
Add terms surrounded by zero together in matrix
Posted:
Jul 31, 2005 1:10 AM


On 29 Jul 2005, at 06:42, mchangun@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

