Re: optimization

*To*: mathgroup at smc.vnet.net*Subject*: [mg96534] Re: [mg96523] optimization*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Mon, 16 Feb 2009 06:54:12 -0500 (EST)*Reply-to*: hanlonr at cox.net

A simplification u = Array[a, {3, 2}]; MapThread[Equal, {Map[Apply[Plus, #] &, u], Table[1, {Length[u]}]}] == Thread[Total /@ u == 1] True Another simplification largo = Flatten[u]; MapThread[Greater, {largo, Table[0, {Length[largo]}]}] == Thread[largo > 0] True Fix error Flatten[Thread[Total /@ u == 1]; Thread[largo > 0]] {a(1,1)>0,a(1,2)>0,a(2,1)>0,a(2,2)>0,a(3,1)>0,a(3,2)>0} Note that the semi-colon suppresses all of the first expression (i.e., the constraints that the rows sum to 1). You intended Join[Thread[Total /@ u == 1], Thread[largo > 0]] Nested tables can be written as a single Table Table[Table[a[k, i], {k, 1, 5}], {i, 1, 3}] == Table[a[k, i], {i, 3}, {k, 5}] True You don't need to Flatten prior to using Total Total[Flatten[u]] == Total[u, 2] True You have a circular definition for u (u defined in terms of u) paso2[individuo_, centro_, m_] := Module[ {largo, a, u}, u = Array[a, {Length[individuo], Length[centro]}]; largo = Flatten[u]; NMinimize[{Total[ Table[ EuclideanDistance[individuo[[k]], centro[[i]]]*u[[k, i]]^m, {i, Length[centro]}, {k, Length[individuo]}], 2], Join[Thread[Total /@ u == 1], Thread[largo > 0]]}, largo]] Bob Hanlon ---- Francisco Gutierrez <fgutiers2002 at yahoo.com> wrote: ============= Dear Friends:I have the following optimization function: paso2[individuo_,centro_,m_]:=Module[{largo,u=Array[u,{Length[individuo],Length[centro]}]},largo=Flatten[u];NMinimize[{Total[Flatten[Table[Table[EuclideanDistance[individuo[[k]],centro[[i]]]*u[[k,i]]^m,{k,1,Length[individuo]}],{i,1,Length[centro]}]]], Flatten[MapThread[Equal,{Map[Apply[Plus,#]&,u],Table[1,{Length[u]}]}];MapThread[Greater,{largo,Table[0,{Length[largo]}]}]]},largo]] in Mathematica 7. However, this function is obviously not capturing the restriction that u[i,j]+u[i,k]==1For example, u[1,1]+u[1,2] should sum up to 1this restriction corresponding to the following piece of code: Flatten[MapThread[Equal,{Map[Apply[Plus,#]&,u],Table[1,{Length[u]}]}] How can I solve this?In general, what I am doing wrong so as to not repeat the error?ThanksFrancisco Guti=E9rrez --- On Thu, 2/12/09, Albert Retey <awnl at gmx-topmail.de> wrote: From: Albert Retey <awnl at gmx-topmail.de> Subject: [mg96534] [mg96523] [mg96389] Re: Reposted, Reformatted Re: "mapping" functions over l= ists, again! To: mathgroup at smc.vnet.net Date: Thursday, February 12, 2009, 6:42 AM Hi, > This is a repost of an earlier post, as there were problems with the emai= l > formatting for some people. I have composed this one in Notepad, and cut = and > pasted to Outlook. Hope it works better. it does work better :-) > This is a list of lists, the lowest level lists containing pairs of {real= , > complex}. The individual lists are not all the same length, and the total > number of lists can vary, and I need to preserve the list structure. > > So I think the most succinct way of expressing my problem is, what form d= oes > fxn take if I want to Map it across my lists of {real,complex} so that it > returns {fxn1[real],fxn2[complex]} or even {real,fxn[complex]}? If I understand correctly what you try to achieve I think the easiest solution will be to use replacement only, it will preserve the structure and can be set up to handle the pairs as you want. Basically I see two approaches, which also can be combined. The first one looks simpler, but only works if the structure of your data doesn't change (the sizes of the list of list can of course be arbitrary): Replace[shortTestdata, {r_, c_} :> {f1[r], f2[c]}, {2}] the trick is to restrict the replacement to the correct level. If you are working with data where the level at which the pairs appear can be arbitrary, you can pick the parts to be treated with a more elaborated rule, like this: shortTestdata /. {r_ /; Element[r, Reals],c_ /; Element[c, Complexes]}:> {f1[r], f2[c]} (/. is short for ReplaceAll). This version should work with arbitrary structured lists of arbitrary depths. I think using the Element-Function to decide what kind of number you are looking at is in your case closer to what you want to achieve than to look at the Head, which has some implications, as you have learned from other posts. Finally you could combine the two approaches: Replace[shortTestdata, {r_ /; Element[r, Reals],c_ /; Element[c, Complexes]}:> {f1[r], f2[c]}, {2}] this could make sense if you have a well structured list, but the pairs could be either of the form {reals,reals}, {real,complex}, {complex,complex} and you want to treat these cases differently. You should be aware that other approaches might be faster, but these will for sure be more elaborate and error prone than the code above. hth, albert

**Re: Shortest Path Problem**

**Stirling 1st problem in Infinite sums**

**optimization**

**Re: Re: optimization**