An inverse Fibonacci algorithm?
Since OP has asked about matrix solution not involving any floating point computations, here it is. We can achieve O(logn) complexity this way, assuming numeric operations have O(1) complexity.
Let's take 2x2 matrix A having following structure
1 1
1 0
Now consider vector (8, 5), storing two consecutive fibonacci numbers. If you multiply it by this matrix, you'll get (8*1 + 5*1, 8*1 + 5*0) = (13, 8) - the next fibonacci number.
If we generalize, A^n * (1, 0) = (f(n), f(n - 1)).
The actual algorithm takes two steps.
- Calculate
A^2,A^4,A^8, etc. until we pass desired number. - Do a binary search by
n, using calculated powers ofA.
On a side note, any sequence of the form f(n) = k1*f(n-1) + k2*f(n-2) + k3*f(n-3) + .. + kt*f(n-t) can be presented like this.
Wikipedia gives the result as
n(F) = Floor[ Log(F Sqrt(5) + 1/2)/Log(Phi)]
where Phi is the golden ratio.
If you can easily interpret F(n) in binary,
You may be suspicious of the constants 1.7 and 1.1. These work because d*1.44042009041 + C never gets very close to an integer.
I can post a derivation tomorrow if there is interest.
Here is a table with n = 2 through 91, which shows the formula result before flooring:
n formula w/o floor F(n) F(n) in binary
2 2.540 1 1
3 3.981 2 10
4 4.581 3 11
5 5.421 5 101
6 6.862 8 1000
7 7.462 13 1101
8 8.302 21 10101
9 9.743 34 100010
10 10.343 55 110111
11 11.183 89 1011001
12 12.623 144 10010000
13 13.223 233 11101001
14 14.064 377 101111001
15 15.504 610 1001100010
16 16.104 987 1111011011
17 17.545 1597 11000111101
18 18.385 2584 101000011000
19 19.825 4181 1000001010101
20 20.425 6765 1101001101101
21 21.266 10946 10101011000010
22 22.706 17711 100010100101111
23 23.306 28657 110111111110001
24 24.147 46368 1011010100100000
25 25.587 75025 10010010100010001
26 26.187 121393 11101101000110001
27 27.028 196418 101111111101000010
28 28.468 317811 1001101100101110011
29 29.068 514229 1111101100010110101
30 30.508 832040 11001011001000101000
31 31.349 1346269 101001000101011011101
32 32.789 2178309 1000010011110100000101
33 33.389 3524578 1101011100011111100010
34 34.230 5702887 10101110000010011100111
35 35.670 9227465 100011001100110011001001
36 36.270 14930352 111000111101000110110000
37 37.111 24157817 1011100001001111001111001
38 38.551 39088169 10010101000111000000101001
39 39.151 63245986 11110001010000111010100010
40 40.591 102334155 110000110010111111011001011
41 41.432 165580141 1001110111101000110101101101
42 42.032 267914296 1111111110000000110000111000
43 43.472 433494437 11001110101101001100110100101
44 44.313 701408733 101001110011101010010111011101
45 45.753 1134903170 1000011101001010011111110000010
46 46.353 1836311903 1101101011100111110010101011111
47 47.193 2971215073 10110001000110010010010011100001
48 48.634 4807526976 100011110100011010000101001000000
49 49.234 7778742049 111001111101001100010111100100001
50 50.074 12586269025 1011101110001100110011100101100001
51 51.515 20365011074 10010111101110110010110100010000010
52 52.115 32951280099 11110101100000011001010000111100011
53 53.555 53316291173 110001101001111001100000101001100101
54 54.396 86267571272 1010000010101111100101010110001001000
55 55.836 139583862445 10000001111111110110001011011010101101
56 56.436 225851433717 11010010010101110010110110001011110101
57 57.276 365435296162 101010100010101101001000001100110100010
58 58.717 591286729879 1000100110101011011011110111110010010111
59 59.317 956722026041 1101111011000001000100111001011000111001
60 60.157 1548008755920 10110100001101100100000110001001011010000
61 61.598 2504730781961 100100011100101101100101101010100100001001
62 62.198 4052739537881 111010111110011010000110011011101111011001
63 63.038 6557470319842 1011111011011000111101100000110010011100010
64 64.478 10610209857723 10011010011001100001110010100010000010111011
65 65.078 17167680177565 11111001110100101001011110101000010110011101
66 66.519 27777890035288 110010100001110001011010001001010011001011000
67 67.359 44945570212853 1010001110000010110100101111110010101111110101
68 68.800 72723460248141 10000100010010001000000000000111101001001001101
69 69.400 117669030460994 11010110000010011110100110000101111111001000010
70 70.240 190392490709135 101011010010100100110100110001101101000010001111
71 71.681 308061521170129 1000110000010111000101001100010011100111011010001
72 72.281 498454011879264 1110001010101011101011110010100001001111101100000
73 73.121 806515533049393 10110111011000010110000111110110100110111000110001
74 74.561 1304969544928657 100101000101101110011100110001010110000110110010001
75 75.161 2111485077978050 111100000000110001001101110000001010111101111000010
76 76.602 3416454622906707 1100001000110011111101010100001100001000100101010011
77 77.442 5527939700884757 10011101000111010000111000010001101100000010100010101
78 78.042 8944394323791464 11111110001101110000100010110011001101000111001101000
79 79.483 14472334024676221 110011011010101000001011011000100111001001001101111101
80 80.323 23416728348467685 1010011001100010110001111101111000000110010000111100101
81 81.764 37889062373143906 10000110100110111110011011000111100111111011010101100010
82 82.364 61305790721611591 11011001110011010100101010110110101000101101011101000111
83 83.204 99194853094755497 101100000011010010011000101111110010000101000110010101001
84 84.644 160500643816367088 1000111010001101100111110000110100111001010110001111110000
85 85.244 259695496911122585 1110011010100111111010110110110011001001111111000010011001
86 86.085 420196140727489673 10111010100110101100010100111101000000011010101010010001001
87 87.525 679891637638612258 100101101111011101011101011110011011001101010100010100100010
88 88.125 1100087778366101931 111101000100010011000000000110000011010000101001100110101011
89 89.566 1779979416004714189 1100010110011110000011101100100011110011101111101111011001101
90 90.406 2880067194370816120 10011111111000000011011101101010100001101110100111100001111000
91 91.846 4660046610375530309 100000010101011110011111011001111000000001100100101011101000101
'
Measuring memory usage by counting unbounded words is sort of silly, but as long as that's the model, there's an O(log n) time, O(1) word solution similar to Nikita Rybak's that in essence computes n via its Zeckendorf representation, which is based on the Fibonacci numbers (YO DAWG).
Define the matrix
1 1
A = ,
1 0
which satisfies
F(m + 1) F(m)
A^m = .
F(m) F(m - 1)
Instead of the sequence A^(2^k), we're going to use the sequence A^F(k). The latter sequence has the property that we can move forward with a matrix multiply
A^F(k + 1) = A^F(k - 1) * A^F(k)
and backward with a matrix inverse and multiplication
A^F(k - 1) = A^F(k + 1) (A^F(k))^-1,
so we can build a bidirectional iterator with only eight six twelve words assuming we store everything as rationals (to avoid assuming the existence of a unit-cost divide). The rest is just adapting this O(1)-space algorithm for finding a Zeckendorf representation.
def zeck(n):
a, b = (0, 1)
while b < n:
a, b = (b, a + b)
yield a
n1 = a
while n1 < n:
a, b = (b - a, a)
if n1 + a <= n:
yield a
n1 += a
a, b = (b - a, a)
>>> list(zeck(0))
[0]
>>> list(zeck(2))
[1, 1]
>>> list(zeck(12))
[8, 3, 1]
>>> list(zeck(750))
[610, 89, 34, 13, 3, 1]