1. Trees Thm 2.1. (Cayley 1889) There are nn-2 different labeled trees
on n vertices. Or it can be stated as: the complete
graph Kn has nn-2spanning trees.

2. A spanning tree of a connected graph is a subgraph that is a tree.
Every tree with at least 2 vertices has at least 2 vertices of degree 1. 6 8 4 5 7 1 2 10 3 9
Consider the sequence:

3. Proof 1 (Prufer):
Given a spanning tree T in Kn, let T1=T and generate a sequence of trees T1, T2,…, Tn-1 and two sequences of vertices as follows: Given Ti with n-i+1 vertices, i=1..n-1, let xi be the least indexed degree 1 vertex of Ti and delete xi and its incident edge {xi, yi} from Ti to obtain Ti+1 on n-i vertices.
Thus T can be encoded as (y1,…, yn-2).
Claim that the encoding is 1-1 and onto. Note that there are nn-2 such encodings.
First we have yn-1 = n, since every tree has at least 2 vertices of degree 1 and n will never be the least indexed degree 1 vertex.

4. Second, xk,…, xn-1 and n are the vertices of Tk.
Third, {xi, yi}, i = k, …, n-1, are exactly the edges of Tk in some order.
v occurs among y1,…, yn-2 degT(v) -1 times.
Similarly, a vertex v in Tk occurs among yk,…, yn-2 is its degree in Tk less 1.
The degree-1 vertices of Tk are those elements of V not in {x1,…, xk-1 } and {yk,…, yn-1 }. This implies that xk is the least degree-1 vertex of Tk not in the above sets.
x1 is the least element of V not in the encoding of T and we can uniquely determine xk from the encoding and x1,…, xk-1. □

5. D has 2 trees rooted at 1 and n and a number (say k) of circuits.
Proof 2:
Consider a mapping f from {2, 3,...,n-1} to {1, 2,…, n}. There are nn-2 such mappings f.
Construct a digraph D on the vertices 1 to n by defining the edge set {(i, f(i)): i=2,..n-1}.
D has 2 trees rooted at 1 and n and a number (say k) of circuits.
The rightmost element in the i-th component, denoted ri, is its minimal element (and li is its left neighbor). 21 1 4 5 3 7 12 19 6 20 16 9 17 10 11 2 13 14 8 15 18

6. The circuits are ordered by r1< r2 < … < rk.
Proof 2:
The circuits are ordered by r1< r2 < … < rk.
We obtain a tree by adding the edges {1, l1}, {r1, l2}, …, {rk-1, lk}, {rk, n} and deleting the edges {ri, li}.
Let r0=1 and ri be the minimal number on the path from ri-1 to n. By traversing the path from 1 to n, we can recover the function f.  1 21 4 5 3 7 12 19 6 20 16 9 17 10 11 2 13 14 8 15 18

7. Proof 3: Recall the multinomial coefficient.
Denote the number of labeled trees with n vertices and degree sequence d1,...,dn by t(n; d1,...,dn) which is 0 if any di is 0.
Replace n by n-2, k by n,
ri by di-1 and xi by 1.

8. t(n;d1,...,dn): # of spanning trees
t(3; 1,2,1)=t(3; 2,1,1)=t(3;1,1,2)=1
t(4;2,1,1,1)=?
t(4;3,1,1,1)= t(3;2,1,1) =1
t(4;2,2,1,1)= t(3;2,1,1) + t(3;1,2,1) 3 4 1 2 3 4

9. Proof 3:

10. If there are k components, then how many
edges are there?
A graph with no polygon (cycle) as subgraph is called a forest. Each component of a forest is a tree.
A weighted graph is a graph together with a function associated a real number c(e) to each edge e, c is called the weighted function.
Given a weighted connected graph G, define the cost of a spanning tree T of G as
c(T)=
A spanning tree may represent a network of cities where c({x, y}) is the cost of erecting a telephone line joining cities x and y.

11. Minimum spanning tree problem is to find a spanning tree with the smallest weight.
Greedy algorithm: Input G: weighted and connected.
1. At each iteration, we have a set {e1,…, ei} of i independent edges, which has n-i components.
2. If i < n-1, let ei+1 be an edge with ends in different components and whose cost is minimum.
3. Stop when n-1 edges are chosen.

12. Thm With e1,…, en-1 chosen as above, the spanning tree T0 has the property that c(T0)  c(T) for any spanning tree T.
Pf: Let T has edges a1,…an-1 with the weight increasing.
Claim that c(ei)  c(ai) for i=1,…, n-1.
If it is false, then there is some k such that
c(ek) > c(ak)  c(ak-1) …. c(a1).
Since none of ak,…,a1 was chosen at the point when ek was chosen, each of them must have both ends in the same component of G{e1,…, ek-1}. The # of components of G{a1,…,ak}  the # of components of G{e1,…,ek-1}, i.e., n-k+1. It contradicts that G{a1,…ak } is a forest. 

13. The operation of BFS on an undirected graph1 1 v w x y v w x y Q Q s w r 1 1
(c) r s t u
(d) r s t u 1 2 1 2 1 2 2 1 2 v w x y v w x y Q r t x Q t x v 1 2 2 1 2 2
(e) r s t u
(f) r s t u 1 2 3 1 2 3 2 1 2 2 1 2 3 v w x y v w x y Q x v u Q v u y 2 2 3 2 3 3

14. (g) r s t u
(h) r s t u 1 2 3 1 2 3 2 1 2 3 2 1 2 3 v w x y v w x y Q u y y 3 3 3
(i) r s t u 1 2 3 2 1 2 3 v w x y Q

15. Breadth-first search：
Initially, vertices are colored white.
Discovered vertices are colored green.
When done, discovered vertices are colored black.
d[u] stores the distance from s to u.
is a predecessor to u on its shortest path.
Q is a first-in first-out queue.

16. Depth First Search: (a) (b) 1/ 1/ 2/ (c) (d) 1/ 2/ 1/ 2/ 3/ 4/ 3/ u v
ancestor descendent
Discovery time
(a) u v
(b) u v 1/ 1/ 2/ x y x y
(c)
(d) u v u v 1/ 2/ 1/ 2/ 3/ 4/ 3/ x y x y

17. (e) (f) (g) 1/ 2/ 1/ 2/ 1/ 2/ 4/ 3/ 4/5 3/ 4/5 3/6 (h) (i) (j) 1/ 2/7u v
(f) u v
(g) u v 1/ 2/ 1/ 2/ 1/ 2/ B B B 4/ 3/
4/5 3/
4/5
3/6 x y x y x y
finish time
(h)
(i) u v
(j) u v u v 1/
2/7 1/
2/7
1/8
2/7 B B F B F
4/5
3/6
4/5
3/6
4/5
3/6 y x y x y

18. Since G is finite, inductively, we can show y is a descendent of x.
Proposition 2.3. If vertices x and y are adjacent in G, then one of the them is a descendent of the other in any DFS tree T.
Pf:
Suppose x is discovered at time k and earlier than y in the DFS. Let u be the next discovered vertex after x. If u=y then we are done. If not then y is still waiting to be discovered. u x
Since G is finite, inductively, we can show y is a descendent of x. y

19. Isthmus (or bridge) is an edge in a graph whose deletion results in a disconnected graph.
Proposition 2.4. Let {x, y} be an edge (not isthmus) of T in G. Let x be the parent of y. Then there is an edge in G but not in T joining some descendent a of y and some ancestor b of x. b x y a
Assigning a direction to each edge of an undirected graph G is called an
orientation of G.
A walk in a digraph may be called strong when each edge is traversed
by the walk according to its direction.
A digraph is strongly connected if for any two vertices x and y, there is a
strong walk from x to y.

20. Thm 2.5. G: finite, connected, no isthmus.
Then G admits a strong orientation.
Pf. Goal: Construct a digraph D from G by choosing a direction for each edge.
Find a DFS tree T and number the vertices from v0. Let e={vi, vj} be an edge of G with i < j. If e is in T, direct it from vi to vj , else from vj to vi.
Need to show that D is strongly connected. There is a strong walk from v0 to any vertex x. It remains to show there is a strong walk from x to v0.
By induction on the distance between x and v0 , and Prop 2.4. x v0