RE: RE: Re: doing things on a procedural way and doing them on a functional way
- To: mathgroup at smc.vnet.net
- Subject: [mg47010] RE: [mg46991] RE: [mg46988] Re: doing things on a procedural way and doing them on a functional way
- From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
- Date: Sat, 20 Mar 2004 03:50:22 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
A remark to add... >-----Original Message----- >From: Wolf, Hartmut To: mathgroup at smc.vnet.net >Sent: Friday, March 19, 2004 7:36 AM >To: mathgroup at smc.vnet.net >Subject: [mg47010] [mg46991] RE: [mg46988] Re: doing things on a procedural way >and doing them on a functional way > > > >>-----Original Message----- >>From: Paul Abbott [mailto:paul at physics.uwa.edu.au] To: mathgroup at smc.vnet.net >To: mathgroup at smc.vnet.net >>Sent: Thursday, March 18, 2004 10:38 AM >>To: mathgroup at smc.vnet.net >>Subject: [mg47010] [mg46991] [mg46988] Re: doing things on a procedural >>way and doing them on a functional way >> [...] >> >>I can't help but feel that there should be an elegant way to do this >>using Distribute ... >> >>Cheers, >>Paul >> [...] > >You're completely right, >the solution, I published yesterday, also works with >distribute (instead of Outer): > >In[32]:= nocnac[{arg_}, op_] := {arg} >In[33]:= >nocnac[{args__}, op_] := > Flatten[ReplaceList[{args}, {a__, b__} :> > Distribute[op[nocnac[{a}, op], nocnac[{b}, op]], List]], 1] > > [...] Effectively both solutions are the same, as for Distribute you always can find an equivalent expression with Outer, e.g.: In[164]:= Distribute[List[{1, 2}, {{3, 4}, 5}, 6], List] Out[164]= {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}} ...this is by definition the same as In[170]:= Distribute[List[{1, 2}, {{3, 4}, 5}, {6}], List] Out[170]= {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}} ...functionally equivalent to In[166]:= Flatten[Outer[List, {1, 2}, {{3, 4}, 5}, {6}, 1], 2] Out[166]= {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}} You may have observed that my first solution using Outer doesn't work with List, whereas that one above with Distribute does. This is not a problem of Outer, I just have been a bit careless with levels (at Flatten and Outer). This code also works with List: In[143]:= nocnac[{arg_}, op_] := {arg} nocnac[{args__}, op_] := Flatten[ReplaceList[{args}, {a__, b__} :> Outer[op, nocnac[{a}, op], nocnac[{b}, op], 1] ], 2] > > >In[39]:= Length[nocnac[Range[#], op]] & /@ Range[10] >Out[39]= >{1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862} > > [...] > >I don't know, but assume that this can be expressed simply >through some sort of combinatorical numbers. Yes, indeed, it is given by In[171]:= Binomial[2 #, #]/(# + 1) & /@ Range[0, 9] Out[171]= {1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862} ...and the numbers c[n_] := Binomial[2n, n]/(n + 1) are called the Catalan numbers. See e.g. http://www.research.att.com/~njas/sequences/ and my observation... > >In[57]:= nc[#]/nc[# - 1] & /@ Range[2, 30] // N >Out[57]= >{1., 2., 2.5, 2.8, 3., 3.14286, 3.25, 3.33333, 3.4, > 3.45455, 3.5, 3.53846, 3.57143, 3.6, 3.625, 3.64706, > 3.66667, 3.68421, 3.7, 3.71429, 3.72727, 3.73913, > 3.75, 3.76, 3.76923, 3.77778, 3.78571, 3.7931, 3.8} > [...] > >The conjecture of course is, that asymtotically nc[n] ~ 4^n. > is nothing else than the simple fact In[182]:= Limit[2n (2n - 1)/(n (n + 1)), n -> Infinity] Out[182]= 4 -- Hartmut Wolf