Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: generalized foldlist problem - part 2

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69422] Re: generalized foldlist problem - part 2
  • From: Arkadiusz.Majka at gmail.com
  • Date: Tue, 12 Sep 2006 06:52:30 -0400 (EDT)
  • References: <7BEBBEE71B48844B82F6D39CC9A3ABF00153C2DE@OCE1B045> <edm0pm$cgf$1@smc.vnet.net>

Andrzej,

Apologies po replying so late. Everything is ok - your code is correct,
I have not found other bugs. It is also ten times faster than mine....
:)

Thank you very much,

Arek


Andrzej Kozlowski napisal(a):
> Indeed, there was another bug. The new code is:
>
> GF[P_]:= Module[ { M=P}, Do[ If[ And @@ Thread[ Total[ Take[ M,s]]
> <1], Continue[], M = NestWhile[
> Join[ ( PadRight[ #1, Length[#1] + 1] & ) /@
> Take[ #1, s-1], ( PadLeft[ #1, Length[#1] + 1] & ) /@
> Take[ #1, s -1- Dimensions[#1][ [1]]]] & , M,
> ! And @@ Thread[ Total[ #1[ [ Range[s], Range[ Length[
> Flatten[ Split[ #1[ [s]]] /. { l___,
> { (0)..}} :> {l}]]]]]] < 1] & ]],
> { s, Dimensions[M][ [1]]}]; M]
>
>
> It works in your example:
>
> list1= { 0.5,0.7,0.4};
> list2= { 2,2,4};
> p = GFMatrix[list1, list2]
> {{0.5, 0.5, 0, 0, 0, 0}, {0, 0.7, 0.7, 0, 0, 0}, {0, 0, 0.4, 0.4,
> 0.4, 0.4}}
> GF[p]
> {{0.5, 0.5, 0, 0, 0, 0, 0, 0}, {0, 0, 0.7, 0.7, 0, 0, 0, 0}, {0, 0,
> 0, 0, 0.4, 0.4, 0.4, 0.4}}
> Total[%]
> {0.5, 0.5, 0.7, 0.7, 0.4, 0.4, 0.4, 0.4}
>
> Please test it and let me know if you find any other bugs.
>
> Andrzej Kozlowski
>
>
> On 5 Sep 2006, at 10:05, Majka Arkadiusz wrote:
>
> > Andrzej!
> >
> > The very good new is that this is what I expected! Thank you very
> > much.
> > There is one , I hope small problem. Your program does not work for
> > any
> > list1.
> >
> > You can check
> >
> > List1={0.5,0.7,0.4}
> > List2={2,2,4}
> >
> > GFMatrix is ok. But GF seems to drop in infinite loop - calculation
> > doesn't
> > finish. I made many tests and I find only one pattern: everything
> > is ok.
> > when elements of list1 are small. But I don't know how small - it
> > looks like
> > that a partial sum of their elements is not allowed to exceed 1 ,
> > but maybe
> > I'm wrong.
> >
> >
> > Any ideas?
> >
> > Arek
> >
> >
> > -----Original Message-----
> > From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl]
To: mathgroup at smc.vnet.net
> > Subject: [mg69422] Re:  Re: generalized foldlist problem - part 2
> >
> > (tm) Pro* I
> > think I have corrected GF also, (sooner than I expected):
> >
> >
> >
> > GF[P_] := Module[{M = P}, Do[If[And @@ Thread[Total[
> >      Take[M, s]] < 1], Continue[], M = NestWhile[
> >        Join[(PadRight[#1, Length[#1] + 1] & ) /@
> >           Take[#1, s - 1], (PadLeft[#1, Length[#1] + 1] & ) /@
> >           Take[#1, s - 1 - Dimensions[#1][[1]]]] & , M,
> >         ! And @@ Thread[Total[#1[[All, Range[Length[
> >                 Flatten[Split[#1[[s]]] /. {l___,
> >                     {(0) ..}} :> {l}]]]]]] < 1] & ]],
> >      {s, Dimensions[M][[1]]}]; M]
> >
> >
> > It works at least in the case of your first example. Please test it
> > and let
> > me know if you find any problems. I really cannot spare have the
> > time to do
> > it myself.
> >
> > By the way, even if this is finally what you desire, I can see that
> > the code
> > could be made a more efficient. However, I think it is efficient
> > enough, and
> > as I wrote above I should not be spending too much time on this so
> > I hope
> > other people will look at this carefully or propose their own
> > alternative
> > code.
> >
> > Andrzej Kozlowski
> >
> > On 4 Sep 2006, at 20:09, Andrzej Kozlowski wrote:
> >
> >> O.K., I see. I think, GFMatrix needs only a slight correction:
> >>
> >> GFMatrix[list1_,
> >>     list2_] :=
> >> Transpose[DeleteCases[Transpose[Normal[SparseArray[Flatten[Table[If
> >> [j <=
> >>        i + list2[[i]] - 1, {i, j} -> list1[[i]], {i, j} -> 0], {
> >>             i, Length[list2]}, {j, i, Total[list2] }], 1]]]],
> >> {Repeated[0]}]]
> >>
> >>
> >> I see the problem with GF too, but I have not got the time to
> >> correct it right now, so I will probably send you corrected version
> >> tomorrow.
> >>
> >> Andrzej
> >>
> >>
> >>
> >>
> >> On 4 Sep 2006, at 16:39, Majka Arkadiusz wrote:
> >>
> >>> Yes, I read it.
> >>>
> >>> Take:
> >>>
> >>> List1={0.4,0.2,0.9}
> >>> List2={8,1,7}
> >>>
> >>> What I want to mean: 0.4 appears 8 times, 0.2 - 1 time, 0.9 -
> >>> seven times.
> >>> And my matrix reads
> >>>
> >>> {{0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0},{0,0.2,0,0,0,0,0,0,0},
> >>> {0,0,0.9,0.9,0.9,
> >>> 0.9,0.9,0.9,0.9}}
> >>>
> >>> Your solution is bounded to dim=3. Could you , please enlarge it ?
> >>>
> >>> Moreover, for your your solution which is
> >>>
> >>> 0.4	0.4	0.4	0	0
> >>> 0	0.2	0	0	0
> >>> 0	0	0.9	0.9	0.9
> >>>
> >>> we apply GF
> >>>
> >>> and get
> >>>
> >>> 0.4	0.4	0.4	0	0	0
> >>> 0	0	0.2	0	0	0
> >>> 0	0	0	0.9	0.9	0.9
> >>>
> >>> This is not what I want.
> >>>
> >>> I want you to
> >>>
> >>> 1) keep the first row - because all elements are <=1
> >>> 2) keep the the first two rows - because first row + second row is
> >>> <=1
> >>> (suming along columns)
> >>> 3) you can't keep first three rows (whole matrix in this example)
> >>> because
> >>> suming along the third column gives 1.3 what is >1. So I have to
> >>> shift it on
> >>> the right of one position and the third row appears to be
> >>>
> >>> 0	0	0	0	0.9	0.9	0.9
> >>>
> >>> Now the solution is
> >>>
> >>> 0.4	0.4	0.4	0	0
> >>> 0	0.2	0	0	0
> >>> 0	0	0	0.9	0.9	0.9
> >>>
> >>> Now its enough to fill right corner by zero's and Plus@@%
> >>>
> >>>
> >>> Another example, how you procedures should work
> >>>
> >>> list1 = {0.8, 0.1, 0.2}
> >>> list2 = {8, 3, 3}
> >>>
> >>> GFMatrix[list1, list2] gives
> >>>
> >>>  0.8	0.8	0.8	0.8	0.8	0.8	0.8	0.8
> >>> 0	0.1	0.1	0.1	0	0	0	0
> >>> 0	0	0.2	0.2	0.2	0	0	0
> >>>
> >>> And GF[%] gives
> >>>
> >>> 0.8	0.8	0.8	0.8	0.8	0.8	0.8	0.8
> >>> 0	0.1	0.1	0.1	0	0	0	0
> >>> 0	0	0	0	0.2	0	0	0
> >>>
> >>> One more example
> >>>
> >>> List1={0.8,0.3,0.9}
> >>> List2={3,3,3}
> >>>
> >>> GF
> >>>
> >>> 0.8	0.8	0.8	0	0
> >>> 0	0	0	0.3	0.3	0.3	0
> >>> 0	0	0	0	0	0	0.9	0.9	0.9
> >>>
> >>> One more
> >>>
> >>> List1={0.5,0.5,0.5}
> >>> List2={3,3,3}
> >>>
> >>> GF
> >>>
> >>> 0.5	0.5	0.5
> >>> 0	0.5	0.5	0.5
> >>> 0	0	0	0.5	0.5	0.5
> >>>
> >>> I hope its clear but I am ready to clarify as long as we get what
> >>> I desire
> >>> :)
> >>>
> >>> Arek
> >>>
> >>> -----Original Message-----
> >>> From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl]
To: mathgroup at smc.vnet.net
> >>> Subject: [mg69422] Re:  Re: generalized foldlist problem - part 2
> >>>
> >>> Hmm... Did you really have a look at the solutions I posted? Here
> >>> are my two
> >>> functions:
> >>>
> >>> GFMatrix[list1_, list2_] := Normal[SparseArray[Flatten[Table[If[j
> >>> <= i +
> >>> list2[[i]]*-1, {i, j} -> list1[[i]], {i, j} -> 0], {i, Length
> >>> [list2]}, {j,
> >>> i, Length[list1] + i - 1}], 1]]]
> >>>
> >>> GF[P_]:= Module[ { M=P}, Do[ M = NestWhile[ Join[ ( PadRight[ #1,
> >>> Length[#1]
> >>> + 1] & ) /@ Take[ #1, s], ( PadLeft[ #1, Length[#1] + 1] & ) /@
> >>> Take[ #1, s
> >>> - Dimensions[#1][ [1]]]] & , M, ! And @@ Thread[ Total[ #1[ [ All,
> >>> Range[
> >>> Length[ Flatten[ Split[ #1[ [s]]] /. { l___, { (0)..}} :>
> >>> {l}]]]]]] < 1] &
> >>> ], { s, Dimensions[M][ [1]]}]; M]
> >>>
> >>> Now, let's consider your example:
> >>>
> >>> list1 = Table[Random[], {10}];
> >>> list2 = Table[Random[Integer, {1, 5}], {10}]
> >>>
> >>> Here is the solution:
> >>>
> >>> In[32]:=
> >>> m=GFMatrix[list1,list2];
> >>>
> >>> In[31]:=
> >>> Total[GF[m]]//Timing
> >>>
> >>> Out[31]=
> >>> {0.015873 Second,
> >>> {0.518177,0.518177,0.518177,0.518177,0.548916,0.588317,
> >>>
> >>> 0.588317,0.588317,0.997371,0.997371,0.997371,0.997371,0.997371,0.534
> >>> 6
> >>> 07,
> >>>
> >>> 0.534607,0.534607,0.534607,0.78564,0.251032,0.251032,0.535924,0.5359
> >>> 2
> >>> 4,
> >>>
> >>> 0.284892,0.284892,0.412818,0.127926,0.127926,0.127926,0.818314,0.690
> >>> 3
> >>> 88,
> >>>      0.690388,0.690388,0.690388,0.776409,0,0,0,0,0,0,0,0,0}}
> >>>
> >>> This is pretty fast, and I think satisfies your requirements. By
> >>> contrast,
> >>> your own solution does not work at all on my machine, producing
> >>> just a bunch
> >>> of error messages. Perhaps the solution is in some way not what
> >>> you wanted,
> >>> but then it would be helpful if you could point out what is wrong
> >>> with it.
> >>>
> >>> Andrzej Kozlowski
> >>>
> >>>
> >>>
> >>>
> >>>
> >>> On 4 Sep 2006, at 09:47, Arkadiusz Majka wrote:
> >>>
> >>>> Thanks for all answers. Andrzej presented "full solution" - and
> >>>> he is
> >>>> right - that's what I wanted to get. (Of course, I didn't take me a
> >>>> lot of time to get it from Jens or others answers)
> >>>>
> >>>> In the code below you will see what I want to do next with our
> >>>> matrix
> >>>> - it is explained below.
> >>>>
> >>>> So lets take
> >>>>
> >>>> list1=Table[Random[],{10}];
> >>>> list2=Table[Random[Integer,{1,5}],{10}]
> >>>>
> >>>> lets use Anrzeje's solution
> >>>>
> >>>> ma=GFMatrix[list1,list2]
> >>>>
> >>>> and now my code that manipulates ma in the way explained below
> >>>> (it is
> >>>> not allowed that Plus@@ma[[i]] is >1 for every i. )
> >>>>
> >>>> The code is the following
> >>>>
> >>>> moje[T_] := Module[{i, matrix, dl, ile, ile1, ile22, ile33},
> >>>>     dl[i_] := First[Flatten[Position[Plus @@ Take[matrix[i], i +
> >>>>     2], _?(# > 1 &)]] /. {} -> {0}];
> >>>>     matrixNew[i_] := ReplacePart[matrix[i], matrix[
> >>>>                 i][[i + 2]] /. {0 -> list1[[i + 2]]}, i + 2];
> >>>>     ile2[i_] := If[dl[i] > 0, Last[Flatten[Position[Plus @@ Take[
> >>>>     matrixNew[i], i + 2], _?(# > 1 &)]] /. {} -> {0}], 0];
> >>>>     ile33[i_] := ile2[i] - dl[i] + If[dl[i] > 0, 1, 0];
> >>>>     matrix[0_] := ma;
> >>>>     matrix[i_] := matrix[i] = Flatten[Join[{Map[PadRight[#,
> >>>>     Dimensions[Map[PadRight[#,
> >>>>       Length[#] + ile33[i - 1], 0,
> >>>>         ile33[i - 1]] &, Drop[matrix[i - 1], i]]][[2]]] &, Take[
> >>>>                   matrix[i - 1], i]], Map[PadRight[#, Length[#] +
> >>>> ile33[i -
> >>>>                 1], 0, ile33[i - 1]] &, Drop[matrix[i - 1],
> >>>> i]]}], 1];
> >>>>     matrix[T]]
> >>>>
> >>>> and solution
> >>>>
> >>>> moje[Length[list1] - 1]
> >>>>
> >>>> This algorithm is very brutal and very time consuming. Program
> >>>> repeats
> >>>> calculations etc. Do you have any ideas how to improve it or
> >>>> rebuild
> >>>> it entirely?
> >>>>
> >>>> p.s. I made many tests using my code and it is correct, i.e it
> >>>> generates "solutions" that i expect.
> >>>>
> >>>> p.s2. The algorithm tries to simulate agents (say computing jobs)
> >>>> that
> >>>> arrive to a service (say computing element). "Power" of a
> >>>> service is
> >>>> <= one. "Power" of agents <=1 - so sometimes more than one agent
> >>>> can
> >>>> occupy  a service , but their sum must by <= 1. Agents come in to a
> >>>> service following the rule FCFS (first come first serve) - it means
> >>>> tha tall agents must patiently waiting in the queue before
> >>>> execution.
> >>>>
> >>>> Have a beautiful week,
> >>>>
> >>>> Arek
> >>>>
> >>>>
> >>>>
> >>>> Andrzej Kozlowski napisal(a):
> >>>>> On 1 Sep 2006, at 23:41, Andrzej Kozlowski wrote:
> >>>>>
> >>>>>>
> >>>>>> On 1 Sep 2006, at 11:41, Andrzej Kozlowski wrote:
> >>>>>>
> >>>>>>>
> >>>>>>> On 31 Aug 2006, at 09:38, Arkadiusz Majka wrote:
> >>>>>>>
> >>>>>>>> Dear All,
> >>>>>>>>
> >>>>>>>> Thank all of you for your help regarding my "generalized fold
> >>>>>>>> list
> >>>>>>>> problem", that sounds
> >>>>>>>>
> >>>>>>>>>> I have two list
> >>>>>>>>>
> >>>>>>>>> list1=={a,b,c,d,e}
> >>>>>>>>> list2=={3,2,5,1,6}
> >>>>>>>>>
> >>>>>>>>> and I want to apply a modified version of FoldList to list1 in
> >>>>>>>>> the following way: list2 indicates that element a appears
> >>>>>>>>> only 3
> >>>>>>>>> times (if space enough) beginning from the beginning of the
> >>>>>>>>> list
> >>>>>>>>> , element b appears 2 times, c - 5 times , etc.
> >>>>>>>>>
> >>>>>>>>> So the output should be
> >>>>>>>>>
> >>>>>>>>> GeneralizedFoldList[list1,list2]=={a,a+b,a+b+c,c+d,c+e}
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>>
> >>>>>>>> Now, I still need your help. What I want to do now, and what I
> >>>>>>>> have
> >>>>>>>> problem with, is to add a constraint to the algorithm, i.e
> >>>>>>>> every
> >>>>>>>> element in my GeneralizedFoldList must be less than one.
> >>>>>>>> The following example says what to do if it is not.
> >>>>>>>>
> >>>>>>>> Lets take two lists
> >>>>>>>>
> >>>>>>>> list1=={0.9,0.8,0.7}
> >>>>>>>> list2=={3,3,3}
> >>>>>>>>
> >>>>>>>> All your algorithms use PadRight (you pad 0's). So the
> >>>>>>>> following
> >>>>>>>> matrix
> >>>>>>>> is built
> >>>>>>>>
> >>>>>>>> {{0.9, 0.9, 0.9, 0, 0}, {0, 0.8, 0.8, 0.8, 0}, {0, 0, 0.7, 0.7,
> >>>>>>>> 0.7}}
> >>>>>>>>
> >>>>>>>> and we add elements along colums and obtain {0.9, 1.7, 2.4,
> >>>>>>>> 1.5,
> >>>>>>>> 0.7}
> >>>>>>>>
> >>>>>>>> The first element is less than 1 so it's ok. The second is > 1
> >>>>>>>> what I
> >>>>>>>> need to avoid. I want to avoid it by shifting the nonzero
> >>>>>>>> elements of
> >>>>>>>> the second and third row of above matrix of two positions:
> >>>>>>>> {0,0,0,0.8,0.8,0.8,0}, {0,0,0,0,0.7,0.7,0.7}.
> >>>>>>>>
> >>>>>>>> I go on with suming along columns and discover that
> >>>>>>>> everything is
> >>>>>>>> fine
> >>>>>>>> until I have to add 0.8 and 0.7 what is >1. So I repeat the
> >>>>>>>> procedure
> >>>>>>>> by shfting hte third row of the number of position that is
> >>>>>>>> needed to
> >>>>>>>> ensure that my sum is <1.
> >>>>>>>>
> >>>>>>>> Finally I obtain {{0.9, 0.9, 0.9, 0, 0, 0, 0, 0,
> >>>>>>>> 0},{0,0,0,0.8,0.8,0.8,0,0,0},{0,0,0,0,0,0,0.7,0.7,0.7}}
> >>>>>>>> and Plus@@% is what I desire.
> >>>>>>>>
> >>>>>>>> Thanks in advance! I hope you will manage :)
> >>>>>>>>
> >>>>>>>> Arek
> >>>>>>>>
> >>>>>>>
> >>>>>>>
> >>>>>>> Here is a (not very pretty) code that should do what you want
> >>>>>>> starting with a matrix of the above kind:
> >>>>>>>
> >>>>>>>
> >>>>>>> GF[P_]:==Module[{M==P},Do[M == NestWhile[
> >>>>>>>       Join[(PadRight[#1, Length[#1] + 1] & ) /@
> >>>>>>>          Take[#1, s], (PadLeft[#1, Length[#1] + 1] & ) /@
> >>>>>>>          Take[#1, s - Dimensions[#1][[1]]]] & , M,
> >>>>>>>        !And @@ Thread[Total[#1[[All,Range[Length[
> >>>>>>>                Flatten[Split[#1[[s]]] /. {l___,
> >>>>>>>                    {(0)..}} :> {l}]]]]]] < 1] & ],
> >>>>>>>     {s, Dimensions[M][[1]]}]; M]
> >>>>>>>
> >>>>>>>
> >>>>>>> Let's verify that it works. First, starting with your original
> >>>>>>> case:
> >>>>>>>
> >>>>>>>
> >>>>>>> In[8]:==
> >>>>>>> M=={{0.9,0.9,0.9,0,0},{0,0.8,0.8,0.8,0},{0,0,0.7,0.7,0.7}};
> >>>>>>>
> >>>>>>>
> >>>>>>> In[9]:==
> >>>>>>> GF[M]//InputForm
> >>>>>>>
> >>>>>>> Out[9]//InputForm==
> >>>>>>> {{0.9, 0.9, 0.9, 0, 0, 0, 0, 0, 0},
> >>>>>>> {0, 0, 0, 0.8, 0.8, 0.8, 0, 0, 0},
> >>>>>>> {0, 0, 0, 0, 0, 0, 0.7, 0.7, 0.7}}
> >>>>>>>
> >>>>>>> Now let's change M somewhat:
> >>>>>>>
> >>>>>>> In[10]:==
> >>>>>>> M=={{0.9,0.9,0.1,0,0},{0,0.1,0.8,0.3,0},{0,0,0.7,0.9,0.7}};
> >>>>>>>
> >>>>>>>
> >>>>>>> In[11]:==
> >>>>>>> GF[M]//InputForm
> >>>>>>>
> >>>>>>> Out[11]//InputForm==
> >>>>>>> {{0.9, 0.9, 0.1, 0, 0, 0, 0, 0}, {0, 0, 0.1, 0.8, 0.3, 0,
> >>>>>>>    0, 0}, {0, 0, 0, 0, 0, 0.7, 0.9, 0.7}}
> >>>>>>>
> >>>>>>> So you only need to combine this with your favourite code from
> >>>>>>> part 1.
> >>>>>>>
> >>>>>>> (PS. In answering part 1 I obviously misunderstood your
> >>>>>>> question, but
> >>>>>>> I think that was mainly because I do not think your function
> >>>>>>> is in
> >>>>>>> any way a "generalisation of Fold", or even of Fold
> >>>>>>> [Plus,element,list]. As far as I can see it does not really
> >>>>>>> perform
> >>>>>>> any kind of "folding".
> >>>>>>>
> >>>>>>> Andrzej Kozlowski
> >>>>>>>
> >>>>>>
> >>>>>>
> >>>>>> In fact I still do not fully understand your original question
> >>>>>> and,
> >>>>>> in particular, if the example you originally posted (and
> >>>>>> which, if
> >>>>>> any, of the solutions various people provided) were correct.
> >>>>>> Just in
> >>>>>> case, here is a "solution" to your original question, which I
> >>>>>> think
> >>>>>> is different from any of the ones I have seen:
> >>>>>>
> >>>>>> GFMatrix[list1_,
> >>>>>>      list2_] :== Normal[SparseArray[Flatten[Table[If[j =B2 i +
> >>>>>> list2
> >>>>>> [[i]] - 1,
> >>>>>> Rule[{i, j}, list1[[i]]], Rule[{i,
> >>>>>>             j}, 0]], {i, Length[list2]}, {j, i, Length[list1] +
> >>>>>> i -
> >>>>>> 1}], 1]]]
> >>>>>>
> >>>>>> GFMatrix builds the matrix, which, I think, you need in the
> >>>>>> second
> >>>>>> part of the question. The code makes no use of PadLeft or
> >>>>>> PadRight.
> >>>>>> However, note that in your original example, the output is not
> >>>>>> the
> >>>>>> same as you gave as the correct one, although the output below
> >>>>>> seems
> >>>>>> to me to fit better your verbal description of what you wanted
> >>>>>> done:
> >>>>>>
> >>>>>>
> >>>>>> list1=={a,b,c,d,e};
> >>>>>> list2=={3,2,5,1,6};
> >>>>>>
> >>>>>>
> >>>>>> GFMatrix[list1,list2]
> >>>>>>
> >>>>>>
> >>>>>> {{a, a, a, 0, 0, 0, 0, 0, 0},
> >>>>>>    {0, b, b, 0, 0, 0, 0, 0,
> >>>>>>     0}, {0, 0, c, c, c, c, c,
> >>>>>>     0, 0}, {0, 0, 0, d, 0, 0,
> >>>>>>     0, 0, 0}, {0, 0, 0, 0, e,
> >>>>>>     e, e, e, e}}
> >>>>>>
> >>>>>>
> >>>>>> Total[%]
> >>>>>>
> >>>>>>
> >>>>>> {a,a+b,a+b+c,c+d,c+e,c+e,c+e,e,e}
> >>>>>>
> >>>>>> This is longer than the output you gave as the right solution,
> >>>>>> which
> >>>>>> was only a part of the above, namely {a,a+b,a+b+c,c+d,c+e}.
> >>>>>>
> >>>>>> Andrzej Kozlowski
> >>>>>>
> >>>>>>
> >>>>>
> >>>>> Oops, the code I copied and pasted in did not come out the way it
> >>>>> should have done. It shoudlhave been:
> >>>>>
> >>>>> GFMatrix[list1_, list2_] := Normal[SparseArray[Flatten[Table[If
> >>>>> [j <=
> >>>>> i + list2[[i]] - 1, {i, j} -> list1[[i]], {i, j} -> 0], {i, Length
> >>>>> [list2]}, {j, i, Length[list1] + i - 1}], 1]]]
> >>>>>
> >>>>> Andrzej Kozlowski
> >>>>
> >>


  • Prev by Date: Re: StylePrint and 2D strings
  • Next by Date: Re: Boolean algebra
  • Previous by thread: Re: Re: generalized foldlist problem - part 2
  • Next by thread: the ellipse and the circle