expression.infiltrate

$\newcommand{\Ed}{\mathsf{E}} \newcommand{\Fd}{\mathsf{F}} \newcommand{\Gd}{\mathsf{G}} \newcommand{\infiltrate}{\mathbin{\uparrow}} \newcommand{\shuffle}{\mathbin{\between}}$ Create the infiltration product of two expressions. In a way the infiltration product combines the conjunction (synchronized product) and the shuffle product.

See also:

• automaton.infiltrate
• expression.conjunction
• expression.shuffle

Examples

import vcsn
c = vcsn.context('lal_char, seriesset<lal_char, z>')
exp = c.expression
c

$\{\ldots\}\to\mathsf{Series}[\{\ldots\}\to\mathbb{Z}]$

The following simple example aims at emphasizing that the transitions of the infiltration combine those of the shuffle and the conjunction products.

x = exp("<x>a"); x

$ \left\langle x \right\rangle \,a$

y = exp("<y>a"); y

$ \left\langle y \right\rangle \,a$

conj = x & y ; conj

$ \left\langle x \, y \right\rangle \,a$

shuff = x.shuffle(y) ; shuff

$ \left\langle x \right\rangle \,a \between \left\langle y \right\rangle \,a$

inf = x.infiltrate(y) ; inf

$ \left\langle x \right\rangle \,a \uparrow \left\langle y \right\rangle \,a$

inf.automaton() == (conj + shuff).automaton()

True

Associativity

This operator is associative: $(\Ed \infiltrate \Fd) \infiltrate \Gd \equiv \Ed \infiltrate (\Fd \infiltrate \Gd)$, i.e., both expressions denote the same series.

x = exp('<x>a', 'trivial')
y = exp('<y>a', 'trivial')
z = exp('<z>a', 'trivial')

x.infiltrate(y).infiltrate(z).automaton()

x.infiltrate(y.infiltrate(z)).automaton()