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# Pattern matching more than once
*To*: mathgroup@smc.vnet.net
*Subject*: [mg11991] Pattern matching more than once
*From*: asari@math.uiuc.edu (ASARI Hirotsugu)
*Date*: Fri, 17 Apr 1998 03:40:27 -0400
*Organization*: University of Illinois at Urbana-Champaign
I have been stuck with the following problem for about 10 days. I can't
think of an elegant solution.
Input: Two lists, from some universal set Output: Many "spliced" lists
obtained from the input.
e.g. {{a,b,c},{d,e,f}} --> {{a,b,c},{d,e,f}} (no common element)
{{a,b,c},{d,b,f}} --> {{a,b,c},{d,b,f},{a,b,f},{d,b,c}}
{{a,b,c,d,e},{f,b,g,d,h}} -->
{{a,b,c,d,e},{a,b,c,d,h},{a,b,g,d,e},{a,b,g,d,h},
{f,b,c,d,e},{f,b,c,d,h},{f,b,g,d,e},{f,b,g,d,h}}
I will regard the input as list of chains (totally ordered set) coming
from a partially ordered set, and I would like to extend these chains
as much as possible. If there is at most one common element in the
inputs, the following will do:
extendList[{list1:{pre1___,x_,post1___},list2:{pre2___,x_,post2___}}]:=
{list1,list2,{pre1,x,post2},{pre2,x,post1}};
The problem is that when the lists have more than one common element,
the second common element will not be considered at all. I suppose I
could just write, for instance,
extendList2[{list1:{pre1___,x_,mid1___,y_,post1___},
list2:{pre2___,x_,mid2___,y_,post2___}}]:=
Union[
{list1,{pre1,x,mid1,y,post2},{pre1,x,mid2,y,post1},
{pre1,x,mid2,y,post2},list2,{pre2,x,mid1,y,post1},
{pre2,x,mid1,y,post2},{pre2,x,mid2,y,post1}}]; But this is very ugly.
I suppose I could write some For[] loop through
Intersection[list1,list2], but I would rather avoid it if I could.
Any help would be appreciated.
--
ASARI Hirotsugu // http://www.math.uiuc.edu/~asari/
finger://math.uiuc.edu/asari // ph://ns.uiuc.edu/asari
"We are what we pretend to be, so we must be careful
about what we pretend to be." --Kurt Vonnegut
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