Computational Processes Mathematics for Computer Science MIT 6.042J/18.062J Computational Processes Copyright © Albert Meyer, 2002. Prof. Albert Meyer & Dr. Radhika Nagpal

Die Hard Picture source: http://movieweb.com/movie/diehard3/

Die Hard Supplies: 3 Gallon Jug 5 Gallon Jug Water

Die Hard Transferring water: 3 Gallon Jug 5 Gallon Jug

Die Hard Transferring water: 3 Gallon Jug 5 Gallon Jug

Psychopath’s challenge: Disarm bomb by putting 4 gallons 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?

Die Hard Work it out now!

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

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

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

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

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

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

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

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

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?

Die Hard Work it out now!

State machine: Step by step procedure, possibly responding to input.

The state graph of a 99-bounded counter: State machines The state graph of a 99-bounded counter: start state overflow 1 2 99

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

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

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

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

State machines Die Hard Transitions: Fill the little jug: Fill the big jug: Empty the little jug: Empty the big jug:

5. Pour from little jug into big jug: If no overflow, then State machines 5. Pour from little jug into big jug: If no overflow, then

5. Pour from little jug into big jug: If no overflow, then b + l  5 State machines 5. Pour from little jug into big jug: If no overflow, then b + l  5

(b,l)  5. Pour from little jug into big jug: (for l >0) State machines 5. Pour from little jug into big jug: (for l >0) If no overflow, then …, (b,l)  otherwise.

5. Pour from little jug into big jug: (for l >0) State machines 5. Pour from little jug into big jug: (for l >0) 6. Pour from big jug into little jug. Likewise.

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

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

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

The Robot The robot is on a grid. y 2 1 x 0 1 2 3

The robot can move diagonally. y 2 1 x 0 1 2 3

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

Robot Invariant NO! P((x,y)) ::= x+y is even

Robot Invariant NO! P((x,y)) ::= x+y is even P((0,0)) is true.

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

So all positions (x,y) reachable by robot 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.

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

GCD correctness Example: GCD(414,662)

GCD correctness Example: GCD(414,662) = GCD(662,414) since 414 (mod 662) = 414

GCD correctness Example: GCD(414,662) = GCD(662,414) since 414 mod 662 = 414 = GCD(414,248) since 662 mod 414 = 248

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 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 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 = 2

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 = 2 = GCD(2,0) since 82 mod 2 = 0

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 = 2 = GCD(2,0) since 82 mod 2 = 0 Return 2.

P((x,y)) ::= [gcd(a,b) = gcd(x,y)]. GCD correctness The Invariant is P((x,y)) ::= [gcd(a,b) = gcd(x,y)].

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

Transitions: (x,y)  (y, rem(x,y)) GCD correctness Transitions: (x,y)  (y, rem(x,y))

Transitions: (x,y)  (y, rem(x,y)) GCD correctness Transitions: (x,y)  (y, rem(x,y)) Invariant holds by Lemma: gcd(x,y) = gcd(y,rem(x,y)), for y  0.

Conclusion: on termination GCD correctness Conclusion: on termination x equals gcd(a,b).

Conclusion: on termination 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).

Class Problems

Robert Floyd Picture source: http://www.stanford.edu/dept/news/report/news/november7/floydobit-117.html

A derived variable v is simply a function Derived variables A derived variable v is simply a function that associate value with state. Die Hard Example: total amount of water in the two jugs

For GCD, already have varables Derived variables For GCD, already have varables x, y.

Proof of GCD termination: Derived variables Proof of GCD termination: Show that variable y is strictly decreasing and natural number valued.

Termination follows by least number principle: Derived variables Termination follows by least number principle: y must take a least value, and thus the algorithm is stuck.

Strictly Decreasing Variable 16 12 8 4 State

Weakly Decreasing Variable 16 12 8 4 State

Weakly Decreasing Variable We used to call weakly decreasing variables “nonincreasing” variables. OK terminology but remember: nondecreasing is NOT EQUAL to “not decreasing.”

There are 5 boys and 5 girls Stable Marriage There are 5 boys and 5 girls 2 3 4 5 A B D E

Preferences: Boys Girls 1: CBEAD A: 35214 2 : ABECD B : 52143 Stable marriage Preferences: Boys Girls 1: CBEAD A: 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415

Stable marriage Boys 1: CBEAD 2 : ABECD 3 : DCBAE 4 : ACDBE 5 : ABDEC

Boys 1: CBEAD 2 : ABECD 3 : DCBAE 4 : ACDBE 5 : ABDEC Marry 1 with C: Stable marriage Boys 1: CBEAD 2 : ABECD 3 : DCBAE 4 : ACDBE 5 : ABDEC Marry 1 with C: C 1

Only 4 boys and 4 girls left. Boys 2 : ABED 3 : DBAE 4 : ADBE 5 : ABDE Stable marriage Only 4 boys and 4 girls left. Boys 2 : ABED 3 : DBAE 4 : ADBE 5 : ABDE 2 3 4 5 A B D E

A problem: Rogue couples Stable marriage A problem: Rogue couples 1 C 4 B Girl C likes like boy 4 better Boy 4 like girl C better

The mating algorithm – day by day Stable marriage The mating algorithm – day by day

The mating algorithm – day by day Stable marriage The mating algorithm – day by day Morning: boys serenade favorite girls

The mating algorithm – day by day Stable marriage The mating algorithm – day by day Morning: boys serenade favorite girls Afternoon: girl rejects all but best Serenader

The mating algorithm – day by day Stable marriage The mating algorithm – day by day Morning: boys serenade favorite girls Afternoon: girl rejects all but best Serenader Evening: rejected boys cross off girls

Termination condition: stop when no rejection is possible. Stable marriage Termination condition: stop when no rejection is possible. The algorithm terminates. Everybody ends up married. The resulting marriages are stable.

Infinitely Wide Trees … …

Infinitely Wide Trees … …

Infinitely Wide Trees

Infinitely Wide Trees

Tutorial Problem 1