Older blog entries for maragato (starting at number 33)

That’s how you make atheists

A recent Pew poll in the US shows that agnostics and atheists seem to know more about religion than the rest of the population. That’s not really a big surprise as previous similar polls shows similar results.

But even if it’s not a big surprise, it can still seem strange to many. To me, the best explanation I’ve been able to come to is beautifully summarized by American Atheists’ president Dave Silverman, for the New York Times:

I have heard many times that atheists know more about religion than religious people. Atheism is an effect of that knowledge, not a lack of knowledge. I gave a Bible to my daughter. That’s how you make atheists.

CNN has a test with 10 questions from the survey. I can’t tell if those are the easiest ones or randomly picked, but I found it extremely easy. Got 10 out of 10 right.

Syndicated 2010-09-28 15:44:26 from Roberto Teixeira's blog

Euler 15 in Python

This one isn’t even funny…

Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20×20 grid?

Your first thought would be to generate the routes, but for a 20×20 grid, those amount to BILLIONS and you’d try to do it recursively too! Forget it.

But if you have some CompSci-level math background, though, you’ll remember this one–reading The Art of Computer Programming, Vol. 4 also helps. It’s a matter of combinatorics and if we take w for the width and h for the height, all we need to do is calculate,

and that’s it. I didn’t even write a program for this, instead using Python’s interactive shell, but just for the same of completeness, here’s a “program” to calculate the possibles paths in a 20×20 grid.

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import math
w = h = 20
print math.factorial(w + h) / (math.factorial(w) * math.factorial(h))

That’s it.

Syndicated 2010-08-07 22:39:27 from Roberto Teixeira's blog

Euler 11 in Python

Project Euler’s problem #11 statement goes:

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is .

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?

This one is remarkably easy but also was quite fun. I think it’s because it reminds me of the kind of work we’d do during our Algorithms classes during my first year in college. And so this one goes to my Algorithms teacher, Ricardo Vargas Dornelles—best teacher I’ve ever had too.

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import operator
numbers = [[8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8],
[49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0],
[81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65],
[52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91],
[22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],
[24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50],
[32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],
[67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],
[24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],
[21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95],
[78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92],
[16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57],
[86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58],
[19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40],
[4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66],
[88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69],
[4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],
[20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16],
[20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54],
[1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48]]
 
maxp = 0
for row in range(0,16):
        for col in range(0,16):
                maxp = max(maxp, reduce(lambda x,y: x*y, \
                                        [n for n in numbers[row][col:col+4]]))
                maxp = max(maxp, numbers[row][col] * numbers[row + 1][col + 1] * \
                                 numbers[row+2][col+2] * numbers[row+3][col+3])
                maxp = max(maxp, reduce(lambda x,y: x*y, \
                                        [n[col] for n in numbers[row:row+4]]))
                if col > 3:
                        maxp = max(maxp, numbers[row][col] * \
                                         numbers[row + 1][col - 1] * \
                                         numbers[row+2][col-2] * \
                                         numbers[row+3][col-3])
 
print maxp

As well, this runs in 0.020s, so not bad at all.

Syndicated 2010-08-06 17:19:56 from Roberto Teixeira's blog

Euler 10 in C++

The other night I took on Project Euler’s problem #10. The problem statement went like this:

The sum of the primes below 10 is .

Find the sum of all the primes below two million.

I went with Python and you can find the result here. I just rewrote the exact same algorithm using C++ and Boost, just for fun.

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#include <iostream>
#include <cmath>
#include <boost/dynamic_bitset.hpp>
using namespace std;
 
unsigned long long prime_sum(int limit)
{
        if (limit < 2) return 0;
 
        long sieve_size = ceil(limit / 2);
        boost::dynamic_bitset<> sieve(sieve_size);
        unsigned long long sum = 2;
 
        sieve.flip();
 
        for (unsigned int i = 0; i < floor(sqrt(limit) / 2); i++) {
                if (!sieve[i])
                        continue;
                for (int j = (i * (i + 3) * 2) + 3; j < sieve_size; j += (i * 2) + 3)
                        sieve[j] = 0;
        }
 
        for (unsigned int i = 0; i < sieve_size; i++)
                if (sieve[i])
                        sum += (i * 2) + 3;
 
        return sum;
}
 
 
int main(void)
{
        cout << prime_sum(2000000) << endl;
        return 0;
}

Runs in 0.019s.

Syndicated 2010-08-06 15:46:32 from Roberto Teixeira's blog

Euler 10 in Python

I decided to take on Project Euler’s problem #10. Its statement goes like this:

The sum of the primes below 10 is .

Find the sum of all the primes below two million.

My first attempt used brute force and testing primality using only previously found primes:

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primes = []
 
def is_prime(n):
        if not (n < 2 or any(n % x == 0 
           for x in filter(lambda x: x < math.ceil(n ** 2), primes))):
                primes.append(n)
                return True
        return False
 
sum_primes = 0
n = 1   
while n < 2000000:
    if (is_prime(n)):
        sum_primes += n
    n += 1  
print sum_primes

It worked if you consider an execution time of over 7 minutes reasonable. Clearly a bad solution. So I decided to rethink it a bit and tried again. This time, I decided to use a little creative thinking. For every number I tested for primality, I took the time to mark its multiples as not primes (when I first test, say, 3, I already know that will not be a prime numbers, so I don’t have to test their primality later.)

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max_primes = 2000000
numbers = [0] * max_primes
 
def is_prime(n):
        if n <= 2:
                return True
        if numbers[n] == 1:
                return False
        c = 2
        m = 0
        while True:
                m = n * c
                if m < max_primes:
                        numbers[m] = 1
                else:
                        break
                c+= 1
 
        #if any(n % x == 0 for x in xrange(2, int(n ** 0.5) + 1)):
        i = 2
        while i < int(n** 0.5 + 1):
                if n % i == 0:
                        return False
                i += 1
 
        numbers[n] = 1
        return True
 
def sum_primes():
        soma = 0
        for i in range (2, max_primes):
                if is_prime(i):
                        soma += i
        return soma
 
print sum_primes()

This is a much better algorithm and it gave the correct result in 54 seconds. Project Euler has a “one-minute rule”, which state that all its problems should be solvable “on a modestly powered computer in less than one minute,” which means the solution applies. Still, it’s a brute force solution and I wanted a better one.

So today I picked my copy of Concrete Mathematics and read about primes. Honestly, I’ve been reading more about primes lately than I even thought I would.

Anyways, Knuth—yes, I know the book has three authors, but I can’t help thinking of it as a Knuth book—talks about several strategies to test primality, but it also mentions sieve algorithms that are used to generate lists of prime numbers. Exactly what I wanted!

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def prime_list(limit):
        if limit < 2:
                return []
        sieve_size = limit / 2
        sieve = [True] * sieve_size
 
        for i in range(0, int(limit ** 0.5) / 2):
                if not sieve[i]:
                        continue
                for j in range((i * (i + 3) * 2) + 3, sieve_size, (i * 2) + 3):
                        sieve[j] = False
 
        primes = [2]
        primes.extend([(i * 2) + 3 for i in range(0, sieve_size) if sieve[i]])
 
        return primes
 
print reduce(lambda x,y: x+y, prime_list(2000000))

Et voilà! Correct result in 0.456s! The code is based on the description of the Sieve of Eratosthenes found in Concrete Mathematics. Also interesting to note, my idea in the second algorithm above is actually the basis of this sieve algorithm, I was just thinking “backwards.”

Syndicated 2010-08-06 01:07:00 from Roberto Teixeira's blog

Euler 9 in C

So I recently wrote an ugly solution to Project Euler’s problem #9. It was written in Python and even though every other Project Euler solution I wrote ran in less than one second, this one took over 18 seconds to get to the answer. Obviously it’s my fault for using a purely brute force algorithm.

Anyway, boredom is a great motivator and I just rewrote the exact same algorithm in C, just to see how much faster it would be.

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#include <stdio.h>
 
int is_triplet(int a, int b, int c)
{
        return (((a < b) && (b < c)) && ((a * a + b * b) == (c * c)));
}
 
int main(void)
{
        for (int a = 0; a < 1000; a++)
                for (int b = a + 1; b < 1000; b++)
                        for (int c = b + 1; c < 1000; c++)
                                if ((a + b + c == 1000) && is_triplet(a, b, c)) {
                                        printf("%d %d %d = %d\n", a, b, c,
                                               a * b * c);
                                        return 0;
                                }
}

It’s the exact same algorithm, but instead of 18 seconds, it ran in 0.072s.

Update: And here’s the version suggested by Gustavo Niemeyer:

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#include <stdio.h>
 
int is_triplet(int a, int b, int c)
{
        return ((a * a + b * b) == (c * c));
}
 
int main(void)
{
        for (int a = 1; a < 1000; a++)
                for (int b = a + 1; b < (1000 - a) / 2; b++) {
                        int c = 1000 - a - b;
                        if (is_triplet(a, b, c)) {
                                printf("%d %d %d = %d\n", a, b, c,
                                       a * b * c);
                                return 0;
                        }
                }
}

It now runs in 0.002s

Syndicated 2010-08-04 00:43:42 from Roberto Teixeira's blog

Euler 9

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #9 statement is –

A Pythagorean triplet is a set of three natural numbers, , for which,

For example, .

There exists exactly one Pythagorean triplet for which .

Find the product .

Butt ugly, super slow, brute-force solution:

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from sys import exit
 
def is_triplet(a, b, c):
        if a < b and b < c:
                return a ** 2 + b ** 2 == c ** 2
 
for a in range(0, 1000):
        for b in range(a + 1, 1000):
                for c in range(b + 1, 1000):
                        if a + b + c == 1000 and is_triplet(a, b, c):
                                print a * b * c
                                exit(0)

I’ll have to rethink this, as it’s really inefficient. As it is, it runs in 18.034s!

Updated: take a look at another take at this problem and algorithm.

Updated 2010/08/04: Gustavo Niemeyer pointed an obvious optimization to the algorithm above (written in C)–the innermost loop is unnecessary. I rewrote it in Python to see the difference:

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from sys import exit
 
def is_triplet(a, b, c):
        return (a ** 2 + b ** 2 == c ** 2)
 
for a in range(1, 1000):
        for b in range(a + 1, (1000 - a) / 2):
                c = 1000 - a - b
                if is_triplet(a, b, c):
                        print "%d * %d * %d = %d" % (a, b, c, a * b * c)

From 18s to 0.076s. As well, following Eduardo Habkost’s suggestion, I used psyco and execution time went down to 0.023s.

Syndicated 2010-08-03 17:45:17 from Roberto Teixeira's blog

Euler 6

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #6 statement is –

The sum of the squares of the first ten natural numbers is,

The square of the sum of the first ten natural numbers is,

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is .

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Can’t be more straightforward than that, really.

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sqr_sum = 0
num_sum = 0
for i in range(1,100 + 1):
        num_sum += i
        sqr_sum += i**2
 
num_sum = num_sum**2
print sqr_sum - num_sum

I’m pretty sure Niemeyer could rewrite this in one line, but whatever. This runs in 0.015s.

Syndicated 2010-08-03 17:35:28 from Roberto Teixeira's blog

Euler 5

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #5 statement is –

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

My first attempt was using brute force, but it didn’t go very well. It took a considerable amount of time, so I decided to think about the problem a bit. In the end, it boils down to the a relation between the lesser common multiplier and the greater common divisor.

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def gcd(a, b):
        if b== 0: return a
        return gcd(b, a % b)
 
def lcm(a, b):
        return a * b / gcd(a, b)
 
if __name__ == "__main__":
        b = 2
        for i in range(3,20):
                b = lcm(b, i)
        print b

While the original code took several seconds before I killed it, this runs in 0.015s.

Syndicated 2010-08-03 17:30:36 from Roberto Teixeira's blog

Because acme hurt my pride[1] by making fun of the still inexisting KDE interface for Apt, I've started creating KApt as a kpart last night. Quite straightforward actually. It already can query packages and update package lists. I just forgot to put it on CVS, but I'll do it tonight.

[1] it's a joke, Arnaldo :-)