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MathGroup Archive 2005

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Re: Set of strings reducing problem (Again!)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg60373] Re: [mg60339] Set of strings reducing problem (Again!)
  • From: "Edson Ferreira" <edsferr at uol.com.br>
  • Date: Wed, 14 Sep 2005 03:27:46 -0400 (EDT)
  • References: <200509131007.GAA09839@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Well,

It was easier than I thought it would be!!

Instead of:

ncl = StringJoin @@@ (
cl //. {x___List,
{a___, p_String, c___},
{a___, q_String, c___},
y___List} :>
{x, {a, p + q /. r, c},
y} /; StringQ[p + q /.
r])

I put:

ncl = StringJoin @@@ (
cl //. {x___List,
{a___, p_String, c___},
z___List,
{a___, q_String, c___},
y___List} :>
{x, {a, p + q /. r, c}, z, y} /; StringQ[p + q /. r])

It worked flawlessly!!!!!! Perfect!

Now I see that Pattern Matching is a VERY powerful tool in Mathematica!

Thanks folks!!!

Edson Ferreira
Mechanical Engineer
São Paulo - Brazil

----- Original Message ----- 
From: "Edson Ferreira" <edsferr at uol.com.br>
To: mathgroup at smc.vnet.net
Subject: [mg60373] [mg60339] Set of strings reducing problem (Again!)


>
> Dear Math Gurus,
>
> Sometime ago I have posted a problem here and got some very clever
> solutions which I would like to thank.
>
> Currently I'm working with one solution I received, but It didn't make
> the "complete" job. The strange behaviour is that for some cases it
> works the way I want.
>
> Let's post the problem (in italic) again (hope you remember):
>
> Let's say a have a list of "n" equal length strings:
>
> L={"11111111",
>   "11112111",
>   "1111X111",
>       ...
>       ...
>       ...
>   "21122211"}
>
> The characters used in strings are only "1", "X", "2", "U", "M", "D" and
>
> "T".
>
> What I want is a reduced set of strings (with all the resulting strings
> with the same length as all the original ones).
>
> The rule to "join" two strings is the following:
>
> If one string is different from the other by just one character then
> take the characters that are different and apply the rule bellow:
>
> "1" + "X" = "D"
> "1" + "2" = "M"
> "1" + "U" = "T"
> "X" + "2" = "U"
> "X" + "M" = "T"
> "2" + "D" = "T"
>
> For example, suppose I have these two elements in the list : "11112111"
> and "1111X111"
>
> The rule will transform these two strings into one : "1111U111"
>
> After all the possible transformations (always using two strings with
> only one different character and resulting another string) I will obtain
> a reduced set of strings.
>
> How can I do that with mathematica??
>
> I guess the first step is a function to identify is two strings are
> different by just one ccharacter.
> A loop then search in the set for any ocurrences of that and apply all
> possible transformations until we can't get any redution.
>
> For instance : {"T11TTTTT"} would be generated from a set with 729
> unique strings !!!!!
>
> Some Observations:
>
> - The strings are all unique.
> - The different character, can ocurr at any position.
> - The length of all strings is the same.
> -All I want is a reduced set of string, thus when you compare two
> strings and you find they differ by only one character, you ELIMINATE
> the original ones and ADD the NEW one to the original set.
> -The rules I gave are the only ones.
> -It's impossible to encounter something like "1111M111" + "11112111"
> -The two strings that will be "joined" have only one different character
> at same position!
> -You can't join "21111111" vs. "11111112" (characters at different
> positions)
>
> The clever solution I received for a set L is the following:
>
> L = {"111", "11X", "112",
>    "1X1", "1XX", "1X2", "121",
>    "12X", "122"};
>
> cl = Characters /@ L;
>
> r = Dispatch[{"1" + "X" -> "D",
>     "1" + "2" -> "M",
>     "1" + "U" -> "T",
>     "X" + "2" -> "U",
>     "X" + "M" -> "T",
>     "2" + "D" -> "T"}];
>
> ncl = StringJoin @@@(
>   cl //. {x___List,
>      {a___, p_String, c___},
>      {a___, q_String, c___},
>      y___List} :>
>     {x, {a, p + q /. r, c},
>       y} /; StringQ[p + q /.
>        r])
>
> Which evaluates {"1TT"}
>
> Now finally I can explain the problem (see the code bellow):
>
> In[1]:=
> Unprotect[D];
> In[2]:=
> U={"2","X"};
> In[3]:=
> M={"1","2"};
> In[4]:=
> D={"1","X"};
> In[5]:=
> T={"1","2","X"};
> In[6]:=
> L=Flatten[Outer[StringJoin,T,T,T,D]];
> In[7]:=
> L = Select[L, Count[Characters[#], "1"] > 1 &];
> In[8]:=
> cl = Characters /@ L;
> In[9]:=
> r = Dispatch[{"1" + "X" -> "D",
> "1" + "2" -> "M",
> "1" + "U" -> "T",
> "X" + "2" -> "U",
> "X" + "M" -> "T",
> "2" + "D" -> "T"}];
> In[10]:=
> ncl = StringJoin @@@ (
> cl //. {x___List,
> {a___, p_String, c___},
> {a___, q_String, c___},
> y___List} :>
> {x, {a, p + q /. r, c},
> y} /; StringQ[p + q /.
> r])
> Out[10]=
> {11TD, 121D, 12U1, 1X1D, 1XU1, 211D, 21U1, 2U11, X11D, X1U1, XU11}
>
>
> If look at some pairs of positions in the result you see that some
> combinations were not done: 2+4, 3+5, 6+9, 7+10 and 8+11
>
> The correct result is: {11TD,1U1D,1UU1,U11D,U1U1,UU11}
>
> Now, finally, the question:
>
> Why this code does not perform all possible combinations???
>
> It looks, as Stephen Layland once said, that this code will only find
> sets of strings that differ by one character that are
> right next to each other...
>
> Any solution????
>
> Sorry for the looooooooooong history...
>
> Thanks!!!!!!!!!
>
> Edson
>
>
> 


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