[ventana]

A fun quote from the article, discussing a basic Fibonacci recursive implementation:

> Each call branches into two more calls, so the total number of calls grows as O(2ⁿ).

Well, no, not really. If anyone bothers counting how many recursive calls are actually made, the result is far from powers of two:

   n | result | # of calls
   1 |      1 |          1
   2 |      1 |          3
   3 |      2 |          5
   4 |      3 |          9
   5 |      5 |         15
   6 |      8 |         25
   7 |     13 |         41
   8 |     21 |         67
   9 |     34 |        109
  10 |     55 |        177
  11 |     89 |        287
  12 |    144 |        465
  13 |    233 |        753
  14 |    377 |       1219
  15 |    610 |       1973
  16 |    987 |       3193
  17 |   1597 |       5167
  18 |   2584 |       8361
  19 |   4181 |      13529
  20 |   6765 |      21891

A curious person will then calculate the actual ratio:

   n | result | # of calls |              ratio
   1 |      1 |          1 |                  1
   2 |      1 |          3 |                  3
   3 |      2 |          5 | 1.6666666666666667
   4 |      3 |          9 |                1.8
   5 |      5 |         15 | 1.6666666666666667
   6 |      8 |         25 | 1.6666666666666667
   7 |     13 |         41 |               1.64
   8 |     21 |         67 | 1.6341463414634145
   9 |     34 |        109 |  1.626865671641791
  10 |     55 |        177 | 1.6238532110091743
  11 |     89 |        287 | 1.6214689265536724
  12 |    144 |        465 | 1.6202090592334495
  13 |    233 |        753 | 1.6193548387096774
  14 |    377 |       1219 | 1.6188579017264275
  15 |    610 |       1973 | 1.6185397867104183
  16 |    987 |       3193 | 1.6183476938672072
  17 |   1597 |       5167 | 1.6182273723770748
  18 |   2584 |       8361 | 1.6181536675053223
  19 |   4181 |      13529 | 1.6181078818323167
  20 |   6765 |      21891 | 1.6180796806859339

and will notice that it gets close to φ = (1 + √5) / 2 ≈ 1.618033989, which makes the number of recursive calls O(φⁿ), which is much more fun than O(2ⁿ).

[Chinjut]

Yes, because the number of calls in this setup is 2 * the next result - 1, and the Fibonacci sequence itself grows at this Θ(φⁿ) rate.

[veltas]

Really it's worse than exponential, because the size of the input is not n, it's the number of bits needed to store n i.e. log n. So as the number of bits k grow, it's growing phi^(2^k).

[recursive]

Big O analysis never implies size in bits. It's just often done. In this case, n is just the numerical value of the input, so I don't think this is correct.