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May 18, 2000, 3:00:00 AM5/18/00

to 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? :-)

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

May 18, 2000, 3:00:00 AM5/18/00

to 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".

<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

May 19, 2000, 3:00:00 AM5/19/00

to

François Pinard wrote:

>

> Hi, people. Here is a Fractran machine, with a demo program.

>

> Hi, people. Here is a Fractran machine, with a demo program.

How long does the demo take to halt ?

BB

May 19, 2000, 3:00:00 AM5/19/00

to

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:

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/

May 19, 2000, 3:00:00 AM5/19/00

to 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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