In 1969 George Spencer-Brown (GSB) published "Laws of Form" which presented a logical system based on a single action, a distinction, that is both an operation and a value. This notebook describes a Python implementation that mimics the Laws of Form notation and uses it to develop a model of computer circuits.
See The Markable Mark.
(()) =
()() = ()
A((B)) = AB
A() = ()
A(AB) = A(B)
I call these three laws the Bricken Basis after William Bricken who figured out that the third law is complete with the other two. GSB had the first two laws and "Each Way" as the basis. (TODO: Find and include the references for all this.)
(If anything here is unclear read The Markable Mark. George Burnett-Stuart has done a fantastic job there explaining the Laws of Form.)
We can use data structures made solely out of Python frozenset and string objects to represent the forms of the Laws of Form notation. I'm going to use the terms "expression" and "form" interchangably in this document.
class Form(frozenset):
def __str__(self):
# Because frozenset is immutable, and the contents are all string or frozenset,
# we can cache the string repr of a form.
try:
return self._str
except AttributeError:
self._str = '(%s)' % ' '.join(sorted(map(str, self)))
return self._str
__repr__ = __str__
def F(*terms):
'''Create a Form from terms.'''
return Form([
term if isinstance(term, (basestring, Form)) else F(*term)
for term in terms
])
Define a few variable names.
a, b, c = 'abc'
Some examples of forms.
A = F(a, b, c)
A
B = F(a, (b, (c,)))
B
Forms like a b c must be enclosed in a pair of nested containers like so (( a b c )), this lets us treat them as a single (Python) object without inverting the logical value of the form.
C = F((a, b, c))
C
Duplicate terms in a form are automatically removed by frozenset.
F(a, (b,), a, (b,))
Order is irrelevant, again due to frozenset.
F(b, a, c) == F(a, b, c)
It's prefectly okay to create forms out of other forms (not just strings.)
F(A, (B, (C,)), a)
Mark = F()
Mark
There is no way to represent Void directly in a programming language so we have to use the simplest Void-valued form instead.
Void = F(Mark)
Void
We can use a Python dict as a context or environment that supplies values (Mark or Void) for the names in a form.
env = dict(a=Mark, b=Mark, c=Mark)
reify(form, environment) Function¶Given forms with string variable names in them we want to be able to substitute values from an environment. If these values are Mark or Void the result will be a pure arithmentic form.
def reify(form, environment):
if isinstance(form, basestring):
return environment.get(form, form)
return Form(reify(inner, environment) for inner in form)
for form in (A, B, C):
print form, u'⟶', reify(form, env)
void(form) Function¶Once the forms have been rendered to pure arithmetic we can use the void() function to find the value of each expression.
def void(form):
return any(not void(i) for i in form)
The void() function returns a Boolean value (Python True or False), for convenience let's write a function that returns the Mark or Void value of a form.
def value_of(form, m=Mark, v=Void):
return (m, v)[void(form)]
Now we can use the void() function (by way of value_of()) to calculate the base value of each expression structure.
for form in (A, B, C):
arith = reify(form, env)
print form, u'⟶', arith, u'⟶', value_of(arith)
For $n$ variables there are $2^n$ possible assignments of the two values of Mark and Void. If we generate environments that each contain one of the possible assignments of names to the base value we can evaluate an expression containing those names and compute its value.
from itertools import product, izip
BASE = Void, Mark
def environments_of_variables(*variables):
universe = [BASE] * len(variables)
for values in product(*universe):
yield dict(izip(variables, values))
envs = list(environments_of_variables(*'abc'))
envs
This is a bit hard to read, so let's define a helper function to convert an environment to a string format.
def format_env(env, m='()', v=' '):
return ' '.join((v, m)[not env[k]] for k in sorted(env))
# Note that Mark is an empty frozenset so in a Boolean context in Python it is False,
# likewise Void is a set with one member, so Python considers it True in a Boolean context.
# The `not` in the expression is just to force such a Boolean context, and we compensate
# by putting `v` in the zero-is-False position in the indexed tuple.
Now we can print out the environments in a table. Notice that it looks just like a list of the eight three-bit binary numbers.
print 'i a b c i in Binary'
for i, env in enumerate(envs):
print i, format_env(env, v='--'), '%3s' % (bin(i)[2:],)
Let's pick one of the expressions and iterate through the environments showing the result of reifying that expression in that environment.
print B
print '-----------'
for i, env in enumerate(envs):
e = reify(B, env)
print i, format_env(env, v='--'), u'⟶', e, u'⟶', value_of(e, m='()', v='')
Let's render the above as a Truth Table.
def truth_table_3(expression):
print expression
print ' a b c | Value'
print '---------+------'
for E in envs:
e = reify(expression, E)
print format_env(E), '|', value_of(e, m='()', v='')
truth_table_3(B)
This makes it clear that each expression in Laws of Form calculus is describing a digital Boolean circuit. The names are its inputs and its Void/Mark value is its output. Each boundary is a multi-input NOR gate, known as the Peirce arrow or Quine dagger (See Sheffer stroke and NOR gate.) Instead of two Boolean values there is only one value and non-existance.
In order to work with expressions as digital circuits, let's define some helper functions that will create logic circuits out of simpler forms. The names of the functions below reflect the choice of Mark as Boolean True but this is just a convention.
nor = lambda *bits: F(*bits)
or_ = lambda *bits: F(bits)
and_ = lambda *bits: Form(F(bit) for bit in bits)
nand = lambda *bits: nor(and_(*bits))
nxor = eqiv = lambda a, b: F((a, (b,)), ((a,), b))
xor = lambda a, b: F(nxor(a, b))
# To build logical expressions with Void as Boolean True use these functions.
anti_nor = nand
anti_or = and_
anti_and = or_
anti_nand = nor
anti_eqiv = xor
anti_xor = eqiv
Some examples:
a, b, c = 'abc'
some_expressions = (
nor(a, b, c),
or_(a, b, c),
and_(a, b, c),
nand(a, b, c),
xor(a, b),
eqiv(a, b),
xor(a, xor(b, c)),
)
for expression in some_expressions:
print expression
And let's rewrite the truth_table_3() function to make it work for any number of variables.
def yield_variables_of(expression):
'''Yield all string members of an expression.'''
if isinstance(expression, basestring):
yield expression
else:
for inner in expression:
for leaf in yield_variables_of(inner):
yield leaf
def collect_names(expression):
'''Return a set of the variables mentioned in an expression.'''
return set(yield_variables_of(expression))
def truth_table(expression):
'''Print a truth table for an expression.'''
names = sorted(collect_names(expression))
header = ' ' + ' '.join(names)
n = 1 + len(header)
header += ' | Value'
print expression
print header
print '-' * n + '+------'
for env in environments_of_variables(*names):
e = reify(expression, env)
print format_env(env), '|', ['()', ''][void(e)]
We can use this truth_table() function to examine the expressions we created above.
truth_table(nor(a, b, c))
truth_table(or_(a, b, c))
truth_table(and_(a, b, c))
truth_table(xor(a, b))
truth_table(eqiv(a, b))
truth_table(xor(a, xor(b, c)))
E1 = and_(
or_(and_(a, b), and_(b, c), and_(c, a)), # Any two variables...
nand(a, b, c) # ...but not all three.
)
truth_table(E1)
This is a brute-force SAT solver that doesn't even bother to stop once it's found a solution.
Sometimes we will have a function for which we know the behavior (truth table) but not an expression and we want the expression. For example, imagine that we didn't just create the expression for this table:
a b c | Value
---------+------
|
() |
() |
() () | ()
() |
() () | ()
() () | ()
() () () |
To write an expression for this table, first we should understand that each row can be represented as an expression.
⟶ ( a b c )
() ⟶ ( a b (c))
() ⟶ ( a (b) c )
() () ⟶ ( a (b) (c))
() ⟶ ((a) b c )
() () ⟶ ((a) b (c))
() () ⟶ ((a) (b) c )
() () () ⟶ ((a) (b) (c))
Each of the above expressions will be true (Mark-valued) for only one possible combination of the three input variables. For example, let's look at the sixth expression above:
e6 = F((a,), b, (c,))
truth_table(e6)
To make an expression that is Mark-valued for just certain rows of the table, pick those rows' expressions,
() () | ( a (b) (c))
() () | ((a) b (c))
() () | ((a) (b) c )
And write them down as terms in an OR expression:
E = (a(b)(c)) ((a)b(c)) ((a)(b)c)
In conventional notation this is called Disjunctive normal form:
E = (¬a ∧ b ∧ c) ∨ (a ∧ ¬b ∧ c) ∨ (a ∧ b ∧ ¬c)
Here it is in action:
e4 = ( a, (b,), (c,))
e6 = ((a,), b, (c,))
e7 = ((a,), (b,), c )
E2 = or_(e4, e6, e7)
truth_table(E2)
Note that the expression E2 above is equivalent to the ealier expression E1 that has the same truth table, in other words:
((((((a) (b)) ((b) (c)) ((c) (a))))) ((((a) (b) (c)))))
equals
(((a (b) (c)) ((a) b (c)) ((a) (b) c)))
We can demonstrate this equivalence by evaluating the expression formed by eqiv() from these two.
For every environment (from the set of possible values for the variables) if both expressions have the same value when evaluated then the eqiv() of those expressions will be Mark-valued (true in our chosen context.)
truth_table(eqiv(E1, E2))
The truth table above shows that the equivalence expression is true (Mark-valued by our current convention) for all possible assignments of Mark/Void to the three variables a, b, and c. This indicates that the expression is a tautology.
If you have two binary digits ("bits") and you are interested in the (binary) sum of these digits you will need two circuits, one for the "ones place" and one for the "twos place" or "carry bit".
Consider:
a b | c s
----+----
0 0 | 0 0
0 1 | 0 1
1 0 | 0 1
1 1 | 1 0
Treating each output column ('c' for carry, 's' for sum) as a single expression, it's easy to see that the carry bit is just AND and the sum bit is just XOR of the two input bits.
a, b = 'ab'
half_bit_adder = {
'Sum': xor(a, b),
'Carry': and_(a, b),
}
for name, expr in half_bit_adder.items():
print name
truth_table(expr)
print
In order to add two multi-bit binary numbers we need adder circuits that are designed to work with three input bits: the two bits to add together and a carry bit from the previous addition:
a b Cin Sum Cout
0 0 0 | 0 0
0 0 1 | 1 0
0 1 0 | 1 0
0 1 1 | 0 1
1 0 0 | 1 0
1 0 1 | 0 1
1 1 0 | 0 1
1 1 1 | 1 1
Looking back at our table of three-variable expressions:
⟶ ( a b c )
() ⟶ ( a b (c))
() ⟶ ( a (b) c )
() () ⟶ ( a (b) (c))
() ⟶ ((a) b c )
() () ⟶ ((a) b (c))
() () ⟶ ((a) (b) c )
() () () ⟶ ((a) (b) (c))
We can easily determine expressions for sum and carry:
Sum = (a b (c)) (a (b) c) ((a) b c) ((a) (b) (c))
Cout = (a (b) (c)) ((a) b (c)) ((a) (b) c) ((a) (b) (c))Sum = F(( (a, b, (c,)), (a, (b,), c), ((a,), b, c), ((a,), (b,), (c,)) ),)
Carry = F(( (a, (b,), (c,)), ((a,), b, (c,)), ((a,), (b,), c), ((a,), (b,), (c,)) ),)
print 'Sum'
truth_table(Sum)
print
print 'Carry'
truth_table(Carry)
Let's make a full_bit_adder() function that can define new expressions in terms of variables (or expressions) passed into it.
def full_bit_adder(a, b, c):
return (
F(( (a, b, (c,)), (a, (b,), c), ((a,), b, c), ((a,), (b,), (c,)) ),),
F(( (a, (b,), (c,)), ((a,), b, (c,)), ((a,), (b,), c), ((a,), (b,), (c,)) ),),
)
Now we can chain it to make a set of circuits that define together an eight-bit adder circuit with carry.
sum0, cout = full_bit_adder('a0', 'b0', 'Cin')
sum1, cout = full_bit_adder('a1', 'b1', cout)
sum2, cout = full_bit_adder('a2', 'b2', cout)
sum3, cout = full_bit_adder('a3', 'b3', cout)
sum4, cout = full_bit_adder('a4', 'b4', cout)
sum5, cout = full_bit_adder('a5', 'b5', cout)
sum6, cout = full_bit_adder('a6', 'b6', cout)
sum7, cout = full_bit_adder('a7', 'b7', cout)
Unfortunately, the sizes of the resulting expression explode:
map(len, map(str, (sum0, sum1, sum2, sum3, sum4, sum5, sum6, sum7, cout)))
We could also use the definitions from the Wikipedia article:
S = A ⊕ B ⊕ C
Cout = (A ⋅ B) + (Cin ⋅ (A ⊕ B))def full_bit_adder(a, b, c):
return (
xor(xor(a, b), c),
or_(and_(a, b), and_(c, xor(a, b))),
)
sum0, cout = full_bit_adder('a0', 'b0', 'Cin')
print 'Sum'
truth_table(sum0)
print
print 'Carry'
truth_table(cout)
sum1, cout = full_bit_adder('a1', 'b1', cout)
sum2, cout = full_bit_adder('a2', 'b2', cout)
sum3, cout = full_bit_adder('a3', 'b3', cout)
sum4, cout = full_bit_adder('a4', 'b4', cout)
sum5, cout = full_bit_adder('a5', 'b5', cout)
sum6, cout = full_bit_adder('a6', 'b6', cout)
sum7, cout = full_bit_adder('a7', 'b7', cout)
The sizes of these expression are much more tractable:
map(len, map(str, (sum0, sum1, sum2, sum3, sum4, sum5, sum6, sum7, cout)))
The Form Python datastructure is based on frozenset so duplicate terms are automatically removed and order of terms is irrelevant just as we would prefer. But we want to be able to automatically simplify forms beyond just that. Ideally, we would like a function that applies the rules of the calculus automatically:
A((B)) = AB
A() = ()
A(AB) = A(B)
I'm going to specify the behaviour of the desired function in a unittest.
import unittest
Let's deal with three easy cases first: string, the Mark, and the Void. The simplify() function should just return them unchanged.
class UnwrapTest0(unittest.TestCase):
def testMark(self):
self.assertEqual(Mark, simplify(Mark))
def testVoid(self):
self.assertEqual(Void, simplify(Void))
def testLeaf(self):
self.assertEqual('a', simplify('a'))
def simplify(form):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
if __name__ == '__main__':
unittest.main(argv=['ignored', 'UnwrapTest0'], exit=False)
(a)¶A single string in a form (a) should also be returned unchanged:
class UnwrapTest1(unittest.TestCase):
def testNegatedLeaf(self):
a = nor('a')
self.assertEqual(a, simplify(a))
def simplify(form):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
# We know it's a Form and it's not empty (else it would be the Mark and
# returned above.)
# Let's just recurse.
return Form(simplify(inner) for inner in form)
if __name__ == '__main__':
unittest.main(argv=['ignored', 'UnwrapTest1'], exit=False)
So far, so good. But what about ((a))? This should be returned as just a. And ((a b)) should remain ((a b)) because we can't represent just a b as a single Python object, so we have to retain the outer pair of containers to hold them without inverting the Mark/Void value (if we just used one container.)
class UnwrapTest2(unittest.TestCase):
def testUnwrapLeaf(self):
'''((a)) = a'''
a = or_('a')
self.assertEqual('a', simplify(a))
def testDoNotUnwrapTwoLeaves(self):
'''((a b)) = ((a b))'''
a = or_('a', 'b')
self.assertEqual(a, simplify(a))
def simplify(form):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
# We know it's a Form and it's not empty (else it would be the Mark and
# returned above.)
# Let's just recurse.
result = Form(simplify(inner) for inner in form)
# Check for ((a)) and return just a.
# If there is more than one item in the inner container ((a b..))
# then we must keep the outer containers.
if len(result) == 1:
inner, = result # inner = (a)
if isinstance(inner, Form) and len(inner) == 1:
a, = inner
return a
return result
if __name__ == '__main__':
unittest.main(argv=['ignored', 'UnwrapTest2'], exit=False)
Does it work for (((a))) = (a) and ((((a)))) = a and so on?
class UnwrapTest3(unittest.TestCase):
def testMultiUnwrapLeaf(self):
A = 'a'
B = nor(A)
a = nor(B)
self.assertEqual(A, simplify(a))
a = nor(a)
self.assertEqual(B, simplify(a))
a = nor(a)
self.assertEqual(A, simplify(a))
a = nor(a)
self.assertEqual(B, simplify(a))
a = nor(a)
self.assertEqual(A, simplify(a))
a = nor(a)
self.assertEqual(B, simplify(a))
def testMultiDoNotUnwrapTwoLeaves(self):
e = F('a', 'b')
f = a = nor(e)
self.assertEqual(f, simplify(a))
a = nor(a)
self.assertEqual(e, simplify(a))
a = nor(a)
self.assertEqual(f, simplify(a))
a = nor(a)
self.assertEqual(e, simplify(a))
a = nor(a)
self.assertEqual(f, simplify(a))
a = nor(a)
self.assertEqual(e, simplify(a))
# Technically, several of the tests above are redundant,
# I'm not willing to figure out the right point ot stop
# right now, so I just do extra tests.
if __name__ == '__main__':
unittest.main(argv=['ignored', 'UnwrapTest3'], exit=False)
But now let's trick our function, it can't handle (a ((b c))) = (a b c) yet. This is going to require an auxiliary helper function that is similar to simplify() but that yields terms into an outer context.
class UnwrapTest4(unittest.TestCase):
def testMultiUnwrapLeaf(self):
a, b, c = 'abc'
f = F(a,((b, c),))
e = F(a, b, c)
self.assertEqual(e, simplify(f))
def testMulti_blah_Leaf(self):
a, b, c = 'abc'
f = F(a,(((b, c),),),)
e = F(a, (b, c))
self.assertEqual(e, simplify(f))
def testMulti_blah_blah_Leaf(self):
a, b, c, d = 'abcd'
f = F(a,((((b, c),), d),))
e = F(a, b, c, d)
self.assertEqual(e, simplify(f))
def simplify(form):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
# We know it's a Form and it's not empty (else it would be the Mark and
# returned above.)
result = []
for inner in simplify_gen(form): # Use the generator instead of recursing into simplify().
result.append(inner)
result = Form(result)
# Check for ((a)) and return just a.
# If there is more than one item in the inner container ((a b..))
# then we must keep the outer containers.
if len(result) == 1:
inner, = result # inner = (a)
if isinstance(inner, Form):
if len(inner) == 1:
a, = inner
return a
else:
# len(inner) cannot be 0, because that means form is Void
# and would already have been returned.
assert len(inner) > 1, repr(inner)
# What to do here?
# We cannot yield the items in inner into the containing context
# because we don't have it (or even know if it exists.)
# Therefore we need a different simplify() generator function that yields
# the simplified contents of a form, and we have to call that instead
# of recurring on simplify() above.
pass
return result
def simplify_gen(form):
for inner in form:
inner = simplify(inner)
# Now inner is simplified, except for ((a b...)) which simplify() can't handle.
# Three easy cases, strings, Mark, or Void.
if isinstance(inner, basestring):
yield inner
continue
if inner == Mark:
yield inner
assert False # The simplify() function will not keep iterating after this.
return # Partial implementation of ()A = ().
if inner == Void:
continue # Omit Void. Implementation of (()) = .
# We know it's a Form and it's not empty (else it would be the Mark and
# yielded above.)
# Check for ((...)) and return just ... .
if len(inner) > 1: # (foo bar)
yield inner
continue
assert len(inner) == 1, repr(inner) # Just in case...
inner_inner, = inner
if isinstance(inner_inner, Form): # inner_inner = (...)
for inner_inner_inner in inner_inner:
yield inner_inner_inner
continue
#else: # inner_inner = foo ; inner = (foo)
yield inner
if __name__ == '__main__':
unittest.main(argv=['ignored', 'UnwrapTest4'], exit=False)
If the Mark occurs in a sub-form it should Occlude all sibling sub-forms, rendering its container form Void.
class MarkTest0(unittest.TestCase):
def testMarkOccludes0(self):
a, b, c = 'abc'
f = F(a, (), b, c)
self.assertEqual(Void, simplify(f))
def testMarkOccludes1(self):
a, b, c = 'abc'
f = F(a, (b, c, ()))
e = F(a)
self.assertEqual(e, simplify(f))
def testMarkOccludes2(self):
a, b, c = 'abc'
f = F(a, (b, ((), c)))
e = F(a, (b,))
self.assertEqual(e, simplify(f))
def simplify(form):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
# We know it's a Form and it's not empty (else it would be the Mark and
# returned above.)
result = []
for inner in simplify_gen(form):
if inner == Mark:
return Void # Discard any other inner forms, form is Void.
result.append(inner)
result = Form(result)
# Check for ((a)) and return just a.
# If there is more than one item in the inner container ((a b..))
# then we must keep the outer containers.
if len(result) == 1:
inner, = result # inner = (a)
if isinstance(inner, Form):
if len(inner) == 1:
a, = inner
return a
return result
if __name__ == '__main__':
unittest.main(argv=['ignored', 'MarkTest0'], exit=False)
So we have (()) = -- and ()A = () what about A(AB) = A(B)?
class PervadeTest0(unittest.TestCase):
def testPervade0(self):
a = 'a'
f = F(a, (a,))
self.assertEqual(Void, simplify(f))
def testPervade1(self):
a = 'a'
f = F(((a,),), (a,))
self.assertEqual(Void, simplify(f))
def testPervade2(self):
a = 'a'
b = 'b'
f = F(a, (b, (a,)))
e = F(a)
self.assertEqual(e, simplify(f))
def simplify(form, exclude=None):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
# We know it's a Form and it's not empty (else it would be the Mark and
# returned above.)
if exclude is None:
exclude = set()
new_stuff = form - exclude
exclude = exclude | new_stuff
result = []
for inner in simplify_gen(new_stuff, exclude):
if inner == Mark:
return Void # Discard any other inner forms, form is Void.
result.append(inner)
result = Form(result)
# Check for ((a)) and return just a.
# If there is more than one item in the inner container ((a b..))
# then we must keep the outer containers.
if len(result) == 1:
inner, = result # inner = (a)
if isinstance(inner, Form):
if len(inner) == 1:
a, = inner
return a
return result
def simplify_gen(form, exclude):
for inner in form:
inner = simplify(inner, exclude)
# Now inner is simplified, except for ((a b...)) which simplify() can't handle.
# Three easy cases, strings, Mark, or Void.
if isinstance(inner, basestring):
yield inner
continue
if inner == Mark:
yield inner
assert False # The simplify() function will not keep iterating after this.
return # Partial implementation of ()A = ().
if inner == Void:
continue # Omit Void. Implementation of (()) = .
# We know it's a Form and it's not empty (else it would be the Mark and
# yielded above.)
# Check for ((...)) and return just ... .
if len(inner) > 1: # (foo bar)
yield inner
continue
assert len(inner) == 1, repr(inner) # Just in case...
inner_inner, = inner
if isinstance(inner_inner, Form): # inner_inner = (...)
for inner_inner_inner in inner_inner:
yield inner_inner_inner
continue
#else: # inner_inner = foo ; inner = (foo)
yield inner
if __name__ == '__main__':
unittest.main(argv=['ignored', 'PervadeTest0'], exit=False)
TODO set up Hypothesis to generate test cases...
# Run ALL the tests!
if __name__ == '__main__':
unittest.main(argv=['ignored'], exit=False)
truth_table(F((a, b), (a, (b,))))
The form says, "if b then a else a". I'll come back to the interpretation of "Each-Way" as an if-then-else statement later.
The thing to note here is that the value for a can be a whole expression which appears twice in the new form: once next to b and once next to (b).
In the first case we can remove any occurances of b from the a next to it
b (...(b c (d ...)))
b (...( c (d ...)))
and in the second case we can change any occurances of b to the Mark.
(b)(...(b c (d ...)))
(b)((b)(b c (d ...)))
(b)(...(b (b) c (d ...)))
(b)(...(b ( ) c (d ...)))
(b)(...( ( ) ))
(b)(... )
We can send (b) into the form until it reaches and b, at which point b(b) becomes () and sweeps out any siblings rendering its containing form Void.
For the first case we can use simplify() and pass in b as a member of the exclude set.
A = F(a, (b, (c,)))
A
simplify(A, {b})
with_mark¶In the second case (b)...b... = (b)...()... we can modify the simplify() function to accept a name that it should treat as Mark-valued.
class MarkitTest0(unittest.TestCase):
def testMarkit0(self):
a, b, c = 'abc'
f = F(a, (b, (c,)))
e = F(a)
self.assertEqual(e, simplify(f, with_mark=b))
def simplify(form, exclude=None, with_mark=None):
# Three easy cases, for strings, Mark, or Void, just return it.
if isinstance(form, basestring) or form in BASE:
return form
# We know it's a Form and it's not empty (else it would be the Mark and
# returned above.)
if exclude is None:
exclude = set()
new_stuff = form - exclude
exclude = exclude | new_stuff
result = []
for inner in simplify_gen(new_stuff, exclude, with_mark):
if inner == Mark or inner == with_mark:
return Void # Discard any other inner forms, form is Void.
result.append(inner)
result = Form(result)
# Check for ((a)) and return just a.
# If there is more than one item in the inner container ((a b..))
# then we must keep the outer containers.
if len(result) == 1:
inner, = result # inner = (a)
if isinstance(inner, Form):
if len(inner) == 1:
a, = inner
return a
return result
def simplify_gen(form, exclude, with_mark):
for inner in form:
inner = simplify(inner, exclude, with_mark)
# Now inner is simplified, except for ((a b...)) which simplify() can't handle.
# Three easy cases, strings, Mark, or Void.
if isinstance(inner, basestring):
yield inner
continue
if inner == Mark or inner == with_mark:
yield Mark
assert False # The simplify() function will not keep iterating after this.
return # Partial implementation of ()A = ().
if inner == Void:
continue # Omit Void. Implementation of (()) = .
# We know it's a Form and it's not empty (else it would be the Mark and
# yielded above.)
# Check for ((...)) and return just ... .
if len(inner) > 1: # (foo bar)
yield inner
continue
assert len(inner) == 1, repr(inner) # Just in case...
inner_inner, = inner
if isinstance(inner_inner, Form): # inner_inner = (...)
for inner_inner_inner in inner_inner:
if inner_inner_inner == with_mark:
yield Mark
assert False # The simplify() function will not keep iterating after this.
return # Never reached, could delete this line.
yield inner_inner_inner
continue
#else: # inner_inner = foo ; inner = (foo)
yield inner
if __name__ == '__main__':
unittest.main(argv=['ignored', 'MarkitTest0'], exit=False)
simplify(A, with_mark=b)
Now we can create a new form that is equivalent to A.
def each_way(form, name):
return simplify(F(
( name , form),
((name,), simplify(form, with_mark=name)),
))
Ab = each_way(A, b)
print A, '=', Ab
In this particular case the original form A was so simple that the new version Ab is actually a bit larger. With a large expression to start with the form after simplification would (usually) be smaller.
Recall our original expressions for the sum and carry bits of a full-bit adder circuit, derived from the truth tables:
Sum = F(( (a, b, (c,)), (a, (b,), c), ((a,), b, c), ((a,), (b,), (c,)) ),)
Carry = F(( (a, (b,), (c,)), ((a,), b, (c,)), ((a,), (b,), c), ((a,), (b,), (c,)) ),)
Sum
Carry
And the expressions derived from the definitions on Wikipedia:
Sum, Carry = full_bit_adder(a, b, c)
Sum
Carry
We can use the each_way() function to look for simpler equivalent forms by, for example, iterating though the names and trying it with each. Try the following cells with both versions of the Sum and Carry above.
print Sum
for name in (a, b, c, a, b, c):
Sum = each_way(Sum, name)
print Sum
print Carry
for name in (a, b, c, a, b, c):
Carry = each_way(Carry, name)
print Carry
Let's redefine the full_bit_adder() function with the smallest version of each above.
def full_bit_adder(a, b, c):
'''From the truth table.'''
return (
F(( (a, b, (c,)), (a, (b,), c), ((a,), b, c), ((a,), (b,), (c,)) ),),
F(( (a, (b,), (c,)), ((a,), b, (c,)), ((a,), (b,), c), ((a,), (b,), (c,)) ),),
)
# Sizes: [63, 327, 1383, 5607, 22503, 90087, 360423, 1441767, 1441773]
def full_bit_adder(a, b, c):
'''Simplest forms from above.'''
return (
F( (((b,), (c,)), (a,), (b, c)), (((b,), c), ((c,), b), a) ),
F( (((a,), (c,)), b), ((b,), a, c) ),
)
# Sizes: [57, 177, 417, 897, 1857, 3777, 7617, 15297, 7653]
def full_bit_adder(a, b, c):
'''Based on the definitions from Wikipedia.'''
return (
xor(xor(a, b), c), # ((((((((a) b) ((b) a)))) c) (((((a) b) ((b) a))) (c))))
or_(and_(a, b), and_(c, xor(a, b))), # ((((((((a) b) ((b) a)))) (c)) ((a) (b))))
)
# Sizes: [67, 159, 251, 343, 435, 527, 619, 711, 371]
def full_bit_adder(a, b, c):
'''Based on the definitions from Wikipedia.'''
return (
simplify(xor(xor(a, b), c)),
simplify(or_(and_(a, b), and_(c, xor(a, b)))),
)
# Sizes: [59, 135, 211, 287, 363, 439, 515, 591, 311]
In this case, the version from the definitions does much better than the other two.
Sum, Carry = full_bit_adder(a, b, c)
truth_table(Sum)
print
truth_table(Carry)
print
sum0, cout = full_bit_adder('a0', 'b0', 'Cin')
sum1, cout = full_bit_adder('a1', 'b1', cout)
sum2, cout = full_bit_adder('a2', 'b2', cout)
sum3, cout = full_bit_adder('a3', 'b3', cout)
sum4, cout = full_bit_adder('a4', 'b4', cout)
sum5, cout = full_bit_adder('a5', 'b5', cout)
sum6, cout = full_bit_adder('a6', 'b6', cout)
sum7, cout = full_bit_adder('a7', 'b7', cout)
print map(len, map(str, (sum0, sum1, sum2, sum3, sum4, sum5, sum6, sum7, cout)))
This is something of an Interlude, we aren't going to use it below, but it's too cool to omit mention.
We can use the simplify() function to create a more efficient SAT solver along the lines of the DPLL algorithm.
It works by selecting a name from the form, and simplifying the form with that name first as Void then as Mark, then recursing with the new form and the next name. If the resulting simplified form becomes the Mark then our choices (of assigning Void or Mark to the names selected so far) constitute a "solution" to the original form. That is, if we reify() the form with the environment returned by the dpll() function the result will be Mark-valued.
def dpll(E, partial=None, unit=True):
if partial is None:
partial = {}
else:
partial = partial.copy() # so we can backtrack later..
if isinstance(E, basestring):
partial[E] = Mark
return partial
if unit:
E = assign_unit_clauses(E, partial)
if not E:
return partial
if Mark in E:
return
v = next_symbol_of(E)
partial[v] = Void
res = dpll(simplify(E, {v}), partial, unit)
if res is not None:
return res
partial[v] = Mark
return dpll(simplify(E, with_mark=v), partial, unit)
def assign_unit_clauses(E, partial):
'''
Find and assign values to an "unit clauses" in the form, simplifying as you go.
A unit clause is a bare name or a negated name: a or (a), for these clauses we
can set them to Void or Mark, respectively, to contibute to making their containing
Form the Mark.
'''
on, off, E = find_units(E)
while on or off:
while on:
if on & off: return Void
term = first_of(on)
partial[term] = Mark
ON, OFF, E = find_units(simplify(E, with_mark=term))
on |= ON
on.remove(term)
off |= OFF
while off:
if on & off: return Void
term = first_of(off)
partial[term] = Void
ON, OFF, E = find_units(simplify(E, {term}))
off |= OFF
off.remove(term)
on |= ON
return E
def next_symbol_of(E):
if isinstance(E, basestring):
return E
for it in E:
return next_symbol_of(it)
raise Exception("no more symbols")
def find_units(E):
'''
Return two sets and a possibly-reduced E. The literals in the first
set must be Void and those in the second must be set Mark to have the
entire expression become Void.
'''
on, off, poly = set(), set(), set()
for clause in E:
if isinstance(clause, basestring):
off.add(clause)
continue
if len(clause) != 1:
poly.add(clause)
continue
(n,) = clause # Unwrap one layer of containment.
if isinstance(n, basestring):
on.add(n)
else:
poly.add(clause)
return on, off, Form(poly)
# Return any item from a form.
first_of = lambda form: next(iter(form))
A = F(a, (b, (c,)))
truth_table(A)
solution = dpll(A)
arith = reify(A, solution)
print
print 'A solution:', solution
print
print 'Reifies to', arith, u'⟶', value_of(arith)
dpll_iter()¶We can write a generator version of the function that keeps looking for solutions after the first.
def dpll_iter(E, partial=None, unit=True):
if partial is None:
partial = {}
else:
partial = partial.copy() # so we can backtrack later..
if isinstance(E, basestring):
partial[E] = Mark
yield partial
return
if unit:
E = assign_unit_clauses(E, partial)
if not E:
yield partial
return
if Mark in E:
return
v = next_symbol_of(E)
partial[v] = Void
for res in dpll_iter(simplify(E, {v}), partial, unit):
yield res
partial[v] = Mark
for res in dpll_iter(simplify(E, with_mark=v), partial, unit):
yield res
for solution in dpll_iter(A):
print (solution)
Sum, Carry = full_bit_adder(a, b, c)
for solution in dpll_iter(Sum, unit=False):
r = reify(Sum, solution)
print (solution), r, '=', simplify(r)
for solution in dpll_iter(Carry, unit=False):
r = reify(Carry, solution)
print (solution), r, '=', simplify(r)
Notice that the reified form still has c in it but that doesn't prevent the simplify() function from reducing the form to the Mark. This should be the case for all solutions generated by the dpll_iter() function.
for solution in dpll_iter(cout):
print simplify(reify(cout, solution)),
Interesting! Some solutions do not simplify() completely in one go. The form (((a5) a5)) is Mark-valued:
(((a5) a5))
((( ) a5))
((( ) ))
( )
()E = F(((a,), a))
print E
print simplify(E)
Something to keep in mind.
sum0, cout = full_bit_adder('a0', 'b0', 'Cin')
sum1, cout = full_bit_adder('a1', 'b1', cout)
sum2, cout = full_bit_adder('a2', 'b2', cout)
sum3, cout = full_bit_adder('a3', 'b3', cout)
Put the circuit expressions into a handy dictionary, and we are ready to add some numbers.
CIRCUITS = {
'sum0': sum0,
'sum1': sum1,
'sum2': sum2,
'sum3': sum3,
'cout': cout,
}
A bunch of crufty junk to print out a nice truth table with the columns arranged to make it (relatively) easy to see the addition.
def stringy_env(env):
return {
k: value_of(e, v='--', m='()')
for k, e in env.items()
}
INPUTs = ('Cin', 'a3', 'a2', 'a1', 'a0', 'b3', 'b2', 'b1', 'b0')
OUTPUTs = ('cout', 'sum3', 'sum2', 'sum1', 'sum0')
format_string = '%(' + ')s %('.join(INPUTs) + ')s | %(' + ')s %('.join(OUTPUTs) + ')s'
print 'Ci|a3 a2 a1 a0|b3 b2 b1 b0 | Co s3 s2 s1 s0'
results = {}
for env in environments_of_variables(*INPUTs):
for name, expression in CIRCUITS.items():
results[name] = value_of(reify(expression, env))
env.update(results)
print format_string % stringy_env(env)
That was a bit steep, let's formalize it and make it a little easier to work with.
First let's have a register of named values:
R = {name: Void for name in 'Cin a3 a2 a1 a0 b3 b2 b1 b0 Cout'.split()}
Let's have a program of named expressions that give new values when evaluated in terms of the current values in R (this is just the same CIRCUITS, but feeding back the results into the "b" bits):
P = {
'b0': sum0,
'b1': sum1,
'b2': sum2,
'b3': sum3,
'Cout': cout,
}
One cycle of the machine means to evaluate each named expression in the program with the current values in the register.
make_reify_reducer = lambda env: lambda form: value_of(reify(form, env))
def cycle(program, register):
rr = make_reify_reducer(register)
return {bit: rr(expression) for bit, expression in program.iteritems()}
With all the register values at "zero" (Void) nothing happens.
R.update(cycle(P, R))
R
Let's make a nice display function to inspect our little adder computer.
def show_as_int(*names):
'''
Return a function that converts a sequence of
named bits (as Void/Mark values) into a integer.
'''
def inner(register):
i, n = 0, 1
for name in names:
if not register[name]:
i += n
n <<= 1
return i
return inner
a_register = show_as_int('a0', 'a1', 'a2', 'a3')
b_register = show_as_int('b0', 'b1', 'b2', 'b3')
def show_computer_state(R):
print 'a: %-3i b: %-3i Cin: %-3i Cout: %-3i' % (
a_register(R),
b_register(R),
int(not R['Cin']),
int(not R['Cout']),
)
show_computer_state(R)
Let's set one bit to true (Mark-valued in the chosen convention. We could have Void be true but we would have to form the circuit expressions differently.)
R['a0'] = Mark
Now let's count to twenty.
for _ in range(20):
show_computer_state(R)
R.update(cycle(P, R))
You can see that at the sixteenth step the "Cout" carry bit is true and the count cycles back to zero.
# Reset the register.
R = {name: Void for name in 'Cin a3 a2 a1 a0 b3 b2 b1 b0 Cout'.split()}
# Count by three.
R['a0'] = R['a1'] = Mark
# Print out twenty cycles.
for _ in range(20):
show_computer_state(R)
R.update(cycle(P, R))
You can see that the "b" bits are indeed counting by threes: 0, 3, 6, 9, 12, 15 & carry, 2, 5, 8, 11, 14 & carry, 1, 4, 7, 10, 13 & carry, 0, 3, 6, 9, ...
This is my basic model for computation: A register, a program, and a cycle function. Note that reducing the form on each cycle isn't necessary, we can run the cycles and just reify() without reducing and we get new circuits that define bits in terms of the register values N cycles in the past.
# Universe
U = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
# Register.
R = {name: Void for name in U}
# A program to XOR each bit with its neighbor, wrapping at the edge.
P = {
name: xor(name, U[(U.index(name) + 1) % len(U)])
for name in U
}
def show(reg):
'''Simple visualization of the register.'''
print ''.join('.0'[not reg[name]] for name in U)
# Set one "bit" in the register.
R[U[-1]] = Mark
# Run through some cycles.
for _ in range(100):
show(R)
R.update(cycle(P, R))
Before building larger "computers" I want to switch to a more efficient implementation based on a register as a set of names that are currently Mark-valued, and a set_solve() function that evaluates a form in terms of such a set, and assuming all other names are Void-valued.
def set_solve(form, marks):
'''
Given a form and a set of names that are Marks assume all other names
in the form are Void and reduce to basic value (Mark or Void.)
'''
return (
(Void, Mark)[form in marks]
if isinstance(form, basestring)
else _set_solve(form, marks)
)
def _set_solve(form, marks):
for inner in form:
if isinstance(inner, basestring):
if inner in marks:
return Void
continue
if not _set_solve(inner, marks): # Mark
return Void
return Mark
A = F(a, (b, (c,)))
print A
print set_solve(A, {a})
print set_solve(A, {b})
print set_solve(A, {c})
To calculate the new R first collect all the names in R that are not mentioned in P (and so cannot be set to Void by it) then add the names evaluated by solving P's expressions with the marks in R.
def set_cycle(R, P):
return R.difference(P).union(
signal
for signal, expression in P.iteritems()
if not set_solve(expression, R)
)
# Universe
U = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
# Register.
R = {U[-1]}
# A program to XOR each bit with its neighbor, wrapping at the edge.
P = {
name: xor(name, U[(U.index(name) + 1) % len(U)])
for name in U
}
def show(reg):
'''Simple visualization of the register.'''
print ''.join('.0'[name in reg] for name in U)
# Run through some cycles.
for _ in range(100):
show(R)
R = set_cycle(R, P)
def set_show_as_int(*names):
def inner(register):
i, n = 0, 1
for name in names:
if name in register:
i += n
n <<= 1
return i
return inner
def ifte(predicate, true, false):
return F(
((predicate,), true),
( predicate , false),
)
E = ifte(a, b, c)
truth_table(E)
If a is Mark-valued the value of the whole form is that of b, but if a is Void-valued the value of the whole form is that of c.
w/ a = ()
((( a) b) ( a c))
(((()) b) (() c))
(( b) (() ))
(( b) )
b
w/ a =
(((a) b) (a c))
((( ) b) ( c))
((( ) ) ( c))
( ( c))
cdef flip_flop(q, reset, set_):
return F(reset, (q, set_))
q, r, s = 'qrs'
E = flip_flop(q, r, s)
truth_table(E)
This is a form that can be used in a circuit to "remember" a value.
w/ r = ()
((q s) r)
((q s) ())
( ())
w/ s = (), r = ___
((q s) r)
((q ()) )
(( ()) )
( )
w/ s = ___, r = ___
((q s) r)
((q ) )
q
If both are Void then the form is just q, if r is Mark then the form is Void, otherwise if s is Mark the form becomes Mark. This is called a "flip-flop" circuit, and it comprises a simple machine to remember one bit.
Consider a simple computer:
U = q, r, s = 'qrs'
P = {
q: flip_flop(q, r, s),
r: Void,
s: Void,
}
R = set() # Initially all signals are off, Void-valued.
bools_of = lambda universe: lambda register: (name in register for name in universe)
def simple_show(universe):
B = bools_of(universe)
def _shower(register):
print ''.join('.0'[b] for b in B(register))
return _shower
show = simple_show(U)
show(R)
def SET(R):
R |= {s}
show(R)
return set_cycle(R, P)
def RESET(R):
R |= {r}
show(R)
return set_cycle(R, P)
print U
show(R) ; R = set_cycle(R, P)
show(R) ; R = set_cycle(R, P)
show(R) ; R = set_cycle(R, P)
R = SET(R)
show(R) ; R = set_cycle(R, P)
show(R) ; R = set_cycle(R, P)
R = SET(R)
show(R) ; R = set_cycle(R, P)
show(R) ; R = set_cycle(R, P)
R = RESET(R)
show(R) ; R = set_cycle(R, P)
show(R) ; R = set_cycle(R, P)
R = RESET(R)
show(R) ; R = set_cycle(R, P)
show(R) ; R = set_cycle(R, P)
You can see that q is stable unless s or r set or reset it.
We can use the system we have developed so far to build addressable RAM.
# Width in bits of the DATA registers.
WIDTH = 2
# Width in bits of the ADDR registers.
LENGTH = 3 # Actual length 2**LENGTH
P = {}
We'll assume a single WRITE bit that sets a RAM location determined by the ADDR sub-register to the contents of the DATA sub-register.
WRITE = 'WRITE'
# Address register.
ADDR = ['addr_%i' % i for i in range(LENGTH)]
ADDR
# Data register.
DATA = ['a%i' % i for i in range(WIDTH)]
DATA
We can recognize which RAM location we want to address by using the expressions for each row from above to create a predicate for each location that depends on the ADDR bits.
def make_ram(program, addrs, data, width, write):
RAM = []
for addr, env in enumerate(environments_of_variables(*addrs)):
addr_predicate = F((name,) if env[name] else name for name in env)
predicate = and_(write, addr_predicate)
for bit in range(width):
ram = 'RAM_%i_%i' % (addr, bit)
RAM.append(ram)
program[ram] = simplify(ifte(predicate, data[bit], ram))
return RAM
RAM = make_ram(P, ADDR, DATA, WIDTH, WRITE)
RAM
P
E = P['RAM_0_0']
E
Note that if the WRITE bit is unset then each RAM bit is just set to itself.
simplify(E, exclude={'WRITE'})
But if the WRITE bit is set then each RAM location still depends on the -per-location-predicate.
simplify(E, with_mark='WRITE')
def make_accumulator(program, addrs, data, width, read, alts):
addr_predicates = [
F((name,) if env[name] else name for name in env)
for env in environments_of_variables(*addrs)
]
for bit in range(width):
a = data[bit]
alt = alts[bit]
foo = Void
for addr, predicate in enumerate(addr_predicates):
ram = 'RAM_%i_%i' % (addr, bit)
foo = (ifte(predicate, ram, foo))
program[a] = ifte(read, foo, alt)
p = {}
make_accumulator(p, ADDR, DATA, WIDTH, 'READ', DATA)
p
E = p[DATA[0]]
E
simplify(E, with_mark='READ')
simplify(E, {'READ'})
each_way(E, 'a0')
each_way(E, 'WRITE')
each_way(E, 'addr_0')
each_way(E, 'a0')
simplify(each_way(E, 'WRITE'), with_mark='WRITE')
, Sorting Networks for routing, more basic functions.
Eventually: Orchestration with Joy.
The rule A(AB) = A(B) is the powerhouse of LoF.
w/ A = ()
A(AB) = A(B)
()(()B) = ()(B)
() = ()
w/ A =
A(AB) = A(B)
(B) = (B)
Be aware of the recursive nature of this rule:
A(...(...(A B)))
A(.A.(...(A B)))
A(.A.(.A.(A B)))
A(.A.(.A.( B)))
A(.A.(...( B)))
A(...(...( B)))
There is this too:
(A)(...(...(... A B)))
(A)((A)(...(... A B)))
(A)((A)((A)(... A B)))
(A)((A)((A)((A) A B)))
(A)((A)((A)(( ) A B)))
(A)((A)(...(( ) )))
(A)(...(... ))
Summarized:
(A)(...(...(... A )))
(A)(...(...(... () )))
(A)(...(... ))Given a string form of an arithmetic expression (in other words a string that consists of only balanced pairs of parentheses) we can reduce it to its Mark/Void value by substitution to a fixed-point.
# Translate the LoF Arithmetic rules to string substitution rules.
reduce_string = lambda s: (
s
.replace('()()', '()')
.replace('(())', '')
)
def to_fixed_point(initial_value, F, limit=10000):
'''Do value = F(value) until value == F(value).'''
next_value = F(initial_value)
while next_value != initial_value:
if not limit: # A safety mechanism. Bail out after N iterations.
raise RuntimeError('Reached limit of allowed iterations without finding fixed point.')
limit -= 1
initial_value = next_value
next_value = F(initial_value)
return initial_value
from operator import add
def dyck_language(left, right=0, t='', A='(', B=')', op=add):
'''Yield balanced pairs of A and B.'''
if not (left or right):
yield t
return
if left > 0:
for i in dyck_language(left - 1, right + 1, op(t, A), A, B, op):
yield i
if right > 0:
for i in dyck_language(left, right - 1, op(t, B), A, B, op):
yield i
for s in dyck_language(5):
print s, u'⟶', to_fixed_point(s, reduce_string)
void() and mark()¶The void() and mark() functions can be defined recursively in terms of each other. Note that void() uses any() while mark() uses all(). These functions implement a depth-first search. If we used versions of any() and all() that evaluated their arguments in parallel void() could return after the True result while mark() depends on all terms's results so its runtime will be bound by term with the greatest runtime.
def void(form):
return any(mark(i) for i in form)
def mark(form):
return all(void(i) for i in form)
for form in BASE:
for func in (void, mark):
print form, 'is', func.__name__, '?', func(form)
The Void and the Mark are not Boolean true and false. Rather they correspond to non-existance and existance. When translating a traditional logical statement into Laws of Form the first thing we must do is choose which mapping we would like to use: true = Mark or true = Void. The notation works the same way once the translation is made, so the only real criteria for choosing is which gives the smaller form.
If you examine the truth tables for the basic forms above in light of this, you can see that each defines two logical functions depending on whether you treat Void as true and Mark as false, or Void as false and Mark as true.
For example, the juxtaposition of two terms next to each other a b corresponds to OR if Mark is true, and to AND if Void is true. Likewise, the form ((a)(b)) is AND if Mark is true and OR if Void is true.
Consider:
(A ∧ ¬B) ∨ (C ∧ D)
(This reads "(A and not B) or (C and D)" in case you have a hard time remembering what the symbols mean like I do.)
If we choose Mark to be true then the form is:
((A) B) ((C)(D))
If we choose Void to be true then the form is:
((A (B)) (C D))
As I said above, the notation works the same way either way, so once the translation is made you can forget about the Boolean true/false and just work with the Laws of Form rules. Logic is containers.
Consider also the De Morgan dual of the original statement:
¬((¬A ∨ B) ∧ (¬C ∨ ¬D))
If we choose Mark to be true then the form is:
(( ((A) B) ((C)(D)) ))
The outer pair of containers can be deleted leaving the same form as above:
((A) B) ((C)(D))
Likewise, if we choose Void to be true then the form is:
((((A)) (B)) (((C)) ((D))))
Again, A((B)) => AB reduces this form to the same one above:
((A (B)) (C D))
In the Laws of Form there are no De Morgan Dual statements. If you translate a logic statement and its dual into Laws of Form notation they both reduce to the same form.
To me this is a clear indication that the Laws of Form are superior to the traditional notation involving ¬ ∨ ∧, etc. LoF replaces all the symbols with just names and boundaries. The Laws of Form are not dualistic, they work directly in terms of existance and non-existance. Duality only comes in when you interpret the forms as Boolean statements and have to pick a mapping.
from collections import Counter
histo = Counter(yield_variables_of(cout))
histo
#import pprint as pp
#E = cout
#for var, count in histo.most_common():
# print 'stan', var, count
# E = to_fixed_point(simplify(standardize(E, var)), simplify)
# print len(str(E))
# pp.pprint(dict(Counter(yield_variables_of(E))))
# print '------'
Rather than manually calling standard_form() let's define a function that reduces a form to a (hopefully) smaller equivalent form by going through all the variables in the form and using standard_form() with each. Along with clean and unwrap we can drive an expression to a fixed point.
def STAN(form):
for var, _ in Counter(yield_variables_of(form)).most_common():
form = to_fixed_point(simplify(standardize(form, var)), simplify)
return form
sum0, cout = full_bit_adder('a0', 'b0', 'Cin')
sum1, cout = full_bit_adder('a1', 'b1', cout)
sum2, cout = full_bit_adder('a2', 'b2', cout)
sum3, cout = full_bit_adder('a3', 'b3', cout)
map(len, map(str, (sum0, sum1, sum2, sum3, cout)))
sum0, cout = full_bit_adder('a0', 'b0', 'Cin')
sum0 = to_fixed_point(sum0, STAN)
cout = to_fixed_point(cout, STAN)
sum1, cout = full_bit_adder('a1', 'b1', cout)
sum1 = to_fixed_point(sum1, STAN)
cout = to_fixed_point(cout, STAN)
sum2, cout = full_bit_adder('a2', 'b2', cout)
sum2 = to_fixed_point(sum2, STAN)
cout = to_fixed_point(cout, STAN)
sum3, cout = full_bit_adder('a3', 'b3', cout)
sum3 = to_fixed_point(sum3, STAN)
cout = to_fixed_point(cout, STAN)
map(len, map(str, (sum0, sum1, sum2, sum3, cout)))
It would be useful and fun to write a simple search algorithm that tried different ways to reduce a form to see if it could find particulaly compact versions.
Let's generate the expressions for the next two output bits, and the carry bit.
The sum3 bit expression is pretty big.
sum3
But it's only about 1/9th of size of the previous version (which was 9261.)
len(str(sum3))
Let's simplify the first one manually just for fun:
(((((())) (())) ((()))))
(( ) ) ( )
( )
Sure enough, it reduces to Mark after just a few applications of the rule (()) = __ (the underscores indicates the absence of any value, aka Void.) We could also just delete variables that are Void in the original expression:
((((a)b)(c)))
(( ) )( )
( )
C = F((a, b))
for form in (A, B, C):
arth = reify(form, env)
print form, u'⟶', arth, u'⟶', value_of(arth)
print A
Aa = simplify(A, {a})
print a, Aa
Aab = simplify(Aa, {b})
print a, b, Aab
Aabc = simplify(Aab, {c})
print a, b, c, Aabc
print a, Aa
Aab = simplify(Aa, with_mark=b)
print a, F(b), Aab
print a, Aa
Aac = simplify(Aa, with_mark=c)
print a, F(c), Aac
print a, Aa
Aac = simplify(Aa, {c})
print a, c, Aac
from collections import Counter
histo = Counter(yield_variables_of(sum7))
histo
len(str(sum7))
Immediately we can just call simplify() until it stops shinking the expression.
s7 = simplify(sum7)
len(str(s7))
s7 = simplify(s7)
len(str(s7))
Once was enough (we should consider adding a call to simplify() in the full_bit_adder() function.)
Let's try using each_way() with the most common names in the form.
s7 = simplify(each_way(s7, 'a0'))
len(str(s7))
s7 = simplify(s7)
len(str(s7))
Counter(yield_variables_of(s7))
s7 = sum7
#for name, count in histo.most_common():
# s7 = simplify(each_way(s7, name))
# print len(str(s7))
def super_simple(form):
return to_fixed_point(form, simplify)
len(str(sum7))
s7 = super_simple(sum7)
len(str(s7))
s7 = sum7
#for name, count in histo.most_common():
# s7 = super_simple(each_way(s7, name))
# print len(str(s7))
print ' '.join(name[:2] for name in sorted(R))
for _ in range(20):
print format_env(R), '=', b_register(R)
R.update(cycle(P, R))