSA: • Notes:– machine is non-

deterministic• see an i

• we don’t know if it’s the “i” in rise or formalize

– we can name the states anything we want

• here: states are named after the portion of the suffix recognized so far

s

i

iz

ize

z

i

e

a-z

Exercise 1

• Step 1: Draw the FSA: • Step 2: Write the Prolog

• Hint: use the underscore variable to simulate the range a-z

s

i

iz

ize

z

i

e

a-z

Exercise 1

• Step 1: Draw the FSA: • Step 2: Write the Prolog

s([_|L]) :- s(L).

s([i|L]) :- i(L).

i([z|L]) :- iz(L).

i([e|L]) :- ize(L).

ize([]).

s

i

iz

ize

z

i

e

a-z

String Transitions

[last lecture]

• all machines have had just a single character label on the arc

• so if we allow strings to label arcs– do they endow the FSA with any

more power?

b

• Answer: No– because we can always convert a

machine with string-transitions into one without

abb

a b b

Exercise 2

• Let’s simplify the FSA for Exercise 1 using string transitions

• Give the modified Prolog form

s

i

iz

ize

z

i

e

a-z

s

ize

ize

a-z

Exercise 2

• Give the modified Prolog form

s([_|L]) :- s(L).s([i,z,e|L]) :- ize(L).ize([]).

s([_|L]) :- s(L).s([i,z,e]).

s

i

iz

ize

z

i

e

a-z

s

ize

ize

a-z

Homework Question 1

• Part 1: (8pts)• Modify your machine so that it accepts

– formalize– modernize– summarize– concretize– sterilize

• but rejects– *summaryize– *concreteize– *sterileize

• Part 2: (2pts)– explain what rule of English are we trying to encode here?

Empty Transitions

[last lecture]• how about allowing the empty

character? – i.e. go from x to y without seeing a input

character– does this endow the FSA with any more

power?

• Answer: No– because we can always convert a

machine with empty transitions into one without

x y

a

b

a

b

b

a

b>

a b

NDFSA

• Deterministic FSA– deterministic

• it’s clear which state we’re always in• deterministic = no choice

• NDFSA– ND = non-deterministic

• i.e. we could be in more than one state• non-deterministic choice point

– example:• see an “a”, either in state 2 or 3 next

s x

y

aa

b

b

>

> 1 2a

a

3b

>

NDFSA

• NDFSA are generally not more powerful than regular (deterministic) FSA

• because

• we can transform any NDFSA into an equivalent FSA

• trick:– (set of states construction)– construct new machine with

states = set of possible states of the old machine

1 2a

a

3b

>

{1}>

a{1}> {2,3}

{3}ba

{1}> {2,3}

{3}ba

{1}> {2,3}

1 2,3a

3b

2,32>

Homework Question 2

• (10pts)• Part 1: (8pts)• Transform the NDFSA shown

on the right into a deterministic FSA

• Give the Prolog implementation of the machine

• Part 2: (2pts)– what does this machine do?

– i.e. what is the language it accepts?

q4>

Summary

• Total: 20 pts

• Question 1: 10pts

• Question 2: 10pts