GNOME in a 22” screen

Posted by amrlima on Janeiro 24, 2010 | Comments (0) | Trackbacks (0)

Just a quik post to show how awsome GNOME looks in a 22” screen with 1920×1080 :D .

GNOME in a 22'' screen

A small gift to myself before worst days arrive.

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Categories: Pessoal, gnome

Another C mathematical program

Posted by amrlima on Janeiro 15, 2010 | Comments (2) | Trackbacks (1)

Disclaimer: For the non geeky programming lovers out there, sorry for spamming you with this bunch of incomprehensible lines of text!

The first time I wrote some kind of program was back in high school with a Ti80 calculator which had some sort of BASIC programming language. After reading the manual I spent hours making small mathematical programs to speedup my problem solving study. I must say the programs were not that useful to math because in the tests I had of course to show all the calculation I did… Then I forgot most things about programming… Until some years ago :) .

Anyways, what does this have to do with the code below? Almost nothing! The code bellow calculates the mode of n numbers provided by a user. It dynamically creates an n size array and checks with value is seen more often. I still haven’t done a python version this time, if someone wan’ts to do it in python or any other language, please do! :D

Here it goes.

#include <stdio.h>
#include <stdlib.h>
 
/*Create a n sized array*/
int* makearray(int n)
{
    int* v;
    int i;
 
    v = (int*) malloc(n*sizeof(int)); /*Alocates n int numbers*/
    printf("Array elements:\n");
    for (i = 0; i < n; i++){ /*Fill the array with user input*/
        scanf("%d", &v[i]);
    }
 
    return v;    
}
 
/*Get the size of an array*/
int arraylen(int v[])
{
    int i=0;
    while(v[i]) i++;
    return i;
}
 
/*Count how many times a given element is repeated in an array*/
int countelem(int v[], int num)
{
    int i, n=0;
 
    for (i=0; i < arraylen(v); i++){
        if (v[i] == (int) num){
            n++;
        }
    }
    return n;
}
 
int main()
{
    int n, i, count = 0, mode;
    int* v;
 
    printf("Size of the array?\n");
    /*We don't wan't anyone creating a oversized array: DANGER'*/
    /*So we limit it's' size. Here I randomly defined 100*/
    do{
        scanf("%d", &n);
 
    } while(n > 100);
 
    v = makearray(n);
    /*Update the wich value is repeated more ofted -> Mode*/
    for (i = 0; i < n; i++){
        if (countelem(v, v[i]) > count){
            count = countelem(v, v[i]); /*Update the value of the maximum number of repeats*/
            printf("%d\n", count);
            mode = v[i]; /*Update mode*/
        } 
 
 
    }
 
    printf("Mode: %d\n", mode);
    return 0;
 
}

EDIT: There was a little bug with my code, it’s fixed now :) .

One of my colegues made a python version :D . Hope he doesn’t mind if I publish it :) 

#!/bin/sh python
#-*- coding: utf-8 -*-
 
 
def makearray(v):
  """Creates the dictionary with the elements input by the user
  """
  print "Array elements (end with empty line):"
  while(1):
    s = raw_input("-> ")    # get an element
    if not s:
      break
    try:
      i = int(s)
      if v.has_key(i):      # if already exists, increment frequency
        v[i] += 1
      else:                 # if not, create it
        v[i] = 1
    except:                 # error handling
       print "not an integer"
 
 
#-----------------------------------------------------------------------
# program
 
# instead of a vector or array, we do this with a dictionary
# that way we can keep the frequency count with each entry
v = {}
makearray(v)
 
# get the mode
if len(v):
  mode = 0
  maxfreq = 0
 
  for i in v.keys():
    if v[i] > maxfreq:
      mode = i
      maxfreq = v[i]
 
  print "Mode %d, with frequency %d" % (mode, maxfreq)
 
else:
  print "Empty array"

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Categories: Programação

Prime factors code in C and Python

Posted by amrlima on Janeiro 14, 2010 | Comments (0) | Trackbacks (0)

Today someone spoke about prime factors and I thought it was something interesting to code. I needed to practice my C so I’ve coded it in C. Here it goes:

#include <stdio.h>
#define VMAX 25
 
int main()
{
    /*Prime numbers < 100 */
    int primes[VMAX] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
    int factors[VMAX] = {}; /*Prime factors*/
    int repeat[VMAX] = {}; /*Store times each factor repeats*/
    int a, i, n=0;
 
    printf("Type a number to find it's prime factors: \n");
    scanf("%d", &a);
 
    /*Start to divide the number by prime numbers while a is not 1*/
    for (i=0; i<VMAX; i++){
        while ((a != 1) && (a % primes[i] == 0)){ 
 
            a = a / primes[i];
            factors[i] = primes[i]; /*The prime factor is the one being used*/
            n++; /*Count how many times each prime is used*/
            } 
        if ((a % primes[i] == 0) ) n = 0; 
        repeat[i] = n;
        n = 0; /*Reinitialize counter*/
 
    }
    printf("Prime factors:\n");
    for (i=0; i<VMAX; i++){
 
        /*Print all factors in the array != 0*/
        if (factors[i] !=0){
            /*Since x ^ 0 = 1, we have to print x ^ 1 instead*/
            if (repeat[i] == 0) printf("%d ^ 1\n", factors[i]);
            else printf("%d ^ %d\n", factors[i], repeat[i]);
        }    
    }       
 
    return 0;
 
}

I really like python but unfortunately I haven’t had the time to code anything with python. Now it’s learning C , then C++ and so on. But this time I wanted to code this in python also. I’m a bit out of practice and my brain was thinking in C. But here it goes.

#! /usr/bin/env/python
VMAX = 25
 
#Prime numbers < 100
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
factors = [0]*VMAX #Prime factors
repeat = [0]*VMAX #Store times each factor repeats
i = 0
n = 0
k = 0
 
a = raw_input("Type a number to find it's prime factors:")
 
#Start to divide the number by prime numbers while a is not 1
for i in range(VMAX): 
 
    while ((a != 1) & (int (a) % primes[i] == 0)):
 
        a = int (a) / primes[i]
 
        print primes[i]
        factors[i] = primes[i] # The prime factor is the one being used
 
        n = n + 1 #*Count how many times each prime is used
        print n
    if (int (a) % primes[i] == 0):
        n = 0 #Reinitialize counter
    repeat[k] = n
    n = 0
    k = k + 1
 
print "Prime factors:"
 
#Print all factors in the array != 0
for i in range(VMAX):
 
    if (factors[i] !=0):
    #Since x ^ 0 = 1, we have to print x ^ 1 instead
        if (repeat[i] == 0):
            print factors[i]
        else: print "%s ^ %s " % (factors[i], repeat[i])

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Breathing more now, but not for long

Posted by amrlima on Janeiro 13, 2010 | Comments (3) | Trackbacks (0)

The last few days have been just a tiny bit more quiet and I can finally have *some rest at night. I like being busy but the last weeks have been ridiculous. In a few days work will catch up and take my time away again. Plus I have to study some more for the exams. The good part is that I have some work already done so I think I won’t have to work till very late.

Apart from work and boring stuff, I went with the flow and finally signed up for facebook. I was quite septic but I must admit it was a pleasant surprise.

I’ve discovered, thanks to a University college in facebook a really great site to discover music:  thesixtyone . Check it out.

I’m trying out gnome-shell with the new message tray, pretty cool :) Check it out as well! I just hope Xorg doesn’t lock on me with KMS…

Bye!

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