# Fractran machine!

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### François Pinard

ungelesen,
18.05.2000, 03:00:0018.05.00
an pytho...@python.org
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.

"""\
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()
(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))
try:
fractran.run()
finally:
fractran.status()

class Fractran:

self.program = []
for numerator, denominator in fractions:
self.program.append((long(numerator), long(denominator)))

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

ungelesen,
18.05.2000, 03:00:0018.05.00
an Python Mailing List
Hi All--

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".

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

ungelesen,
19.05.2000, 03:00:0019.05.00
an
François Pinard wrote:
>
> Hi, people. Here is a Fractran machine, with a demo program.

How long does the demo take to halt ?

BB

### Bernhard Herzog

ungelesen,
19.05.2000, 03:00:0019.05.00
an
Boris Borcic <bor...@geneva-link.ch> writes:

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:

(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

ungelesen,
19.05.2000, 03:00:0019.05.00
an Boris Borcic
Boris Borcic <bor...@geneva-link.ch> écrit:

> 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. :-)

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