Programming

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 #8 statement is —

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

This is another very straightforward problem. My solution —

number = '73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450'

max_prod = 0
for i in range(1, len(number) - 4):
        prod = reduce(lambda x,y: x*y, [int(s) for s in number[i:i+5]])
        if prod > max_prod:
                max_prod = prod

print max_prod

This ran in 0m0.023s.

Project Euler’s Problem #7 statement is —

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10001st prime number?

It’s a straightforward problem to solve using brute force, but I went with a mixed approach doing some math instead.

<span class='line'><span class="c">#!/usr/bin/python</span>
</span><span class='line'><span class="kn">import</span> <span class="nn">math</span>
</span><span class='line'>
</span><span class='line'><span class="k">def</span> <span class="nf">is_prime</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
</span><span class='line'>    <span class="k">return</span> <span class="ow">not</span> <span class="p">(</span><span class="n">n</span> <span class="o"><</span> <span class="mi">2</span> <span class="ow">or</span> <span class="nb">any</span><span class="p">(</span><span class="n">n</span> <span class="o">%</span> <span class="n">x</span> <span class="o">==</span> <span class="mi">0</span> <span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="nb">xrange</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="nb">int</span><span class="p">(</span><span class="n">n</span> <span class="o">**</span> <span class="mf">0.5</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)))</span>
</span><span class='line'>
</span><span class='line'><span class="n">currentMax</span> <span class="o">=</span><span class="mi">0</span>
</span><span class='line'><span class="n">primes</span> <span class="o">=</span> <span class="mi">1</span>
</span><span class='line'><span class="n">counter</span> <span class="o">=</span> <span class="mi">3</span>
</span><span class='line'>
</span><span class='line'><span class="k">while</span> <span class="p">(</span><span class="n">primes</span> <span class="o"><</span> <span class="mi">10001</span><span class="p">):</span>
</span><span class='line'>    <span class="k">if</span> <span class="p">(</span><span class="n">is_prime</span><span class="p">(</span><span class="n">counter</span><span class="p">)):</span>
</span><span class='line'>        <span class="n">currentMax</span> <span class="o">=</span> <span class="n">counter</span>
</span><span class='line'>        <span class="n">primes</span><span class="o">+=</span><span class="mi">1</span>
</span><span class='line'>        <span class="n">counter</span><span class="o">+=</span><span class="mi">2</span>
</span><span class='line'>
</span><span class='line'><span class="k">print</span> <span class="n">currentMax</span>
</span>

I initially used a dumb brute-force which ended up running fast enough. But reading about primes, I came across this property by which if a number cannot be divided by any number smaller than sqrt(n), it is a prime. That required significantly less tests to be performed inside is_prime, but then again, it had to perform a square root calculation for every number we need to test. In the end, the algorithm above performed slightly better (0.02s to be exact.)

Updated: Thanks to Eduardo Habkost for pointing out the fact that I had mistakenly pasted the wrong version of the script. I fixed the text to reflect that.

Euler 8

Euler 7