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Math::NumSeq::FibonacciRepresentations -- count of representations by sum of Fibonacci numbers
use Math::NumSeq::FibonacciRepresentations; my $seq = Math::NumSeq::FibonacciRepresentations->new; my ($i, $value) = $seq->next;
This is the Fibonacci representations function R(i) which is the number of ways i can be represented as a sum of distinct Fibonacci numbers,
1, 1, 1, 2, 1, 2, 2, 1, 3, 2, 2, 3, 1, 3, 3, 2, 4, 2, 3, 3, ... starting i=0
For example R(11)=3 because 11 is a sum of Fibonacci numbers in the following three different ways
11 = 8 + 3 R(11)=3 sums = 8 + 2 + 1 = 5 + 3 + 2 + 1
The pattern in the values can be seen by arranging them in rows of an array.
1 i=0 to 0 1 1 i=0 to 1 1 1 i=1 to 2 1 2 1 i=2 to 4 1 2 2 1 i=4 to 7 1 3 2 2 3 1 i=7 to 12 1 3 3 2 4 2 3 3 1 i=12 to 20 1 4 3 3 5 2 4 4 2 5 3 3 4 1 i=20 to 33 1 4 4 3 6 3 5 5 2 6 4 4 6 2 5 5 3 6 3 4 4 1 i=33 to 54 F[y]-1 to F[y+1]-1
There are Fibonacci F[y-1] many values in each row, running from i=F[y]-1 to i=F[y+1]-1. Notice the ranges overlap so each "1" at the right hand end is a repeat of the "1" at the left end. There's just a single 1 in the sequence for each block.
New values in row y are the sum of adjacent values in row y-2, or equivalently a pattern of duplications and sums from row y-1. For example in the third row the "2" has duplicated to be 2,2, then in the fourth row the adjacent 1,2 values sum to give new "3". The row y-2 is a kind of Fibonacci style delay to when values are summed, resulting in F many values in each row.
See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence classes.
$seq = Math::NumSeq::FibonacciRepresentations->new ()
Create and return a new sequence object.
$value = $seq->ith($i)
$i'th value of the sequence.
($v0, $v1) = $seq->ith_pair($i)
Return two values
ith($i+1)from the sequence. As described below ("Ith Pair") two values can be calculated with the same work as one.
$bool = $seq->pred($value)
Return true if
$valueoccurs in the sequence, which means simply integer
ith_pair() the two sequence values at an arbitrary i,i+1 can be calculated from the Zeckendorf form of i+1 (see "Zeckendorf Base" in Math::NumSeq::Fibbinary) as per
Jean Berstel, "An Exercise on Fibonacci Representations", Theoretical Informatics and Applications, volume 35, 2001, pages 491-498. http://archive.numdam.org/numdam-bin/fitem?id=ITA_2001__35_6_491_0
Berstel uses a state machine which results in matrices based on runs of consecutive 0-bits in the Zeckendorf form. If making the Zeckendorf breakdown by some sort of logarithm or high binary bit then that would be Fibonacci number indices and their differences are the run lengths.
If making the Zeckendorf breakdown by individual Fibonacci number comparisons to give the fibbinary form then it's convenient to take each bit individually rather than in runs. In the following algorithm, runs of 0-bits do "r += rprev" every second time and hence become "r += rprev*floor(run/2)" which is per the M matrices of Berstel.
rprev = 1 = R(0) will become R[i] r = 1 = R(1) will become R[i+1] zeros = 0 how many consecutive bit=0 seen for each bit of fibbinary(i+1) from high to low if bit=0 then if zeros odd then r += rprev zeros ^= 1 if bit=1 then if zeros odd then rprev += r else rprev = r zeros = 0 result R[i]=rprev, R[i+1]=r
The loop maintains r=R[bits] where "bits" is those bits of fibbinary(i+1) which have been processed so far. rprev is the immediately preceding sequence value.
The loop action is to append either a 0-bit or 1-bit Zeckendorf term to the bits processed so far and step r,rprev accordingly.
"zeros" is the count of 0-bits seen since the last 1-bit. This is kept only modulo 2 since the test is just for odd or even run of zeros, not the full count.
For a 1-bit the zeros count becomes 0 since there are now no 0-bits since the last 1-bit seen. In the Zeckendorf form a 1-bit always has a 0-bit immediately below it. That bit can be worked into the bit=1 case,
if bit=1 then if zeros odd then rprev += r else rprev = r next lower bit of fibbinary(i+1) is 0-bit, skip it zeros = 1
This is as if the bit=0 code is done immediately after the bit=1. zeros=0 is even after the bit=1 so there's no change to r,rprev by the bit=0 code, simply skip the 0-bit and record zeros=1.
When calculating Fibonacci numbers for fibbinary(i+1) it's desirable to use integers <=i only so as not to overflow a finite number type. This can be done by finding the biggest Fibonacci f1<=i and subtracting it before doing the +1, giving i+1-f1 without overflow. If the biggest Fibonacci <=i+1 is in fact f2=f1+f0<=i+1 then will have i+1-f1 >= f0 and should subtract that too for i+1-(f1+f0). The loop begins at the f1 bit in this case, or at the f0 bit if not.
When the high 1-bit is handled like this to avoid overflow the second highest bit is always a 0-bit the same as above. So the loop can begin one lower, so f0 if f2 was subtracted or the Fibonacci below f0 if not. Initial zero=1 to record the 0-bit skipped.
f1<=i and f1+f0<=i+1 only occurs when i+1=f1+f0 exactly, so it has all 0-bits. This can be treated explicitly by floor(count/2) which is the 0-bit cases in the loop. i+1=Fibonacci won't occur very often, but returning count/2 is about the amount code as an i-=f0 and setup to loop from f1.
The effect of the algorithm in each case is to descend through the array above ("Array") by taking or not taking mediant r+rprev or duplication rprev=r. This can be compared to the Stern diatomic sequence calculation which goes by taking or not taking the mediant, no duplicating rprev=r case.
The Fibonacci representations sequence contains the Stern diatomic sequence (eg. Math::NumSeq::SternDiatomic) as a subset, per
Marjorie Bicknell-Johnson, "Stern's Diatomic Array Applied to Fibonacci Representations", Fibonacci Quarterly, volume 41, number 2, May 2003, pages 169-180.
Taking the R(i) at indices i for which i in Zeckendorf form uses only even Fibonaccis gives the Stern diatomic sequence.
These indices have fibbinary value (Math::NumSeq::Fibbinary) with 1-bits only at even bit positions (counting the least significant bit position as 0 and going up from there). Even positions are either 0 or 1. Odd positions are always 0. The highest bit is always a 1-bit and it must be at an even position.
fibbinary(i) = 10a0b0c...0z ^ ^ ^ ^ even bits a,b,c,etc 0 or 1, odd bits always 0
In the "Ith Pair" calculation above this kind of i always has an odd number of 0-bits between each 1-bit. So the 1-bit step is always rprev+=r, and the 0-bit step at the even positions is r+=prev. Those two steps are the same as the Stern diatomic calculation per "Ith Pair" in Math::NumSeq::SternDiatomic.
Copyright 2014 Kevin Ryde
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