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Topic: Strange Recursion (Read 6978 times) 

ThudnBlunder
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Strange Recursion
« on: Oct 5^{th}, 2003, 7:14am » 
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Define F(n) thus: F(n) = n  F(F(n1)); F(0) = 0 What is F(100000000)?


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towr
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Re: Strange Recursion
« Reply #1 on: Oct 5^{th}, 2003, 10:17am » 
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I get ::61803399:: It seems f(n)/n converges to a rather interesting number..

« Last Edit: Oct 5^{th}, 2003, 10:29am by towr » 
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Barukh
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Re: Strange Recursion
« Reply #2 on: Oct 6^{th}, 2003, 2:00am » 
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I got the same answer as towr; and the same interesting number... Also, it has something to do with a very famous number sequence. I believe I can give a proof. Very nice problem, THUDandBLUNDER!


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towr
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Re: Strange Recursion
« Reply #3 on: Oct 6^{th}, 2003, 5:26am » 
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I have to be honest though, I looked up the sequence, and how to get the answer more easily. I had first tried mathematica, but it wanted to do it the hard way which took too long. And programming it more efficiently would take too much memory (even though it's then linear in both time and space, which isn't overall bad). hint: You can use the Zeckendorf representation of the number Of course you can get the answer even more easily by approximation, once you know what F(n)/n converges to.

« Last Edit: Oct 6^{th}, 2003, 5:26am by towr » 
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Barukh
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Re: Strange Recursion
« Reply #4 on: Oct 6^{th}, 2003, 11:22am » 
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on Oct 6^{th}, 2003, 5:26am, towr wrote: Exactly! That's how I did it. In fact: if Z is the Zeckendorf Representation (ZR) of n, then the ZR of F(n) is Z shifted one position to the right. The steps of the proof are as follows: 1) F(n+1)  F(n) = 0 or 1. 2) If F(n) = k, then F(n+k) = n. on Oct 6^{th}, 2003, 5:26am, towr wrote:And programming it more efficiently would take too much memory (even though it's then linear in both time and space, which isn't overall bad). 
 I think, at least time should be logarithmic in n.

« Last Edit: Oct 6^{th}, 2003, 11:22am by Barukh » 
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towr
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Re: Strange Recursion
« Reply #5 on: Oct 6^{th}, 2003, 12:59pm » 
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on Oct 6^{th}, 2003, 11:22am, Barukh wrote: I think, at least time should be logarithmic in n. 
 You have to create the whole table (up to some point), and look up the two values in each step. to create the table up to some n: int F[n] F[0] = 0; for(int i=1; i < n; i++) F[i] = i  F[F[i1]]; This is linear, the same number of operations in each step; which is much better than the recursive function: int f(int i) { if (!i) return 0; else return i  f(f(i1)); } this is exponential, I think. The evaluation tree grows fast with i. (But I haven't done a proper analysis, suffice it to say it's much worse) Of course I'm open to suggestions if you can get a better performance than the linear algorithm.. (Not including the approxitive method that gives it in constant time, which isn't trivial to prove correct)


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Barukh
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Re: Strange Recursion
« Reply #6 on: Oct 7^{th}, 2003, 3:32am » 
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on Oct 6^{th}, 2003, 12:59pm, towr wrote:Of course I'm open to suggestions if you can get a better performance than the linear algorithm.. 
 Here's the version that does it in logarithmic time (and space) for a given number n: Code:int F(n) { int f[2*log(n)]; int F_n, i; f[0] = 0; f[1] = 1; for (i = 1; f[i] + f[i1] < n; i++) f[i+1] = f[i1] + f[i]; for (F_n = 0; n; i) if (n >= f[i]) { F_n += f[i1]; n = f[i]; i; } return F_n; } 
 on Oct 6^{th}, 2003, 12:59pm, towr wrote: (Not including the approxitive method that gives it in constant time, which isn't trivial to prove correct) 
 Again, I believe I've got the proof.

« Last Edit: Oct 11^{th}, 2003, 5:31am by Barukh » 
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towr
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Re: Strange Recursion
« Reply #7 on: Oct 7^{th}, 2003, 5:34am » 
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on Oct 7^{th}, 2003, 3:32am, Barukh wrote: Here's the version that does it in logarithmic time (and space) for a given number n: 
 Nice. I hadn't thought about implementing it that way. Even though that's how I solved it originally Quote:Again, I believe I've got the proof. 
 I believe you, but I'm sure it's not trivial (well, not to me anyway)


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Barukh
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Re: Strange Recursion
« Reply #8 on: Oct 9^{th}, 2003, 10:16am » 
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Oops... I've found a flaw in the proof I believed I had! So, currently I don't have a proof. I also corrected the code .


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Barukh
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Re: Strange Recursion
« Reply #9 on: Oct 11^{th}, 2003, 6:18am » 
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At last! Here's the proof. Denote f_{k} the kth Fibonacci number. Then the following statement answers the question: If n = f_{k} + f_{m} + … + f_{s} is represented as a sum of distinct Fibonacci numbers, 2 [le] k < m < … < s, then F(n) = f_{k1} + f_{m1} + … + f_{s1}. The proof is comprised of a number of steps. 1) For every n > 0, F(n) – F(n1) = 0, or F(n) – F(n1) = 1. Use induction by n: a. F(1) – F(0) = 1. b. F(n+1) – F(n) = F(F(n1))  F(F(n)) + 1. Now, if F(n) = F(n1), then F(n+1) – F(n) = 1; if F(n) = k = F(n1) + 1, then F(n+1) – F(n) = F(k) – F(k+1) + 1 = 0 or 1. 2) For every n > 0, if F(n) = k, then F(n+k) = n. Again, use induction by n: a. F(1) = 1, and F(2) = 1. b. Let F(n+1) = k. According to 1), F(n) = k or k1. If F(n) = k, then F(n+k) = n, and F(n+k+1) = n+k+1 – F(n) = n+1. If F(n) = k1, then F(n+k1) = n, so that F(n+k) = n+k – F(n) = n, and F(n+k+1) = n+k+1 – F(n) = n+1. 3) For every k [ge] 2, F(f_{k}) = f_{k1}. This is proved using 2) and induction by k. 4) For every pair 2 [le] k < m, F(f_{k} + f_{m}) = f_{k1} + f_{m1}. Use induction on k: a. For every m > 2, F(f_{m} + f_{2}) = F(f_{m} + 1) = f_{m} + 1 – F(f_{m1}) = f_{m} + 1  f_{m2} = f_{m1} + 1 = f_{m1} + f_{1} (uses 3) two times). b. Assume it’s true upto k, then for every m > k, F(f_{k} + f_{m}) = f_{k1} + f_{m1}, and according to 2): F(f_{k} + f_{m} + f_{k1} + f_{m1}) = F(f_{k+1} + f_{m+1}) = f_{k} + f_{m}. 5. Generalize 4) to a sum of arbitrary number of f_{k}'s.

« Last Edit: Oct 11^{th}, 2003, 6:22am by Barukh » 
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