Re: Re: Increasing scattered subsequence

• To: mathgroup at smc.vnet.net
• Subject: [mg80113] Re: [mg80086] Re: Increasing scattered subsequence
• From: Fred Simons <f.h.simons at tue.nl>
• Date: Mon, 13 Aug 2007 04:26:32 -0400 (EDT)
• References: <fa11b990e477.46baf2d3@bgu.ac.il> <D370EB89-ACFD-4FFF-B4E7-84A25FF267C5@mimuw.edu.pl> <C82B7D4D-6C0B-47E1-8603-DEE030A6E387@mimuw.edu.pl> <f69ab100e04a.46bc3a23@bgu.ac.il> <28688943.1186820838583.JavaMail.root@m35> <op.twwnxqppqu6oor@monster.gateway.2wire.net> <200708121116.HAA27746@smc.vnet.net>

```Here is another oneliner for this problem(?):

In[4]:= lst = {9, 2, 10, 3, 14, 9};

In[5]:= n = -\[Infinity];
Reap[If[# > n, Sow[#]; n = #] & /@ lst][[2, 1]]

Out[6]= {9, 10, 14}

Fred Simons
Eindhoven University of Technology

Andrzej Kozlowski wrote:
> (...)
>>> me that make sme suspicious.  What do you say "you need to use
>>> recursion and pattern matching, Select and Join"? This sounds to me
>>> like some sort of test problem so I have decided to answer it but
>>> without using any of these functions (although it may not be the
>>> simplest way to do this). So here is my answer:
>>>
>>>   ls = {9, 2, 10, 3, 14, 9};
>>>
>>> Reverse[Last[Last[Reap[NestWhile[With[{a = First[Ordering[#, -1]]},
>>> Sow[#[[a]]]; Take[#, a - 1]] &,ls,Length[#] > 0 &]]]]]
>>>
>>> {9, 10, 14}
>>>
>>>

```

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