Concrete mathematics : a foundation for computer science

7.3 SOLVING RECURRENCES 329
And with chutzpah and luck we might even have been able to smoke out
the constant $. But it sure is simpler and more reliable to have generating
functions as a tool.
Example 3: Mutually recursive sequences.
Sometimes we have two or more recurrences that depend on each other.
Then we can form generating functions for both of them, and solve both by
a simple extension of our four-step method.
For example, let’s return to the problem of 3 x n domino tilings that we
explored earlier this’ chapter. If we want to know only the total number of
ways, Ll,, to cover a 3 x n rectangle with dominoes, without breaking this
number down into vertical dominoes versus horizontal dominoes, we needn’t
go into as much detail as we did before. We can merely set up the recurrences
uo = 1 , Ul =o; vo = 0, v, =l;
u, =2v,-, fl.lnp2, vn = LLl + vn4 ) for n 3 2.
Here V, is the number of ways to cover a 3 x n rectangle-minus-corner, using
(3n - 1)/2 dominoes. These recurrences are easy to discover, if we consider
the possible domino configurations at the rectangle’s left edge, as before. Here
are the values of U, and V,, for small n:
nlO1234 5 6 7
\ ,r
i \ (7.34)
Let’s find closed forms, in four steps. First (Step l), we have
U, = 2V,-1 + U-2 + [n=Ol , vll = b-1 +v,-2,
for all n. Hence (Step 2),
U(z) = ZzV(zj + z%l(z)+l , V(z) = d(z) + z2V(z)
Now (Step 3) we must solve two equations in two unknowns; but these are
easy, since the second equation yields V(z) = zU(z)/(l - 2’); we find
l-22
z
U(z) = --. V(z] =
l-422 +24' 1 - 422 + 24
(We had this formula for U(z) in (7.10), but with z3 instead of z2. In that
derivation, n was the number of dominoes; now it’s the width of the rectangle.)
The denominator 1 - 4z2 + z4 is a function of z2; this is what makes
UI~+J = 0 and V2, = 0, as they should be. We can take advantage of this