Re: A programming puzzle.
- To: mathgroup at smc.vnet.net
- Subject: [mg60853] Re: [mg60810] A programming puzzle.
- From: <bsyehuda at gmail.com>
- Date: Fri, 30 Sep 2005 03:57:31 -0400 (EDT)
- References: <200509290941.FAA01111@smc.vnet.net>
- Reply-to: bsyehuda at gmail.com
- Sender: owner-wri-mathgroup at wolfram.com
I do not know what you call an "Elegant solution", since if you already have
a way of doing this, why would you need another one?
Anyway
s[n_Integer]:=Sum[a[i]b[i],{i,n}]
generates the input sum
and
makeMyList[x_] := (Plus @@@ Rest[FoldList[Flatten[{List[#1, #2]}] &, {}, \
#[[1]]]].Subtract @@@ Partition[#[[2]], 2, 1, {
1, 1}, 0]) &@(x /. Times -> List /. Plus -> List)
will work on it to give you the desired result
that is
makeMyList[s[5]]
will produce
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]
and check by Simplify[%]
and get back
a[1] b[1] + a[2] b[2] + a[3] b[3] + a[4] b[4] + a[5] b[5]
I do not find it elegant, do you?
yehuda
On 9/29/05, jackgoldberg at comcast.net <jackgoldberg at comcast.net> wrote:
>
> 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
>
>
- References:
- A programming puzzle.
- From: jackgoldberg@comcast.net
- A programming puzzle.