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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg103232] Re: [mg103181] Re: how to get the longest ordered sub sequence of a
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Fri, 11 Sep 2009 05:26:03 -0400 (EDT)
  • References: <h87pgp$5gt$1@smc.vnet.net> <200909101118.HAA17784@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

That's the longest CONSECUTIVE ordered subsequence... and here's a  
slightly different code for it:

list = {3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 7, 9, 8, 9};
sp = Split[list, OrderedQ[{#1, #2}] &];
Last@SortBy[sp, Length]

{3, 5, 7, 9}

Bobby

On Thu, 10 Sep 2009 06:18:19 -0500, Valeri Astanoff <astanoff at gmail.com>  
wrote:

> On 9 sep, 10:37, a boy <a.dozy.... at gmail.com> wrote:
>> how to get a (strict or not-strict)decreasing sub sequence of a list?
>>                                     =
>          ----------------
>>
>> increasing                               =
>     ?
>
> Good day,
>
> Just an example of what you can do :
>
> In[1]:= list = {3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 7, 9, 8, 9};
>
> In[2]:= sp = Split[list, OrderedQ[{#1, #2}] &]
> Out[2]= {{3}, {1, 4}, {1, 5, 9}, {2, 6}, {5}, {3, 5, 7, 9}, {8, 9}}
>
> In[3]:= pos = Ordering[sp, 1, (Length[#1] >= Length[#2]) &] // First
> Out[3]= 6
>
> In[4]:= sp[[pos]]
> Out[4]= {3, 5, 7, 9}
>
> hth
>
> --
> Valeri Astanoff
>


-- 
DrMajorBob at yahoo.com


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