automaton.evaluate.ipynb

Kernel: Python 3

automaton.evaluate(w)

Evaluates the weight of the given word through the automata.

Preconditions:

• w must be a valid word in the labelset.

• automaton must not have spontaneous cycles.

Examples

import vcsn

a = vcsn.context('lal(ab), b').de_bruijn(2)
a

a.evaluate('b')

$\bot$

a.evaluate('bbabb')

$\top$

You can also write automaton('word') to evaluate a word:

a('bbabb')

$\top$

a = vcsn.Z.expression('(<2>C+c)los(<3>e(s+<4>d)+ing)').standard()
a.shortest(10)

$\left\langle 24\right\rangle \mathit{Closed} \oplus \left\langle 6\right\rangle \mathit{Closes} \oplus \left\langle 12\right\rangle \mathit{closed} \oplus \left\langle 3\right\rangle \mathit{closes} \oplus \left\langle 2\right\rangle \mathit{Closing} \oplus \mathit{closing}$

a.evaluate('Closing')

$2$

a.evaluate('close') # but not enough

$0$

All automaton types are supported, evaluate is not limited to free labelsets. For instance, with word-labeled automata:

a = vcsn.context('law,q').expression('<2>(ab(<3>cd)*(ef))<5>', "associative").automaton()
a

a.evaluate('abcdef')

$30$

a.evaluate('abcdcdef')

$90$

Polynomials are also supported:

p = vcsn.context('law,q').polynomial('<2>abcdef+abcdcdef')
a.evaluate(p)

$150$

Spontaneous transitions are allowed:

%%automaton -s a
context = "wordset<char_letters(abcdef)>, q"
$ -> 0 <2>
0 -> 1 \e
1 -> 1 <3>cd
1 -> 2 \e
2 -> $ <5>

a.evaluate("cd")

$30$

a.evaluate("")

$10$

Tuplesets are fully supported:

ctx = vcsn.context('lat<lan, lan>, zmin')
a = ctx.expression('(<0>(a|a+b|b))* (<1>[^]|\e + <1>\e|[^] + <2>(a|[^a]+b|[^b])){*}').automaton()
a #Compute the LCS distance between two words

a.evaluate("aaa|ab")

$3$

a.evaluate("aba|ab")

$1$