import urllib, json

import sys

import tweepy

from tweepy import OAuthHandler

def twitter_fetch(screen_name = "BBCNews",maxnumtweets=10):

'Fetch tweets from @BBCNews'

# API described at https://dev.twitter.com/docs/api/1.1/get/statuses/user_timeline

consumer_token = '' #substitute values from twitter website

consumer_secret = ''

access_token = ''

access_secret = ''

auth = tweepy.OAuthHandler(consumer_token,consumer_secret)

auth.set_access_token(access_token,access_secret)

api = tweepy.API(auth)

#print api.me().name

#api.update_status('Hello -tweepy + oauth!')

for status in tweepy.Cursor(api.user_timeline,id=screen_name).items(2):

print status.text+'\n'

if __name__ == '__main__':

twitter_fetch('BBCNews',10)

YOu can obtain these tokens by registering your app with twitter(mandatory) - Keep them private.

References:

http://pythonhosted.org/tweepy/html/api.html

# Technical Diversity

## Wednesday, June 5, 2013

## Thursday, March 1, 2012

### Hailstone sequences - Easy one

Problem

=================================================

Start with x,

if x is even --> x becomes half

if x is off --> x becomes triple plus one

Sequence is believed to end with '1' all the time (hailstone sequence)

=================================================

Solution

============

I decided to put my newly learnt skills of Python to use for this easy one. :)

Here it goes:

=================================================

Start with x,

if x is even --> x becomes half

if x is off --> x becomes triple plus one

Sequence is believed to end with '1' all the time (hailstone sequence)

=================================================

Solution

============

I decided to put my newly learnt skills of Python to use for this easy one. :)

Here it goes:

*def hailstone_seq(x):*

print x,

while x > 1:

if x % 2 == 0:

x = x/2

else:

x = 3*x + 1

print x,print x,

while x > 1:

if x % 2 == 0:

x = x/2

else:

x = 3*x + 1

print x,

## Friday, September 30, 2011

### Hamming Numbers

This exercise is taken from www.programmingpraxis.com

=========================================================

=========================================================

=========================================================

=========================================================

*"The sequence of Hamming numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, … (A051037) consists of all numbers of the form 2i·3j·5k where i, j and k are non-negative integers. Edsger Dijkstra introduced this sequence to computer science in his book A Discipline of Programming, and it has been a staple of beginning programming courses ever since. Dijkstra wrote a program based on three axioms:*

Axiom 1: The value 1 is in the sequence.

Axiom 2: If x is in the sequence, so are 2 * x, 3 * x, and 5 * x.

Axiom 3: The sequence contains no other values other than those that belong to it on account of Axioms 1 and 2.

Your task is to write a program to compute the first thousand terms of the Hamming sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below."Axiom 1: The value 1 is in the sequence.

Axiom 2: If x is in the sequence, so are 2 * x, 3 * x, and 5 * x.

Axiom 3: The sequence contains no other values other than those that belong to it on account of Axioms 1 and 2.

Your task is to write a program to compute the first thousand terms of the Hamming sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below."

=========================================================

**Here is my C solution**=========================================================

#include#include int ham_num(int x) { int h = x; while (h && (h%2 == 0)) h = h/2; while (h && (h%3 == 0)) h = h/3; while (h && (h%5 == 0)) h = h/5; /* * h will be 1, when it is ham num, else it will be * something else - example h =10 */ return h; } int main(int argc, char *argv[]) { int x = 1, cnt = 1000; printf("Hamming Numbers : \n"); if (argc > 1) cnt = atoi(argv[1]); while (cnt) { if (ham_num(x) == 1) { printf("%d ",x); cnt--; } x++; } printf("\n"); }

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