Concrete mathematics : a foundation for computer science

312 GENERATING FUNCTIONS
Now we have three equations in three unknowns (U, V, and A). We can solve
them by first solving for V and A in terms of U, then plugging the results
into the equation for U:
v = (I - Q)-ml, A = (I-g)-‘ou;
u = I + B(l-B,)-‘ml + B(I- gyou + pJu
And the final equation can be solved for U, giving the compact formula
u = 1 - B(l-@)-‘[I -
I
B(I-gJ-‘o - R’
(7.8)
This expression defines the infinite sum U, just as (7.4) defines T.
The next step is to go commutative. Everything simplifies beautifully
when we detach all the dominoes and use only powers of II and =:
u =
1
1 - O&(1 - ,3)-~’ - Po(l - ,3)-l - ,3
l-o3
= (I- ,3)2 -20%;
(1 - c33)-’
-
= l-202o(1
- &:I+
1 2020 404 02
=m+
80603 ~-
(1 - ,3)3 + (1 - ,3)5 + (1 - ,3)7 +...
= t (m;2k)2’.,,2kak+h.
k,m>O
(This derivation deserves careful scrutiny. The last step uses the formula
(1 - ,)-2k--1 = Em (m+mZk)Wm, identity (5.56).) Let’s take a good look at
the bottom line to see what it tells us. First, it says that every 3 x n tiling
uses an even number of vertical dominoes. Moreover, if there are 2k verticals,
there must be at least k horizontals, and the total number of horizontals must
be k + 3m for some m 3 0. Finally, the number of possible tilings with 2k
verticals and k + 3m horizontals is exactly (“i2k)2k.
We now are able to analyze the 3 x 4 tilings that left us doubtful when we
began looking at the 3 x n problem. When n = 4 the total area is 12, so we
need six dominoes altogether. There are 2k verticals and k + 3m horizontals,
I /earned in another
class about “regular
expressions.” If I’m
not mistaken, we
can write
u = (LB,*0
+BR*o+H)*
in the language of
regular expressions;
so there must be
some connection
between regular
expressions and generating
functions.