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- Find an when a 0 = 0 and an+1 = 3an + 1 when n ≥ 0. Answer: Let A(x) =
n≥ 0 anx n, the generating function for the ans. Multiply both sides of the recursion by xn+1^ and sum:
A(x) − a 0 = 3xA(x) +
∑^ ∞
n=
xn+1^ = 3A(x) + x 1 − x
Hence
A(x) = x (1 − x)(1 − 3 x)
1 − 3 x
1 − x
(3n/2)xn^ −
(1/2)xn.
Thus an = (3n^ − 1)/2.
- For the graph shown here, find two different lineal spanning trees with A as root vertex Answer: There six possible trees. For all of them, A has three principal subtrees, each of which is a path. One tree consists of F alone. Another consists of EG or GE. The third consists of CBD, CDB, DBC, or DCB.
- (a) A strictly decreasing function from k to n is given by f (i) = k+2−i for 1 ≤ i ≤ k. Prove that its rank is k. (n is much larger than k.)
Answer: Recall that RANK(f ) =
∑^ k
i=
f (i) − 1 k + 1 − i
. Since the binomial coeffi-
cients are
(k+2−i− 1 k+1−i
(k+1−i k+1−i
= 1, the sum is k. (b) A strictly decreasing function from k to n is given by f (i) = k+3−i for 1 ≤ i ≤ k. Prove that its rank is k(k + 3) 2
Answer: As in (a), but the binomial coefficients are now
(k+2−i k+1−i
= k + 2 − i. Hence the terms in the sum for the rank go from k + 1 down to 2. We use induction on k to prove that the sum is the given answer. When k = 1, the result is true. For k, the formula and the induction hypothesis give
Sum = (k + 1) + (k − 1)((k − 1) + 3) 2
k(k + 3) 2
- If T is an unlabeled binary RP-tree with more than one leaf, let T 1 and T 2 be its left and right principal subtrees and let |T | be the number of leaves in T. Using Exercise 9.1.12(c) or otherwise, prove: (You do NOT need to prove Exercise 9.1.12.) When n ≥ 2 over half the unlabeled binary RP-trees with n leaves have either |T 1 | = 1 or |T 2 | = 1.
Answer: When n = 2, the result is true. When n > 2, the number of trees with
|T 1 | = 1 is b 1 bn− 1 = bn− 1 and likewise for |T 2 | = 1. The the fraction of trees with either |T 1 | = 1 or |T 2 | = 1 is 2bn− 1 /bn. From the exercise, (n + 1)bn+1 = 2(2n − 1)bn and so 2 bn− 1 bn
2 nbn− 1 nbn
2 nbn− 1 2(2(n − 1) − 1)bn− 1
n 2 n − 3
2 − 3 /n
- An oriented simple graph is a simple graph which has been converted to a digraph by assigning an orientation to each edge. Answer: This problem appeared on the first hour exam. (a) Prove that the number of n-vertex oriented simple graphs is 3(
n 2 ) . (b) State and prove a formula for the number of n-vertex oriented simple graphs that have exactly q edges.
- Let Bn be the number of partitions of n. It was proved in the text that
Bn+1 =
∑^ ∞
k=
n k
Bk for n ≥ 0 provided B 0 = 1. Although the sum is infinite, only
finitely many terms are nonzero since
(n k
= 0 if k > n. Prove by induction that
Bn =
e
∑^ ∞
i=
in i! for n ≥ 0 , where 0^0 equals 1.
Hints: Remember that
i=
k=0(anything) =^
k=
i=0(same thing). You may find (j + 1)m/j! = (j + 1)m+1/(j + 1)! useful in your proof. Answer: The case n = 0 is straightforward: ∑^ ∞
i=
i^0 i!
∑^ ∞
i=
i! = e.
For induction, we use the recursion for Bn+1 and the formula for Bk:
Bn+1 =
∑^ ∞
k=
n k
Bk =
∑^ ∞
k=
n k
e
∑^ ∞
i=
ik i!
=
e
∑^ ∞
i=
i!
∑^ ∞
i=
n k
ik
e
∑^ ∞
i=
i! (1 + i)n,
where the last equality is due to the binomial theorem. We want this to equal
1 e
∑^ ∞
i=
in+ i!
(b) State and prove a formula for Q(x) =
∑^ ∞
n=
qnxn. Answer: If k is odd, the kth part can appear any number of times, so we get 1 + xk^ + (xk)^2 + · · · = (1 − xk)−^1. Thus
Q(x) =
k odd
1 − xk^
(c) Prove that Q(x) = D(x) and hence qn = dn as follows: Multiply D(x) by F (x) =
∏^ ∞
k=
1 − xk
and combine the two factors having the same power of x.
Divide the result by F (x) and show that this is the same as the Q(x) you obtained in (b). Answer: Proceeding as instructed:
D(x)F (x) =
k=
(1 + xk)
k=
(1 − xk)
∏^ ∞
k=
(1 − x^2 k) =
k even
(1 − xk)
and so
D(x)F (x) F (x)
k even
(1 − xk) ∏
all k
(1 − xk)
k odd
(1 − xk)
