1. Is There Something to CS Theory?
Alan Kaylor Cline
July 10, 2009

2. Can one do computer science without doing programming?

3. Can one do computer science
without doing programming?
No and Yes

4. How many games are required to determine the winner of
Problem:
How many games are required to determine the winner of
a single elimination tournament with 18 teams?
Single Elimination Tournament: a tournament in which each game involves exactly two teams and after which exactly one is the winner of the game and one is the loser. The tournament proceeds until all but one team has lost.

5. The Pigeonhole Principle

6. The Pigeonhole Principle
Statement
Children’s Version: “If k > n, you can’t stuff k pigeons in n holes without having at least two pigeons in the same hole.”

7. The Pigeonhole Principle
Statement
Children’s Version: “If k > n, you can’t stuff k pigeons in n holes without having at least two pigeons in the same hole.”
Smartypants Version: “No injective function exists mapping a set of higher cardinality into a set of lower cardinality.”

8. The Pigeonhole Principle
Example
Twelve people are on an elevator and they exit on ten different floors. At least two got of on the same floor.

9. For a real number x, the ceiling(x) equals
The ceiling function:
For a real number x, the ceiling(x) equals
the smallest integer greater than or equal to x
Examples:
ceiling(3.7) = 4
ceiling(3.0) = 3
ceiling(0.0) = 0
If you are familiar with the truncation function, notice that
the ceiling function goes in the opposite direction –
up not down.
If you owe a store 12.7 cents and they make you pay 13 cents,
they have used the ceiling function.

10. The Extended (i.e. coolguy)
Pigeonhole Principle
Statement
Children’s Version: “If you try to stuff k pigeons in n holes there must be at least ceiling (n/k) pigeons in some hole.”

11. The Extended (i.e. coolguy)
Pigeonhole Principle
Statement
Children’s Version: “If you try to stuff k pigeons in n holes there must be at least ceiling (n/k) pigeons in some hole.”
Smartypants Version: “If sets A and B are finite and f:A B, then there is some element b of B so that cardinality(f -1(b)) is at least ceiling (cardinality(B)/ cardinality(A).”

12. The Extended (i.e. coolguy) Pigeonhole Principle
Example
Twelve people are on an elevator and they exit on five different floors. At least three got off on the same floor.
(since the ceiling(12/5) = 3)

13. The Extended (i.e. coolguy) Pigeonhole Principle
Example
Twelve people are on an elevator and they exit on five different floors. At least three got off on the same floor.
(since the ceiling(12/5) = 3)
Example of even cooler “continuous version”
If you travel 12 miles in 5 hours, you must have traveled at least 2.4 miles/hour at some moment.

14. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.

15. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.

16. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies. F F F

17. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies. E E E

18. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.

19. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.
How would you solve this?
You could write down every possible
acquaintanceship relation.

20. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.
How would you solve this?
You could write down every possible
acquaintanceship relation.
There are 15 pairs of individuals.
Each pair has two
possibilities: friends or enemies.
That’s 215 different relations.

21. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.
How would you solve this?
You could write down every possible
acquaintanceship relation.
There are 15 pairs of individuals.
Each pair has two
possibilities: friends or enemies.
That’s 215 different relations.
By analyzing one per minute,
you could prove this in 546 hours.

22. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.
Could the pigeonhole
principle be applied to this?

23. Application 1: Among any group of six acquaintances
there is either a subgroup
of three mutual friends or three mutual enemies.
Could the pigeonhole
principle be applied to this?
I am glad you asked.
Yes.

24. Begin by choosing one person:*
Begin by choosing one person: *

25. * * Begin by choosing one person: Five acquaintances remain
These five must fall into two classes:
friends and enemies

26. * * Begin by choosing one person: Five acquaintances remain
These five must fall into two classes:
friends and enemies
The extended pigeonhole principle
says that at least three must be
in the same class -
that is: three friends or three enemies

27. Suppose the three are friends of :*
Suppose the three are friends of : * ? ? ?

28. * * Suppose the three are friends of :
Either at least two of the three are friends of each other… * ? ?

29. * * Suppose the three are friends of :
Either at least two of the three are friends of each other… * ? ?
In which case we have three mutual friends.

30. * * Suppose the three are friends of :
Either at least two of the three are friends of each other…
or none of the three are friends *

31. * * Suppose the three are friends of :
Either at least two of the three are friends of each other…
or none of the three are friends *
In which case we have three mutual enemies.

32. Similar argument if we suppose the three are enemies of .*
Similar argument if we suppose the three are enemies of . * ? ? ?

33. Application 2:
Given twelve coins – exactly eleven of which have equal weight
determine which coin is different and whether it is
heavy or light in a minimal number of weighings using a
three position balance. H

34. OUR SOLUTION

35. Application 3: In any sequence of n2+1 distinct integers, there is
a subsequence of length n+1 that is either
strictly increasing or strictly decreasing
n=2: 3,5,1,2, ,3,5,4,1
n=3: 2,5,4,6,10,7,9,1,8, ,1,6,3,8,9,2,4,5,7
n=4: 7,9,13,3,22,6,4,8,25,1,2,16,19,26,
10,12,15,20,23,5,24,11,14,21,18,17

36. Application 3: In any sequence of n2+1 distinct integers, there is
a subsequence of length n+1 that is either
strictly increasing or strictly decreasing
n=2: 3,5,1,2, ,4,5,3,1
n=3: 2,5,4,6,10,7,9,1,8, ,1,6,3,8,9,2,4,5,7
n=4: 7,9,13,3,22,6,4,8,25,1,2,16,19,26,
10,12,15,20,23,5,24,11,14,21,18,17

37. Application 3: In any sequence of n2+1 distinct integers, there is
a subsequence of length n+1 that is either
strictly increasing or strictly decreasing
Idea: Could we solve this by considering cases?
For sequences of length 2: 2 cases
For sequences of length 5: 120 cases
For sequences of length 10: 3,628,800 cases
For sequences of length 17: cases
For sequences of length 26: cases
For sequences of length 37: cases

38. The Limits of Computation
Speed: speed of light = m/s
Distance: proton width = 10 –15 m
With one operation being performed in
the time light crosses a proton
there would be operations per second.
Compare this with current serial processor
speeds of operations per second

39. The Limits of Computation
With one operation being performed in
the time light crosses a proton
there would be operations per second.
Big Bang: 14 Billion years ago
… that’s seconds ago
So we could have done
operations since the Big Bang.
So we could not have proved this (using enumeration)
even for the case of subsequences of length 7 from
sequences of length 37.

40. But with the pigeon hole principle we can prove it in two minutes.

41. We can apply the pigeon hole principle to a sorting problem

42. We can apply the pigeon hole principle to a sorting problem
How many comparisons are required to sort n distinct elements?

43. Think of a sorting algorithm as determining
in which permutation that an array is presented

44. is presented as the permutation: 3, 1, 4, 2, 5
Think of a sorting algorithm as determining
in which permutation that an array is presented
The array:
2, -5, 6, 0, 12
is presented as the permutation:
3, 1, 4, 2, 5

45. is presented as the permutation: 3, 1, 4, 2, 5
Think of a sorting algorithm as determining
in which permutation that an array is presented
The array:
2, -5, 6, 0, 12
is presented as the permutation:
3, 1, 4, 2, 5
Thus the problem of sorting is equivalent to
determining in which of the
n! permutations an array is presented

46. In terms of functions, the sorting algorithm is a function with
n! possible different inputs that must act differently on each

47. In terms of functions, the sorting algorithm is a function with
n! possible different inputs that must act differently on each
Think of the various permutations as leaves at the bottom
of a binary tree with comparisons
at the internal nodes of the tree.
A tree of height k (i.e. with at most k comparisons to reach any leaf)
has at most 2k leaves.

48. In terms of functions, the sorting algorithm is a function with
n! possible different inputs that must act differently on each
Think of the various permutations as leaves at the bottom
of a binary tree with comparisons
at the internal nodes of the tree.
A tree of height k (i.e. with at most k comparisons to reach any leaf)
has at most 2k leaves.
Thus, we must have

49. In terms of functions, the sorting algorithm is a function with
n! possible different inputs that must act differently on each
Think of the various permutations as leaves at the bottom
of a binary tree with comparisons
at the internal nodes of the tree.
A tree of height k (i.e. with at most k comparisons to reach any leaf)
has at most 2k leaves.
Thus, we must have
or equivalently

50. We have shown that the minimum number of comparisons for a comparison based sorting algorithm is

51. Have you seen this presented as
We have shown that the minimum number of comparisons for a comparison based sorting algorithm is
Have you seen this presented as

52. Have you seen this presented as
We have shown that the minimum number of comparisons for a comparison based sorting algorithm is
Have you seen this presented as
For large values of n these are close,
but they are certainly not identical

53. vs.
vs. mergesort

54. What is the volume of the largest sphere that can be fit
Problem:
What is the volume of the largest
sphere that can be fit
between a unit radius cylinder
adjoining a floor and a wall?

55. What is the volume of the largest sphere that can be fit
Problem:
What is the volume of the largest
sphere that can be fit
between a unit radius cylinder
adjoining a floor and a wall?
so the volume is
Radius is

56. How should we calculate
if we use

57. How should we calculate
Some alternatives:
if we use
expression
computed as
relative error
double precision error multiple
2.51%
103.
5.95%
137.
15.47%
339. %
71,791.
1.05%
35.
0.44%
18.
0.15% 1.

58. How should we calculate?
Some alternatives:
expression
computed as
relative error
double precision error multiple
2.51%
103.
5.95%
137.
15.47%
339. %
71,791.
1.05%
35.
0.44%
18.
0.15% 1.

59. How should we calculate?
Some alternatives:
expression
computed as
relative error
double precision error multiple
2.51%
103.
5.95%
137.
15.47%
339. %
71,791.
1.05%
35.
0.44%
18.
0.15% 1.

60. double precision error multiple
How should we calculate ?
Some alternatives:
expression
computed as
relative error
double precision error multiple
2.51%
103.
5.95%
137.
15.47%
339. %
71,791.
1.05%
35.
0.44%
18.
0.15% 1.

61. An easy way to approximate pi is to construct a sequence of inscribed
regular polygons of 2n sides
We have:
and for n = 2, 3, … :

94. algebraically equivalent
Let’s use this
algebraically equivalent
version of the formula for :
and for n = 2, 3, … :

96. .

97. .

101. The steps of scientific computation
Observation of nature
Construction of a mathematical model
Selection of a computational method to solve the mathematical formulation
Program the method
Execute the program and display the results
Interpret the results and compare to observations
(Possibly) Refine