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

20. 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}

21. 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

22. 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

23. 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)

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

25. 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

26. 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

27. (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.

28. 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.

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

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

31. 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

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

33. The robot can move diagonally. y2 1 x

34. 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

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

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

37. 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

38. 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.

39. 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.

40. GCD correctness
Example: GCD(414,662)

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

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

43. 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

44. 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

45. 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

46. 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

47. 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.

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

49. 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.

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

51. 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.

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

53. 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).

54. Class Problems

55. Robert Floyd
Picture source:

56. 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

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

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

59. 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.

60. Strictly Decreasing Variable16 12 8 4
State

61. Weakly Decreasing Variable16 12 8 4
State

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

63. There are 5 boys and 5 girls
Stable Marriage
There are 5 boys and 5 girls
A B D E

64. 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

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

66. 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

67. 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
A B D E

68. 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

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

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

71. 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

72. 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

73. 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.

74. Infinitely Wide Trees … …

75. Infinitely Wide Trees … …

76. Infinitely Wide Trees

77. Infinitely Wide Trees

78. Tutorial Problem 1