http://blog.ossme.net Its easier than getting stuck in Tue, 16 Jun 2009 08:17:35 +0000 http://backend.userland.com/rss092 en It has been a long, long tie since I posted. I have a fair few more Project Euler solutions to post, and I will post them soon. Work has got in the way of my Erlanging. I'm gutted. I'd ditch Java in a flash for paid Erlang work. Anyway. I start ... http://blog.ossme.net/2009/06/collective-inteligence-in-erlang/ It has been a while since I Euler'd but I have been mega depressed what with working in Sheffield and trying to retro-fit web security as an after thought to a web app that I was thrown at last minute...so at home, to cheer up, it is Euler time again. I've ... http://blog.ossme.net/2009/03/euler-22-name-scores-in-a-list/ The problem is: The following iterative sequence is defined for the set of positive integers: n n/2 (n is even) n 3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 40 20 10 5 16 8 ... http://blog.ossme.net/2008/12/problem-14-find-the-longest-sequence/ This is not a tough problem for lisp or erlang. Some bright spark in the forum simply states that he cut and paste the list and wrapped it with (+ ) and had the result. I think the size of the numbers presents problems for some languages and the optimization ... http://blog.ossme.net/2008/12/problem-13-work-out-the-first-ten-digits-of-the-sum-of-the-following-one-hundred-50-digit-numbers/ This is so similar to problem 16. The problem is: n! means n (n 1) ... 3 2 1 Find the sum of the digits in the number 100! Which is cool 'cos I have learned what n! means. My code is: prob20(N) -> L = lists:seq(N, 1, -1), NBang ... http://blog.ossme.net/2008/12/problem-20-sum-of-the-digits-in-100/ I realised that you can sort the Project Euler problems by difficulty. So I did. Which is why the order gets a bit odd from now onwards. I have no time to do Euler problems so I am cherry picking what I can fit into the spare 5 and 10 ... http://blog.ossme.net/2008/12/problem-16-sum-of-digits/ I haven't posted for a while. I'm contracting for a VoD producing company pathologically averse to normal quality measures which has taken all (ALL) my time. Anyway, the problem:The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes ... http://blog.ossme.net/2008/12/problem-10-sum-of-the-primes/ So I read a comment on reddit about how bad my maths was (and how badly I abuse Erlang too). I started reading Algorithms in a Nutshell as I figured I might have enough maths now to understand it. Well I don't really but I did understand the Big O ... http://blog.ossme.net/2008/11/problem-1-again/ Project Euler problem 9 is back in proper maths mode with this beauty: A Pythagorean triplet is a set of three natural numbers, a b c, for which, a2 + b2 = c2 For example, 32 + 42 = 9 + 16 = 25 = 52. There exists exactly one Pythagorean triplet ... http://blog.ossme.net/2008/11/problem-9-pythagorean-triplets/ Problem 8 in Project Euler is not so much a maths problem as a text manipulation one (isn't it?): 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 Again it just makes sense to me to brute force it. That is try every sequence of 5 digits numbers and ... http://blog.ossme.net/2008/11/euler-8-text-handling-is-not-a-strong-suit/