Sony Ericsson has some of the best designs for phones, just have a look at this one:-
Thursday, September 20, 2007
Saturday, September 01, 2007
Friday, August 03, 2007
Project Euler #39
its definitely brute force and im sure there is a better way to get the answer, but here we go
Problem:-
f p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p 1000, is the number of solutions maximised?
My Solution:-
import math, time
t = time.time()
maximum = {}
for a in xrange(1,500):
for b in xrange(1,500):
c = math.sqrt(pow(a,2)+pow(b,2))
p = a+b+c
if p == int(p) and p < 1000:
if maximum.has_key(p):
maximum[p] += 1
else:
maximum[p] = 1
e = maximum.keys()
e.sort(cmp = lambda a,b: cmp(maximum[a],maximum[b]))
print e[-1]
print time.time()-t
Friday, July 13, 2007
Namespaces for PHP
Thursday, July 05, 2007
Project Euler
Here is the solution to one of the problems
How Many Lychrel Numbers are below 10,000 ? http://projecteuler.net/index.php?section=problems&id=55
my solution in python, which takes abt 39 seconds:-
def lychrel(x):
i = 0
while i < 50:
x += int(str(x)[::-1])
if str(x) == str(x)[::-1]:
return False
i += 1
return True
print len ([x for x in range(1,10000) if lychrel(x)])
Saturday, May 26, 2007
Microsoft Popfly
The site is in alpha, and there are definitely some issues that I have come across. But the UI is amazing, it works with firefox ( big plus ! ).
I will be messing around with this and see what mashups I can come up with. Please feel free to add me as a friend ( http://www.popfly.ms/users/Kashif ) and I will post invites for users as soon as I get them :)
Sunday, May 06, 2007
Vice Funds
Now for anyone who still reads this would you invest in it ? Abhi , I think I already know your answer :)
Saturday, February 24, 2007
Roman Numeral Converter
In this challenge, you will be given a number in roman numeral form and must print out its integer value.
( from codegolf)
My latest solution at 129 bytes:-
r,l,s={'M':1000,'D':500,'C':100,'L':50,'X':10,'V':5,'I':1},1000,0
for j in raw_input():s+=r[j]-(0,l*2)[r[j]>l];l=r[j]
print s
Any other ideas ?
Update:
After manzo's comment, I searched a little and came up with this. Right now its at 119 bytes
r,l,s=dict(M=1000,D=500,C=100,L=50,X=10,V=5,I=1),1000,0
for j in raw_input():s+=r[j]-(0,l*2)[r[j]>l];l=r[j]
print s
Saturday, January 06, 2007
Colors for Visual Studio
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
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, December 02, 2006
Thursday, October 05, 2006
Cant argue with this
Saturday, September 23, 2006
Learning 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]
Thursday, August 31, 2006
Friday, August 25, 2006
Friday, June 16, 2006
Struts so far has been awesome
Tuesday, May 30, 2006
Lets see here
Michael awakens and finding himself in the most awkward position ever ( you have to hear the story from him for the full effect )
Awesome Career Fair
We all passed OS ( this one was too close )
PGH 547 is being remodelled ( and not the way we thought it would be, this one not so much of a highlist )
Completing the SE project 30 minutes before the demo
At the moment this is all that I remember , and good luck to everyone who will be starting their internship tomorrow.
