Re: Re: 2D pattern matching

*To*: mathgroup at smc.vnet.net*Subject*: [mg77437] Re: [mg77264] Re: [mg77140] 2D pattern matching*From*: János <janos.lobb at yale.edu>*Date*: Fri, 8 Jun 2007 05:41:02 -0400 (EDT)*References*: <200706051024.GAA00465@smc.vnet.net> <3789605.1181136409894.JavaMail.root@m35> <op.ttilxgo2qu6oor@monster.ma.dl.cox.net>

Of course if he is changing that 2x2 with a nxm where either n or m is not 2, then of course he will be in trouble if he wants to do the change for every 2.2 at once. As a newbie I had just 2x2-s in my mind :) Thanks, J=E1nos P.S. What has happened to DrBob ? Did he go to the Army ? :) On Jun 6, 2007, at 3:33 PM, DrMajorBob wrote: > That solution indicates, initially, the need for an inverse to > Partition[x_,{2,2},{1,1}] and similar things. > > After changing one of the 2x2's, though, the "blown-up" matrix > isn't in the domain of that inverse. > > Bobby > > On Wed, 06 Jun 2007 06:03:31 -0500, J=E1nos <janos.lobb at yale.edu> = wrote: > >> >> On Jun 5, 2007, at 6:24 AM, alexxx.magni at gmail.com wrote: >> >>> hi everybody, >>> >>> do you know how to do a pattern match involving 2D matrices? >>> If e.g. you have a list >>> >>> a = {{0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 1, 0}, = {0, >>> 0, 0, 1, 1, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}} >>> >>> i.e. >>> 0 0 0 0 0 0 >>> 0 1 0 0 0 0 >>> 0 1 0 1 1 0 >>> 0 0 0 1 1 0 >>> 0 0 0 0 0 0 >>> >>> is it possible for me to write a rule by which I am able to find the >>> 2x2 square of 1's, and replace it with something else? >>> >>> I studied all I could on patterns, but didnt find an answer... >>> >>> thanks for ANY help >>> >>> Alessandro Magni >>> >> >> Here is a newbie bit-blow-up approach: >> >> In[2]:= >> b = Partition[a, {2, 2}, >> {1, 1}] >> Out[2]= >> {{{{0, 0}, {0, 1}}, >> {{0, 0}, {1, 0}}, >> {{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}}, >> {{{0, 1}, {0, 1}}, >> {{1, 0}, {1, 0}}, >> {{0, 0}, {0, 1}}, >> {{0, 0}, {1, 1}}, >> {{0, 0}, {1, 0}}}, >> {{{0, 1}, {0, 0}}, >> {{1, 0}, {0, 0}}, >> {{0, 1}, {0, 1}}, >> {{1, 1}, {1, 1}}, >> {{1, 0}, {1, 0}}}, >> {{{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}, >> {{0, 1}, {0, 0}}, >> {{1, 1}, {0, 0}}, >> {{1, 0}, {0, 0}}}, >> {{{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}, >> {{0, 0}, {0, 0}}}} >> >> /If you do a MatrixForm on b you will able to see where that block >> is and how you should get the staring coordinates:)/ >> >> Then >> >> In[4]:= >> Position[b, {{1, 1}, {1, 1}}] >> Out[4]= >> {{3, 4}} >> >> gives the upper left position of the block in b and that is the same >> as in a. /I leave the proof of it to the experts :)/ >> >> Replacing now in a is easy because you know how many cells in right >> and down do you want to replace from the found starting position. >> >> >> J=E1nos >> >> >> --------------------------------------------- >> "Its like giving a glass of ice water to somebody in Hell" >> Steve Jobs about iTunes on Windows >> >> >> >> > > > > -- > DrMajorBob at bigfoot.com

**References**:**2D pattern matching***From:*"alexxx.magni@gmail.com" <alexxx.magni@gmail.com>