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

2. Die Hard
Picture source:

3. Die Hard
Supplies:
3 Gallon Jug
5 Gallon Jug
Water

4. Die Hard
Transferring water:
3 Gallon Jug
5 Gallon Jug

5. Die Hard
Transferring water:
3 Gallon Jug
5 Gallon Jug

6. 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?

7. Die Hard
Work it out now!

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

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

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

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

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

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

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

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

16. 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?

17. Die Hard
Work it out now!

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

19. 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

20. 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

21. 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)

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

23. (b,l)  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
b + l  5
otherwise
(b,l) 

24. 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.

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

26. 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

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

28. The robot can move diagonally. y2 1 x

29. 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

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

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

32. 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.

33. 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.

34. 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.

35. 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.

36. 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).

37. Class Problems

38. Robert Floyd
Picture source: