Fractran machine!
François Pinard
discover what it does. You might prefer to read the code and guess? :-)
# Versions: Pascal in 1985, C++ in 1989, Python in 2000.
"""\
Interpreter for the Fractran machine.
The Fractran machine (probably invented by John Conway) holds an accumulator,
which is an integer of unlimited length. It also holds a finite program
expressed as a list of rational numbers, or fractions.
Each computation cycle works as follow. Find the first fraction for which
the denominator exactly divides the accumulator. If there is no such
fraction, the machine halts. Multiply the accumulator by this fraction.
Also, whenever the result is an exact power of two, output the exponent.
"""
def main():
fractran = Fractran()
fractran.load_program((17, 91),
(78, 85),
(19, 51),
(23, 38),
(29, 33),
(77, 29),
(95, 23),
(77, 19),
(1, 17),
(11, 13),
(13, 11),
(15, 14),
(15, 2),
(55, 1))
fractran.load_accumulator(2)
try:
fractran.run()
finally:
fractran.status()
class Fractran:
def load_program(self, *fractions):
self.program = []
for numerator, denominator in fractions:
self.program.append((long(numerator), long(denominator)))
def load_accumulator(self, accumulator):
self.accumulator = long(accumulator)
def run(self):
count = self.step_count = 0
accumulator = self.accumulator
while 1:
count = count + 1
for numerator, denominator in self.program:
quotient, remainder = divmod(accumulator, denominator)
if remainder == 0:
accumulator = quotient * numerator
break
if remainder != 0:
break
if accumulator & accumulator - 1 == 0:
print ('Printing %d after %d steps'
% (length(accumulator)-1, count))
self.step_count = count
self.accumulator = accumulator
def status(self):
print 'Stopping after %d steps' % self.step_count
print ' with accumulator =', self.accumulator
def length(number):
count = 0
while number > 256:
count = count + 8
number = number >> 8
while number > 0:
count = count + 1
number = number >> 1
return count
if __name__ == '__main__':
main()
--
François Pinard http://www.iro.umontreal.ca/~pinard
Ivan Van Laningham
François Pinard wrote:
>
> Hi, people. Here is a Fractran machine, with a demo program. Try it to
> discover what it does. You might prefer to read the code and guess? :-)
>
> # Versions: Pascal in 1985, C++ in 1989, Python in 2000.
>
That seems good for this year's motto: "Python in 2000".
<don't-tread-on-me>-ly y'rs,
Ivan;-)
----------------------------------------------
Ivan Van Laningham
Axent Technologies, Inc.
http://www.pauahtun.org
http://www.foretec.com/python/workshops/1998-11/proceedings.html
Army Signal Corps: Cu Chi, Class of '70
Author: Teach Yourself Python in 24 Hours
Boris Borcic
>
> Hi, people. Here is a Fractran machine, with a demo program.
How long does the demo take to halt ?
BB
Bernhard Herzog
I don't claim to have understood how the demo program for the fractran
machine works, but it seems pretty obvious that it will never halt.
From the doc string:
Each computation cycle works as follow. Find the first fraction for
which the denominator exactly divides the accumulator. If there is no
such fraction, the machine halts. ...
And the fractran program:
fractran.load_program((17, 91),
(78, 85),
(19, 51),
(23, 38),
(29, 33),
(77, 29),
(95, 23),
(77, 19),
(1, 17),
(11, 13),
(13, 11),
(15, 14),
(15, 2),
(55, 1))
Note that the last item has a denominator of 1, so there will always be
a fraction whose denominator exactly divides the accumulator.
Another observation: If a program has a fraction whose denomintor is 1,
it must be the last one in the program. Fractions after that will never
be used.
--
Bernhard Herzog | Sketch, a drawing program for Unix
her...@online.de | http://sketch.sourceforge.net/
François Pinard
> How long does the demo take to halt ?
Once it finishes printing the whole series of prime numbers! Yet, it is
likely that you will run out of memory before it does. :-)