1. Please elaborate with your own illustrations!
Sets and Counting
JA Ritcey
EE 416
Revised Oct
Please elaborate with your own illustrations!

2. Elementary Set Theory
Universe – superset containing any element of interest. This is the complement of the empty set.
Subset – collection of none, some, or all elements
Set operations – complement, union, intersection
Cardinality – the number of elements in a set, size
Cardinality can be finite, countable, or uncountable
Venn diagrams are useful pictorial representations
Set operations can be carried out in MATLAB!

3. Set Operations U = { u_1, … , u_N } with card(U) = N
Count by placing elements of U in one-to-one correspondence with the natural numbers
You should be able to carry out set operations by taking unions, intersections, and complements.
You should be able to handle both lists and geometric regions.

4. Disjoint or Mutually Exclusive Sets
2 sets a disjoint or mutually exclusive if they have no common elements
disjoint means the intersection is the empty set
Disjoint sets are simpler than those with overlap
A collection of sets is mutually disjoint if no two intersect. Together they partition the union.
Partition – a collection of disjoint subsets that is complete and together yields the entirety

5. Partitions
A set S is partitioned by a collection of disjoint subsets that together encompass all of S
The days of the week are partitioned into work days and weekend
The numbers {1,2,3, …} are partitioned into odd and even
Partitions are used to when something is held invariant over each subset

6. Sensor Nodes Partitioned by Radius from Acces Point

7. Cartesian Products Given 2 sets A = { a_i } B = { b_j } create the
Ordered pair c = (a_i, b_j )
C = AxB is called the Cartesian product
|C| = |AxB| = |A|x|B| . This is called the multplication principle in counting (combinatorics)
Select one from column A and 1 from column B
This can be extended to ordered J-tuples
C= A_1 x … x A_J the J-fold product
A vector can be created this way

8. Power Sets Power Set Pwr(S) is the set of all subsets of any S
If a set S is finite, say |S| = N, then
|Pwr(S)| = 2^N
To see this equate each member of Pwr(S) with the binary string (b_1, b_2, … , b_N ).
The b_i in (0,1) label whether element s_i is in the particular subset.
There are 2^N binary strings of length N
If the cardinality, or size, of A is countable, the power set has uncountable cardinality

9. Properties of the Power Set
Pwr(S) has some closure properties
A collection of sets C is closed under set operations:
(1) If A in C, A^c in C (closed under complements)
(2) If A_1, …A_N in C, their union is in C
Often this last property is extended to countable unions. If A_1, A_2,… in C, their union is in C
Collections of sets with this property are called algebras and sigma-algebras.
When S is uncountable Pwr(S) is too big to assign probabilities to everything – a subtle point

10. Generating some small power sets
function powerset(n);
%function powerset(n);
n1=n-1;
for knt = 0:2^n1-1
C= bitget(knt,n1:-1:1);
disp([knt C])
end

11. Counting Finite Sets
We have already done some counting, using the Multiplications Principle
Cartesian product card |C| = |AxB| = |A| |B|
All pairs must be possible
The size of A and B can be different, but finite
There are 52 = 4x13 cards in a deck
There are 100 = 2x50 US Senators
Lets call a set S an N-set when |S|=N

12. Ordered Selection w/ Repetition
Given a base N-set S
In how many ways can we select k objects
Assume an ordered selected With replacement. So that repetition is possible
We can do this is N^k ways by the multiplication principle
Just equate the selection of an object with an ordered k-tuple The elements are independently selected and each drawn from an N-set (repetitions)

13. Tabular View Repetition With Without W/ Order N^k N!/(N-k)!
WO/ Order (N+k-1)!/ (N-1)! k! N!/(N-k)!k!
(N+k-1 choose k) (N choose k)
It is easier to remember these, when they are considered as a group

14. Ordered Selection W/Replacement
Each Selection is a row
Possible Choices along columns
7^12 = 7 x7 …x7
Cartesian Product
Illustrates the Multiplication Principle 1 2 3 4 5 6 7 8 9 10 11 12

15. Examples Cards = 4 suits x 13 face values = 52 cards
You can place r balls into n cells in n^r ways
n^r = n x n … x n
Harder: How many ways can r flags be put on n flagpoles?
Ans: n x (n+1) x (n+2) x … x (n+r-1)
Why: The second flag can be put above or below the first. So the number of choices increases

16. Ordered Selection wo/ Repetition
Given a base N-set S
In how many ways can we select k objects
Assume an ordered selected without replacement. So that repetition is not possible
We can do this is (N)_k = N(N-1) …(N-k+1) ways, again by the multiplication principle
Just equate the selection of an object with an ordered k-tuple The elements are each drawn from sets of decreasing size
When k=N, we can select in N! ways

17. Tabular View Repetition With Without W/ Order N^k N!/(N-k)!
WO/ Order (N+k-1)!/ (N-1)! k! N!/(N-k)!k!
(N+k-1 choose k) (N choose k)
It is easier to remember these, when they are considered as a group

18. Factorial Interpretation
The factorial n! = n(n-1)(n-2) …(3)(2)(1) 0!=1
It gets very big very fast,
n! asy sqrt(2pi)n^(n+1/2)exp(-n)
Then (N)(N-1)…(N-k+1) = N! / (N-k)!
Best ways to compute the factorial is through the Gamma Function (gamma.m) More later.
You can permute (reorder) an N-tuple in N! ways

19. Factorial Growth

20. Stirling’s Formula

21. Matlab M-File functioncompareGamma(M); m = [1:M]';half=1/2;
fact = gamma(m+1);
plot( m, fact, 'b*-');grid
title('m! = Gamma(m+1) - Grows very fast');
xlabel('m');ylabel('G(m)');
%try logs to compress the dynamic range
figure(2)
subplot(121);
stir =
sqrt(2*pi).*m.^(m+half).*exp(-m);
semilogy( m, fact, 'b*-',m,stir,'ro');grid
subplot(122);
RE = abs(stir-fact)./fact;
semilogy(m,RE,'r*');grid
title('Relative Error');
figure(1)

22. Unordered selection wo/ Repetition
What if we select without regard to the order.
Solve this by reconsidering ordered selection wo/ replacement of k objects from N-set
Each such selection resulted a k-tuple that indexed the selection a_i is that object selected on the i^th turn
But a k-tuple can be reordered (permuted) in k! ways
Therefore we need to reduce our result to
N(N-1)…(N-k+1)/k! = N! / (N-k)! K!
Combinations - binomial coefficient ( N choose k )

23. Tabular View Repetition With Without W/ Order N^k N!/(N-k)!
WO/ Order (N+k-1)!/ (N-1)! k! N!/(N-k)!k!
(N+k-1 choose k) (N choose k)
It is easier to remember these, when they are considered as a group

24. Computing Binomial Coefficient
{ n choose k } = n! / (n-k)! k!
But n! = gamma(n+1) = G(n+1) (gamma.m)
{ n choose k } = G(n+1)/G(n+1-k) G(k+1)
Even better to use loggamma = log(Gamma(m))
{ n choose k } =
exp( gammln(n+1) -gammaln(n+1-k) -gammaln(k+1) )

25. More Binomial Coefficient
binomialcoef(n) Think (1+x)^n. The coefficients:
1 1
Computation – recursion, approximate, logGamma

26. Matlab Code function binomialcoef(n);
%generates an array of binomial coefficients
k = [0:n];
lgGn = gammaln(n+1);
lgGnk = gammaln(n+1-k);
lgGk = gammaln(k+1);
bico = exp( lgGn -lgGnk -lgGk );
bico = floor(bico);
fprintf('%5.0f', bico);

27. Unordered w/ Repetition
This is the most complicated situation so far
Thinks for a long time, then come up with a
Clever way to represent a selection bars & balls
4 objects choose 2  o||o| select object 1 and 3
Another choice is |||oo select object 4 twice
To select k objects from N-set
Use N-1 bars and k balls  string of |||oo
But these can be drawn in ( N+k-1 choose k) ways
(N+k-1)!/(N-1)!k!

28. Tabular View Repetition With Without W/ Order N^k N!/(N-k)!
WO/ Order (N+k-1)!/ (N-1)! k! N!/(N-k)!k!
(N+k-1 choose k) (N choose k)
It is easier to remember these, when they are considered as a group