(setq lst1 '((A 6) (B 1) (C 2) (D 1) (E 4)))
(setq lst2 '((A 3) (B 1) (C 1) (D 1) (E 3)))
> ((A 3) nil c nil E)
(setq lst1 '((A 6) (B 1) (C 2) (D 2) (E 4)))
(setq lst2 '((A 9) (B 2) (C 2) (D 1) (E 5))) > ((A 3) (B 1) (C 0) nil (E 1))
(nil '((A 9) (B 2) (C 2) (D 1) (E 5)))
(setq lst1 '((A 6) (B 1) (C 2) (D 2) (E 4)))
(setq lst2 '((A 15) (B 3) (C 4) (D 5) (E 9))) > ((A 3) (B 1) (C 0) (D 1) (E 1))
(('((A 6) (B 1) (C 2) (D 2) (E 4)) 2) ' ((A 3) (B 1) (C 0) (D 1) (E 1)))
Simultaneous existence,No nil
A 15-2*6=3
B 3-2*1=1
C 4-2*2=0
D 5-2*2=1
E 9-2*4=1
A 15-2*6=3
B 3-2*1=1
C 4-2*2=0
D 5-2*2=1
E 9-2*4=1
I think I see now but how do you then decode the result? And what does Run Length Encoding have to do with this operation?
How does: ((A 3) (B 1) (C 0) (D 1) (E 1))
get back to: ((A 6) (B 1) (C 2) (D 2) (E 4))
or: ((A 15) (B 3) (C 4) (D 5) (E 9))