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 April 28th, 2014, 01:06 PM #1 Member   Joined: Jul 2010 Posts: 83 Thanks: 2 Unique Integer Mapping with Phi and the Golden Angle One of the most interesting properties of the golden ratio is it's supreme irrationality above all others. Of course, the golden angle is essentially nothing more than a restatement of phi so the same holds true for it as well. Now it may not seem from the outset that this property could possibly provide any means to, say, produce pseudorandom sequences of integers but in fact it can be done and the key turns out to be none other than the fibonacci sequence. Consider the following: Given some N select any greater consecutive sequence of fibonacci numbers {F1, F2} and calculate G1 = {N*F2 mod F1}. G1 is, in a sense, the fractional (non-integer) result of multiplying N with an approximation of the golden ratio; the conjecture here that G1 for any given N < F1 is unique. Similarly, given some N select any greater consecutive sequence of fibonacci numbers {F1, F2, F3} then calculate G2 = {N*F1 mod F3}. In this case, G2 rather represents the fractional (non-integer) result of multiplying N with an approximation of the golden *angle* (and likewise conjectured unique). To help visualize what's happening here, imagine a circle divided into F3 pieces. G2 then is the "position" on this circle obtained when N is multiplied by the golden angle and then truncated to fit "onto" the circle. The uniqueness of this mapping perhaps explains why nature has employed it all along in the placement of leaves and so forth; non-overlap is kept to the most optimal level. But is it really true that this mapping is unique? Empirical evidence certainly points to such a conclusion, although a proof as yet seems elusive; in any case, the risk of it being false appears negligible at worst. Only time will tell, of course. Cheers! April 28th, 2014, 01:25 PM   #2
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 Originally Posted by Sebastian Garth Given some N select any greater consecutive sequence of fibonacci numbers {F1, F2} and calculate G1 = {N*F2 mod F1}. G1 is, in a sense, the fractional (non-integer) result of multiplying N with an approximation of the golden ratio; the conjecture here that G1 for any given N < F1 is unique.
Let me see if I understand the conjecture. Pick any consecutive Fibonacci numbers F1 < F2. Then there are no 0 <= m < n < F1 with m*F2 = n*F2 (mod F1). Is this correct? April 28th, 2014, 01:56 PM   #3
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 Originally Posted by CRGreathouse Let me see if I understand the conjecture. Pick any consecutive Fibonacci numbers F1 < F2. Then there are no 0 <= m < n < F1 with m*F2 = n*F2 (mod F1). Is this correct?
Correct. April 28th, 2014, 08:17 PM #4 Member   Joined: Jul 2010 Posts: 83 Thanks: 2 Also note that any other irrational is bound to have more near "collisions" with unit fractions than phi (just consider it's continued fraction expansion). So output correlations between inputs M and N which share common divisors or what have you is kept to a minimum. April 29th, 2014, 06:49 AM   #5
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 Originally Posted by Sebastian Garth Correct.
So you're trying to prove that F1 does not divide (n-m)*F2, with 0 <= m < n < F1. The sequence of Fibonacci numbers is a gcd sequence, so gcd(F1, F2) is Fibonacci number at the gcd of their indices. Since they're consecutive the gcd of their indices is 1, so gcd(F1, F2) = F_1 = 1, and so F1 divides (n-m)*F2 if and only if F1 divides n-m. But this is between 1 and F1-1 and so F1 cannot divide it. QED. April 29th, 2014, 09:01 AM   #6
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 Originally Posted by CRGreathouse So you're trying to prove that F1 does not divide (n-m)*F2, with 0 <= m < n < F1. The sequence of Fibonacci numbers is a gcd sequence, so gcd(F1, F2) is Fibonacci number at the gcd of their indices. Since they're consecutive the gcd of their indices is 1, so gcd(F1, F2) = F_1 = 1, and so F1 divides (n-m)*F2 if and only if F1 divides n-m. But this is between 1 and F1-1 and so F1 cannot divide it. QED.
Ah, well done! Tags angle, golden, integer, mapping, phi, unique ### how many golden angles fit in circle

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