Anonymous user
Talk:Zeckendorf number representation: Difference between revisions
Talk:Zeckendorf number representation (view source)
Revision as of 02:12, 28 November 2012
, 11 years agono edit summary
(→Consensus on the sequence: no duplicate 1s, no required method) |
No edit summary |
||
| (10 intermediate revisions by 6 users not shown) | |||
Line 14:
: The sequence should be specified as 1,2,3,5,8..., and a Zeckendorf representation of a non-negative integer n is n expressed as the sum of non-consecutive terms in that sequence. This is sufficient and unambiguous: every n >= 0 has a unique such representation, and vice versa. I don't think the task should specify ''how'' one derives such a summation from n; listing a method as a hint, fine, putting it in the spec as if it's required, no. --[[User:Ledrug|Ledrug]] 23:47, 11 October 2012 (UTC)
:: I agree with these points. I'll add that it seems most sites say Zeckendorf representation only specifies a sum, and does not specify the binary place value coding that turns 3 + 1 into 101. I kind of like the binary coding but I think it might be good to describe it as an additional encoding on top of Zeckendorf representation. [http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibrep.html#fibbase Dr Ron Knott's site] calls this the "Fibonacci base system." Other links:
::* [http://mathworld.wolfram.com/ZeckendorfRepresentation.html Mathworld] hints at the binary encoding by using 0 or 1 as a multiplier of Fibonacci terms.
::* [http://mathworld.wolfram.com/ZeckendorfsTheorem.html Z's theorem at MathWorld]
::* [http://en.wikipedia.org/wiki/Zeckendorf%27s_theorem Wikipedia] is careful to say the theorem applies to ''distinct'' Fibonacci numbers (my emphasis.)
::* [http://oeis.org/A003714 OEIS A003714], "Fibbinary numbers".
:: —[[User:Sonia|Sonia]] 00:32, 12 October 2012 (UTC)
:: Oh wait, there's [http://oeis.org/A014417 OEIS A014417] "Representation of n in base of Fibonacci numbers" which is even better. —[[User:Sonia|Sonia]] 00:39, 12 October 2012 (UTC)
:::I've rewritten the task to be based on distinct Fibonacci numbers, and the task now references the OEIS sequence as the desired result without specifying the algorithm. I don't think it will be any great hardship to the current entries to switch to this approach, since it merely involves starting the Fibonacci sequence at a point where it leaves out the first 1. --[[User:TimToady|TimToady]] 05:16, 12 October 2012 (UTC)
::::Okay, I think I've patched everything to be consistent with the new task description, but please double-check my work to see if I've done something stupid in your favorite language. --[[User:TimToady|TimToady]] 06:27, 12 October 2012 (UTC)
:::::Much thanks guys, the fixes are great. I am currently trying not to get a full blown cold, which appears to be what's affecting my concentration. --[[User:Paddy3118|Paddy3118]] 18:35, 12 October 2012 (UTC)
: It should be noted that this task can be solved without using a Fibonacci sequence. See the '''generic''' example in the REXX section. -- [[User:Gerard Schildberger|Gerard Schildberger]] 21:34, 23 October 2012 (UTC)
:: That doesn't make sense at all. Try the sequence 1, 2, 4 ... 2^n, how do you make a representation that's garanteed to have no 11s? --[[User:Ledrug|Ledrug]] 23:13, 23 October 2012 (UTC)
::: I don't understand "try the sequence". Do you want those numbers expressed as Zeckendorf numbers (instead of 0, 1, 2, 3, 4, 5, ..., 20)? The '''generic''' REXX example doesn't ''use'' a sequence at all. -- [[User:Gerard Schildberger|Gerard Schildberger]] 23:22, 23 October 2012 (UTC)
:::: Never mind, I thought "solved without using a Fibonacci sequence" meant "using some other sequence in its place", which obviously was my misunderstanding. --[[User:Ledrug|Ledrug]] 00:29, 24 October 2012 (UTC)
: Also, the '''general''' REXX solution solves the task for '''N''' Zeckendorf numbers, which merely generates a Fibonacci sequence whose last numer is greater or equal to '''N''' (instead of hard coding six Fibonacci numbers). I think it'd be a better task to list up to '''N''' Zeckendorf numbers, with 20 being the default. That way, no short cuts would be used in the examples and thereby hiding the limitations of the programs. -- [[User:Gerard Schildberger|Gerard Schildberger]] 21:34, 23 October 2012 (UTC)
==Perl 6, wrong fib sequence==
Line 20 ⟶ 42:
:By the way, according to your own criteria, the Python solutions are incorrect for 1, since using the 1,1 in the reversed order greedily should produce '10', not '1'. You shouldn't have to special-case 1 like that. --17:59, 11 October 2012 (UTC)
::More to the point, if you don't introduce a special discontinuity like that at 1, then the 1,1 and 1,2 solutions are isomorphic modulo the presence or absence of a trailing 0. Though I should probably not have said "incorrect" above; as Sonia points out, it's ambiguously specified. But I'd very much prefer to disambiguate the rule by saying something like "Using all the numbers in the sequence less than or equal to the number." This is in keeping with the greediness and the reversal of the sequence. Alternately I would not be adverse to going back to a formulation that uses "All the positive Fibonacci numbers in order." That is, the 1,2 solution. Which most of the implementations chose back when it was still unspecified as 1,1... --[[User:TimToady|TimToady]] 18:11, 11 October 2012 (UTC)
== J wiki edit trouble ==
It's supposed to show as
<lang J>
' '&~:Filter@:":@:#:@:#.@:((|. fib 2+i.8) e. fsum)&.>i.3 7
┌──────┬──────┬──────┬──────┬──────┬──────┬──────┐
│0 │1 │10 │100 │101 │1000 │1001 │
├──────┼──────┼──────┼──────┼──────┼──────┼──────┤
│1010 │10000 │10001 │10010 │10100 │10101 │100000│
├──────┼──────┼──────┼──────┼──────┼──────┼──────┤
│100001│100010│100100│100101│101000│101001│101010│
└──────┴──────┴──────┴──────┴──────┴──────┴──────┘
</lang>
--LambertDW 02:12, 28 November 2012 (UTC)
| |||