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MathGroup Archive 2005

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Re: Add terms surrounded by zero together in matrix

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59149] Re: [mg59123] Add terms surrounded by zero together in matrix
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 30 Jul 2005 01:25:29 -0400 (EDT)
  • References: <200507290442.AAA03344@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 29 Jul 2005, at 06:42, mchangun at gmail.com wrote:

> Hi All,
>
> I think this is a rather tough problem to solve.  I'm stumped and  
> would
> really appreciated it if someone can come up with a solution.
>
> What i want to do is this.  Suppose i have the following matrix:
>
> 0       0       0       1       0
> 0       0       1       2       0
> 0       0       0       2       1
> 1       3       0       0       0
> 0       0       0       0       0
> 0       0       0       0       0
> 0       0       1       1       0
> 5       0       3       0       0
> 0       0       0       0       0
> 0       0       0       3       1
>
> I'd like to go through it and sum the elements which are surrounded by
> zeros.  So for the above case, an output:
>
> [7 4 5 5 4]
>
> is required.  The order in which the groups surrounded by zero is
> summed does not matter.
>
> The elements are always integers greater than 0.
>
> Thanks for any help!
>
>


O.K., Here is a solution. I think the algorithm is rather nice but  
the implementation certainly isn't, with a nasty procedural Do loop,  
nested Blocks  etc, but I can't really afford the time to try to make  
it nicer. Perhaps someone else will.

Here is the function:


SumsOfTermsSurroundedByZero[AA_] :=
   Block[{MakeNames, A = AA, p = First[Dimensions[AA]], q = Last 
[Dimensions[AA]]},
    MakeNames[1, 1] := A[[1,1]] = Unique[z]*A[[1,1]]; MakeNames[1,  
i_] :=
      A[[1,i]] = Block[{d}, If[(d = Variables[A[[1,i - 1]]]) != {},  
First[d]*A[[1,i]],
         Unique[z]*A[[1,i]]]]; MakeNames[i_, 1] :=
      A[[i,1]] = Block[{d}, Which[(d = Variables[A[[i - 1,1]]]) != {},
         First[d]*A[[i,1]], (d = Variables[A[[i - 1,2]]]) != {}, First 
[d]*A[[i,1]],
         True, Unique[z]*A[[i,1]]]]; MakeNames[i_, q] :=
      A[[i,q]] = Block[{d}, Which[(d = Variables[A[[i - 1,q - 1]]]) ! 
= {},
         First[d]*A[[i,5]], (d = Variables[A[[i - 1,q]]]) != {}, First 
[d]*A[[i,q]],
         (d = Variables[A[[i,q - 1]]]) != {}, First[d]*A[[i,q]], True,
         Unique[z]*A[[i,q]]]]; MakeNames[i_, j_] :=
      A[[i,j]] = Block[{d}, Which[(d = Variables[A[[i - 1,j - 1]]]) ! 
= {},
         First[d]*A[[i,j]], (d = Variables[A[[i - 1,j]]]) != {}, First 
[d]*A[[i,j]],
         (d = Variables[A[[i - 1,j + 1]]]) != {}, First[d]*A[[i,j]],
         (d = Variables[A[[i,j - 1]]]) != {}, First[d]*A[[i,j]], True,
         Unique[z]*A[[i,j]]]]; Do[MakeNames[i, j], {i, p}, {j, q}];
     Cases[List @@ Plus @@ Flatten[A], _?NumericQ, Infinity]]

Here is your matrix defined using proper Mathematica syntax:

AA = {{0,0 , 0 , 1, 0}, {0, 0 , 1 , 2, 0}, {0, 0, 0, 2 , 1}, {1, 3,  
0 , 0 , 0}, {0,
     0, 0, 0 , 0}, {0 , 0, 0 , 0, 0}, {0, 0 , 1 , 1, 0}, {5 , 0, 3,  
0 , 0}, {0,
      0, 0, 0, 0}, {0, 0, 0, 3, 1}}

And here is the solution:

In[3]:=
SumsOfTermsSurroundedByZero[AA]

Out[3]=
{7,4,5,5,4}


I have not tested it on other examples but your own but it should  
work in all cases.

Andrzej Kozlowski



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