JA Ritcey EE416-2013 Please elaborate with your own illustrations! Sets and Counting JA Ritcey EE416-2013 Please elaborate with your own illustrations!

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!

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. Often simpler using indicator functions I_A (x) You should be able to handle both lists and geometric regions.

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

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

Sensor Nodes Partitioned by Radius from Acces Point

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

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

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

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 0 0 0 0 0 1 0 0 0 1 2 0 0 1 0 3 0 0 1 1 4 0 1 0 0 5 0 1 0 1 6 0 1 1 0 7 0 1 1 1 8 1 0 0 0 9 1 0 0 1 10 1 0 1 0 11 1 0 1 1 12 1 1 0 0 13 1 1 0 1 14 1 1 1 0 15 1 1 1 1

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

Selection w/ Replacement Given a base N-set S In how many ways can we select k objects Assume a selection With Replacement. So that repetition is possible. We can do this is N^k ways by the multiplication principle N^k = NxNx…xN k-times 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)

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

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

Selection WO/ Replacement Given a base N-set S In how many ways can we select k objects Assume a Selection 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

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

Factorial Growth

Stirling’s Formula

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)

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 )

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) )

More Binomial Coefficient binomialcoef(n) Think (1+x)^n. The coefficients: 1 1 1 2 1 1 3 3 1 1 3 5 3 1 1 4 9 9 4 1 1 6 15 19 15 6 1 (1+x)^3 = 1x^0 + 3x^1 +3x^2 +1x^3 for example Computation – recursion, approximate, logGamma

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);

Unordered w/ Repetition This is the most complicated situation so far Think, 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! = ( N+k-1 choose N-1)

Tabular View Repetition With Without W/ Order N^k (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