06-18-2014, 12:50 PM | #421 |
Zageron E. Tazaterra
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Re: The Project Euler thread
Haha
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06-20-2014, 04:29 PM | #422 |
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Re: The Project Euler thread
Site is up, limited form
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06-26-2014, 02:56 AM | #423 |
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Re: The Project Euler thread
ETA
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06-26-2014, 06:27 AM | #424 |
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Re: The Project Euler thread
who knows
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08-8-2014, 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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08-8-2014, 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; 08-8-2014 at 07:28 AM.. |
08-8-2014, 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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08-8-2014, 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.
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08-8-2014, 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_(n-1) + p_(n-2) and q_n = a_n*q_(n-1) + q_(n-2) 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_(k-1) + p_(k-2) ------------------------------------- ( a_k + 1/a_(k+1) )*q_(k-1) + q_(k-2) = a_k*p_(k-1) + p_(k-2) + p_(k-1)/a_(k+1) ---------------------------------------- a_k*q_(k-1) + q_(k-2) + q_(k-1)/a_(k+1) = p_k + p_(k-1)/a_(k+1) ---------------------- q_k + q_(k-1)/a_(k+1) = a_(k+1)*p_k + p_(k-1) ---------------------- a_(k+1)*q_k + q_(k-1) = 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_(n-1) - q_n*p_(n-1) = (-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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08-8-2014, 03:28 PM | #430 |
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Re: The Project Euler thread
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08-16-2014, 07:34 AM | #431 |
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Re: The Project Euler thread
aaaaaaaaaaaaaaaaaaand live
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08-19-2014, 04:08 AM | #432 |
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Re: The Project Euler thread
\o/
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02-19-2015, 02:32 AM | #433 |
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Re: The Project Euler thread
thread revive
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03-7-2015, 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; 03-7-2015 at 07:36 PM.. |
03-7-2015, 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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03-7-2015, 08:57 PM | #436 |
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Re: The Project Euler thread
Sort by difficulty instead
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03-7-2015, 08:59 PM | #437 | |
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Re: The Project Euler thread
I tried 206 earlier and failed.
Quote:
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03-7-2015, 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; 03-7-2015 at 09:06 PM.. |
03-12-2015, 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 1-12, 14, 16, and 20. I'll probably work on 13 now.
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03-13-2015, 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; 03-13-2015 at 10:25 PM.. |
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