code-golf

The puzzle A little puzzle I heard while I was in high school went something like this... The questioner would ask me to give him a number; On hearing the number, the questioner would do some sort of transformation on it repeatedly (for example, he might say ten is three ) until eventually arriving at the number 4 (at which point he would finish with four is magic ). Any number seems to be transformable into four eventually, no matter what. The goal was to try to figure out the transformation function and then be able to reliably proctor this puzzle yourself. The solution The transformation

The goal Today's Code Golf challenge is to create a regex parser in as few characters as possible. The syntax No, I'm not asking you to match Perl-style regular expressions. There's already a very reliable interpreter for those, after all! :-) Here's all you need to know about regex syntax for this challenge: A term is defined as a single literal character, or a regular expression within grouping parentheses () . The * (asterisk) character represents a Kleene star operation on the previous TERM. This means zero or more of the previous term, concatenated together. The + (plus) character

I just came across this little problem on UVA's Online Judge and thought, that it may be a good candidate for a little code-golf. The problem: You are to design a program to assist an architect in drawing the skyline of a city given the locations of the buildings in the city. To make the problem tractable, all buildings are rectangular in shape and they share a common bottom (the city they are built in is very flat). The city is also viewed as two-dimensional. A building is specified by an ordered triple (Li, Hi, Ri) where Li and Ri are left and right coordinates, respectively, of building i

This should be easy. Here's my array (rather, a method of generating representative test arrays): >>> ri = numpy.random.randint >>> ri2 = lambda x: ''.join(ri(0,9,x).astype('S')) >>> a = array([float(ri2(x)+ '.' + ri2(y)) for x,y in ri(1,10,(10,2))]) >>> a array([ 7.99914000e+01, 2.08000000e+01, 3.94000000e+02, 4.66100000e+03, 5.00000000e+00, 1.72575100e+03, 3.91500000e+02, 1.90610000e+04, 1.16247000e+04, 3.53920000e+02]) I want a list of strings where '\n'.join(list_o_strings) would print: 79.9914 20.8 394.0 4661.0 5.0 1725.751 391.5 19061.0 11624.7 353.92 I want to space pad to the left and

Challenge The shortest program by character count that accepts standard input of the form X-Y R , with the following guarantees: R is a non-negative decimal number less than or equal to 8 X and Y are non-negative angles given in decimal as multiples of 45° ( 0 , 45 , 90 , 135 , etc.) X is less than Y Y is not 360 if X is 0 And produces on standard output an ASCII "arc" from the starting angle X to the ending angle Y of radius R , where: The vertex of the arc is represented by o Angles of 0 and 180 are represented by - Angles of 45 and 225 are represented by / Angles of 90 and 270 are

The Challenge The shortest program by character count that outputs the n-bit Gray Code . n will be an arbitrary number smaller than 1000 100000 (due to user suggestions) that is taken from standard input. The gray code will be printed in standard output, like in the example. Note : I don't expect the program to print the gray code in a reasonable time ( n=100000 is overkill); I do expect it to start printing though. Example Input : 4 Expected Output : 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000 gnibbler Python - 53 chars n=1<<input() for x in range(n):print

Write a program that takes 3 integers separated by spaces and perform every single combination of addition, subtraction, multiplication and division operations possible and display the result with the operation combination used. Example: $./solution 1 2 3 Results in the following output 1+2+3 = 6 1-2-3 = -4 1*2*3 = 6 1/2/3 = 0 (integer answers only, round up at .5) 1*2-3 = -1 3*1+2 = 5 etc... Order of operation rules apply, assume there will be no parenthesis used i.e. (3-1)*2 = 4 is not a combination, although you could implement this for "extra credit" For results where a divide by 0 occurs