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Math::NumSeq::Fibbinary -- without consecutive 1-bits
use Math::NumSeq::Fibbinary; my $seq = Math::NumSeq::Fibbinary->new; my ($i, $value) = $seq->next;
This sequence is the fibbinary numbers
0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, ... starting i=0
being integers which have no adjacent 1-bits when written in binary, taken in ascending order.
i fibbinary fibbinary (decimal) (binary) --- --------- -------- 0 0 0 1 1 1 2 2 10 3 4 100 4 5 101 5 8 1000 6 9 1001 7 10 1010 8 16 10000 9 17 10001
For example at i=4 fibbinary is 5. The next fibbinary is not 6 or 7 because they have adjacent 1-bits (110 and 111), the next without adjacent 1s is 8 (100).
The two highest bits must be "10...", they cannot be "11...". So there's effectively a block of 2^k values (not all used) followed by a gap of 2^k values, etc.
The least significant bit of each fibbinary is the Fibonacci word sequence, per Math::NumSeq::FibonacciWord.
All numbers without adjacent 1-bits can also be generated simply by taking the binary expansion and changing each "1" to "01", but that doesn't given them in ascending order the way the fibbinary here does.
The bits of the fibbinary values encode Fibonacci numbers used to represent i in Zeckendorf style Fibonacci base. In the Zeckendorf base system an integer i is a sum of Fibonacci numbers,
i = F(k1) + F(k2) + ... F(kn) k1 > k2 > ... > kn
Each k is chosen as the highest Fibonacci less than the remainder at that point. For example, reckoning the Fibonaccis as F(0)=1, F(1)=2, etc,
20 = 13+5+1 = F(5)+F(3)+F(0)
The k's are then assembled as 1-bits in binary to encode this sum in an integer,
fibbinary(20) = 2^5 + 2^3 + 2^0 = 41
The gaps between Fibonacci numbers means that after subtracting F(k) the next cannot be F(k-1), it must be F(k-2) or less. For that reason there's no adjacent 1-bits in the fibbinary numbers.
The connection between no adjacent 1s and the Fibonacci sequence can be seen by considering values with high bit 2^k. The further bits in it cannot have 2^(k-1) but only 2^(k-2), so effectively the number of new values are not from the previous k-1 but the second previous k-2, the same way as the Fibonacci sequence adds not the previous term (which would by double) but the one before instead.
See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence classes.
Move the current i so
$valueon the next call. If
$valueis not in the sequence then move so as to return the next higher value which is.
$value = $seq->ith($i)
$i'th fibbinary number.
$bool = $seq->pred($value)
Return true if
$valueis a fibbinary number, which means that in binary it doesn't have any consecutive 1-bits.
$i = $seq->value_to_i($value)
$i = $seq->value_to_i_ceil($value)
$i = $seq->value_to_i_floor($value)
Return the index i of
$valueis not in the sequence then
value_to_i_ceil()returns the i of the next higher value which is, or
value_to_i_floor()the i of the next lower value.
$i = $seq->value_to_i_estimate($value)
Return an estimate of the i corresponding to
For a given fibbinary number, the next fibbinary is +1 if the lowest bit is 2^2=4 or more. If however the low bit is 2^1=2 or 2^0=1 then the run of low alternating ...101 or ...1010 must be cleared and the bit above set. For example 1001010 becomes 1010000. All cases can be handled by some bit twiddling
# value=fibbinary filled = (value >> 1) | value mask = ((filled+1) ^ filled) >> 1 next value = (value | mask) + 1
value = 1001010 filled = 1101111 mask = 1111 next = 1010000
"filled" means trailing ...01010 has the zeros filled in to ...01111. Then those low ones can be extracted with +1 and XOR (the usual trick for getting low ones). +1 means the bit above the filled part is included so 11111, but a shift drops back to "mask" just 01111. OR-ing and incrementing then clears those low bits and sets the next higher bit to make ...10000.
This works for any fibbinary input, both odd "...10101" and even "...1010" endings and also zeros "...0000". In the zeros case the result is just a +1 for "...0001", and that includes input value=0 giving next=1.
The i'th fibbinary number can be calculated as per "Zeckendorf Base" above. Reckoning the Fibonacci numbers as F(0)=1, F(1)=2, F(2)=3, F(3)=5, etc,
find the biggest F(k) <= i subtract i -= F(k) fibbinary result += 2^k repeat until i=0
To find each F(k)<=i either just work downwards through the Fibonacci numbers, or the Fibonaccis grow as (phi^k)/sqrt(5) with phi=(sqrt(5)+1)/2 the golden ratio, so an F(k) could be found by a log base phi of i. Or taking log2 of i (the bit length of i) might give 2 or 3 candidates for k. Calculating log base phi is unlikely to be faster, but log 2 high bit might quickly go to a nearly-correct place in a table.
Testing for a fibbinary value can be done by a shift and AND,
is_fibbinary = ((value & (value >> 1)) == 0)
Any adjacent 1-bits overlap in the shift+AND and come through as non-zero.
& operator converts NV float to UV integer. If an NV value is an integer but bigger than a UV then bits will be lost to the
&. Conversion to
Math::BigInt or similar is necessary to preserve the full value.
Floats which are integers but bigger than an UV might be of interest, or it might be thought any float means rounded-off and therefore inaccurate and not of interest. The current code has some experimental automatic BigInt conversion which works for floats and for BigFloat or BigRat integers too, but don't rely on this quite yet. (A BigInt input directly is fine of course.)
In a fibbinary value each bit becomes a Fibonacci F[i] to add to make i, as per "Zeckendorf Base" above.
If a number is not a fibbinary then the next lower fibbinary can be had by finding the highest 11 pair and changing it and all the bits below to 101010...etc. For example 10011001 is not a fibbinary and must change down to 10010101, ie. the 11001 reduces to 10101, that being the biggest fibbinary no-adjacent-1s which is 10xxx.
bits 2^k from high to low if bit set if prev bit set too then treat remainder as 010101... else i += F[k]
If working downwards adding F[k] values then it's easy enough to notice an adjacent 11 pair. An alternative would be to find all 11 pairs by bit-twiddling per "Predicate" above and the highest 1-bit (if any) of those is the place to mangle.
In general i grows as a power of phi=1.618 and the values grow as a power of 2. So an estimate can be had from
value = 2^k i = F[k+1] ~= phi^(k+1) / (phi + 1/phi) ~= C * phi^k where C=phi/(phi + 1/phi) log(i/C)/log(phi) ~= log(value)/log(2) i_estimate = C * value^(log(phi)/log(2))
The power log(phi)/log(2)=0.694 reduces the value to give an i approximation. That power can also be approximated by shifting off the least significant 1-0.694=0.306 of the bits of the value. This is fast and may be enough accuracy for a bigint.
highbitpos of value i_estimate = value >> floor(highbitpos * 0.306)
Copyright 2011, 2012, 2013, 2014 Kevin Ryde
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