# Problem with algorithm

Discussion in 'Python' started by Jia Lu, Apr 13, 2007.

1. ### Jia LuGuest

Hi all.
I want to create a large list like:

aaaa ~ zzzz

Is there any good algorithm to do this?

Thanx

Jia Lu

Jia Lu, Apr 13, 2007

2. ### Guest

On Apr 12, 10:16ï¿½pm, "Jia Lu" <> wrote:
> Hi all.
>  I want to create a large list like:
>
> aaaa ~ zzzz
>
> Is there any good algorithm to do this?

Sure.
test = '01'

for m in test:
for n in test:
for o in test:
for p in test:
print m+n+o+p

## 0000
## 0001
## 0010
## 0011
## 0100
## 0101
## 0110
## 0111
## 1000
## 1001
## 1010
## 1011
## 1100
## 1101
## 1110
## 1111

Now just change test='01' to test='abcdefghijklmnopqrstuvwxyz'.

>
> Thanx
>
> Jia Lu

, Apr 13, 2007

3. ### Charles SandersGuest

wrote:
> On Apr 12, 10:16ï¿½pm, "Jia Lu" <> wrote:
>> Hi all.
>> ï¿½I want to create a large list like:
>>
>> aaaa ~ zzzz
>>
>> Is there any good algorithm to do this?

>
> Sure.
> test = '01'
>
> for m in test:
> for n in test:
> for o in test:
> for p in test:
> print m+n+o+p

[snip]

Forgive any silly mistakes I have made (I've been teaching
myself python for about 1 week) but there is a moderately
well known algorithm for this that extends to arbitrary
lengths of both the list of alternatives and the length
of the required output, and avoids deeply nested loops.
I know that it is no better for small and constant output
lengths, but for longer lengths or if the output length
can vary it should be better. There is a similar algorithm
if duplicates are not allowed (ie abcd ... wxyz).

My attempt at a python translation of the algorithm:

def m_from_n ( v, m ):
"""
Print all combinations of m things from v[0] ... v[n-1],
duplicates OK. Yields a list.
"""
x = [0] * m
while True:
yield [ v for i in x ]
i = m - 1
while i>=0 and x==len(v)-1:
x = 0
i = i - 1
if i >= 0:
x = x + 1
else:
return

for y in m_from_n( "xyz", 2 ):
print ''.join(y)

xx
xy
xz
yx
yy
yz
zx
zy
zz

for y in m_from_n( [0,1], 3 ):
print y

[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]

for y in m_from_n( "abcdefghijklmnopqrstuvwxyz", 4 ):
print ''.join(y)

should more or less do what you want.

Charles

Charles Sanders, Apr 13, 2007
4. ### Paul RubinGuest

Charles Sanders <> writes:
> Forgive any silly mistakes I have made (I've been teaching
> myself python for about 1 week) but there is a moderately
> well known algorithm for this that extends to arbitrary
> lengths of both the list of alternatives and the length
> of the required output, and avoids deeply nested loops.

s = "abcd"

def a(n):
if n==0:
yield ''
return
for c in s:
for r in a(n-1):
yield c+r

print list(a(3))

Paul Rubin, Apr 13, 2007
5. ### Charles SandersGuest

Paul Rubin wrote:
[snip]
>
> def a(n):
> if n==0:
> yield ''
> return
> for c in s:
> for r in a(n-1):
> yield c+r
>
> print list(a(3))

Of course, obvious in retrospect, recursion instead of iteration.
I have yet to completely wean myself off Fortran style thinking.

Charles

Charles Sanders, Apr 13, 2007
6. ### Jia LuGuest

> for m in test:
> for n in test:
> for o in test:
> for p in test:
> print m+n+o+p

But if I consider about a combination of over 26 letter's list just
like:
"abcdefssdzxcvzxcvzcv"
"asllxcvxcbbedfgdfgdg"
......

Need I write 26 for loops to do this?

Thanx

Jia LU

Jia Lu, Apr 13, 2007
7. ### azraelGuest

I think that this would be very silly to do. bad kung foo. The
recoursion technique would be more satisfying. You sholud consider
that this would take about 4 lines to write. Also be avare of the
default recoursion depth in python wich is 1000. you can get and set
the recoursion limit hrough importing the "sys" module and using
getrecoursionlimit() and setrecoursionlimit().

On Apr 13, 9:16 am, "Jia Lu" <> wrote:
> > for m in test:
> > for n in test:
> > for o in test:
> > for p in test:
> > print m+n+o+p

>
> But if I consider about a combination of over 26 letter's list just
> like:
> "abcdefssdzxcvzxcvzcv"
> "asllxcvxcbbedfgdfgdg"
> .....
>
> Need I write 26 for loops to do this?
>
> Thanx
>
> Jia LU

azrael, Apr 13, 2007

On Apr 13, 8:16 am, "Jia Lu" <> wrote:
> > for m in test:
> > for n in test:
> > for o in test:
> > for p in test:
> > print m+n+o+p

>
> But if I consider about a combination of over 26 letter's list just
> like:
> "abcdefssdzxcvzxcvzcv"
> "asllxcvxcbbedfgdfgdg"
> .....
>
> Need I write 26 for loops to do this?
>
> Thanx
>
> Jia LU

Try this: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/502199

You could then write something like:

import string
for thiscomb in comb2( *([string.lowercase]*26) ):
...

Mind you, it generates a lot of combinations.

9. ### Michael HoffmanGuest

azrael wrote:
> I think that this would be very silly to do. bad kung foo. The
> recoursion technique would be more satisfying. You sholud consider
> that this would take about 4 lines to write. Also be avare of the
> default recoursion depth in python wich is 1000. you can get and set
> the recoursion limit hrough importing the "sys" module and using
> getrecoursionlimit() and setrecoursionlimit().

Well, you'd have to spell sys.getrecursionlimit() correctly, but yes

At least in the past, raising the recursion limit past a certain point
would result in the CPython interpreter crashing, so it's not completely
scalable.
--
Michael Hoffman

Michael Hoffman, Apr 13, 2007
10. ### azraelGuest

sorry for the bad grammar. I didn't investigate the StackLess Python,
but as I have been reading about it (so if it was correct), the
recursionlimit should not be the problem using StackLess Python.
>From my expirience with python and recursions, it works well to the

depth of about 200 to 500 (depending od algorithm and purpose). I
think that in this case it should work well with about 500. If you
need a bigger string, then lett it repeat and merge the different
strings.
You could also generate multidimensional hash.

Best Regards

On Apr 13, 2:24 pm, Michael Hoffman <> wrote:
> azrael wrote:
> > I think that this would be very silly to do. bad kung foo. The
> > recoursion technique would be more satisfying. You sholud consider
> > that this would take about 4 lines to write. Also be avare of the
> > default recoursion depth in python wich is 1000. you can get and set
> > the recoursion limit hrough importing the "sys" module and using
> > getrecoursionlimit() and setrecoursionlimit().

>
> Well, you'd have to spell sys.getrecursionlimit() correctly, but yes
>
> At least in the past, raising the recursion limit past a certain point
> would result in the CPython interpreter crashing, so it's not completely
> scalable.
> --
> Michael Hoffman

azrael, Apr 13, 2007
11. ### Steve HoldenGuest

Jia Lu wrote:
>> for m in test:
>> for n in test:
>> for o in test:
>> for p in test:
>> print m+n+o+p

>
> But if I consider about a combination of over 26 letter's list just
> like:
> "abcdefssdzxcvzxcvzcv"
> "asllxcvxcbbedfgdfgdg"
> .....
>
> Need I write 26 for loops to do this?
>
> Thanx
>
> Jia LU
>

Your new example uses 20-byte strings anyway, so to produce those using
the specified method you would need 20 nested for loops, not 26.

I'm pretty sure you could give a separate name to each atom ont he known
universe with a scheme like this. Do you really need 20-byte strings?

regards
Steve
--
Steve Holden +44 150 684 7255 +1 800 494 3119
Holden Web LLC/Ltd http://www.holdenweb.com
Skype: holdenweb http://del.icio.us/steve.holden
Recent Ramblings http://holdenweb.blogspot.com

Steve Holden, Apr 13, 2007
12. ### Paul McGuireGuest

On Apr 13, 8:53 am, Steve Holden <> wrote:
> Jia Lu wrote:
> >> for m in test:
> >> for n in test:
> >> for o in test:
> >> for p in test:
> >> print m+n+o+p

>
> > Thanx for your anwser.
> > But if I consider about a combination of over 26 letter's list just
> > like:
> > "abcdefssdzxcvzxcvzcv"
> > "asllxcvxcbbedfgdfgdg"
> > .....

>
> > Need I write 26 for loops to do this?

>
> > Thanx

>
> > Jia LU

>
> Your new example uses 20-byte strings anyway, so to produce those using
> the specified method you would need 20 nested for loops, not 26.
>
> I'm pretty sure you could give a separate name to each atom ont he known
> universe with a scheme like this. Do you really need 20-byte strings?
>
> regards
> Steve
> --
> Steve Holden +44 150 684 7255 +1 800 494 3119
> Holden Web LLC/Ltd http://www.holdenweb.com
> Skype: holdenweb http://del.icio.us/steve.holden
> Recent Ramblings http://holdenweb.blogspot.com- Hide quoted text -
>
> - Show quoted text -

If you just expand the length to five million* or so, one of those
strings will contain all the works of Shakespeare.

-- Paul
* ref: Project Gutenberg - http://www.gutenberg.org/etext/100 -
unzipped plaintext is ~5.3Mb

Paul McGuire, Apr 13, 2007
13. ### Jia LuGuest

> If you just expand the length to five million* or so, one of those
> strings will contain all the works of Shakespeare.

Oops, you have this formula in math?

Actually I want to scan a range of network for some certain files.

Jia Lu, Apr 13, 2007
14. ### Paul McGuireGuest

On Apr 13, 9:27 am, "Jia Lu" <> wrote:
> > If you just expand the length to five million* or so, one of those
> > strings will contain all the works of Shakespeare.

>
> Oops, you have this formula in math?
>
> Actually I want to scan a range of network for some certain files.

Sorry, Jia Lu, I don't. I was actually just joking, alluding to the
old saying that goes "if you had an infinite number of monkeys typing
randomly on an infinite number of typewriters, they will eventually
type out the works of Shakespeare." "Typewriters"! who uses
typewriters any more?!

-- Paul

Paul McGuire, Apr 13, 2007
15. ### Michael BentleyGuest

On Apr 13, 2007, at 9:19 AM, Paul McGuire wrote:

> If you just expand the length to five million* or so, one of those
> strings will contain all the works of Shakespeare.

Not likely, even with a tiny sampling of the works of Shakespeare:

#

import string
import random

def main(bardText, maxTries=5000000):
tries = 0
while tries < maxTries:
tries += 1
attempt = []
for letter in bardText.lower():
if random.choice(
string.lowercase[:26]
+ string.punctuation
+ ' '
) == letter:
attempt.append(letter)
else:
break
if len(attempt) >= 4:
print '%d: %s' % (
tries,
''.join(attempt)
)

if __name__ == "__main__":
main("Alas, poor Yorick!")

Michael Bentley, Apr 13, 2007
16. ### Paul McGuireGuest

On Apr 13, 10:22 am, Michael Bentley <>
wrote:
> On Apr 13, 2007, at 9:19 AM, Paul McGuire wrote:
>
> > If you just expand the length to five million* or so, one of those
> > strings will contain all the works of Shakespeare.

>
> Not likely, even with a tiny sampling of the works of Shakespeare:
>
> #
>
> import string
> import random
>
> def main(bardText, maxTries=5000000):
> tries = 0
> while tries < maxTries:
> tries += 1
> attempt = []
> for letter in bardText.lower():
> if random.choice(
> string.lowercase[:26]
> + string.punctuation
> + ' '
> ) == letter:
> attempt.append(letter)
> else:
> break
> if len(attempt) >= 4:
> print '%d: %s' % (
> tries,
> ''.join(attempt)
> )
>
> if __name__ == "__main__":
> main("Alas, poor Yorick!")

5000000 << infinity

Keep tryin'!

Also, the OP's technique was not doing random string permutations, but
generating an exhaustive list of all possible sequences from aaa... to
zzz... . So I think the works of Shakespeare are *bound* to be in
there somewhere.

For proof, here's an extract from my sample code from running this
exhaustive program with length=14:

....
ALASPOORYORICG
ALASPOORYORICH
ALASPOORYORICI
ALASPOORYORICJ
ALASPOORYORICK
ALASPOORYORICL
ALASPOORYORICM
ALASPOORYORICN
ALASPOORYORICO
....

-- Paul
(too late for April 1, unfortunately)

Paul McGuire, Apr 13, 2007
17. ### Carsten HaeseGuest

On Fri, 2007-04-13 at 10:22 -0500, Michael Bentley wrote:
> On Apr 13, 2007, at 9:19 AM, Paul McGuire wrote:
>
> > If you just expand the length to five million* or so, one of those
> > strings will contain all the works of Shakespeare.

>
> Not likely, even with a tiny sampling of the works of Shakespeare:

Actually, the OP seems to be interested in generating *all* strings of
length N. If you generate the set of *all* strings of 5 million
characters length, at least one of them will contain all works of
Shakespeare. That statement is utterly true and utterly impractical,
which is, of course, the point of Paul's joke.

-Carsten

Carsten Haese, Apr 13, 2007
18. ### Paul McGuireGuest

On Apr 13, 10:49 am, Carsten Haese <> wrote:
> On Fri, 2007-04-13 at 10:22 -0500, Michael Bentley wrote:
> > On Apr 13, 2007, at 9:19 AM, Paul McGuire wrote:

>
> > > If you just expand the length to five million* or so, one of those
> > > strings will contain all the works of Shakespeare.

>
> > Not likely, even with a tiny sampling of the works of Shakespeare:

>
> Actually, the OP seems to be interested in generating *all* strings of
> length N. If you generate the set of *all* strings of 5 million
> characters length, at least one of them will contain all works of
> Shakespeare. That statement is utterly true and utterly impractical,
> which is, of course, the point of Paul's joke.
>
> -Carsten

But even random typing will *eventually* get there (where "eventually"
= several gazillion times the age of the universe) - see
http://en.wikipedia.org/wiki/Infinite_monkey_theorem.

-- Paul
If I see farther, it is because I stand on the shoulders of an
infinite number of monkeys.

Paul McGuire, Apr 13, 2007
19. ### Paul McGuireGuest

On Apr 13, 8:53 am, Steve Holden <> wrote:
>
> I'm pretty sure you could give a separate name to each atom ont he known
> universe with a scheme like this. Do you really need 20-byte strings?
>

Steve,

Based on the Wikipedia article's estimate of 10**79 atoms in the
observable universe (is that all?), we would need a string of about 57
characters long to give each one a separate name.

(And I'll bet you've typed on an old Royal or two in your time...)

-- Paul

Paul McGuire, Apr 13, 2007
20. ### Paul McGuireGuest

On Apr 13, 10:41 am, "Paul McGuire" <> wrote:
> On Apr 13, 10:22 am, Michael Bentley <>
> wrote:
>
>
>
>
>
> > On Apr 13, 2007, at 9:19 AM, Paul McGuire wrote:

>
> > > If you just expand the length to five million* or so, one of those
> > > strings will contain all the works of Shakespeare.

>
> > Not likely, even with a tiny sampling of the works of Shakespeare:

>
> > #

>
> > import string
> > import random

>
> > def main(bardText, maxTries=5000000):
> > tries = 0
> > while tries < maxTries:
> > tries += 1
> > attempt = []
> > for letter in bardText.lower():
> > if random.choice(
> > string.lowercase[:26]
> > + string.punctuation
> > + ' '
> > ) == letter:
> > attempt.append(letter)
> > else:
> > break
> > if len(attempt) >= 4:
> > print '%d: %s' % (
> > tries,
> > ''.join(attempt)
> > )

>
> > if __name__ == "__main__":
> > main("Alas, poor Yorick!")

>
> 5000000 << infinity
>
> Keep tryin'!
>
> Also, the OP's technique was not doing random string permutations, but
> generating an exhaustive list of all possible sequences from aaa... to
> zzz... . So I think the works of Shakespeare are *bound* to be in
> there somewhere.
>
> For proof, here's an extract from my sample code from running this
> exhaustive program with length=14:
>
> ...
> ALASPOORYORICG
> ALASPOORYORICH
> ALASPOORYORICI
> ALASPOORYORICJ
> ALASPOORYORICK
> ALASPOORYORICL
> ALASPOORYORICM
> ALASPOORYORICN
> ALASPOORYORICO
> ...
>
> -- Paul
> (too late for April 1, unfortunately)- Hide quoted text -
>
> - Show quoted text -

And apologies to the OP for beating a dead horse into the ground.

-- Paul

Paul McGuire, Apr 13, 2007