Re: The Project Euler thread
Haha
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Re: The Project Euler thread
Site is up, limited form
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Re: The Project Euler thread
ETA
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Re: The Project Euler thread
who knows
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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. |
Re: The Project Euler thread
If that pattern works, shouldn't you use dynamic programming?
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Re: The Project Euler thread
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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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Re: The Project Euler thread
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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 |
Re: The Project Euler thread
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Re: The Project Euler thread
aaaaaaaaaaaaaaaaaaand live
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Re: The Project Euler thread
\o/
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Re: The Project Euler thread
thread revive
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Re: The Project Euler thread
This thread needs more action
New problem up in 2.5 hours |
Re: The Project Euler thread
I'm too stupid I don't have a sufficient math foundation to do stuff over 50 probably. |
Re: The Project Euler thread
Sort by difficulty instead
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Re: The Project Euler thread
I tried 206 earlier and failed.
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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. |
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. |
Re: The Project Euler thread
Try sorting by difficulty, too -- there are easy / good problems later on in the problem set, too.
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