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Topic: Add terms surrounded by zero together in matrix
Replies: 6   Last Post: Jul 31, 2005 1:37 AM

 Messages: [ Previous | Next ]
 Peter Pein Posts: 1,147 Registered: 5/4/05
Re: Add terms surrounded by zero together in matrix
Posted: Jul 31, 2005 1:37 AM

Andrzej Kozlowski schrieb:
> Here is a slightly improved version. (I just removed unnecessary
> inner Blocks and a unnecessary conversion of the final expression to
> List before applying Cases, which of course is never needed).
>
> Andrzej Kozlowski
>
>
> SumsOfTermsSurroundedByZero[AA_] :=
> Block[{d, 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]] = If[(d = Variables[A[[1, i - 1]]]) != {}, First[d]*A
> [[1, i]],
>
> Unique[z]*A[[1, i]]]; MakeNames[i_, 1] :=
> A[[i, 1]] = 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]] = 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]] = 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[Plus @@ Flatten[A], _?NumericQ, Infinity]]
>
>
>
>
>
> On 29 Jul 2005, at 20:55, Andrzej Kozlowski wrote:
>
>

>>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
>>
>>
>>

>
>

Wow, a nice trick!

I surrounded the array by zeros to get rid of the case differentiations
and applied MakeNames only to positive elements. Additionally I tried to
manage the case of "V"-shaped patterns:

SumsOfTermsSurroundedByZero[AA_]:=
Block[{d,MakeNames,A},
A=Prepend[Flatten[{0,#,0}]&/@AA,Table[0,{2+Length[AA[[1]]]}]];
MakeNames[i_,j_]:=A[[i,j]]*=
If[(d=Variables[A[[{i-1,i},{j-1,j,j+1}]]])==={},
Unique[z],
If[Length[d]>1,
Set[#,First[d]]&/@Rest[d]];
First[d]
];
(*Do[MakeNames[i,j],{i,p},{j,q}];*)
(*Cases[Plus@@Flatten[A],_?NumericQ,Infinity]*)
MakeNames@@@Position[A,_?Positive];
Coefficient[Plus@@Flatten[A],#]&/@Variables[A]
]

Regards,
Peter

--
Peter Pein
Berlin

Date Subject Author
7/30/05 Andrzej Kozlowski
7/31/05 Peter Pein
7/31/05 Andrzej Kozlowski
7/31/05 Simons, F.H.
7/31/05 Andrzej Kozlowski