MATH 310, FALL 2003(Combinatorial Problem Solving)

Lecture 13, Monday, September 29

Example 3: Pitcher Pouring Puzzle We are given three pitchers of

water, of sizes 10 quarts, 7 quarts and 4 quarts.

Initially, the 10-quart is full and the other two pitchers are empty.

Is there a way to pour among pitchers to obtain exactly 2 quarts in one of the pitchers?

The State Space The state space contains the following:

S – set of states s0 – initial state F – final or goal states (could be only one) P - set of rules D – Admissibility function D:P S

{True,False} U – Rule application: U:P S S.

Example 3: Pitcher Pouring Puzzle – Exercise 16(a)

We are given three pitchers of water, of sizes 8 quarts, 5 quarts and 3 quarts.

Initially, the 8-quart is full and the other two pitchers are empty.

Is there a way to pour among pitchers to obtain exactly 4 quarts in one of the pitchers?

Example 3: Pitcher Pouring Puzzle – Exercise 16(a)

We are given three pitchers of water, of sizes 8 quarts, 5 quarts and 3 quarts.

Initially, the 8-quart is full and the other two pitchers are empty.

Is there a way to pour among pitchers to obtain exactly 4 quarts in one of the pitchers?

State Space Here is a part of the

state space for Exercise 16(a).

What is the initial state? What are the final

states? What are the rules? Are there any states

missing? Are there any edges

missing? Find DFS and BFS trees.

800

503 350

530

053

323

233

251

701 710 413

440

620602152

143

Example 4: Jealous Wives Puzzle. Three jealous wives

and three husbands come to a river.

The boat is for two people.

Find a sequence of boat trips that will get 6 people across the river without ever letting any husband to be without his wife in the presence of another wife.

A,a,B,b,C,c

Example 4: Jealous Wives Puzzle. Three jealous wives

and three husbands come to a river.

The boat is for two people.

Find a sequence of boat trips that will get 6 people across the river without ever letting any husband to be without his wife in the presence of another wife.

A,a,B,b,C,c

A Center of a Tree T Let T be an arbitrary

tree and let r be any of its vertices. Let Tr be the tree rooted at r. Vertex r is called a center if the correspondig rooted tree Tr has minimal height. Compare Exercise 18, p. 103.

7

5

4 81

6

23

root 1 2 3 4 5 6 7 8

height 3 4 3 4 4 3 4 2

4.1. Shortest Paths Homework (MATH 310#5M):

• Read 4.2. • Do Exercises 4.1: 1,2,4,6,8,10,12• Volunteers:

• ____________• ____________• Problem: 1.

No Section of Chapter 4 will be on Test 1.No Section of Chapter 4 will be on Test 1.

Dijkstra’s Algorithm At each step of

the algorithm the (green) edge e between a labeled vertex p and unlabeled vertex q is selected in such a way, that

• d(p) + k(e)

is minimal.

a

Labeled verticesb

Unlabeled vertices

p

q

a