code-golf

Post your shortest code, by character count, to check if a player has won, and if so, which. Assume you have an integer array in a variable b (board), which holds the Tic Tac Toe board, and the moves of the players where: 0 = nothing set 1 = player 1 (X) 2 = player 2 (O) So, given the array b = [ 1, 2, 1, 0, 1, 2, 1, 0, 2 ] would represent the board X|O|X -+-+- |X|O -+-+- X| |O For that situation, your code should output 1 to indicate player 1 has won. If no-one has won you can output 0 or false . My own (Ruby) solution will be up soon. Edit : Sorry, forgot to mark it as community wiki. You

The Challenge: Write the shortest program that implements John H. Conway's Game of Life cellular automaton. [ link ] EDIT: After about a week of competition, I have selected a victor: pdehaan , for managing to beat the Matlab solution by one character with perl. For those who haven't heard of Game of Life, you take a grid (ideally infinite) of square cells. Cells can be alive (filled) or dead (empty). We determine which cells are alive in the next step of time by applying the following rules: Any live cell with fewer than two live neighbours dies, as if caused by under-population. Any live

The problem: >>> a = dict(a=1,b=2 ) >>> b = dict( b=3,c=2) >>> c = ??? c = {'a': 1, 'b': 5, 'c': 2} So, the idea is two add to dictionaries by int/float values in the shortest form. Here's one solution that I've devised, but I don't like it, cause it's long: c = dict([(i,a.get(i,0) + b.get(i,0)) for i in set(a.keys()+b.keys())]) I think there must be a shorter/concise solution (maybe something to do with reduce and operator module? itertools?)... Any ideas? Update: I'm really hoping to find something more elegant like "reduce(operator.add, key = itemgetter(0), a+b)". (Obviously that isn't real

To commemorate the public launch of Stack Overflow, what's the shortest code to cause a stack overflow? Any language welcome. ETA: Just to be clear on this question, seeing as I'm an occasional Scheme user: tail-call "recursion" is really iteration, and any solution which can be converted to an iterative solution relatively trivially by a decent compiler won't be counted. :-P ETA2: I've now selected a “best answer”; see this post for rationale. Thanks to everyone who contributed! :-) Patrick All these answers and no Befunge? I'd wager a fair amount it's shortest solution of them all: 1 Not

The challenge: Build an ASCII chart of the most commonly used words in a given text. The rules: Only accept a-z and A-Z (alphabetic characters) as part of a word. Ignore casing ( She == she for our purpose). Ignore the following words (quite arbitary, I know): the, and, of, to, a, i, it, in, or, is Clarification: considering don't : this would be taken as 2 different 'words' in the ranges a-z and A-Z : ( don and t ). Optionally (it's too late to be formally changing the specifications now) you may choose to drop all single-letter 'words' (this could potentially make for a shortening of the