Blowing in the Wind Lecture 23 CS 15-251 root a a root b b b A tree. a methane H H propane H C C H C H H C H H H C H H C H H C H H C H H H H C H H C H H H H H H ethane butane H some saturated hydrocarbons Putting a picture into words… A tree is a connected graph with no cycles. These are not trees… Theorem: Let G be a graph with n nodes and e edges. The following are equivalent: 1. G is a tree (connected, acyclic). 2. Every two nodes of G are joined by a unique path. 3. G is connected and n = e + 1. 4. G is acyclic and n = e + 1. Proof: Show 1  2  3  4  1. (1) (2) G is a tree  u  v, u and v are connected by a unique path contrapositive (not 2) (not 1) Fix u and v. If there is no path between them, then G is not a tree. Let p1 and p2 be two distinct paths from u to v. p2 w w u v w’ p1 p2 Let w be the first point along p1 whose next edge is different along p2 than p1. Let w’ be the first point along p1 after w whose successor is also in p2. p1 w’ A cycle. G is not a tree. (2) (3) Every two vertices of G are connected by a unique path  G is connected and n=e+1 Clearly G is connected. We show that n=e+1 by induction on n. Base cases: n=1 n=2 Assume the claim is true for all graphs with fewer than n nodes. When we remove an edge from G we break it into exactly two connected components. unique path from u to v By induction, each piece has one fewer edge than nodes. (n-2)+1 edges = n-1 edges u v (3) (4) G is connected and n=e+1  G is acyclic and n=e+1 Suppose G has a cycle of length r. The cycle has r nodes and r edges. For each node not on the cycle, we associate one of its edges which brings us closer to the cycle. Each non-cycle node is now associated with a unique non-cycle edge. This contradicts n = e+1. (4)  (1) G is acyclic and n=e+1  G is acyclic and connected If G is not connected, then it has k > 1 connected components G1,G2,…,Gk . Each is acyclic and connected, and therefore a tree. We already know (1)  (4) So n1  e1  1 n2  e2  1 nk  ek  1  k  ni  n    ei   k  e  k  e  1   i 1  i 1  k Question: How many trees are there with four nodes? 2 trees 16 labeled trees a c b d Question: How many labeled graphs on n nodes are there?   2 n 2 n Each of   edges is either 2 present or not. Question How many labeled trees on n nodes are there? n Why? n 2 The proof will use the correspondence principle. Each labeled tree on n nodes corresponds to   n 2 a sequence in 1, 2, , n , i.e., (n-2) numbers each in the range 1..n  . How to make a sequence from a tree. Loop through i from 1 to n-2 Let l be the degree-1 node with the lowest label. Define the ith element of the sequence as the label of the node adjacent to l. Delete the node l from the tree. 8 5 Example: 4 3 1,3,3,4,4,4 1 7 2 6 How to reconstruct the unique tree from a sequence S. Let I  1,2,3, , n empty sequence Loop until S= Let l = smallest # in I but not in S Let s = first label in sequence S •Add edge {l, s} to the tree. •Delete l from I. •Delete s from S. Add edge {l, s} to the tree, where I  l , s 1 1,3,3,4,4,4 2 3 1 3 5 4 3 4 4 4 6 7 8 The problem of counting the number of n-node trees is much harder. It can be handled by Polya enumeration theory.     T ( x )  x e r 1 T xr The coefficient of xk in T(x) is the # of rooted trees on k nodes. A Game: n point sprouts Start with n points on the page. Two players alternate moves. A move consists of drawing and edge between two points and sprouting a new point somewhere along the edge. Restrictions: •Edges can’t cross •No point can have degree greater than three. A Game: n point sprouts You lose if you can’t make a legal move. (You are allowed to place an edge between a point and itself as shown below.) new point Question: Why does n-node sprouts terminate? How many moves might it take? For a given position, define its degree of freedom as 3 # of nodes 2 # of edges Thus, the starting position in n-node sprouts has degree of freedom 3n Each move drops the degree of freedom by 1. (We add 1 node, but 2 edges.) Theorem: The total # of moves in n-node sprouts is  3n Brussels Sprouts Nodes look like this: Sprouts look like this: Questions: Why does it terminate? How many moves? Is there a winning strategy? Each has 4 liberties. There are 4n liberties at start of an n-node game. Each move uses up two more liberties and creates two more. At any point in the game there are 4n liberties. The board starts as 1 region, but it can be split into many. 1 1 2 3 Each time a region is created, a liberty falls inside it. I.e., every region has at least one liberty. If # of regions < # of liberties then there is still at least one legal move to be played. Some region must still have at least two liberties. If # regions = # liberties, then the game is over! Each liberty must be in a separate region, i.e., each region has at least one liberty. Call a move “region creating” if it adds one to the total # of regions. We start with 1 region. We finish when we have 4n regions. Thus: Each game has exactly 4n-1 region creating moves. But what about the moves that don’t create regions? At the start the n +’s can be viewed as n components. Two +’s are in the same component if and only if there is a path from one to the other. A region-creating move must go between +’s in the same component. Thus, a move that does not create a region must merge two components into one. At the start there are n components. At the end there is 1 component. In any game there are exactly n-1 moves that do not create a region. 4n  1  region creating moves  n 1 component merging moves 5n  2 moves per game odd n  first player wins even n  second player wins