I'm not quite sure I understand your encoding scheme.

Specifically, you're encoding into a sequence of characters, but when
decoding, you're not using (8#2)#:a.i.y and it looks like you've
cooked up some mechanism to compensate for how |."1#: behaves ...

In other words, I think that the fibonacci encoding aspect isn't the
problem here.

So I think I would start with either a careful description of your
"encode to character" mechanism (how can I know if you've implemented
what you intended without that? I'd rather not try to reverse engineer
your thoughts  and assumptions if that's not necessary) or I would
start with an implementation which isn't so complicated.

Thanks,

--
Raul

On Thu, Jul 18, 2019 at 10:46 AM 'Jon Hough' via Programming
<[email protected]> wrote:
>
> Positive integers can be encoded using Fibonacci numbers.
> For the sake of easer, assume Fibonacci numbers are 1,2,3,5,... (only one 1).
> We can encode a positive integer uniquely using non-consecutive Fibonacci 
> numbers. For example
>
> 4 = 1 + 3 (note: 1 and 3 a re non-consecutive)
> 5 = 5 (already a Fib number)
> 6 = 1 + 5
> 7 = 2 + 5
> 8 = 8
>
> These representation can be encoded using bits. 1's where a Fibonacci number 
> is in the representation, 0s when not.
> 4 = 1 + 3 = 101
> 5 = 0001
> 6 = 1001
> 7 = 0101
> 8 = 00001
> ...
>
> We can append a 1 on the end of the encoding a a delimiter, so that during 
> decoding we know easily where to stop. This way integers can be packed 
> together.
>
> Here are some J verbs for encoding and decoding positive integers.
>
>
> NB. ===================================================================
> genfibs=: 3 : 0
> r=. 1 2
> i1=. 1
> i2=. 2
> c=. 2
> while. y > c=. c+1 do.
>   t=. i1 + i2
>   i1=. i2
>   i2=. t
>   r=. r, x: t
> end.
> )
>
>
> Fibs=: genfibs 1400
>
> encode=: 3 : 0
> d=. > (>@:,&:>)/ (<@encode1)"0 y
> r=. d,'0' #~ (#d) -~ 8 * >. 8 %~ # d
> pack r
> )
>
> encode1=: 3 : 0
> n=. x: y
> r=. ''
> k=: ''
> fl=. x: Fibs{~ I. Fibs <: y
> i=. <:#fl
> while. n do.
>   r=. r,'1'
>   n=. n- i{fl
>   k=: k,i{fl
>   i=. i-1
>   while. (i >: 0) *. (n<i{fl) do.
>     r=. r,'0'
>     i=. i-1
>   end.
> end.
> (|.r),'1'
> )
>
>
>
> pack=: 3 : 0
> a.{~ #. _8 ] \"."0 y
> )
>
> decode=: 3 : 0
> i=. , {:|."1 (8 # 2) ,:|."1 #: a.i. y
> n=. ''
> while. 1 e. 1 1 E. i do.
>   idx=. {. I. 1 1 E. i
>   n=. n, +/ Fibs {~ I. i {~ i. >: idx
>   i=. (2+idx) }. i
> end.
> n
> )
>
> NB. ===================================================================
>
> The encode verb outputs the Fib-representation as bytes.
>  encode 4
> �
>
> #: a.i. encode 4
> 1 0 1 1 0 0 0 0
>
>
> decode encode 449239438124834384923493383837734733747181x
> 449239438124834384923493383837734733747181
>
>
> decode encode 1 2 3 4 5 6 7
> 1 2 3 4 5 6 7
>
> Fibonacci encoding is not really a good compression encoding, but has some 
> good error handling properties.
> I am not happy with the decode verb, as I wanted to write it without a while 
> loop. Difficult though as there are some
> awkaward consecutive bits that might appear in the encoding, e.g.
> ...0 1 1 1 0...
> Using 1 1 E. with this breaks decoding unless done carefully. Easier just to 
> use a while loop and eat the input every iteration.
>
> refs: https://en.wikipedia.org/wiki/Fibonacci_coding
> https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem
> ----------------------------------------------------------------------
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