Discrete Math For Computing II

Discrete Math For Computing II

Main Text: Topics in enumeration; principle of inclusion

and exclusion, Partial orders and lattices. Algorithmic complexity; recurrence relations, Graph theory.

Prerequisite: CS 2305 "Discrete Mathematics and its Applications"

Kenneth Rosen, 5th Edition, McGraw Hill.

Contact Information

B. PrabhakaranDepartment of Computer ScienceUniversity of Texas at DallasMail Station EC 31, PO Box 830688Richardson, TX 75083Email: praba@utdallas.eduPhone: 972 883 4680; Fax: 972 883 2349URL: http://www.utdallas.edu/~praba/cs3305.htmlOffice: ES 3.706Office Hours: 1-2pm Tuesdays, Thursdays

Other times by appointments through emailAnnouncements: Made in class and on course web page.TA: TBA.

Course Outline

Selected topics in chapters 6 through 9. Chapter 6: Advanced Counting Techniques:

recurrence relations, principle of inclusion and exclusion

Chapter 7: Relations: properties of binary relations, representing relations, equivalence relations, partial orders

Chapter 8: Graphs: graph representation, isomorphism, Euler paths, shortest path algorithms, planar graphs, graph coloring

Chapter 9: Trees: tree applications, tree traversal, trees and sorting, spanning trees

Course ABET Objectives Ability to construct and solve recurrence relations Ability to use the principle of inclusion and exclusion to solve

problems Ability to understand binary relations and their applications Ability to recognize and use equivalence relations and partial

orderings Ability to use and construct graphs and graph terminology Ability to apply the graph theory concepts of Euler and

Hamilton circuits Ability to identify and use planar graphs and shortest path

problems Ability to use and construct trees and tree terminology Ability to use and construct binary search trees

Evaluation

1 Mid-terms: in class. 75 minutes. Mix of MCQs (Multiple Choice Questions) & Short Questions.

1 Final Exam: 75 minutes or 2 hours (depending on class room availability). Mix of MCQs and Short Questions.

2 - 3 Quizzes: 5-6 MCQs or very short questions. 15-20 minutes each.

Homeworks/assignments: 3 or 4 spread over the semester.

Homeworks

Each homework will be for 10 marks. Homeworks Submission:

Submit on paper to TA/Instructor.

Grading

Home works: 5% Quizzes: 15% Mid-term : 40% Final: 40%

Likely Letter Grades

Mostly Relative grading A-, A, A+: 1.2 – 1.4 times class average B-, B, B+: 1 – 1.2 class average C-, C, C+: 0.8 – 1.0 class average D-, D, D+: 0.7 – 0.8 class average F: Below 0.6 times class average Absolute scores will also influence above the ranges.

Schedule

Quizzes: Dates announced in class & web, a week ahead. Mostly before midterm and final.

Mid-term: February 23, 2006 Final Exam: Last day of class (April 20th) OR 11am,

May 1, 2006 (As per UTD schedule) Subject to minor changes Quiz and homework schedules will be announced in

class and course web page, giving sufficient time for preparation.

Cheating

Academic dishonesty will be taken seriously.

Cheating students will be handed over to Head/Dean for further action.

Remember: home works (and exams too!) are to be done individually.

Any kind of cheating in home works/exams will be dealt with as per UTD guidelines.

Chapter 6

Advanced Counting Techniques

§ 6.1: Recurrence Relations

Definition: A recurrence relation for the sequence {an} is an equation expressing an in terms of one or more of the previous terms of the sequence:

a1,a2,a3,…,an-1, with n>=n0, (n0 being a nonnegative integer).

A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.

Recurrence Example Consider the recurrence relation

an = 2an−1 − an−2 (n≥2). Which of the following are solutions?

an = 3n Yes -> 2 [3(n-1)] – 3(n-2)

= 3n => an

an = 2n No -> a0 = 1, a1 = 2, a2 = 4;

a2 = 2a1 – a0 = 2.2 – 1 = 3 ≠ a2

an = 5 Yes -> an = 2.5 – 5 = 5 = an

Initial conditions

The initial conditions (base conditions):

Specify the terms that precede the first term where the recurrence relation takes effect.

i.e., specify a0

Example Applications

Growth of bank accountInitial Amount P0=$10,000

After n Years= Pn

Compound Interest I = 11%Soln: Pn =Pn-1+(I/100)Pn-1=(1.11)Pn-1

Using Iteration we get, Pn =(1.11)n P0

i.e P30 =(1.11)3010,000=$228,922.97

Example Application

Rabbits and Fibonacci NumbersGrowth of rabbit population in which each rabbit yields 1 new one every period starting 2 periods after its birth.

Pn = Pn−1 + Pn−2 (Fibonacci relation)

Classic Tower of Hanoi Example Problem: Get all disks from peg 1 to

peg 2. Only move 1 disk at a time. Never set a larger disk on a smaller one.

Hanoi Recurrence Relation

Let Hn = # moves for a stack of n disks. Optimal strategy:

Move top n−1 disks to spare peg. (Hn−1 moves)

Move bottom disk. (1 move) Move top n−1 to bottom disk. (Hn−1 moves)

Note: Hn = 2Hn−1 + 1

Why is Hn = 2Hn-1 + 1

Only move 1 disk at a time. Never set a larger disk on a smaller one.

Solving Tower of Hanoi RR

Hn = 2 Hn−1 + 1

= 2 (2 Hn−2 + 1) + 1 = 22 Hn−2 + 2 + 1

= 22(2 Hn−3 + 1) + 2 + 1 = 23 Hn−3 + 22 + 2 + 1 … = 2n−1 H1 + 2n−2 + … + 2 + 1

= 2n−1 + 2n−2 + … + 2 + 1 (since H1=1)

= 2n − 1

§6.2: Solving Recurrences

Definition: A linear homogeneous recurrence relation of degree k with constant coefficient is a recurrence relation of the forman = c1an−1 + … + ckan−k,where the ci are all real, and ck ≠ 0.

The solution is uniquely determined if k initial conditions a0…ak−1 are provided

§6.2: Solving Recurrences.. Linear?: Right hand side is sum of

multiples of previous terms. Homogenous?: No terms that are NOT

multiples of the ajs. Degree?: k-degree since previous k

terms are used.

Solving with const. Coefficients Basic idea: Look for solutions of the form

an = rn, where r is a constant. This requires the characteristic

equation:rn = c1rn−1 + … + ckrn−k, i.e., rk − c1rk−1 − … − ck = 0 (Dividing both sides

by rn-k and subtracting right hand side from left).

The solutions (characteristic roots) can yield an explicit formula for the sequence.

Solving ……

Theorem1: Let c1 and c2 be real numbers.

Suppose that r2 − c1r − c2 = 0 has two distinct roots r1and r2. Then the sequence {an} is a solution of the recurrence relation an = c1an−1 + c2an−2 if and only if an = α1r1

n + α2r2n

for n≥0, where α1, α2 are constants.

Theorem 1: Proof 2 things to prove

Case 1: Roots are r1 & r2, i.e., {an = α1r1n +

α2r2n} an is a solution.

r1 & r2 are roots of r2 − c1r − c2 = 0 r12 =

c1r1 + c2; r22 = c1r2 + c2.

c1an−1 + c2an−2 = c1(α1r1n-1 + α2r2

n-1) + c2(α1r1

n-2 + α2r2n-2)

α1r1n-2 (c1r1 + c2) + α2r2

n-2 (c1r2 + c2)

α1r1n-2 r1

2 + α2r2n-2 r2

2 an

Theorem 1: Proof … 2 things to prove

Case 2: an is a solution an = α1r1n + α2r2

n for some α1 & α2.

an is a solution a0 = C0 = α1 + α2.

C1 = α1r1 + α2r2

α1 = (C1 – C0r2)/(r1- r2) α2 = C0 – α1 = (C0r1 - C1)/(r1- r2) Works only if r1≠ r2

an = α1r1n + α2r2n works for 2 initial conditions Since the initial conditions uniquely

determine the sequence, an = α1r1n + α2r2

n

Example Solve the recurrence an = an−1 + 2an−2 given the

initial conditions a0 = 2, a1 = 7. An = rn rn = rn-1 + 2rn-2 r2 = r +2 Solution: Use theorem 1

c1 = 1, c2 = 2 Characteristic equation:

r2 − r − 2 = 0 Solutions: r = [−(−1)±((−1)2 −

4·1·(−2))1/2] / 2·1= (1±91/2)/2 = (1±3)/2, so r = 2 or r

= −1. So an = α1 2n + α2 (−1)n.

Example Continued… To find α1 and α2, solve the equations for the

initial conditions a0 and a1: a0 = 2 = α120 + α2 (−1)0

a1 = 7 = α121 + α2 (−1)1

Simplifying, we have the pair of equations:2 = α1 + α2

7 = 2α1 − α2which we can solve easily by substitution:

α2 = 2−α1; 7 = 2α1 − (2−α1) = 3α1 − 2; 9 = 3α1; α1 = 3; α2 = -1.

Final answer: an = 3·2n − (−1)n

The Case of Degenerate Roots Theorem2: Let c1 and c2 be real

numbers with c2 ≠ 0. Suppose that r2

− c1r − c2 = 0 has only one root r0.

A sequence {an} is a solution of the recurrence relation an=c1an-1 + c2an-2 if and only if an = α1r0

n + α2nr0n, for all

n≥0, for some constants α1, α2.

k-Linear Homogeneous Recurrence Relations with Constant CoefficientsTheorem3: Let c1,c2,….ck be real numbers. Suppose the C.E.

If this has k distinct roots ri, then the solutions

to the recurrence are of the form: if and only if

for all n≥0, where the αi are constants.

k

iinin aca

1

01

k

i

iki

k rcr

k

i

niin ra

1

Theorem 3: ExampleLet an = 6an-1 – 11an-2 +6an-3;

a0=2,a1=5, & a2=15.

C.E.: r3 – 6r2 + 11r – 6; Roots = 1,2, & 3.Solution: an = α1.1n + α2.2n + α3.3n

a0 = 2 = α1 + α2 + α3

a1 = 5 = α1 + α2.2 + α3.3

a2 = 15 = α1 + α2.4 + α3.9

α1 = 1; α2 = 1; α3 = 2.

an = 1 – 2n + 2.3n

Degenerate t-roots Theorem 4: Suppose there are t roots r1,

…,rt with multiplicities m1,…,mt. mi >= 1 for i = 1…t. Then:

for all n≥0, where all the α are constants.

t

i

ni

m

j

jjin rna

i

1

1

0,

Theorem 4: Example E.g., Roots of C.E. are 2, 2, 2, 5, 5, & 9. Solution: (α1,0 + α1,1n + α1,2n2).2n + (α2,0 + α2,1n).5n +

α3,09n

Linear NonHomogeneous Recurrence Relations with Constant Coefficients

Linear NonHomogeneous RRs with constant coefficients may (unlike LiHoReCoCos) contain some terms F(n) that depend only on n (and not on any ai’s). General form:

an = c1an−1 + … + ckan−k + F(n)

The associated homogeneous recurrence relation(associated LiHoReCoCo).

Solutions of LiNoReCoCos

Theorem 5: If an(p) is a particular

solution to the LiNoReCoCo, then

Then all its solutions are of the form:an = an

(p) + an(h) ,

where an(h) is a solution to the associated

homogeneous RR

)(1

nFacak

iinin

k

iinin aca

1

Theorem 5: Proof {an

(p)} is a particular solution: an

(p) = c1an-1(p) + c2an-2

(p) +..+ckan-k(p) + F(n) (1)

Let bn be a 2nd solution: bn = c1bn-1 + c2bn-2 +..+ckbn-k + F(n) (2)

(2) – (1): bn – an

(p) = c1(bn-1 - an-1(p)) + c2(bn-2 - an-2

(p)) +… + ck(bn-k - an-k

(p))

Hence, {bn – an(p) } is a solution of the

associated homogeneous RR, say {an(h) }. bn

= an(p) + an

(h) .

Example

Find all solutions to an = 3an−1+2n. Which solution has a1 = 3? To solve this LiNoReCoCo, solve its associated

LiHoReCoCo equation: an = 3an−1, and its solutions are an

(h) = α3n, where α is a constant. By Theorem 5, the solutions to the original

problem are all of the form an = an(p) + α3n.

So, all we need to do is find one an(p) that

works.

Trial Solutions Since F(n)=2n, i.e it is linear so a reasonable

trial solution is a linear function in n, say pn = cn + d.Then the equation an = 3an−1+2n becomes,

cn+d = 3(c(n−1)+d) + 2n, (for all n)

Simplifying, we get

(−2c+2)n + (3c−2d) = 0 (collect terms)So c = −1 and d = −3/2.So a(p)

n = −n − 3/2 is a solution.

Finding a Desired Solution From Theorem 5, we know that all general

solutions to our example are of the form:an = −n − 3/2 + α3n.

Solve this for α for the given case, a1 = 3:

3 = −1 − 3/2 + α31

α = 11/6 The answer is an = −n − 3/2 + (11/6)3n

Theorem 6

Suppose {an} satisfies LiNoRR:

And F(n)= ; bt and s are real nos.

When s is not a root of C.E. of the associated LiHoRR, there is a particular solution of the form

)(1

nFacak

iinin

nt

i

itit snb

0

nt

i

itit snp

0

Theorem 6 continue….. When s is a root of this C.E. and its

multiplicity is m, there is a particular solution of the form

This factor nm ensures that the proposed particular solution will not already be a solution of the associated LiHoRR.

nt

i

itit

m snpn

0

Theorem 6: Example

an

= 6an-1

– 9an-2

+ F(n); F(n) = 3n

C.E.: r2 – 6r + 9 => (r-3)2 = 0; r = 3. Solution: By Theorem 6, s = 3,

multiplicity m = 2. Solution is of the form p0n23n, where p0 is polynomial constant.

§6.3: Divide & Conquer R.R.s

Many types of problems are solvable by reducing a problem of size n into some number a of independent subproblems, each of size n/b, where a1 and b>1.

The time complexity to solve such problems is given by a recurrence relation: f(n) = a·f(n/b) + g(n) where g(n) is the extra operations required.

This is called a divide-and-conquer recurrence relation

Divide+Conquer Examples Binary search: Break list into 1 sub-

problem (smaller list) (so a=1) of size n/2 (so b=2). So f(n) = 1.f(n/2)+2 (g(n)=2 constant)

Merge sort: Break list of length n into 2 sublists (a=2), each of size n/2 (so b=2), then merge them, in g(n) = n operations So M(n) = 2M(n/2) + n

Fast Multiplication Example This algorithm splits each of two 2n-bit

integers into two n bits blocks. Thus, from multiplications of two 2n-bit integers, it is reduced to only three multiplications of n bit integers, plus shifts and additions

To find the product ab of two 2n-digit-base 2 numbers, a=(a2n-1a2n-2…a0)2 and b=(b2n-1b2n-2…b0)2, first, we break them in half:

a=2nA1+A0, b=2nB1+B0,

))((2

)12()22(

)(2

)12()22(

))()((2

2

)(22

)2)(2(

1001

00112

10001101

00112

000011111001

00112

001001112

0101

BBAA

BABA

BABABABA

BABA

BABABABABABA

BABA

BABABABA

BBAAab

n

nnn

n

nnn

n

n

nn

nn

Derivation of Fast Multiplication

Zero

(Multiply outpolynomials)

(Factor last polynomial)

Three multiplications, each with n-digit numbers

Recurrence Rel. for Fast Mult.

Notice that the time complexity f(n) of the fast multiplication algorithm obeys the recurrence:

f(2n)=3f(n)+C(n)i.e.,

f(n)=3f(n/2)+C(n)So a=3, b=2.

Time to do the needed adds & subtracts of n-digit and 2n-digitnumbers

Theorem1 Let f(n) = af(n/b) + c whenever n is divisible

by b, where a1, b is an integer greater than 1, and c is a positive real number. Then

if a > 1

if a = 1

Furthermore, when n =bk, where k is a positive integer,

f(n) = C1 + C2.

where C1 = f(1) + c/(a-1) and C2 = -c/(a-1)

)(nf )()(log

log abnOnO

is

abnlog

Theorem1: Proof n = bk; f(n) = akf(1) + a = 1: f(n) = f(1) + ck; Since n = bk, k = logbn f(n) = f(1)

+ clogbn. n is a power of b or not: f(n) is O(logn), when a = 1. a > 1: (from Theorem 1, Section 3.2) f(n) = akf(1) + c(ak – 1)/(a – 1)

= ak[f(1) + c/(a – 1)] – c/(a – 1) = C1nlog

ba + C2, since ak = alogbn = nlogba

f(n) is O(nlogb

a)

1

0

1

0

)1( k

j

jkk

j

j acfaca

MASTER Theorem Let f(n) = af(n/b) + cnd whenever n=bk,

where a1, b>1, and c and d are real number, with c positive and d nonnegative. Then

if a<bd

f(n) is if a=bd

if a>bd

)(

)log(log )(

dnO

ndnOabno

Example

We know that number of comparisons used by the merge sort to sort a list of n elements is less then M(n), where M(n)=2M(n/2) + n

From Master Theorem we find that M(n) is O(nlogn)

§6.4 Generating Functions Generating functions: Used to represent

sequences efficiently by coding the terms of a sequence as coefficients of powers of a variable x in a formal power series

Definition: Let S = {a0, a1, a2, a3, ...} be an (infinite) sequence of real numbers. Then the generating function G(x), of S is the seriesG(x)=a0+a1x+…+akxk+…=

0k

kxa k

Example of GF Let S = {1, 1, 1, 1, ... }. Then

G(x) =1+ x + x2 +...+ xk +... =

if S=1,1,1,1,1,1 By Theorem 1 of Section 3.2, (x6-1)/(x-1) = 1+ x + x2 +...+ x5

G(x) =(x6-1)/(x-1) is the generating function of sequence 1,1,1,1,1,1

0k

kx

Theorems of GF Theorem 1:

Let f(x) = and g(x)= . Then,

f(x) + g(x)= and

f(x)g(x)=

0k

kk xb

0k

kxa k

0

)(k

kk xba k

0 0k

kjk

k

jj xba

Example A expression for the generating function of the

sequence S= {1, 1, 1, 1, 1, 1, ...} is 1/(1- x) 1/(1- x) = 1 + x + x2 + x3 + … Let f(x) = 1/(1-x)2; f(x) = What are the coefficients of ak? From theorem 1, we have (1 + x + x2 + ..) x (1 + x

+ x2 + ..) x

1/(1- x)2 = =

0

)1(k

kxk

0 0

1k

kk

j

x

0k

kxa k

0k

kx

0k

kx

Permutations & Combinations

Permutation: An ordered arrangement of objects from a set of distinct objects. P(n,r): n – distinct objects; r – number of objects in

the permutation P(n,r) = n(n-1)(n-2)…(n-r+1) P(n,r) = n!/(n-r)! Example: Selecting first 3 prize winners from 100

different people 100 x 99 x 98 = 970,200. Combination: Unordered selection of r elements

from a set. C(n,r) = n!/[r!(n-r)!]; n is non-negative, 0 ≤ r ≤ n

Proof: r-permutation is ordered r-combinations.

Permutations & Combinations …

Proof: r-permutation is ordered r-combinations.: P(n,r) = C(n,r) x P(r,r) C(n,r) = P(n,r) / P(r,r) C(n,r) = [n!/(n-r)!]/[r!/(r-r)!] = n!/[r!(n-r)!]

Corollary 1: C(n,r) = C(n,n-r) if n & r are non-negative and r ≤ n.

Proof: C(n,n-r) = n!/[(n-r)!(n- (n-r))!] = n!/[(n-r)!r!] Note: C(n,n) = C(n,0) = 1. Example: Choosing 5 players from 10 member team.

C(10,5) = 10!/(5!5!) = 252.

Binomial Coefficients & Theorem

C(n,r): denoted by n r C(n,r): Binomial coefficient these numbers occur

as co-efficients in the expansion of powers of binomial expressions such as (a+b)n

Binomial Theorem: x, y are variables; n – non-negative integer. (x + y)n = ∑j=0 n C(n,j).xn-j.yj

= C(n,0)xn + C(n,1)xn-1y + C(n,2)xn-2y2 +….C(n,n-1)xyn-1 + C(n,n)yn

Example: (x + y)2 = (x+y)(x+y) = xx + xy + xy + yy = x2 + 2xy + y2

Binomial Coefficients & Theorem

Proof: Count the number of terms of form xn-j.yj Choose xs from the n sums (so that the other

terms are ys) Coefficient of xn-jyj is C(n,n-j) = C(n,j)

Example 1: Expansion of (x + y)4 = ∑j=0 4 C(4,j)x4-jyj

= C(4,0)x4 + C(4,1)x3y + C(4,2)x2y2 + C(4,3)xy3 + C(4,4)y4

= x4 + 4x3y + 6x2y2 + 4xy3 + y4

Example 2: Coeffiecient of x12y13 in (x+y)25

= C(25,13) = 25!/(13! 12!) = 5,200,300

Binomial Theorem: Corollary 1

Corollary 1: n is non-negative ∑k=0 n C(n,k) = 2n

Proof: 2n = (1 + 1)n = ∑k=0 n C(n,k)1k1n-k

= ∑k=0 n C(n,k) Basically, ∑k=0 n C(n,k) is the total number of

subsets of a set with n elements 2n

Binomial Theorem: Corollary 2

Corollary 2: n is positive ∑k=0 n (-1)kC(n,k) = 0 Proof: 0 = 0n = ((-1) + 1)n = ∑k=0 n C(n,k)(-1)k1n-k

= ∑k=0 n C(n,k)(-1)k

C(n,0) + C(n,2) + C(n,4) + .. = C(n,1) + C(n,3) + C(n,5) + …

Binomial Theorem: Corollary 3

Corollary 3: n is positive ∑k=0 n 2kC(n,k) = 3n

Proof: 3n = (1+2)n = ∑k=0 n C(n,k)(2)k1n-k

= ∑k=0 n C(n,k)2k

Permutations With Repetitions

Chapter 4.5 Theorem 1: Number of r-permutations of a set of n

objects with repetitions allowed is nr. Theorem 2: There are C(n+r-1, r) r-combinations

from a set with n elements when repetition of elements is allowed.

Extended Binomial Coefficients Definition: Let u be a real number and k a

non-negative integer. Then the extended binomial coefficient (EBC) is: if k>0,

if k=0.

!/)1)....(1(1

kkuuuuk

e.g:EBC(-2,3) = (-2)(-3)(-4) / 3! = -4. EBC(0.5,3)=(0.5)(-0.5)(-1.5) / 3! = 1/16

Example

11 1 ( 1, ) ,

r rn n rC n r r n r Z

r r

When the top parameter (u) is negative integer, the EBC can be expressed in terms of ordinary binomial coefficients. Using earlier definition: EBC = (-n)(-n-1)….(-n-r+1)/r!

= (-1)rn(n+1)…(n+r-1)/r!= (-1)r(n+r-1)!/(r!*(n-1)!). Therefore,

The Extended Binomial Theorem

Let x be a real number with |x| < 1and let u be a real number. Then

(1+x)u = k

k

uk x

0

Example Find GF for (1+x)-n and (1-x)-n

By EB Theorem, it follows that (1+x)-n =We know that

(1+x)-n =

Replacing x with –x we get(1-x)-n =

0k

knk x

),1()1( rrnCrnr

0

),1()1(k

kk xkknC

0

),1(k

kxkknC

Counting Problems and Generating Functions

Generating Functions can be used to solve a wide variety of counting problems. In particular they can be used to count the number of combinations of various types.

Example Use GF to find the number of ways to select r

objects of n different kinds {ar} if we must select at least one object of each kind?

Chapter 4, theorem. Each of the n kinds of objects contributes the

factor (x+x2+x3…+) to the GF G(x), Hence G(x)= (x+x2+x3…+)n=xn(1+ x+x2+x3…+) =xn/(1-x)n

Using EB Theorem, we haveG(x) = xn.(1-x)-n = xn = xn

G(x)=

0

)(r

rnr x

0

),1(r

rxrrnC

0

),1(r

rnxrrnC

Continue… Substituting t = n+r, we get

G(x)= Replacing t by r as the index of summation,

we getG(x)=

Hence, there are C(r-1,r-n) ways to select r objects of n different kinds if we must select at least one object of each kind.

nt

txnttC ),1(

nr

rxnrrC ),1(

Using GF to Solve Recurrence Relations

We can find the solution to a recurrence relation and its initial conditions by finding an explicit formula for the associated generating function.

Example Solve the R.R., ak=3ak-1 for k=1,2,3… and

initial condition a0=2.

Uses identity1/(1-ax) = Consequently, ak=2 . 3k

1 0

0 0 1 0 01 1 0

00

0 0

3 2.

( ) 3 3 3 ( )

( ) 3 (2 3 )1 3

k k

k k jk k j

k k j

k k k k

k k

a a a

G x a a x a a x a x a x a xG x

aG x a x x

x

0k

kk xa

§6.5 The principle of Inclusion-ExclusionThe principle of Inclusion-Exclusion

If A, B and C are finite sets then |A B| = |A| + |B| – |A B|

Continue…

If A, B and C are finite sets then|A B C| = |A| + |B| + |C| - |A B| - |B C| - |A C| - |A B C |

|A|

|C|

|B|

|A B|

|A C| |B C|

General Inclusion-Exclusion Theorem 1: Let A1,…,An be finite sets.

Then | A1 A2 …… An | =

|...|)1(...|||||| 211

111n

n

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General Theorem: Proof | A1 A2 …… An |: Show an element in the union is

counted only once in the right hand side. Let a be a member of exactly r sets (A1, …). a is counted C(r,1) times by 1st summation, C(r,2) by 2nd

summation of the intersections, and C(r,m) times by mth summation involving m sets Ai.

a is counted exactly C(r,1) – C(r,2) + C(r,3) - … + (-1)r+1C(r,r).

By Corollary 2 of Section 4.4, C(r,0) - C(r,1) + C(r,2) - C(r,3) - … + (-1)rC(r,r) = 0.

C(r,0) = C(r,1) – C(r,2) + C(r,3) - … + (-1)r+1C(r,r). a is counted only once.

§6.6 Applications of Inclusion-Exclusion

Alternative Form: Let Ai be the subset containing the elements that have Property Pi.

Writing these quantities in terms of sets, we have

If the number of elements with none of the properties P1,…,Pn is denoted by N(P1’ P2’…Pn’)

then, N(P1’ P2’…Pn’)=N-

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|...| 21 nAAA

Continue From the inclusion-

exclusion principle, we see thatN(P1’ P2’…Pn’)=N

)...()1(

...)( )(||

21

111

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The Number of Onto Functions Let m and n be positive integers with (m n).

How many ways to assign m different jobs to n different employees if every employee is assigned at least one job?

Assume m = 5, n = 4, following Theorem 1 (next slide): 45 – C(4,1).35 + C(4,2).25 – C(4,3)15 = 240

The Number of Onto Functions Theorem: Let m and n be positive integers with

(m n). Then, there are

nm - C (n,1)·(n -1)m + C (n,2)·(n -2)m+…

+(-1)iC (n,i )·(n-i )m +…+ (-1)n-1C (n,n-1)·1m

onto functions from a set with m elements to a set with n elements

Derangements A derangement is a permutation of objects

that leaves no object in its original position.

Theorem: The number of derangements of a set with n elements is

!

11

!4

1

!3

1

!2

1

!1

11!

nnD n

n

Theorem Proof Property Pi: A permutation that fixes element i. Number of derangements: number of permutations

having none of the properties Pi for i = 1,2,…,n. Dn = N(P1’,P2’,…,Pn’) (from inclusion-exclusion)

= N -

N: number of permutations of n elements. So, N = P(n). Similarly: N(Pi) = P(n-1); N(Pi,Pj) = P(n-2). Inserting these values in the expression for Dn, we get

the formula.

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