Algorithms and Data Structures

N.Wirth. Algorithms and Data Structures. Oberon version 87
f i+1 = f i + f i-1 , f 1 = 1, f 0 = 0
These are the recursive rules (or recurrence relations) defining the Fibonacci numbers:
f = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
Each Fibonacci number is the sum of its two predecessors. As a consequence, the numbers of initial runs
on the two input sequences must be two consecutive Fibonacci numbers in order to make Polyphase work
properly with three sequences.
How about the second example (Table 2.14) with six sequences? The formation rules are easily derived
as
a 4 (L+1) = a 0 (L)
a 3 (L+1) = a 0 (L) + a 4 (L)
a 2 (L+1) = a 0 (L) + a 3 (L)
a 1 (L+1) = a 0 (L) + a 2 (L)
a 0 (L+1) = a 0 (L) + a 1 (L)
= a 0 (L) + a 0 (L-1)
= a 0 (L) + a 0 (L-1) + a 0 (L-2)
= a 0 (L) + a 0 (L-1) + a 0 (L-2) + a 0 (L-3)
= a 0 (L) + a 0 (L-1) + a 0 (L-2) + a 0 (L-3) + a 0 (L-4)
Substituting f i for a 0 (i) yields
f i+1 = f i + f i-1 + f i-2 + f i-3 + f i-4 for i > 4
f 4 = 1
f i = 0 for i < 4
These numbers are the Fibonacci numbers of order 4. In general, the Fibonacci numbers of order p are
defined as follows:
f i+1 (p) = f i (p) + fi-1(p) + ... + f i-p (p) for i > p
f p (p) = 1
f i (p) = 0 for 0 ≤ i < p
Note that the ordinary Fibonacci numbers are those of order 1.
We have now seen that the initial numbers of runs for a perfect Polyphase Sort with N sequences are the
sums of any N-1, N-2, ... , 1 (see Table 2.15) consecutive Fibonacci numbers of order N-2.
L \ N: 3 4 5 6 7 8
1 2 3 4 5 6 7
2 3 5 7 9 11 13
3 5 9 13 17 21 25
4 8 17 25 33 41 49
5 13 31 49 65 81 97
6 21 57 94 129 161 193
7 34 105 181 253 321 385
8 55 193 349 497 636 769
9 89 355 673 977 1261 1531
10 144 653 1297 1921 2501 3049
11 233 1201 2500 3777 4961 6073
12 377 2209 4819 7425 9841 12097
13 610 4063 9289 14597 19521 24097
14 987 7473 17905 28697 38721 48001