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http://programmingpraxis.com

by krat0sprakhar • 14 years ago

109 19 14 years ago

programmingpraxis.com

new_test2• 14 years ago

Another similar excellent resource: http://www.reddit.com/r/dailyprogrammer/

jbranchaud3• 14 years ago

Any idea where so many daily programming problems come from?

new_test2• 14 years ago

http://www.reddit.com/r/dailyprogrammer_ideas

gryphonic• 14 years ago

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...

jschulenklopper• 14 years ago

Obligatory link to a nice and long-running initiative: Project Euler: http://projecteuler.net/.

"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."

chii• 14 years ago

its interesting that all of these are programming "in the small" problems.

Is there a site that gives you common/difficult programming "in the large" problems?

sillysaurus2• 14 years ago

Here are several "real-world" projects in order of increasing difficulty: http://shawnpresser.blogspot.com/2010/10/how-to-become-game-...

I've thought about making this into a series.

VMG• 14 years ago

Looks interesting.

Some suggestions:

* The font needs to be larger

* The color scheme could use a little more contrast

* Pasting source code needs to be easier

Surio• 14 years ago

Agree with OP on all three points. Please look into this.

bduerst• 14 years ago

This is awesome.

eddington• 14 years ago

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)])]

... 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']

peteretep• 14 years ago

And in some rather nicer Perl:

    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;
    }

eg:

    my @solutions = solutions( 5, [1..3] );

JasonFruit• 14 years ago

That's quite pretty Perl. You've lessened my prejudice against the language by writing code that can be read.

finnh• 14 years ago

Is that accurate? This looks like it only checks for sequential runs of two characters apiece.

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:

  grep { ! /(.{2})\1/ }

becomes

  grep { ! /(.{2,})\1/ }

I wonder if it would help performance to say:

  grep { $max = length()/2; $max < 2 || ! /(.{2,$max})\1/ }

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).

notaddicted• 14 years ago

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']

jbranchaud3• 14 years ago

Great resource, I just added this to my list of programming puzzle resources -- http://joshbranchaud.wordpress.com/2011/07/09/my-somewhat-co...

jcl• 14 years ago

You might also want to check out SPOJ, an enormous collection of problems from programming contests, with an online solution evaluator:

http://www.spoj.com/problems/classical/

jbranchaud3• 14 years ago

Thanks for the pointer, that is jam packed!

sillysaurus2• 14 years ago

I used Mathematica to give the solution to Wirth's Problem (http://programmingpraxis.com/2012/12/07/wirth-problem-15-12/).

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.

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