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Re: How to check whether an infinite set is closed under addition?

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  • Subject: [mg124312] Re: How to check whether an infinite set is closed under addition?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 16 Jan 2012 17:14:00 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201201150951.EAA19688@smc.vnet.net> <C9FCBB38-0E20-478B-97BC-BD57313E080F@mimuw.edu.pl>

On 15 Jan 2012, at 12:00, Andrzej Kozlowski wrote:

> Let
> On 15 Jan 2012, at 10:51, Rex wrote:
>
>> Given k positive numbers a_1<a_2<a_3<...<a_k, and all integers greater
>> than a_k, we want to check whether this set {a_1, a_2, a_3,...a_k, a_k
>> + 1, a_k+2, ......} is closed under addition.
>>
>> Is there any easy way to do this? any functions that we could use in
>> Mathematica?
>>
>> Your help will be greatly appreciated.
>>
>>
>
> Lest's call your set {a1,a2,...,a3} "base". Then:
>
> closedQ[base_List] :=
> Complement[Select[Total[Subsets[base, {2}], {2}], # <= Max[base] &],
>   base] == {}
>
> For example:
>
> closedQ[{1, 2, 3}]
>
> True
>
> closedQ[{1, 4, 6, 7}]
>
> False
>
> Andrzej Kozlowski
>

The program above does dome unnecessary comparisons since anything added to the largest element of the base will obviously included in the complete set. If the base is ordered, a better program will be:

closedQ[base_List] :=
 Complement[
   Select[Total[Subsets[Most[base], {2}], {2}], # <= Last[base] &],
   base] == {}

If it is not ordered we could use Sort[base] in place of base.

Andrzej Kozlowski



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