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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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OK, if anyone is still bothering with PE then add me as a friend...
My friend key is 77645475138828_559c6b95618a95bce6341cffa7532d82
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| Tue Oct 25, 2011 12:47 pm |
 
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rustybucket
I haven't seen my friends in so long
Joined: Thu Jun 18, 2009 5:10 pm
Posts: 5836
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Added
I'm 89260527259205_c44a4f6a46d0cb1b56336558f861b57a
_________________Jim
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| Tue Oct 25, 2011 5:09 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Problem 29 - I gave up and brute forced it in the end.
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| Wed Oct 26, 2011 12:16 pm |
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finlay666
Spends far too much time on here
Joined: Thu Apr 23, 2009 9:40 pm
Posts: 4876
Location: Newcastle
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Added too
Mine is
72741547138783_cbda3d0aecb9219d73172fd9ddb124c5
_________________TwitterCharlie Brooker: Macs are glorified Fisher-Price activity centres for adults; computers for scaredy cats too nervous to learn how proper computers work; computers for people who earnestly believe in feng shui.
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| Wed Oct 26, 2011 11:40 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Brute forced 37 in under 2 seconds.
It was an optimised brute force though. Put trivial checks before any other checks (like checking for primes etc...)
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| Thu Oct 27, 2011 11:21 am |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Tried to use my code for 39 to solve 75... It ran for 15 hours... and gave me the wrong answer. 
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| Fri Oct 28, 2011 8:01 am |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Why is it that you can look at a problem for ages and not know where to start then leave it for months and come back and do it first time in 5 mins. Just did problem 35  Runs VERY quickly (not sure how long exactly though, I'd guess less than 0.1 seconds).
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| Mon Dec 19, 2011 11:30 am |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Boom!
107 and then following it very quickly using a very similar algorithm 31.
I was very pleased with 107 as it was all my own work and I didn't research the algorithm's before hand. Both run in around 15 ms.
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| Tue Dec 20, 2011 3:30 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Done 87.
Runs in 0.18 seconds.
3 more to go until level 2!
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| Wed Dec 21, 2011 11:54 am |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Problem 54 done (a bitch to wrote the program for). It's the poker one. How many hands does player 1 win out of 1000 games in a text file. Started writing it and then had to scrap it and start again. Runs in 20ms
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| Wed Dec 21, 2011 3:44 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Problem 149 done! Find the maximum sum of adjacent numbers in a straight line (up, down, left, right and diagonally) from an array of 2000x2000 random numbers. Very pleased as my algorithm solves it in 1.7 seconds using a very efficient algorithm too 1 more to go until lvl 2!
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| Thu Dec 22, 2011 12:50 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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I had a go at 287...
I had a program that worked but I calculated it would take around 500 days to get the answer for N=24.
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| Fri Dec 23, 2011 11:30 am |
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daycoder
Has a life
Joined: Tue Jan 17, 2012 11:41 am
Posts: 2
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 I've just done the first one in Python.
First attempt:
def problem1Version1():
....total = 0
....for n in range(1000):
........if n%3==0:
............total+=n
........elif n%5==0:
............total+=n
....return total
The used List Comprehension using a trick I found ages ago:
def problem1Version2():
....return sum([(n%3==0 or n%5==0) and n or 0 for n in range(1000)])
There's something similar on the Euler forums, but it was doing it in a less tricky way using something I remember reading about, but had forgotten about until reading it on the Euler forums. So I took that and made a new version that take lists, so it works with lots of numbers not just two.
def problem1Version5(target=1000,multiples=[3,5]):
....return sum([n for n in range(target) if len([n for multiple in multiples if n%multiple==0])>0])
But had also earlier made a long-winded version to check whether the numbers were coming out correctly.
def problem1Version4(target=1000,multiples=[3,5]):
....total = 0
....for n in range(target):
........found = False
........for multiple in multiples:
.............found = found or n%multiple==0
........total+=found and n or 0
....return total
--------
Got to start somewhere !
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| Tue Jan 17, 2012 12:08 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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Nice one! Exactly. The whole point of PE is to work through the problems and learn new techniques and tools as you go along. Both from researching different algorithms and from reading how other people have solved the same puzzles you have. When I started out there was no way that I would have been able to solve some of the puzzles I have now solved. It seemed like it almost was impossible. Now I'm currently trying to wrestle with some really tough puzzles (287) but I can kind of see a good way to get into it and make it work.
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| Tue Jan 17, 2012 3:48 pm |
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Fogmeister
I haven't seen my friends in so long
Joined: Thu Apr 23, 2009 7:35 pm
Posts: 6580
Location: Getting there
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 | | | | Fogmeister wrote: Right, I've been working on a Sudoku algorithm for problem 96. I first wrote an algorithm to just fill in the obvious ones (i.e. if 8 different numbers appear in the same column, row or mini grid then the remaining number goes in the current slot). That didn't do much but I kept it. I then implemented a thing that looked at the numbers that could go into the current space. If any of the numbers can't go into another square in the mini grid then place it there. This managed to solve about 10 or 11 out of the 50. I've now implemented a system that checks an assumption of a number. i.e. this slot can take either 4 or 7. Check 4, does that work or does it break the grid? If it works then place 4 and repeat the first set of checks. I've now got a program that solves 43 out of the 50 sudoku grids. The remaining 7 all get broken by the assumption system because they get set so that they become unsolvable  It's good fun though | | | | |
LOL! Just been reading through the thread. Well, this post opened up a can of worms! Due to problem 96 I am now almost finished developing this... http://www.sudokuwiki.org/Sudoku_Solver ... le_Devices It is getting very nice now but also getting to the bit that just requires little tid bits finishing off (i.e. annoying bits rather than proper coding). Can't wait to finish it now though!
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| Tue Jan 17, 2012 4:29 pm |
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