18.314 PSET 5 ANSWER KEY
Problem 1. Prove that in any tree T, any two longest paths cross each other.
Let P1 , P2 be two longest paths that are noncrossing. Since T connected, there exits points x ∈ P1 , y ∈ P2 s.t. xy ∈ E(T ). P1 = Pa xPb , P2 = Pc yPd . Then among
Pa xyPc , Pa xyPd , Pb xyPc , Pb xyPd exits a path that is longer than P1 , P2 .
Problem 2. Find the number of spanning trees for the graph W 6
Special case of Problem 9. Answer : 8100.
Problem 3. Find the probability pn (d), limit.
Using Prufer
code, want to find number of terms of monomial x1d−1 xµ2 2 · · · xµnn in (
(n − 1)n−d−1 .
This is n−2
d−1
n−2
(n − 1)n−d−1
d−1
pn (d) =
nn−2
1
lim pn (d) =
e(d − 1)!
P
xi )n−2 .
Problem 4. Number of trees on n labeled nodes with exactly 4 leaves
Using Prufer code, when di is degree of vertex i,
Y
X
xidi −1 = (
xi )n−2
Lets count the number of terms which exactly four variables have 0 exponents.
n
(n − 4)!
S(n − 2, n − 4)
4
Problem 5. Matchings and matrices
P
Q
det(A − tI) = π∈Sn (−1)sgn(π) ai,π(i)
(1) Terms with π having
k-fixed points correspond to coefficient of tk .
Q
(2) Call π good, if ai,π(i) is nonzero.
(3) Good π cannot have cycles of length ≥ 3.
(4) So n − k should be even, and good π should consist of k-fixed points and (n − k)/2
cycles of length 2.
(5) (n − k)/2 transpositions means that (−1)sgn(π) = (−1)(n−k)/2 .
Combine these facts to get the answer. I gave full credit even if you did not prove the sign
part.
Problem 6. SDR
It is easy to see the following two properties.
(1) Let I be a SDR. If J ⊂ I, then J is a SDR.
Date: September 28, 2007.
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18.314 PSET 5 ANSWER KEY
(2) If I, J are SDR, |I| > |J|, then ∃x ∈ I − J s.t. {x} ∪ J is a SDR.
Hence the SDRs form a matroid, and it was introduced in class that greedy algorithms works
for matroids.
Problem 7. Abel’s Formula
Set k := n − a. Proceed by induction on n. When we differentiate by y, both sides
corresponds to when n − 1. So we only need to show both sides are same for some value of
y, where plugging in y = (−x − n) makes it easy.
Problem 8. (P, w) and trees
From a tree on n + 1 vertices, let n + 1 be the root, and children connected to it be
r1 < r2 < · · · < rk . Draw k-up steps labled r1 , · · · , rk , then one down step. Repeat the
procedure on subtree having rk as root. When finished, repeat on subtree having rk−1 as
root, and so on to r1 . This is easily reversible, and can get a bijeciton between (P, w) and
labled trees on n + 1 vertices.
Problem 9. Spectral Graph theory
Use the fact that this is a product graph of a ring graph and K2 . And number
Q of spanning
trees for a degree d-regular graph with eigenvalues λi of the Laplacian is 1/n (d − λi ). As
long as you seemed to know what is going on, I gave full credit regardless of the calculation.
Problem 10. Km,n
P
v∈V1 (deg(v) − 1) = n + m − 1 − n = m − 1. This is number of times v ∈ V 1 appears in
Prufer coding. So we get nm−1 mn−1 .
Problem 11. Labeling tree
h(v) ways to label a root v. But of all possible labelings, only valid labeling is the one
with root being smallest. Use induction on number of vertices in a tree.
E-mail address: suho@mit.edu