∂RE

Brzozowski’s Derivatives of Regular Expressions

Legend:

∧ intersection
∨ union
∘ concatenation (see below)
¬ complement
ϕ empty set (aka ∅)
λ singleton set containing just the empty string
I set of all letters in alphabet

Derivative of a set R of strings and a string a:

∂a(R)

∂a(a) → λ
∂a(λ) → ϕ
∂a(ϕ) → ϕ
∂a(¬a) → ϕ
∂a(R*) → ∂a(R)∘R*
∂a(¬R) → ¬∂a(R)
∂a(R∘S) → ∂a(R)∘S ∨ δ(R)∘∂a(S)
∂a(R ∧ S) → ∂a(R) ∧ ∂a(S)
∂a(R ∨ S) → ∂a(R) ∨ ∂a(S)

∂ab(R) = ∂b(∂a(R))

Auxiliary predicate function δ (I call it nully) returns either λ if λ R or ϕ otherwise:

δ(a) → ϕ
δ(λ) → λ
δ(ϕ) → ϕ
δ(R*) → λ
δ(¬R) δ(R)≟ϕ → λ
δ(¬R) δ(R)≟λ → ϕ
δ(R∘S) → δ(R) ∧ δ(S)
δ(R ∧ S) → δ(R) ∧ δ(S)
δ(R ∨ S) → δ(R) ∨ δ(S)

Some rules we will use later for “compaction”:

R ∧ ϕ = ϕ ∧ R = ϕ

R ∧ I = I ∧ R = R

R ∨ ϕ = ϕ ∨ R = R

R ∨ I = I ∨ R = I

R∘ϕ = ϕ∘R = ϕ

R∘λ = λ∘R = R

Concatination of sets: for two sets A and B the set A∘B is defined as:

{a∘b for a in A for b in B}

E.g.:

{‘a’, ‘b’}∘{‘c’, ‘d’} → {‘ac’, ‘ad’, ‘bc’, ‘bd’}

Implementation

from functools import partial as curry
from itertools import product

ϕ and λ

The empty set and the set of just the empty string.

phi = frozenset()   # ϕ
y = frozenset({''}) # λ

Two-letter Alphabet

I’m only going to use two symbols (at first) becaase this is enough to illustrate the algorithm and because you can represent any other alphabet with two symbols (if you had to.)

I chose the names O and l (uppercase “o” and lowercase “L”) to look like 0 and 1 (zero and one) respectively.

syms = O, l = frozenset({'0'}), frozenset({'1'})

Representing Regular Expressions

To represent REs in Python I’m going to use tagged tuples. A regular expression is one of:

O
l
(KSTAR, R)
(NOT, R)
(AND, R, S)
(CONS, R, S)
(OR, R, S)

Where R and S stand for regular expressions.

AND, CONS, KSTAR, NOT, OR = 'and cons * not or'.split()  # Tags are just strings.

Because they are formed of frozenset, tuple and str objects only, these datastructures are immutable.

String Representation of RE Datastructures

def stringy(re):
    '''
    Return a nice string repr for a regular expression datastructure.
    '''
    if re == I: return '.'
    if re in syms: return next(iter(re))
    if re == y: return '^'
    if re == phi: return 'X'

    assert isinstance(re, tuple), repr(re)
    tag = re[0]

    if tag == KSTAR:
        body = stringy(re[1])
        if not body: return body
        if len(body) > 1: return '(' + body + ")*"
        return body + '*'

    if tag == NOT:
        body = stringy(re[1])
        if not body: return body
        if len(body) > 1: return '(' + body + ")'"
        return body + "'"

    r, s = stringy(re[1]), stringy(re[2])
    if tag == CONS: return r + s
    if tag == OR:   return '%s | %s' % (r, s)
    if tag == AND:  return '(%s) & (%s)' % (r, s)

    raise ValueError

I

Match anything. Often spelled “.”

I = (0|1)*
I = (KSTAR, (OR, O, l))
print stringy(I)
.

(.111.) & (.01 + 11*)'

The example expression from Brzozowski:

(.111.) & (.01 + 11*)'
   a    &  (b  +  c)'

Note that it contains one of everything.

a = (CONS, I, (CONS, l, (CONS, l, (CONS, l, I))))
b = (CONS, I, (CONS, O, l))
c = (CONS, l, (KSTAR, l))
it = (AND, a, (NOT, (OR, b, c)))
print stringy(it)
(.111.) & ((.01 | 11*)')

nully()

Let’s get that auxiliary predicate function δ out of the way.

def nully(R):
    '''
    δ - Return λ if λ ⊆ R otherwise ϕ.
    '''

    # δ(a) → ϕ
    # δ(ϕ) → ϕ
    if R in syms or R == phi:
        return phi

    # δ(λ) → λ
    if R == y:
        return y

    tag = R[0]

    # δ(R*) → λ
    if tag == KSTAR:
        return y

    # δ(¬R) δ(R)≟ϕ → λ
    # δ(¬R) δ(R)≟λ → ϕ
    if tag == NOT:
        return phi if nully(R[1]) else y

    # δ(R∘S) → δ(R) ∧ δ(S)
    # δ(R ∧ S) → δ(R) ∧ δ(S)
    # δ(R ∨ S) → δ(R) ∨ δ(S)
    r, s = nully(R[1]), nully(R[2])
    return r & s if tag in {AND, CONS} else r | s

No “Compaction”

This is the straightforward version with no “compaction”. It works fine, but does waaaay too much work because the expressions grow each derivation.

def D(symbol):

    def derv(R):

        # ∂a(a) → λ
        if R == {symbol}:
            return y

        # ∂a(λ) → ϕ
        # ∂a(ϕ) → ϕ
        # ∂a(¬a) → ϕ
        if R == y or R == phi or R in syms:
            return phi

        tag = R[0]

        # ∂a(R*) → ∂a(R)∘R*
        if tag == KSTAR:
            return (CONS, derv(R[1]), R)

        # ∂a(¬R) → ¬∂a(R)
        if tag == NOT:
            return (NOT, derv(R[1]))

        r, s = R[1:]

        # ∂a(R∘S) → ∂a(R)∘S ∨ δ(R)∘∂a(S)
        if tag == CONS:
            A = (CONS, derv(r), s)  # A = ∂a(R)∘S
            # A ∨ δ(R) ∘ ∂a(S)
            # A ∨  λ   ∘ ∂a(S) → A ∨ ∂a(S)
            # A ∨  ϕ   ∘ ∂a(S) → A ∨ ϕ → A
            return (OR, A, derv(s)) if nully(r) else A

        # ∂a(R ∧ S) → ∂a(R) ∧ ∂a(S)
        # ∂a(R ∨ S) → ∂a(R) ∨ ∂a(S)
        return (tag, derv(r), derv(s))

    return derv

Compaction Rules

def _compaction_rule(relation, one, zero, a, b):
    return (
        b if a == one else  # R*1 = 1*R = R
        a if b == one else
        zero if a == zero or b == zero else  # R*0 = 0*R = 0
        (relation, a, b)
        )

An elegant symmetry.

# R ∧ I = I ∧ R = R
# R ∧ ϕ = ϕ ∧ R = ϕ
_and = curry(_compaction_rule, AND, I, phi)

# R ∨ ϕ = ϕ ∨ R = R
# R ∨ I = I ∨ R = I
_or = curry(_compaction_rule, OR, phi, I)

# R∘λ = λ∘R = R
# R∘ϕ = ϕ∘R = ϕ
_cons = curry(_compaction_rule, CONS, y, phi)

Memoizing

We can save re-processing by remembering results we have already computed. RE datastructures are immutable and the derv() functions are pure so this is fine.

class Memo(object):

    def __init__(self, f):
        self.f = f
        self.calls = self.hits = 0
        self.mem = {}

    def __call__(self, key):
        self.calls += 1
        try:
            result = self.mem[key]
            self.hits += 1
        except KeyError:
            result = self.mem[key] = self.f(key)
        return result

With “Compaction”

This version uses the rules above to perform compaction. It keeps the expressions from growing too large.

def D_compaction(symbol):

    @Memo
    def derv(R):

        # ∂a(a) → λ
        if R == {symbol}:
            return y

        # ∂a(λ) → ϕ
        # ∂a(ϕ) → ϕ
        # ∂a(¬a) → ϕ
        if R == y or R == phi or R in syms:
            return phi

        tag = R[0]

        # ∂a(R*) → ∂a(R)∘R*
        if tag == KSTAR:
            return _cons(derv(R[1]), R)

        # ∂a(¬R) → ¬∂a(R)
        if tag == NOT:
            return (NOT, derv(R[1]))

        r, s = R[1:]

        # ∂a(R∘S) → ∂a(R)∘S ∨ δ(R)∘∂a(S)
        if tag == CONS:
            A = _cons(derv(r), s)  # A = ∂a(r)∘s
            # A ∨ δ(R) ∘ ∂a(S)
            # A ∨  λ   ∘ ∂a(S) → A ∨ ∂a(S)
            # A ∨  ϕ   ∘ ∂a(S) → A ∨ ϕ → A
            return _or(A, derv(s)) if nully(r) else A

        # ∂a(R ∧ S) → ∂a(R) ∧ ∂a(S)
        # ∂a(R ∨ S) → ∂a(R) ∨ ∂a(S)
        dr, ds = derv(r), derv(s)
        return _and(dr, ds) if tag == AND else _or(dr, ds)

    return derv

Let’s try it out…

(FIXME: redo.)

o, z = D_compaction('0'), D_compaction('1')
REs = set()
N = 5
names = list(product(*(N * [(0, 1)])))
dervs = list(product(*(N * [(o, z)])))
for name, ds in zip(names, dervs):
    R = it
    ds = list(ds)
    while ds:
        R = ds.pop()(R)
        if R == phi or R == I:
            break
        REs.add(R)

print stringy(it) ; print
print o.hits, '/', o.calls
print z.hits, '/', z.calls
print
for s in sorted(map(stringy, REs), key=lambda n: (len(n), n)):
    print s
(.111.) & ((.01 | 11*)')

92 / 122
92 / 122

(.01)'
(.01 | 1)'
(.01 | ^)'
(.01 | 1*)'
(.111.) & ((.01 | 1)')
(.111. | 11.) & ((.01 | ^)')
(.111. | 11. | 1.) & ((.01)')
(.111. | 11.) & ((.01 | 1*)')
(.111. | 11. | 1.) & ((.01 | 1*)')

Should match:

(.111.) & ((.01 | 11*)')

92 / 122
92 / 122

(.01     )'
(.01 | 1 )'
(.01 | ^ )'
(.01 | 1*)'
(.111.)            & ((.01 | 1 )')
(.111. | 11.)      & ((.01 | ^ )')
(.111. | 11.)      & ((.01 | 1*)')
(.111. | 11. | 1.) & ((.01     )')
(.111. | 11. | 1.) & ((.01 | 1*)')

Larger Alphabets

We could parse larger alphabets by defining patterns for e.g. each byte of the ASCII code. Or we can generalize this code. If you study the code above you’ll see that we never use the “set-ness” of the symbols O and l. The only time Python set operators (& and |) appear is in the nully() function, and there they operate on (recursively computed) outputs of that function, never O and l.

What if we try:

(OR, O, l)

∂1((OR, O, l))
                            ∂a(R ∨ S) → ∂a(R) ∨ ∂a(S)
∂1(O) ∨ ∂1(l)
                            ∂a(¬a) → ϕ
ϕ ∨ ∂1(l)
                            ∂a(a) → λ
ϕ ∨ λ
                            ϕ ∨ R = R
λ

And compare it to:

{'0', '1')

∂1({'0', '1'))
                            ∂a(R ∨ S) → ∂a(R) ∨ ∂a(S)
∂1({'0')) ∨ ∂1({'1'))
                            ∂a(¬a) → ϕ
ϕ ∨ ∂1({'1'))
                            ∂a(a) → λ
ϕ ∨ λ
                            ϕ ∨ R = R
λ

This suggests that we should be able to alter the functions above to detect sets and deal with them appropriately. Exercise for the Reader for now.

State Machine

We can drive the regular expressions to flesh out the underlying state machine transition table.

.111. & (.01 + 11*)'

Says, “Three or more 1’s and not ending in 01 nor composed of all 1’s.”

State Machine Diagram

Start at a and follow the transition arrows according to their labels. Accepting states have a double outline. (Graphic generated with Dot from Graphviz.) You’ll see that only paths that lead to one of the accepting states will match the regular expression. All other paths will terminate at one of the non-accepting states.

There’s a happy path to g along 111:

a→c→e→g

After you reach g you’re stuck there eating 1’s until you see a 0, which takes you to the i→j→i|i→j→h→i “trap”. You can’t reach any other states from those two loops.

If you see a 0 before you see 111 you will reach b, which forms another “trap” with d and f. The only way out is another happy path along 111 to h:

b→d→f→h

Once you have reached h you can see as many 1’s or as many 0’ in a row and still be either still at h (for 1’s) or move to i (for 0’s). If you find yourself at i you can see as many 0’s, or repetitions of 10, as there are, but if you see just a 1 you move to j.

RE to FSM

So how do we get the state machine from the regular expression?

It turns out that each RE is effectively a state, and each arrow points to the derivative RE in respect to the arrow’s symbol.

If we label the initial RE a, we can say:

a --0--> ∂0(a)
a --1--> ∂1(a)

And so on, each new unique RE is a new state in the FSM table.

Here are the derived REs at each state:

a = (.111.) & ((.01 | 11*)')
b = (.111.) & ((.01 | 1)')
c = (.111. | 11.) & ((.01 | 1*)')
d = (.111. | 11.) & ((.01 | ^)')
e = (.111. | 11. | 1.) & ((.01 | 1*)')
f = (.111. | 11. | 1.) & ((.01)')
g = (.01 | 1*)'
h = (.01)'
i = (.01 | 1)'
j = (.01 | ^)'

You can see the one-way nature of the g state and the hij “trap” in the way that the .111. on the left-hand side of the & disappears once it has been matched.

from collections import defaultdict
from pprint import pprint
from string import ascii_lowercase
d0, d1 = D_compaction('0'), D_compaction('1')

explore()

def explore(re):

    # Don't have more than 26 states...
    names = defaultdict(iter(ascii_lowercase).next)

    table, accepting = dict(), set()

    to_check = {re}
    while to_check:

        re = to_check.pop()
        state_name = names[re]

        if (state_name, 0) in table:
            continue

        if nully(re):
            accepting.add(state_name)

        o, i = d0(re), d1(re)
        table[state_name, 0] = names[o] ; to_check.add(o)
        table[state_name, 1] = names[i] ; to_check.add(i)

    return table, accepting
table, accepting = explore(it)
table
{('a', 0): 'b',
 ('a', 1): 'c',
 ('b', 0): 'b',
 ('b', 1): 'd',
 ('c', 0): 'b',
 ('c', 1): 'e',
 ('d', 0): 'b',
 ('d', 1): 'f',
 ('e', 0): 'b',
 ('e', 1): 'g',
 ('f', 0): 'b',
 ('f', 1): 'h',
 ('g', 0): 'i',
 ('g', 1): 'g',
 ('h', 0): 'i',
 ('h', 1): 'h',
 ('i', 0): 'i',
 ('i', 1): 'j',
 ('j', 0): 'i',
 ('j', 1): 'h'}
accepting
{'h', 'i'}

Generate Diagram

Once we have the FSM table and the set of accepting states we can generate the diagram above.

_template = '''\
digraph finite_state_machine {
  rankdir=LR;
  size="8,5"
  node [shape = doublecircle]; %s;
  node [shape = circle];
%s
}
'''

def link(fr, nm, label):
    return '  %s -> %s [ label = "%s" ];' % (fr, nm, label)


def make_graph(table, accepting):
    return _template % (
        ' '.join(accepting),
        '\n'.join(
          link(from_, to, char)
          for (from_, char), (to) in sorted(table.iteritems())
          )
        )
print make_graph(table, accepting)
digraph finite_state_machine {
  rankdir=LR;
  size="8,5"
  node [shape = doublecircle]; i h;
  node [shape = circle];
  a -> b [ label = "0" ];
  a -> c [ label = "1" ];
  b -> b [ label = "0" ];
  b -> d [ label = "1" ];
  c -> b [ label = "0" ];
  c -> e [ label = "1" ];
  d -> b [ label = "0" ];
  d -> f [ label = "1" ];
  e -> b [ label = "0" ];
  e -> g [ label = "1" ];
  f -> b [ label = "0" ];
  f -> h [ label = "1" ];
  g -> i [ label = "0" ];
  g -> g [ label = "1" ];
  h -> i [ label = "0" ];
  h -> h [ label = "1" ];
  i -> i [ label = "0" ];
  i -> j [ label = "1" ];
  j -> i [ label = "0" ];
  j -> h [ label = "1" ];
}

Drive a FSM

There are lots of FSM libraries already. Once you have the state transition table they should all be straightforward to use. State Machine code is very simple. Just for fun, here is an implementation in Python that imitates what “compiled” FSM code might look like in an “unrolled” form. Most FSM code uses a little driver loop and a table datastructure, the code below instead acts like JMP instructions (“jump”, or GOTO in higher-level-but-still-low-level languages) to hard-code the information in the table into a little patch of branches.

Trampoline Function

Python has no GOTO statement but we can fake it with a “trampoline” function.

def trampoline(input_, jump_from, accepting):
    I = iter(input_)
    while True:
        try:
            bounce_to = jump_from(I)
        except StopIteration:
            return jump_from in accepting
        jump_from = bounce_to

Stream Functions

Little helpers to process the iterator of our data (a “stream” of “1” and “0” characters, not bits.)

getch = lambda I: int(next(I))


def _1(I):
    '''Loop on ones.'''
    while getch(I): pass


def _0(I):
    '''Loop on zeros.'''
    while not getch(I): pass

A Finite State Machine

With those preliminaries out of the way, from the state table of .111. & (.01 + 11*)' we can immediately write down state machine code. (You have to imagine that these are GOTO statements in C or branches in assembly and that the state names are branch destination labels.)

a = lambda I: c if getch(I) else b
b = lambda I: _0(I) or d
c = lambda I: e if getch(I) else b
d = lambda I: f if getch(I) else b
e = lambda I: g if getch(I) else b
f = lambda I: h if getch(I) else b
g = lambda I: _1(I) or i
h = lambda I: _1(I) or i
i = lambda I: _0(I) or j
j = lambda I: h if getch(I) else i

Note that the implementations of h and g are identical ergo h = g and we could eliminate one in the code but h is an accepting state and g isn’t.

def acceptable(input_):
    return trampoline(input_, a, {h, i})
for n in range(2**5):
    s = bin(n)[2:]
    print '%05s' % s, acceptable(s)
    0 False
    1 False
   10 False
   11 False
  100 False
  101 False
  110 False
  111 False
 1000 False
 1001 False
 1010 False
 1011 False
 1100 False
 1101 False
 1110 True
 1111 False
10000 False
10001 False
10010 False
10011 False
10100 False
10101 False
10110 False
10111 True
11000 False
11001 False
11010 False
11011 False
11100 True
11101 False
11110 True
11111 False

Reversing the Derivatives to Generate Matching Strings

(UNFINISHED) Brzozowski also shewed how to go from the state machine to strings and expressions…

Each of these states is just a name for a Brzozowskian RE, and so, other than the initial state a, they can can be described in terms of the derivative-with-respect-to-N of some other state/RE:

c = d1(a)
b = d0(a)
b = d0(c)
...
i = d0(j)
j = d1(i)

Consider:

c = d1(a)
b = d0(c)

Substituting:

b = d0(d1(a))

Unwrapping:

b = d10(a)

‘’‘

j = d1(d0(j))

Unwrapping:

j = d1(d0(j)) = d01(j)

We have a loop or “fixed point”.

j = d01(j) = d0101(j) = d010101(j) = ...

hmm…

j = (01)*