Thanks, though I think Cliff’s approach is more advanced.
Noticing that your g is applied with ~, you can also use this “hook”
h =: + +
which does x + 2y
Also, you can use the constant noun _1 1, so avoiding the constant function _1
1”_ , as the lhs in this expression:
(_1 1$~]) 5
_1 1 _1 1 _1
So your verb may be modified thus:
4 (h/\.&.|.@, _1 1$~]) 5
4 7 15 29 59 117
Or with another variation for building the list of +-1,
4 (h/\.&.|.@, $&_1 1) 6
4 7 15 29 59 117 235
I wonder if this discussion is what Skip was hoping for?
Cheers,
Mike
Please reply to [email protected].
Sent from my iPad
> On 28 Nov 2018, at 19:26, Nimp O <[email protected]> wrote:
>
> Very interesting mathematical solution Mike.
>
> My initial idea to solve this type of sequence formula was using folds.
> The idea is to define a fold function g and fold it on a list of _1 and 1,
> considering an initial value.
> The result was:
>
> g =. ]+2*[
> 4 ((g~/\.&.|.)@, ]$(_1 1)"_) 5
>
> 4 7 15 29 59 117
>
> I found a bit difficult to work this left foldlist in J. Folds from right
> seem more natural.
>
> This same idea can be implemented in Mathematica as:
>
> In[1]:= FoldList[2 #1 + #2 &, 4, {-1 , 1 , -1 , 1, -1}]
> Out[1]= {4, 7, 15, 29, 59, 117}
> ________________________________
> From: Programming <[email protected]> on behalf of
> 'Mike Day' via Programming <[email protected]>
> Sent: Wednesday, November 28, 2018 2:08 PM
> To: [email protected]
> Subject: Re: [Jprogramming] Recursive verbs
>
> Yes, this is one way to get high terms without having to evaluate all
> their predecessors.
> Bear in mind that it's calculating 2^n twice, and the zero-th term
> should surely be the
> starting value. Also, it assumes that starting value is 4; just change
> "4" to "x" in the
> 13 : ... expression to make it apply for any starting value.
>
> So here's a lovely compact tacit version which addresses those issues,
> at the expense
> of returning the wrong sign for a negative starting value:
>
> fa=: (* ([: | -) 3 %~ [: <: ]) _2&^ NB. with a [: to keep Linda
> happy
> So,
> 4 fa 200x
> 5892106162282964343653861005250929542581410977203573729438379
> cf
> 4 (13 :'x p.~(([:(2&^)>:),.~ _1r3 *([:(2&^)>:) + _1&^)y' ) 199x
> 5892106162282964343653861005250929542581410977203573729438379
>
> If you want a table, the rank needs to be imposed:
> 4 5 fl"0/ 3 4 5
> 29 59 117
> 37 75 149
>
> A possibly easier way to understand why we need ((_2^n) - 1) % 3 is perhaps
> to inspect the general term.
> We have
> a1 = 2a0 - 1 NB. using "Mathematical" notationrather than APL/J
> a2 = 2a1 + 1 = 4a0 - 2 + 1
> a3 = 2a2 - 1 = 8a0 - 4 + 2 - 1
> ...
> an = (2^n).a0 - (2^[n-1]) + (2^[n-2]) ... + (-1)^n
>
> The absolute sum of all terms except the first is | sum {i=0,n-1} _2^i | ,
> which we know from High School maths is | [(_2^n) - 1]%[_2 -1] | ,
> ie | [(_2^n) - 1]% 3 |
>
> If you want the correct sign for negative starting values, use
> (fa * * [) NB. ... which isn't quite so neat.
>
> Cheers,
>
> Mike
> NB - previous correspondence not trimmed
>
>
>> On 28/11/2018 09:49, Linda Alvord wrote:
>> Hi Cliff,
>>
>> It took a while but I got the explicit verb to work.
>>
>> f=: 13 :'4 p.~(([:(2&^)>:),.~ _1r3 *([:(2&^)>:) + _1&^)y'
>>
>> f i.10
>> 7 15 29 59 117 235 469 939 1877 3755
>>
>> f
>> 4 p.~ ([: 2&^ >:) ,.~ _1r3 * ([: 2&^ >:) + _1&^
>>
>> Linda
>>
>> -----Original Message-----
>> From: Programming<[email protected]> On Behalf Of
>> Cliff Reiter
>> Sent: Tuesday, November 27, 2018 12:38 PM
>> To:[email protected]
>> Subject: Re: [Jprogramming] Recursive verbs
>>
>> I like that solution. Here is another approach
>>
>> 4 p.~(2&^@>: ,.~ _1r3 * 2&^@>: + _1&^)i. 10
>> 7 15 29 59 117 235 469 939 1877 3755
>>
>> I found this by composing the linear poly's and noting the constant terms
>> satisfy a two step recursion with eigenvalues 2 and _1.
>>
>> lc=.({.@[+{:@[*{.@]),*&{:
>> lc/\.(7$_1 2,:1 2)
>> _43 128
>> _21 64
>> _11 32
>> _5 16
>> _3 8
>> _1 4
>> _1 2
>>
>>
>>> On 11/27/2018 11:06 AM, Raul Miller wrote:
>>> Sure, ... that's a bit bulkier than I prefer, but it works.
>>>
>>> Here's another alternative:
>>>
>>> evenodd=: , (_1 2,:_1 4)&p.@{:
>>> evenodd evenodd 4
>>> 4 7 15 29 59
>>>
>>> Thanks,
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