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Friday, April 23, 2010

Programming Challenges: The Programming Contest Training Manual

Voy a ir subiendo los ejercicios que vaya resolviendo de "Programming Challenges", los enunciados los voy a poner en ingles como están en el libro para evitar cualquier tipo de problemas en la traducción.
Cualquier bug encontrado o mejora que se pueda hacer al código, se agradecen los comentarios!

Problema 1.6.1: The 3n + 1 Problem

   Consider the following algorithm to generate a sequence of numbers. Start with an
integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this
process with the new value of n, terminating when n = 1. For example, the following
sequence of numbers will be generated for n = 22:
                        22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for
every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000.
   For an input n, the cycle-length of n is the number of numbers generated up to and
including the 1. In the example above, the cycle length of 22 is 16. Given any two
numbers i and j, you are to determine the maximum cycle length over all numbers
between i and j, including both endpoints.

Input
The input will consist of a series of pairs of integers i and j, one pair of integers per
line. All integers will be less than 1,000,000 and greater than 0.

Output
For each pair of input integers i and j, output i, j in the same order in which they
appeared in the input and then the maximum cycle length for integers between and
including i and j. These three numbers should be separated by one space, with all three
numbers on one line and with one line of output for each line of input.

Sample Input
1 10
100 200
201 210
900 1000

Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174

Solución:
def main():
    f = open('input161', 'r')
    input = f.readlines()
    for line in input:
        n, n2 = line.replace('\n', '').split(' ')
        n, n2 = int(n), int(n2)
        maxCycle = 0
        for i in range((n2-n) + 1):
            l = []
            num = n + i
            l.append(num)
            while num != 1:
                num = num/2 if num % 2 == 0 else num*3+1
                l.append(num)
            if len(l) > maxCycle:
                maxCycle = len(l)
        print n, n2, maxCycle

if __name__ == '__main__':
    main()

4 comments:

Anonymous said...

Jejeje Ojala tuvieras el 2.8.2 Poker Hands... en java Saludos Men

Anonymous said...

me parece excelente que estés solucionando los problemas en ese lenguaje jajaja yo uso java, pero te confieso que al ver tus soluciones hasta me han dado ganas de aprender pyphon, en todo caso para tu solución del 3n+1 sería mejor que guardaras en una tabla el numero de cálculos que ya hayas hecho, eso reduciría mucho el tiempo de ejecución.

Diego Sarmentero said...

gracias por la sugerencia!

Dasus said...

Oye yo tambien tenia pensado en guardar en una arreglo el numero de calculos, pero no se que tamaño darle a ese arreglo ?