Preparing frontier information for each node, and applying to path counting in undirected graph

[Johnny_Rotten]

What you describe above, is it the basis for your recent result:

@kuniaki.mukai

% ?- time((between(2, 13, N), findall(I-J, (between(1, N, I), K is I + 1, between(K, N, J)), E),
%	udg_path_count(1-N, E, C), writeln(count_path_of_K(N)=C), fail)).
%@ count_path_of_K(2)=1
%@ count_path_of_K(3)=2
%@ count_path_of_K(4)=5
%@ count_path_of_K(5)=16
%@ count_path_of_K(6)=65
%@ count_path_of_K(7)=326
%@ count_path_of_K(8)=1957
%@ count_path_of_K(9)=13700
%@ count_path_of_K(10)=109601
%@ count_path_of_K(11)=986410
%@ count_path_of_K(12)=9864101
%@ count_path_of_K(13)=108505112
%@ % 1,364,231,392 inferences, 61.385 CPU in 63.662 seconds (96% CPU, 22224211 Lips)
%@ false.

Can it be done with tabling? Or do you throw away certain
non active frontier information in the course of your counting?
I have the feeling tabling is not suited for the best algorithms,

since it lacks a form of automatic forgetting. I am currently
facinated by this forgetting, it has a also a link to streaming,
depending on how much I am allowed to forget, I can replace

two streams by one stream with a sliding window. I can show
that easily for the Pascal triangle example. In the Pascal
triangle the window size is W = 2. The size can be derived

from the constraint C =:= C2+1 here:

@Johnny_Rotten

% rule(+Term, +Term, -Term)
rule(p(R2, C2, M1), p(R2, C, M2), p(R, C, M)) :-
   R is R2+1,
   C =:= C2+1,
   M is M1+M2.