06182014, 12:50 PM  #421 
Zageron E. Tazaterra
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Re: The Project Euler thread
Haha
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06202014, 04:29 PM  #422 
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Re: The Project Euler thread
Site is up, limited form

06262014, 02:56 AM  #423 
I am leonid
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Re: The Project Euler thread
ETA

06262014, 06:27 AM  #424 
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Re: The Project Euler thread
who knows

0882014, 06:42 AM  #425 
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Re: The Project Euler thread
So recently I've gotten more motivation to do smartsy stuff so I picked this up again. (unfortunately we can't have our accounts back yet, if ever)
In any case, I'm currently working on 65. What I don't understand is how someone would go about finding a solution more rigorously.
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0882014, 07:27 AM  #426 
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Re: The Project Euler thread
If that pattern works, shouldn't you use dynamic programming?
Last edited by AutotelicBrown; 0882014 at 07:28 AM.. 
0882014, 07:43 AM  #427 
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Re: The Project Euler thread
That pattern is so simple it only needs like a couple lines of code. But still, what concerns me at the moment is the reasoning for this pattern, an mathematical explanation for how it works, rather than the code/solution, which already works.
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0882014, 07:58 AM  #428 
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Re: The Project Euler thread
Yeah, I noticed that was the point a bit later. Anyway, I'm taking a look into it, I'll let you know if I get something.

0882014, 02:31 PM  #429  
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Re: The Project Euler thread
Quote:
to prove it, we define sequences of numbers using this relation and show that they have the properties we want using induction. so for a given continued fraction [a0; (a1, a2, a3...)] (using the notation in the question) we define p0 = a0 p1 = a0*a1 + 1 q0 = 1 q1 = a1 p_n = a_n*p_(n1) + p_(n2) and q_n = a_n*q_(n1) + q_(n2) we want to prove p_n / q_n = [a0; (a1, a2, ..., a_n)] we use induction. n=0 and n=1 are boring assume true for n=k then [a0; (a1, a2, ..., a_(k+1))] = [a0; (a1, a2, ..., a_k + 1/a_(k+1))] = ( a_k + 1/a_(k+1) )*p_(k1) + p_(k2)  ( a_k + 1/a_(k+1) )*q_(k1) + q_(k2) = a_k*p_(k1) + p_(k2) + p_(k1)/a_(k+1)  a_k*q_(k1) + q_(k2) + q_(k1)/a_(k+1) = p_k + p_(k1)/a_(k+1)  q_k + q_(k1)/a_(k+1) = a_(k+1)*p_k + p_(k1)  a_(k+1)*q_k + q_(k1) = p_(k+1)/q_(k+1) which completes the induction finally we need to prove that these fractions p_n/q_n are irreducible to do this we prove that: p_n*q_(n1)  q_n*p_(n1) = (1)^(n+1) and you do an induction sort of like the previous one. from this it follows instantly that p_n and q_n are coprime, so these recurrences give you the fractions in lowest terms. edit: does anybody know when the fuck i'll be able to log into project euler again because (1) i want to see which problems i've solved and (2) most importantly i want that dark website background back like i've always been used to using, which i can only access while logged in
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0882014, 03:28 PM  #430 
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Re: The Project Euler thread
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08162014, 07:34 AM  #431 
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Re: The Project Euler thread
aaaaaaaaaaaaaaaaaaand live

08192014, 04:08 AM  #432 
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Re: The Project Euler thread
\o/

02192015, 02:32 AM  #433 
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Re: The Project Euler thread
thread revive
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0372015, 07:36 PM  #434 
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Re: The Project Euler thread
This thread needs more action
New problem up in 2.5 hours Last edited by Reincarnate; 0372015 at 07:36 PM.. 
0372015, 08:54 PM  #435 
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Re: The Project Euler thread
I'm too stupid I don't have a sufficient math foundation to do stuff over 50 probably.
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0372015, 08:57 PM  #436 
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Re: The Project Euler thread
Sort by difficulty instead

0372015, 08:59 PM  #437  
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Re: The Project Euler thread
I tried 206 earlier and failed.
Quote:
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0372015, 09:05 PM  #438 
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Re: The Project Euler thread
Find x where x^2 = 1_2_3_4_5_6_7_8_9_0
You know the last blank must be a 0. So now if we divide by 100: x^2/100 = 1_2_3_4_5_6_7_8_9 (x/10)^2 = 1_2_3_4_5_6_7_8_9 Since the righthand number ends in 9, we know x/10 must end in either 3 or 7. That should cut things down some. Last edited by Reincarnate; 0372015 at 09:06 PM.. 
03122015, 10:16 AM  #439 
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Re: The Project Euler thread
Oh hey, I never noticed this thread. I primarily got into programming this/last year and project euler has been fun for me testing out my math skills and my programming skills.
I've solved 112, 14, 16, and 20. I'll probably work on 13 now.
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03132015, 10:25 PM  #440 
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Re: The Project Euler thread
Try sorting by difficulty, too  there are easy / good problems later on in the problem set, too.
Last edited by Reincarnate; 03132015 at 10:25 PM.. 
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