Showing posts with label programming. Show all posts
Showing posts with label programming. Show all posts

Saturday, January 06, 2007

Colors for Visual Studio

Having watched a lot of videos featuring textmate, I really liked the color scheme that was being used by a lot of the videos. After some searching I came across Scott Hanselmans blog where he suggests some colors that bring it quite close.

This is what my current setting looks like:-

Any ideas for the colors of the html tag names other than white ?


Friday, December 08, 2006

Codegolf

Ive been working on this codegolf challenge recently which I found quite fun to do (codegolf == accomplish a challenge in code using least amt of lines )

The challenge in question:-

The game of REVERSE requires you to arrange a list of numbers in numerical order from left to right. To move, you tell the computer how many numbers (counting from the left) to reverse. For example, if the current list is 2 3 4 5 1 6 7 8 9 and you reverse 4, the result will be 5 4 3 2 1 6 7 8 9. Now if you reverse 5, you win.

What we're asking you to do is, given a list of numbers in a random order, produce the moves required to arrange them so they end up in numerical order.

(from codegolf.com)



Now Ive been working on this for a couple of hours and the best I can come up with is this ( My language of choice, python):-


[UPDATE #1]: File Size - 181kb

1) Changed the while loop condition

2)A different way of printing, by adding a comma after the last object one can suppress the new line that is automatically added by print in python

3)Removing unnecessary indentation helped on the file size

4)Didnt need the strip function

import re,sys
n = map(int,re.split(" ",sys.stdin.readline()))
l = len(n)
while l:
m = n.index(max(n[0:l])) + 1
n[:m] = n[m-1::-1]
n[:l] = n[l-1::-1]
print m,l,
l -= 1


[ORIGINAL]


import re,sys
n = map(int,re.split(" ",str(sys.stdin.readline()).strip()))
l = len(n)
while n != sorted(n):
m = n.index(max(n[0:l])) + 1
n[:m] = n[m-1::-1]
n[:l] = n[l-1::-1]
print m,"\n",l
l -= 1


the algorithm works in the following way:-

given a list, get its length

find the index of the maximum value within it, do a reverse from 0 to the index, so now the max is in front (and the move being the index + 1)

then do a reverse again on the length of the list , bringing the max to the end of the list (now the move after that being the length of the list)

decrease the length, so you iterate over a smaller list and repeat the same steps on the smaller section of the list


Anyone have anymore ideas to make this shorter ( either the program itself or the algorithm ) ?

Saturday, September 23, 2006

Learning Python

Just started to pick up on Python,

I am sure everyone who took COSC 2320 with Dr. Anderson remembers the word count program. Well after going through some of the tutorials of python here is the same program in python

Not only is it smaller, but its also easier to understand. Hopefully more to follow


import re
import string

#dictionary to store words and their counts
word_count = {}

#read in text document line by line
for line in open("trial.txt").readlines():

#remove leading and trailing whitespace
line = string.strip(line)

#split the string into words
#based on whitespace, punctuation, digits
for word in re.split("["+string.whitespace+string.punctuation+string.digits+"]",line):

#make the word lower case
word = string.lower(word)

#check if it is actually a word
if re.match("^["+string.lowercase+"]+$",word):

#increment count if true
if word_count.has_key(word):
word_count[word]+=1

#else add entry
else:
word_count[word] = 1

for w in word_count:
print w, ":" ,word_count[w]



Edit: Some more playing around


import re
import string

word_count = {}

text = open("trial.txt").read();

#list of words delimited by whitespace, punctuation and digits
#iterate by words in returned list from split
#lower case all the words in the text
words = re.split("["+string.whitespace+string.punctuation+string.digits+"]",string.lower(text))

#go through the list
for i in range(0,len(words)-1):

#as long as the word in the list is a word and is not already a key
if re.match("^["+string.lowercase+"]+$",words[i]) and not word_count.has_key(words[i]):

#add to the dictionary and get the count from the list
word_count[words[i]] = words.count(words[i])

for w in word_count:
print w,":",word_count[w]