Mathematics for Computer Science MIT 6 042 J18

Mathematics for Computer Science MIT 6. 042 J/18. 062 J Computational Processes Copyright © Albert Meyer, 2002. Prof. Albert Meyer & Dr. Radhika Nagpal March 3, 2002 1

Die Hard Picture source: http: //movieweb. com/movie/diehard 3/ March 3, 2002 2

Die Hard Supplies: 3 Gallon Jug 5 Gallon Jug Water March 3, 2002 3

Die Hard Transferring water: 3 Gallon Jug 5 Gallon Jug March 3, 2002 4

Die Hard Transferring water: 3 Gallon Jug 5 Gallon Jug March 3, 2002 5

Die Hard Psychopath’s challenge: Disarm bomb by putting 4 gallons of water on top, or it will blow up. Question: How to do it? March 3, 2002 6

Die Hard Work it out now! March 3, 2002 7

How to do it Start with empty jugs: (0, 0) 3 Gallon Jug 5 Gallon Jug March 3, 2002 8

How to do it Fill the big jug: (0, 5) 3 Gallon Jug 5 Gallon Jug March 3, 2002 9

How to do it Pour from big to little: (3, 2) 3 Gallon Jug 5 Gallon Jug March 3, 2002 10

How to do it Empty the little: (0, 2) 3 Gallon Jug 5 Gallon Jug March 3, 2002 11

How to do it Pour from big to little: (2, 0) 3 Gallon Jug 5 Gallon Jug March 3, 2002 12

How to do it Fill the big jug: (2, 5) 3 Gallon Jug 5 Gallon Jug March 3, 2002 13

How to do it Pour from big to little: (3, 4) 3 Gallon Jug 5 Gallon Jug Done!! March 3, 2002 14

Die Hard once and for all What if you have a 9 gallon jug instead? 3 Gallon Jug 5 Gallon Jug 9 Gallon Jug March 3, 2002 15

Die Hard once and for all What if you have a 9 gallon jug instead? 3 Gallon Jug 9 Gallon Jug Can you do it? Can you prove it? March 3, 2002 16

Die Hard Work it out now! March 3, 2002 17

State machines State machine: Step by step procedure, possibly responding to input. March 3, 2002 18

State machines The state graph of a 99 -bounded counter: start state 0 1 2 99 overflow States: {0, 1, …, 99, overflow} Transitions: i 99 i+1 0 i < 99 overflow March 3, 2002 19

State machines The state graph of a 99 -bounded counter: start state 0 1 2 99 overflow States: {0, 1, …, 99, overflow} Transitions: overflow March 3, 2002 20

State machines Die hard state machine State = amount of water in the jug: (b, l) where 0 b 3 and 0 l 5. Start state = (0, 0) March 3, 2002 21

State machines Die Hard Transitions: 1. Fill the little jug: 2. Fill the big jug: 3. Empty the little jug: 4. Empty the big jug: March 3, 2002 22

State machines 5. Pour from little jug into big jug: If no overflow, then b+l 5 otherwise (b, l) March 3, 2002 23

State machines 5. Pour from little jug into big jug: (for l >0) 6. Pour from big jug into little jug. Likewise. March 3, 2002 24

State Invariants Die hard once and for all Invariant: P(state) = 3 divides number of gallons in each jug. March 3, 2002 25

State Invariants Proof method (just like induction): 1) Show base case: P(start). 2) Invariant case: if P(q) and q then P(r). r , Conclusion: P holds for all reachable states, including final state (if any). March 3, 2002 26

The Robot y The robot is on a grid. 2 1 0 0 1 2 3 x March 3, 2002 27

The Robot y The robot can move diagonally. 2 1 0 0 1 2 3 x March 3, 2002 28

The Robot y Can it reach from (0, 0) to (1, 0)? ? 2 1 GOAL 0 0 1 2 3 x March 3, 2002 29

Robot Invariant NO! P((x, y)) : : = x + y is even P((0, 0)) is true. Transition adds ± 1 to both x and y March 3, 2002 30

Robot Invariant So all positions (x, y) reachable by robot have x + y even, but 1 + 0 = 1 is odd. Therefore (1, 0) is not reachable. March 3, 2002 31

GCD correctness Computing GCD(a, b) using the Euclidean Algorithm: 1. Let x = a, y = b. 2. If y = 0, return x & terminate. 3. Else set (x: =y, y: =x mod y) simultaneously; 4. Repeat process. March 3, 2002 32

GCD correctness Example: GCD(414, 662) = GCD(662, 414) since 414 mod 662 = 414 = GCD(414, 248) since 662 mod 414 = 248 = GCD(248, 166) since 414 mod 248 = 166 = GCD(166, 82) since 248 mod 166 = 82 = GCD(82, 2) since 166 mod 82 = GCD(2, 0) since 82 mod 2 = 0 Return 2. March 3, 2002 33

GCD correctness The Invariant is P((x, y)) : : = [gcd(a, b) = gcd(x, y)]. P(start, start): at start x = a , y =b. March 3, 2002 34

GCD correctness Transitions: (x, y) (y, rem(x, y)) Invariant holds by Lemma: gcd(x, y) = gcd(y, rem(x, y)), for y 0. March 3, 2002 35

GCD correctness Conclusion: on termination x equals gcd(a, b). Because terminate when y = 0, so x = gcd(x, 0) = gcd(x, y) = gcd(a, b). March 3, 2002 36

Class Problems March 3, 2002 37

Robert Floyd Picture source: http: //www. stanford. edu/dept/news/report/news/november 7/floydobit-117. html March 3, 2002 38