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MathGroup Archive 2002

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Re: Combination/Permutation questions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37462] Re: [mg37436] Combination/Permutation questions
  • From: "Michael Chang" <michael_chang86 at hotmail.com>
  • Date: Thu, 31 Oct 2002 04:41:36 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Andrzej,

Many thanks for your help and solutions!

Actually, for completeness, I believe that for In[3], the input should have 
been

In[3]:= Perms[{a,b,c,d,e},2]

but I think that this is probably clear from Out[3] anyways.

Also, after seeing what *expert* Mathematica programmers do ;), I've found 
that for In[6],

In[6]:=
f[l_List,k_]:=Select[Union[Plus@@@Perms[Join[l,-l],k]],Length[#]==k&]

seems to produce the expected number of sets when using, for example,

In[7]:= f[{a,b,c,d},3]

(A slightly larger set tends to result otherwise, due to the inclusion of 
some sets with Lenth[#]!=k.)

Regards,

Michael

>From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
To: mathgroup at smc.vnet.net
>To: Andrzej Kozlowski <andrzej at tuins.ac.jp>
>CC: michael_chang86 at hotmail.com (Michael Chang), mathgroup at smc.vnet.net
>Subject: [mg37462] Re: [mg37436] Combination/Permutation questions
>Date: Tue, 29 Oct 2002 23:23:21 +0900
>
>I guess I did not exactly answer your last question. To select those  
>expressions that involve just one minus use:
>
>
>l={-a-b,a-b,-a+b,a+b,-a-c,a-c,-b-c,b-c,-a+c,a+c,-b+c,b+c,-a-d,a-d,-b- 
>d,b-d,-c-
>       d,c-d,-a+d,a+d,-b+d,b+d,-c+d,c+d};
>
>
>Select[l,Count[#,-1,2]==1&]
>
>{a-b,-a+b,a-c,b-c,-a+c,-b+c,a-d,b-d,c-d,-a+d,-b+d,-c+d}
>
>
>
>On Tuesday, October 29, 2002, at 05:20 PM, Andrzej Kozlowski wrote:
>
>>
>>On Tuesday, October 29, 2002, at 02:09 PM, Michael Chang wrote:
>>
>>>First, suppose that I have 4 objects, a, b, c, and d, respectively.
>>>How can I generate the permutation set when chosing, say, only *2*
>>>elements.
>>>
>>>In[1]:=  Permutations[{a,b,c,d}]
>>>
>>>gives me the permutation set when choosing *all* 4 objects ... :(
>>
>>
>>In[1]:=
>><<DiscreteMath`Combinatorica`
>>
>>In[2]:=
>>Perms[l_List,k_]:=Flatten[Permutations/@KSubsets[l,k],1]
>>
>>In[3]:=
>>Perms[{a,b,c,d},2]
>>
>>Out[3]=
>>{{a,b},{b,a},{a,c},{c,a},{a,d},{d,a},{a,e},{e,a},{b,c},{c,b},{b,d},{d,b 
>>},{
>>   b,e},{e,b},{c,d},{d,c},{c,e},{e,c},{d,e},{e,d}}
>>
>>
>>
>>
>>
>>>
>>>Second, for the same four objects (a,b,c,d), to generate the
>>>*combination* set (with 2 elements), I use:
>>>
>>>In[2]:=  Needs["DiscreteMath`Combinatorica`"];
>>>In[3]:=  cset=KSubsets[{a,b,c,d},2]
>>>
>>>and this generates the expected 6 combinatorial pairs
>>>({{a,b},{a,c},{a,d},{b,c},{b,d},{c,d}}).
>>>
>>>What I'd like to do next is add each combinatorial set, and I am able
>>>to do this correctly via:
>>>
>>>In[4]:=  Plus@@Transpose[cset]
>>>
>>>to obtain {a+b,a+c,a+d,b+c,b+d,c+d}.
>>
>>Alternatively:
>>
>>In[4]:=
>>cset=KSubsets[{a,b,c,d},2]
>>
>>
>>Out[4]=
>>{{a,b},{a,c},{a,d},{b,c},{b,d},{c,d}}
>>
>>In[5]:=
>>Plus@@@cset
>>
>>Out[5]=
>>{a+b,a+c,a+d,b+c,b+d,c+d}
>>
>>>
>>>My problem lies in the fact that now, I'd like to be able to allow
>>>*each* (a,b,c,d) element to be either plus, or minus (in reality, I'm
>>>trying to generate combinatorial adds/minuses for *functions*
>>>(a,b,c,d)), and to still generate the 'sum' of the cominatorial set.
>>>So for instance, in my current example, for the *first* {a,b}
>>>combinatorial pair, I'd like to be able to generate
>>>
>>>In[5]:=   
>>>Plus@@Transpose[Partition[Flatten[Outer[List,{a,-a},{b,-b}]],2]]
>>>
>>>(which generates {a+b,a-b,-a+b,-a-b}) ... only I'd like this to be
>>>somehow done automatically for *each* combinatorial pair (and, for the
>>>more general case when I generate combinatorial sets involving only
>>>'k' elements).  I've struggled with this for a while, and am only able
>>>to generate such a list *manually* for each of my six original
>>>combinatorial pairs ... a tedious, and somewhat tiresome procedure! :(
>>>  Is there a way of 'easily' doing this?!?
>>
>>For example:
>>
>>In[6]:=
>>f[l_List,k_]:=Rest[Union[Plus@@@Perms[Join[l,-l],k]]]
>>
>>In[7]:=
>>f[{a,b,c,d},2]
>>
>>Out[7]=
>>{-a-b,a-b,-a+b,a+b,-a-c,a-c,-b-c,b-c,-a+c,a+c,-b+c,b+c,-a-d,a-d,-b- 
>>d,b-d,-
>>   c-d,c-d,-a+d,a+d,-b+d,b+d,-c+d,c+d}
>>
>>>
>>>With my newly generated list (having 4*6 'elements'), how can I also
>>>go about finding which expressions involve/use only, say, 1 Minus?  Is
>>>there an 'easy' way of doing this too?  (For instance, I'd like to
>>>then 'filter' for (a-b) and (b-a) ...)  (Perhaps the count of 1 Minus
>>>is a little contrived for this example here, but for the more general
>>>case I'll probably be considering, I might need to filter for, say,
>>>(longer) expressions involving 2 Minuses (say).)
>>
>>
>>In cases like the above the following will work:
>>
>>In[8]:=
>>Count[#,-1,2]&/@%
>>
>>Out[8]=
>>{2,1,1,0,2,1,2,1,1,0,1,0,2,1,2,1,2,1,1,0,1,0,1,0}
>>
>>
>>
>>Andrzej Kozlowski
>>Yokohama, Japan
>>http://www.mimuw.edu.pl/~akoz/
>>http://platon.c.u-tokyo.ac.jp/andrzej/
>>
>>
>Andrzej Kozlowski
>Yokohama, Japan
>http://www.mimuw.edu.pl/~akoz/
>http://platon.c.u-tokyo.ac.jp/andrzej/

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