C[omp]ute

From: "andrew cooke" <andrew@...>

Date: Thu, 19 Jan 2006 18:41:44 -0300 (CLST)

Ha.  Finally something to make me smile after an extremely crappy couple
of days...

I was intrigued by this comment -
http://blogs.msdn.com/code4bill/archive/2005/12/20/505872.aspx - which
refers to the puzzle here -
http://www.microsoft.com/india/code4bill/archives/19_24_dec05.aspx (bottom
of page).

The idea is that you can (completely arbitrarily) assign the weights for a
"binary" digit to be 1,2,3,5,8,13,21... instead of 1,2,4,8,16,32,64,...
where those numbers are from the Fibonacci sequence (each number is the
sum of the previous two; for some reason the initial 0,1 are ignored).

Now since those numbers don't go up as quickly as 2^n you get ambiguities
(duplications?).  The first is 3, which is both 001 and 11 (low bits to
the left).

So.... I downloaded PLT scheme - http://www.plt-scheme.org/ - for no
particular reason except that I was bored with Python and started playing
around.  It took me a while to remember my (flaky) Lisp, but eventually I
got to the point where I had some working code, albeit not very elegant.

First thing I defined was an increment method that takes one of these
numbers and adds 1 to the binary part (see end of post).  Then I plotted
the "fibonacci" values against the "binary" values.  I can't post images
to this blog, but this wasn't that interesting anyway - although it did
appear to be fractal (OK, so that's pretty cool I guess).

Next I plotted how many different representations there were for each
natural number.  That was a much prettier graph, reminding me vaguely of
number spiral - http://www.numberspiral.com/ (just because it was kind-of
pretty and with a lot more arcs/structure than I expected).

Finally - and this is what made me smile - I followed the hint in the post
I linked to first above and looked at the numbers with a single
representation.  Now, I may have a bug, but they appear to be:

  1, 2, 4, 7, 12, 20, 33, 54, 88

Stupid old me, looking at those, thought "bleagh".  Can you see the
pattern dear reader?  Well, despite it being totally obvious I cheated -
http://www.research.att.com/~njas/sequences/?q=1%2C2%2C4%2C7%2C12%2C20%2C33%2C54%2C88
- which tells you soon enough that they're either 1 less than the
Fibonacci series, or the sum of the first n Fibonacci numbers.  Ha!  Sweet
or what?

And.... then I thought, OK, so what was the binary?  At first I thought
"OK, this is obvious - it's all 1s (since it's the sum, right?)".  But
it's not, because they threw away the first 1, remember?  (Check the
sequence above - it's the sums for 0, 1, 1, 2, 3, 5... but the problem
used 1, 2, 3, 5...)

So, any guesses?

I have no idea how this works, but here are some values (low bit to left):

  33 = 1 0 1 0 1 0 1
  54 = 0 1 0 1 0 1 0 1
  88 = 1 0 1 0 1 0 1 0 1

Heh.  Actually, as I type this, I do kind-of see how that works.  Given
that each value is already the sum of previous two...  Hmmm.

Andrew

PS Here's the increment code.  I represented each number by a pair of
lists - one list was the fibonacci numbers, the other was the binary
digits.  I have a strong suspicion this code could be much neater, but
can't see how (excuse the "I can write bad Haskell in any language"
style...):

  (define (increment fib-digits)
    (match fib-digits
      ; digit of zero is easy
      [((n . nn) . (0 . dd)) `((,n . ,nn) . (1 . ,dd))]
      ; for digit of 1 consider progressively longer series
      ; special case of singleton
      [((1) . (1)) '((1 2) . (0 1))]
      ; last two numbers - need to extend the fibonacci series
      [((n1 n2) . (1 1)) (cons (list n1 n2 (+ n1 n2)) '(0 0 1))]
      ; we have at least two more digits so recurse
      [((n . nn) . (1 . dd))
       (match-let (((nn2 . dd2) (increment `(,nn . ,dd))))
         `((,n . ,nn2) . (0 . ,dd2)))]))