For more math oriented challenges (more project euler like), check out the monthly IBM 'I ponder this'. These are usually quite challenging, but the problems are interesting and solutions are posted at the end of the month. http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/pages/in...
"Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems."
Complete solution to today's problem in one line of python
def gen(univ, n=5): return univ if n == 1 else [x+y for y in gen(univ,n-1) for x in univ if all([(x+y)[start:start+upto] != (x+y)[start+upto:start+upto*2] for start in range(len(x+y)) for upto in range(1,(len(x+y)-start+2)/2)])]
sub solutions {
my ( $target_length, $alphabet, $head ) = @_;
$head //= '';
# Base Case our head is the right length
return $head if (length $head) == $target_length;
# General Case
return
map { generate( $target_length, $alphabet, $_ ) }
grep { ! /(.{2})\1/ }
map { $head . $_ }
@$alphabet;
}
Or if the regexp engine manages to do that automatically.
[UPDATE: it is slower to provide your own max length. the perl regexp engine is badass. repeating timings on my mac pro with a target_length of 12 show the plain {2,} version is 15% quicker].
(also, you've a minor typo as you recurse into "generate" instead of "solutions", I assume b/c you renamed the function at the very end to make the eg: code scan better).
Here is an ugly iterative solution, in one lambda:
from itertools import product
f = lambda n: [
s for s in
map(''.join, product("abc", repeat=n))
if not any(a==b
for a,b in
[(s[a:b], s[b:c])
for a,b,c in
[(i,i+w,i+2*w)
for w in range(1,len(s)/2+1)
for i in range(len(s)-2*w+1)]])]
>>> f(5)
['abaca', 'abacb', 'abcab', 'abcac', 'abcba', 'acaba',
'acabc', 'acbab', 'acbac', 'acbca', 'babca', 'babcb',
'bacab', 'bacba', 'bacbc', 'bcaba', 'bcabc', 'bcacb',
'bcbab', 'bcbac', 'cabac', 'cabca', 'cabcb', 'cacba',
'cacbc', 'cbabc', 'cbaca', 'cbacb', 'cbcab', 'cbcac']
19 comments
[ 4.7 ms ] story [ 47.7 ms ] thread"Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems."
Is there a site that gives you common/difficult programming "in the large" problems?
I've thought about making this into a series.
Some suggestions:
* The font needs to be larger
* The color scheme could use a little more contrast
* Pasting source code needs to be easier
def gen(univ, n=5): return univ if n == 1 else [x+y for y in gen(univ,n-1) for x in univ if all([(x+y)[start:start+upto] != (x+y)[start+upto:start+upto*2] for start in range(len(x+y)) for upto in range(1,(len(x+y)-start+2)/2)])]
... god that looks ugly. Call it with
>>> gen(["1", "2", "3"], 5) ['13121', '23121', '21321', '31321', '12321', '12131', '32131', '21231', '31231', '13231', '13212', '23212', '21312', '12312', '32312', '12132', '32132', '23132', '21232', '31232', '31213', '13213', '23213', '12313', '32313', '32123', '13123', '23123', '21323', '31323']
The challenge calls for sequences "such that no two immediately adjacent subsequences are equal".
This, to me, implies subsequences of any length > 1.
For a $target_length of 5, your solution is correct as 2 is the longest repeatable subsequence.
Easy fix, of course:
becomes I wonder if it would help performance to say: Or if the regexp engine manages to do that automatically.[UPDATE: it is slower to provide your own max length. the perl regexp engine is badass. repeating timings on my mac pro with a target_length of 12 show the plain {2,} version is 15% quicker].
(also, you've a minor typo as you recurse into "generate" instead of "solutions", I assume b/c you renamed the function at the very end to make the eg: code scan better).
http://www.spoj.com/problems/classical/
Screenshot: https://dl.dropbox.com/u/315/math/problems/wirth/wirth.png
Output: https://dl.dropbox.com/u/315/math/problems/wirth/wirth_outpu...
Notebook: https://dl.dropbox.com/u/315/math/problems/wirth/wirth.nb
Mathematica code turns out to be surprisingly readable.