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MathGroup Archive 2005

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Re: A programming puzzle.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg60873] Re: A programming puzzle.
  • From: Peter Pein <petsie at dordos.net>
  • Date: Sat, 1 Oct 2005 02:55:48 -0400 (EDT)
  • References: <dhge6k$1kg$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

jackgoldberg at comcast.net schrieb:
> Hello everyone,
> 
> I have a simple problem for which I would like an "elegant" solution.  The problem is to convert the series  
> 
> a1*b1 + a2*b2 + a3*b3 + ... + an*bn
> 
> into the equivalent series 
> 
> a1(b1 - b2) + (a1+a2)(b2 - b3) + ... +(a1 + a2 + ... + an-1)(bn-1 - bn) + 
>          (a1 + a2 + ... + an)bn
> 
> This process, I believe, is called summation-by-parts.  This problem is not hard to do; one simply separates the ai's from the bi's, constructs the "partial" sums for the  ai  series and the differences for the  bi series.  Then a dot product gets the answer.  I am interested in a more elegant solution, if one exists. 
> Taking care of all special cases will probably make any solution rather inelegant, so I am taking the liberty granted to all posers of only allowing series which have at least two terms AND the bi's are not constants and  bi = bj is prohibited unless i = j.  (This is, in fact, the situation I am concerned with.)
> 
> I think some of you will have fun connocting ingenious solutions ...  Let me know if you find any.  Incidentally, timing is not crucial in my applications of "summation-by-parts" nor is storage of intermediate computations a problem.  Thanks!
> 
> Jack Goldberg
> 

Another approach is to expand the series of the product around b1=b2,
b2=b3 and so on (it's ugly, bcause I forced it into one (syntactic) line):

In[1]:=
sumbyparts[pr_]:=
  ((Plus@@(Last/@#)-pr)+Factor[#[[-1,1]]])&@
    FoldList[(Take[Series[#1[[1]],Append[#2,1]][[3]],2]*{1,
              Subtract@@#2})&,{pr},
      Transpose[Drop[(List@@pr)/.x_*y_\[RuleDelayed]y,#]&/@{-1,1}]]

In[2]:= test5=(a/@#).(b/@#)&@Range[5];

In[3]:= sumbyparts[test5]//InputForm
Out[3]//InputForm=
a[1]*(b[1] - b[2]) +
(a[1] + a[2])*(b[2] - b[3]) +
(a[1] + a[2] + a[3])*(b[3] - b[4]) +
(a[1] + a[2] + a[3] + a[4])*(b[4] - b[5]) +
(a[1] + a[2] + a[3] + a[4] + a[5])*b[5]

In[4]:= %//Expand//InputForm
Out[4]//InputForm=
a[1]*b[1] + a[2]*b[2] + a[3]*b[3] + a[4]*b[4] + a[5]*b[5]

In[5]:= sumbyparts[3x+2y+c z]//InputForm
Out[5]//InputForm= 3*(x - y) + 5*(y - z) + (5 + c)*z

In[6]:= %//Expand//InputForm
Out[6]//InputForm= 3*x + 2*y + c*z

Tricky? Maybe... Elegant? NO.

Peter


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