Re: Re: simple iteration question-thanks

*To*: mathgroup at smc.vnet.net*Subject*: [mg89989] Re: [mg89928] Re: simple iteration question-thanks*From*: Francisco Gutierrez <fgutiers2002 at yahoo.com>*Date*: Thu, 26 Jun 2008 04:39:28 -0400 (EDT)*Reply-to*: fgutiers2002 at yahoo.com

Many thanks to Don, Jean Marc, and Oleksandr. Their ideas were very useful to solve my problem. Indeed, it implied using Fors and Ifs. I wonder if it is possible to express all this functionally and not procedurally; I couldn't find a way to do it. Best, Fg --- On Tue, 6/24/08, sashap <pavlyk at gmail.com> wrote: From: sashap <pavlyk at gmail.com> Subject: [mg89989] [mg89928] Re: simple iteration question To: mathgroup at smc.vnet.net Date: Tuesday, June 24, 2008, 5:17 AM On Jun 20, 5:16 am, Francisco Gutierrez <fgutiers2... at yahoo.com> wrote: > Dear Friends: > I have a list of the form list={{3,2,1},{5,6,1},{10,5,1}}, of any length > I have a vector, say vector={1,2,3} > I want to make the following iteration. Take the first component of the > list, make the dot product of it and the vector, and substract a number ( sa > y 50). If the result is bigger or equal than zero, stop and return the vect > or. If it is less than zero, then change the vector in the followig way: = = ve= > ctor+list[[2]], and repeat the test. > Iterate until the result is equal or bigger than zero, or until the list = = en= > ds. > I haven't found a neat way of doing this. Any help is welcome > Francisco > Hey Francisco, here is a short way of doing it: In[59]:= mat = {{3, 2, 1}, {5, 6, 1}, {10, 5, 1}}; vec = {1, 2, 3}; In[61]:= Catch[ Fold[If[#1.#2 > 50, Throw[#1], #1 + mat[[2]]] &, vec, mat]] Out[61]= {6, 8, 4} Oleksandr Pavlyk=0A=0A=0A