You can make it pretty short too, if you allow yourself fix:
rs=1:fix(\f p n-n++f(p++n)p)[1][0]
I came up with this on the train home, but then I realised it was the
same as your solution :)
On 08/11/2007, at 12:57 PM, Alfonso Acosta wrote:
How about this,
infiniteRS :: [Int]
Report from the Rabbit Warren.
Thank you, everybody, for your contribution. The problem was, how to
construct a one-liner which generates the infinite Rabbit Sequence:
10110101101101011010110110101101101011010110110101101011011010110...
This is the result of an *infinite time* evolution of a
rabbit = 1 : concatMap ((0:) . flip replicate 1 . (1+)) rabbit
This nasty acquaintance of mine...
demanded the solution (in Haskell) as a one-liner.
It wasn't you yourself, was it?
No. Please, tell me it isn't so.
Regards,
Yitz
___
Haskell-Cafe
On Thu, 8 Nov 2007 [EMAIL PROTECTED] wrote:
Report from the Rabbit Warren.
Thank you, everybody, for your contribution. The problem was, how to
construct a one-liner which generates the infinite Rabbit Sequence:
10110101101101011010110110101101101011010110110101101011011010110...
This is
On 11/8/07, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I have a vague impression that the solution of Alfonso Acosta:
rs_aa = let acum a1 a2 = a2 ++ acum (a1++a2) a1 in 1 : acum [1] [0]
is also somehow related to this limit stuff, although it is simpler,
and formulated differently.
I
On Thu, 2007-11-08 at 00:56 +0100, [EMAIL PROTECTED]
wrote:
Don't shoot me...
The last exchange with Andrew Bromage made me recall a homework which was
given to some students by a particularly nasty teacher I happen to know.
The question is to generate the whole infinite Rabbit Sequence
Henning Thielemann writes:
And now we have much Haskell code for one sequence to be submitted to the
Online Encyclopedia of Integer Sequences!
Is there anything really there in Haskell?...
Well, if you are interested in something more venerable than rabbits, why
not try the sequence which
Don't shoot me...
The last exchange with Andrew Bromage made me recall a homework which was
given to some students by a particularly nasty teacher I happen to know.
The question is to generate the whole infinite Rabbit Sequence in one
shot (co-recursive, selbstverständlich).
The Rabbit
G'day all.
Quoting [EMAIL PROTECTED]:
This nasty acquaintance of mine asked the students to write down a simple
procedure which generates the sequence after the infinite number of units
of time.
Cool problem! Simple is, of course, in the eye of the beholder.
zipWith (!!) (fix (([1]:).map(=
On 11/8/07, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Would somebody try to solve it, before I unveil the solution? It isn't
difficult.
*** SPOILER WARNING ***
Here's my attempt, which I wrote without peeking:
let fibs' = 1 : 2 : zipWith (+) fibs' (tail fibs')
rabbits = 1 : 0 :
On Thu, Nov 08, 2007 at 12:56:46AM +0100, [EMAIL PROTECTED] wrote:
This nasty acquaintance of mine asked the students to write down a simple
procedure which generates the sequence after the infinite number of units
of time. Of course, any finite prefix of it.
rabbit = let rs = 0 : [x | r - rs,
How about this,
infiniteRS :: [Int]
infiniteRS = let acum a1 a2 = a2 ++ acum (a1++a2) a1 in 1 : acum [1] [0]
it certainly fits in one line but it's not really elegant
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G'day all.
Quoting [EMAIL PROTECTED]:
zipWith (!!) (fix (([1]:).map(= \x-if x==0 then [1] else [1,0]))) [0..]
This was the shortest variant I could manage in the time allotted:
zipWith(!!)(fix(([1]:).map(= \x-1:[0|x==1])))[0..]
Cheers,
Andrew Bromage
Is this what you are looking for:
mrs = [0] : [1] : zipWith (++) (tail mrs) mrs
then you can get the one you want with:
mrs !! index
given a suitable value for index
It seems I didn't read the question carefully - you want the infinite
list.
You can recover the solution from mrs
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