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MathGroup Archive 2002

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Re: Combination/Permutation questions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37442] Re: [mg37436] Combination/Permutation questions
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Wed, 30 Oct 2002 00:50:39 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

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/



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