expression.derivation(label,breaking=False)

Compute the derivation of a weighted expression.

Arguments:

• label: the (non empty) string to derive the expression with.
• breaking: whether to split the result.

See also:

• expression.derived_term
• expression.expansion
• polynomial.split

References:

• lombardy.2005.tcs defines the derivation
• angrand.2010.jalc defines the breaking derivation

Examples

The following function will prove handy: it takes a rational expression and a list of strings, and returns a $\LaTeX$ aligned environment to display nicely the result.

import vcsn
from IPython.display import Latex

def diffs(r, ss):
    eqs = []
    for s in ss:
        eqs.append(r'\frac{{\partial}}{{\partial {0}}} {1}& = {2}'
                   .format(s,
                           r.format('latex'),
                           r.derivation(s).format('latex')))
    return Latex(r'''\begin{{aligned}}
        {0}
    \end{{aligned}}'''.format(r'\\'.join(eqs)))

Classical expressions

In the classical case (labels are letters, and weights are Boolean), this is the construct as described by Antimirov.

b = vcsn.context('lal_char(ab), b')
r = b.expression('[ab]{3}')
r.derivation('a')

$\left(a + b\right)^{2}$

Or, using the diffs function we defined above:

diffs(r, ['a', 'aa', 'aaa', 'aaaa'])

\begin{aligned} \frac{\partial}{\partial a} \left(a + b\right)^{3}& = \left(a + b\right)^{2}\\\frac{\partial}{\partial aa} \left(a + b\right)^{3}& = a + b\\\frac{\partial}{\partial aaa} \left(a + b\right)^{3}& = \varepsilon\\\frac{\partial}{\partial aaaa} \left(a + b\right)^{3}& = \emptyset \end{aligned}

Weighted Expressions

Of course, expressions can be weighted.

q = vcsn.context('lal_char(abc), q')
r = q.expression('(<1/6>a*+<1/3>b*)*')
diffs(r, ['a', 'aa', 'ab', 'b', 'ba', 'bb'])

\begin{aligned} \frac{\partial}{\partial a} \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}& = \left\langle \frac{1}{3}\right\rangle {a}^{*} \, \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}\\\frac{\partial}{\partial aa} \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}& = \left\langle \frac{4}{9}\right\rangle {a}^{*} \, \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}\\\frac{\partial}{\partial ab} \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}& = \left\langle \frac{2}{9}\right\rangle {b}^{*} \, \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}\\\frac{\partial}{\partial b} \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}& = \left\langle \frac{2}{3}\right\rangle {b}^{*} \, \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}\\\frac{\partial}{\partial ba} \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}& = \left\langle \frac{2}{9}\right\rangle {a}^{*} \, \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}\\\frac{\partial}{\partial bb} \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*}& = \left\langle \frac{10}{9}\right\rangle {b}^{*} \, \left( \left\langle \frac{1}{6} \right\rangle \,{a}^{*} + \left\langle \frac{1}{3} \right\rangle \,{b}^{*}\right)^{*} \end{aligned}

And this is tightly connected with the construction of the derived-term automaton.

r.derived_term()

Breaking derivation

The "breaking" derivation "splits" the polynomial at the end.

r = q.expression('[ab](<2>[ab])', 'associative')
r

$\left(a + b\right) \, \left\langle 2 \right\rangle \,\left(a + b\right)$

r.derivation('a')

$ \left\langle 2 \right\rangle \,\left(a + b\right)$

r.derivation('a', True)

$\left\langle 2\right\rangle a \oplus \left\langle 2\right\rangle b$

r.derivation('a').split()

$\left\langle 2\right\rangle a \oplus \left\langle 2\right\rangle b$

Again, this is tightly connected with both flavors of the derived-term automaton.

r.derived_term()

r.derived_term('breaking_derivation')