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Re: how to get the longest ordered sub sequence of a

  • To: mathgroup at smc.vnet.net
  • Subject: [mg103212] Re: [mg103158] how to get the longest ordered sub sequence of a
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Thu, 10 Sep 2009 07:24:15 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

lst = {1, 8, 2, 4, 3, 5};

Select[oss = Select[Subsets[lst], OrderedQ],
 Length[#] == Max[Length /@ oss] &]

{{1, 2, 4, 5}, {1, 2, 3, 5}}

Select[oss = Select[Subsets[lst], OrderedQ[Reverse[#]] &],
 Length[#] == Max[Length /@ oss] &]

{{8, 4, 3}}


Bob Hanlon

---- a boy <a.dozy.boy at gmail.com> wrote: 

=============
Thank all! your answers are right!
However,what I need is the longest not-strict ordered items of a given list
L, not a segment of L. For example,
{1,8,2,4,3,5} -- ascending--> {1,2,4,5}
{1,8,2,4,3,5} --descending-->{8,4,3}

Because when I think this question (
http://groups.google.com/group/comp.soft-sys.math.mathematica/browse_thread/thread/f82401b1a517310c/9ca72a83a2313f50?lnk=gst&q=a.dozy.boy#9ca72a83a2313f50
):
For any integer k and 3^k, suppose 2^j is the closest to 3^k,
Gap[k]=| 3^k-2^j| is the subtraction .

Gap = Function[k, x = k*Log[2, 3];    Min[3^k - 2^Floor[x], 2^Ceiling[x] -
3^k]];
list=Table[{i, Gap[i]}, {i, 1, 5000}]

I want to find a non-strict decreasing items of {Gap[i]} .


On Wed, Sep 9, 2009 at 7:53 PM, Fred Simons <f.h.simons at tue.nl> wrote:

> a boy wrote:
>
>> how to get a (strict or not-strict)decreasing sub sequence of a list?
>>                                             ----------------
>>  increasing                                   ?
>>
>>
>>
>>
> lst=RandomInteger[{1,100}, {5000}];
>
> With[{sublists=Split[lst, #1<#2&]},
>    With[{m=Max[Length /@ sublists]},
>         Select[sublists, Length[#]==m&]]]
>
> {{3,19,22,33,51,66,89,95}}
>
> Fred Simons
> Eindhoven University of Technology
>

--

Bob Hanlon



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