Presentation on theme: "Inclusion-Exclusion Rosen 6.5 & 6.6"— Presentation transcript:
1Inclusion-Exclusion Rosen 6.5 & 6.6 Longin Jan Lateckibasd on Slides byMax Welling, University of California, IrvineVardges Melkonian, Ohio University, Athens
26.5 Inclusion-Exclusion U A B It’s simply a matter of not over-counting the blue area in the intersection.
3Example on Inclusion/Exclusion Rule (2 sets) Question: How many integers from 1 through 100are multiples of 3 or multiples of 7 ?Solution: Let A=the set of integers from 1 through which are multiples of 3;B = the set of integers from 1 through 100which are multiples of 7.Then we want to find n(A B).First note that A B is the set of integersfrom 1 through 100 which are multiples of 21 .n(A B) = n(A) + n(B) - n(A B) (by incl./excl. rule)= – 4 = 43 (by counting the elementsof the three lists)
4Now three Sets area = 4-3=1 U C area = 2-1=1 area = 1 B A Image a blue circle has area 4. The intersections between 2 circles havearea 2 and the intersection between three circles 1. What is the total areacovered?A=4+4+4 – = 12 – = 7.
5Example on Inclusion/Exclusion Rule (3 sets) 3 headache drugs – A,B, and C – were tested on 40 subjects. The results of tests:23 reported relief from drug A;18 reported relief from drug B;31 reported relief from drug C;11 reported relief from both drugs A and B;19 reported relief from both drugs A and C;14 reported relief from both drugs B and C;37 reported relief from at least one of the drugs.Questions:1) How many people got relief from none of the drugs?2) How many people got relief from all 3 drugs?3) How many people got relief from A only?
6Example on Inclusion/Exclusion Rule (3 sets) We are given: n(A)=23, n(B)=18, n(C)=31,n(A B)=11, n(A C)=19, n(B C)=14 ,n(S)=40, n(A B C)=37Q1) How many people got relief from none of the drugs?By difference rule,n((A B C)c ) = n(S) – n(A B C) = = 3SABC
7Example on Inclusion/Exclusion Rule (3 sets) Q2) How many people got relief from all 3 drugs?By inclusion/exclusion rule:n(A B C) = n(A B C)- n(A) - n(B) - n(C)+ n(A B) + n(A C) + n(B C)= 37 – 23 – 18 – = 9Q3) How many people got relief from A only?n(A – (B C)) (by inclusion/exclusion rule)= n(A) – n(A B) - n(A C) + n(A B C)= 23 – 11 – = 2
8The Principle of Inclusion-Exclusion Proof: We show that each element is counted exactly once.Assume element ‘a’ is in r sets out of the n sets A1,...,An.-The first term counts ‘a’ r-times=C(r,1).-The second term counts ‘a’ -C(r,2) times (there are C(r,2) pairs in a set of r elements).-The k’th term counts ‘a’ -C(r,k) times (there are C(r,k) k-subsets in a set of r elements).-...- If k=r then there are precisely (-1)^(r+1) C(r,r) terms.- For k>r ‘a’ is not in the intersection: it is counted 0 times.Total: C(r,1)-C(r,2)+...+(-1)^(r+1)C(r,r)Now use: to show that each element is counted exactly once.
9Applications of Incl.-Excl. We can use inclusion/exclusion to count the number of members of aset that do not have a bunch of properties: P1,P2,...,Pn.Call N(Pi,Pj,Pk,...) the number of elements of a set that do have propertiesPi, Pj, Pk,.... and N the total number of elements in the set.By inclusion/exclusion we then have:Theorem: Let Ai be the subset of elements of a set A that has property Pi.The number of elements in a set A that do not haveproperties P1,...Pn is given then by:
11Examples Compute the number of solutions to x1+x2+x3=11 where x1,x2,x3 non-negative integers and x1 <=3, x2<=4, x4<=6.P1: x1 > 3P2: x2 > 4P3: x3 > 6The solution must have non of the properties P1,P2,P3.The solution of a problem x1+x2+x3=11 with constraints x1 > 3 is solved asfollows:7 more balls4 balls inbasket x1already.Total number of ways:C(7+3-1,7)=36x x x3Therefore: N-N(P1)-N(P2)-N(P3)+N(P1,P2)+N(P2,P3)+N(P1,P3)-N(P1,P2,P3)= C(11+3-1,11) – C(7+3-1,7) – C(6+3-1,6) – C(4+3-1,4) + … – 0.
12Connection with De Morgan’s law So we have 2 ways to solve the last example:x1+x2+x3 = 11 such that non of the following properties hold:P1: x1 > 3P2: x2 > 4P3: x3 > 6or x1+x2+x3=11 such all of the following properties hold:Q1=NOT P1: 0<x1<=3Q2=NOT P2: 0<x2<=4Q3=NOT P3: 0<x3<=6Sometimes this is easier to compute.
13Number of Onto-Functions Onto or surjective functions:A function f from A to B is onto if for every element b in B there is anelement a in A with f(a)=b.If we have m elements in A and n in B, how manyonto functions are there?We want all yi in the range of the function f.Call Pi the property that yi is not in therange of the the function f.Then we are looking for the number of functionsthat has none of the properties P1,...,PnxyfABThere is no element without incoming arrows
14Number of Onto-Functions N(P1’P2’P3’) =N-N(P1)-N(P2) N(P1,P2) -...+(-1)^n N(P1,...,Pn).N: number of function from A B: n^mN(Pi): number of functions that do not have y1 inits range: (n-1)^m.There are n=C(n,1) such terms.N(Pi,Pj): (n-2)^m with C(n,2) terms.Total:n^m – C(n,1)(n-1)^m + C(n,2)(n-2)^m –...+(-1)^(n-1)C(n,n-1)1^m.m=6 and n=3N(P1’P2’P3’) = 3^6 – C(3,1)2^6 + C(3,2)1^6 = 540xyfABThere is no element without incoming arrows