COMPSCI 230 Discrete Mathematics for Computer Science Counting

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COMPSCI 230 Discrete Mathematics for Computer Science

COMPSCI 230 Discrete Mathematics for Computer Science

Counting III Lecture 8 1 X+ 2 X+ 3 X

Counting III Lecture 8 1 X+ 2 X+ 3 X

How many ways to rearrange the letters in the word “CARNEGIEMELLON”? 14 2 positions

How many ways to rearrange the letters in the word “CARNEGIEMELLON”? 14 2 positions of the Ns 12 3 positions of the Es 9 2 positions of the Ls 7! ways to order the rest

Arrange n symbols: r 1 of type 1, r 2 of type 2, …,

Arrange n symbols: r 1 of type 1, r 2 of type 2, …, rk of type k n-r 1 … r 2 n r 1 = = n - r 1 - r 2 - … - rk-1 rk n! (n-r 1)!r 1! (n-r 1 -r 2)!r 2! n! r 1!r 2! … rk! …

CARNEGIEMELLON 14! 2!3!2! = 3, 632, 428, 800

CARNEGIEMELLON 14! 2!3!2! = 3, 632, 428, 800

5 distinct pirates want to divide 20 identical, indivisible bars of gold. How many

5 distinct pirates want to divide 20 identical, indivisible bars of gold. How many different ways can they divide up the loot?

Sequences with 20 G’s and 4 /’s GG/G//GGGGGGGGG/ represents the following division among the

Sequences with 20 G’s and 4 /’s GG/G//GGGGGGGGG/ represents the following division among the pirates: 2, 1, 0, 17, 0 In general, the ith pirate gets the number of G’s after the (i-1)st / and before the ith / This gives a correspondence (bijection) between divisions of the gold and sequences with 20 G’s and 4 /’s

How many different ways to divide up the loot? Sequences with 20 G’s and

How many different ways to divide up the loot? Sequences with 20 G’s and 4 /’s 24 4

How many different ways can n distinct pirates divide k identical, indivisible bars of

How many different ways can n distinct pirates divide k identical, indivisible bars of gold? n+k-1 n-1 = n+k-1 k

Identical/Distinct Dice Suppose that we roll seven dice How many different outcomes are there,

Identical/Distinct Dice Suppose that we roll seven dice How many different outcomes are there, if order matters? 67 What if order doesn’t matter? (E. g. , Yahtzee) 12 7 (Corresponds to 6 pirates and 7 bars of gold)

Multisets A multiset is a set of elements, each of which has a multiplicity

Multisets A multiset is a set of elements, each of which has a multiplicity The size of the multiset is the sum of the multiplicities of all the elements Example: {X, Y, Z} with m(X)=0 m(Y)=3, m(Z)=2 Unary visualization: {Y, Y, Y, Z, Z}

Counting Multisets There number of ways to choose a multiset of size k from

Counting Multisets There number of ways to choose a multiset of size k from n types of elements is: n+k-1 = n-1 k

Taking k items from a set of n Order matters, no repetition: Order matters,

Taking k items from a set of n Order matters, no repetition: Order matters, repetition allowed: Order doesn’t matter, no repetition: n! (n-k)! nk n k Order doesn’t matter, repetition allowed: n+k-1 k

Polynomials Express Choices and Outcomes Products of Sum = Sums of Products ( +

Polynomials Express Choices and Outcomes Products of Sum = Sums of Products ( + + + )( + + )= +

b 3 b 1 b 2 t 1 t 2 b 1 t 1

b 3 b 1 b 2 t 1 t 2 b 1 t 1 b 1 t 2 b 2 t 1 b 2 t 2 b 3 t 1 b 3 t 2 t 1 (b 1+b 2+b 3)(t 1+t 2) = b 1 t 1 + b 1 t 2 + b 2 t 1 + b 2 t 2 + b 3 t 1 + b 3 t 2

What is a Closed Form Expression For ck? (1+X)n = c 0 + c

What is a Closed Form Expression For ck? (1+X)n = c 0 + c 1 X + c 2 X 2 + … + cn. Xn (1+X)(1+X)…(1+X) After multiplying things out, but before combining like terms, we get 2 n cross terms, each corresponding to a path in the choice tree ck, the coefficient of Xk, is the number of paths with exactly k X’s ck = n k

The Binomial Formula (1+X)n n 0 1 = X +…+ Xn 0 1 n

The Binomial Formula (1+X)n n 0 1 = X +…+ Xn 0 1 n Binomial Coefficients binomial expression

The Binomial Formula (1+X)0 = 1 (1+X)1 = 1 + 1 X (1+X)2 =

The Binomial Formula (1+X)0 = 1 (1+X)1 = 1 + 1 X (1+X)2 = 1 + 2 X + 1 X 2 (1+X)3 = 1 + 3 X 2 + 1 X 3 (1+X)4 = 1 + 4 X + 6 X 2 + 4 X 3 + 1 X 4

What is the coefficient of EMSTY in the expansion of (E + M +

What is the coefficient of EMSTY in the expansion of (E + M + S + T + Y)5? 5!

What is the coefficient of EMS 3 TY in the expansion of (E +

What is the coefficient of EMS 3 TY in the expansion of (E + M + S + T + Y)7? The number of ways to rearrange the letters in the word SYSTEMS

What is the coefficient of (X 1 r 1 X 2 r 2…Xkrk) in

What is the coefficient of (X 1 r 1 X 2 r 2…Xkrk) in the expansion of (X 1+X 2+X 3+…+Xk)n? n! r 1!r 2!. . . rk!

Power Series Representation n (1+X)n = k=0 “Product form” or “Generating form” = k=0

Power Series Representation n (1+X)n = k=0 “Product form” or “Generating form” = k=0 n k Xk Xk For k>n, n k “Power Series” or “Taylor Series” Expansion =0

By playing these two representations against each other we obtain a new representation of

By playing these two representations against each other we obtain a new representation of a previous insight: n (1+X)n = k=0 n Let x = 1, 2 n = k=0 n k Xk n k The number of subsets of an n-element set

By varying x, we can discover new identities: n (1+X)n = k=0 n Let

By varying x, we can discover new identities: n (1+X)n = k=0 n Let x = -1, 0= k=0 n Equivalently, k odd n k Xk n (-1)k k n = k even n k

The number of subsets with even size is the same as the number of

The number of subsets with even size is the same as the number of subsets with odd size

n (1+X)n = k=0 n k Xk Proofs that work by manipulating algebraic forms

n (1+X)n = k=0 n k Xk Proofs that work by manipulating algebraic forms are called “algebraic” arguments. Proofs that build a bijection are called “combinatorial” arguments

n k odd n k n = k even n k Let On be

n k odd n k n = k even n k Let On be the set of binary strings of length n with an odd number of ones. Let En be the set of binary strings of length n with an even number of ones. We gave an algebraic proof that On = En

A Combinatorial Proof Let On be the set of binary strings of length n

A Combinatorial Proof Let On be the set of binary strings of length n with an odd number of ones Let En be the set of binary strings of length n with an even number of ones A combinatorial proof must construct a bijection between On and En

An Attempt at a Bijection Let fn be the function that takes an n-bit

An Attempt at a Bijection Let fn be the function that takes an n-bit string and flips all its bits fn is clearly a one-to-one and onto function for odd n. E. g. in f 7 we have: . . . but do even n work? In f 6 we have 0010011 1101100 1001101 0110010 110011 001100 101010 010101 Uh oh. Complementing maps evens to evens!

A Correspondence That Works for all n Let fn be the function that takes

A Correspondence That Works for all n Let fn be the function that takes an n-bit string and flips only the first bit. For example, 0010011 101001101 0001101 110011 010011 101010 001010

n (1+X)n = k=0 n k Xk The binomial coefficients have so many representations

n (1+X)n = k=0 n k Xk The binomial coefficients have so many representations that many fundamental mathematical identities emerge…

The Binomial Formula (1+X)0 = 1 (1+X)1 = 1 + 1 X (1+X)2 =

The Binomial Formula (1+X)0 = 1 (1+X)1 = 1 + 1 X (1+X)2 = 1 + 2 X + 1 X 2 (1+X)3 = 1 + 3 X 2 + 1 X 3 (1+X)4 = 1 + 4 X + 6 X 2 + 4 X 3 + 1 X 4 Pascal’s Triangle: kth row are coefficients of (1+X)k Inductive definition of kth entry of nth row: Pascal(n, 0) = Pascal (n, n) = 1; Pascal(n, k) = Pascal(n-1, k-1) + Pascal(n-1, k)

“Pascal’s Triangle” 0 =1 0 1 =1 0 2 =1 0 3 =1 0

“Pascal’s Triangle” 0 =1 0 1 =1 0 2 =1 0 3 =1 0 1 =1 1 2 =2 1 3 =3 1 2 =1 2 3 =3 2 • Al-Karaji, Baghdad 953 -1029 • Chu Shin-Chieh 1303 • Blaise Pascal 1654 3 =1 3

Pascal’s Triangle “It is extraordinary 1 how fertile in properties the 1 1 triangle

Pascal’s Triangle “It is extraordinary 1 how fertile in properties the 1 1 triangle is. 1 2 1 Everyone can try his 1 3 3 1 hand” 1 4 6 4 1 1 1 5 6 10 15 10 20 5 15 1 6 1

Summing the Rows n 2 n = n k 1 =1 1 + 1

Summing the Rows n 2 n = n k 1 =1 1 + 1 =2 1 + 2 + 1 =4 1 + 3 + 1 =8 1 + 4 + 6 + 4 + 1 = 16 1 + 5 + 10 + 5 + 1 = 32 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 k=0

Odds and Evens 1 1 1 1 2 3 4 5 6 1 3

Odds and Evens 1 1 1 1 2 3 4 5 6 1 3 6 10 15 1 1 4 10 20 1 5 15 1 6 1 + 15 + 1 = 6 + 20 + 6 1

Summing on 1 st Avenue n 1 1 1 1 2 3 4 5

Summing on 1 st Avenue n 1 1 1 1 2 3 4 5 6 1 15 1 4 10 1 5 15 1 6 i=1 1 6 20 i = i=1 3 10 n 1 i = n+1 2 1

Summing on kth Avenue 1 1 1 1 6 2 4 6 1 4

Summing on kth Avenue 1 1 1 1 6 2 4 6 1 4 10 20 i=k 1 3 10 15 1 3 5 n 1 5 15 1 6 1 i = n+1 k

Fibonacci Numbers 1 =2 1 1 =3 = 5 1 2 1 =8 1

Fibonacci Numbers 1 =2 1 1 =3 = 5 1 2 1 =8 1 3 3 1 = 13 1 1 1 4 5 6 6 10 15 4 10 20 1 5 15 1 6 1

Sums of Squares 1 1 1 1 2 2 2 3 2 2 4

Sums of Squares 1 1 1 1 2 2 2 3 2 2 4 5 6 1 3 2 6 10 15 1 2 1 4 10 20 2 1 5 15 1 6 1

Al-Karaji Squares 1 1 =1 2 +2 1 =4 1 1 3 +2 3

Al-Karaji Squares 1 1 =1 2 +2 1 =4 1 1 3 +2 3 1 1 4 +2 6 5 +2 10 6 +2 15 4 10 20 =9 1 5 15 = 16 1 1 6 = 25 1 = 36

Pascal Mod 2

Pascal Mod 2

All these properties can be proved inductively and algebraically. We will give combinatorial proofs

All these properties can be proved inductively and algebraically. We will give combinatorial proofs using the Manhattan block walking representation of binomial coefficients

How many shortest routes from A to B? A B 10 5

How many shortest routes from A to B? A B 10 5

Manhattan jth street 4 There are j+k k 3 2 1 0 0 1

Manhattan jth street 4 There are j+k k 3 2 1 0 0 1 2 kth avenue 3 4 shortest routes from (0, 0) to (j, k)

Manhattan Level n 4 There are 3 2 n k 1 0 0 1

Manhattan Level n 4 There are 3 2 n k 1 0 0 1 2 kth avenue 3 4 shortest routes from (0, 0) to (n-k, k)

Manhattan Level n 4 3 There are 2 1 n k 0 0 1

Manhattan Level n 4 3 There are 2 1 n k 0 0 1 2 kth avenue 3 4 shortest routes from (0, 0) to level n and kth avenue

Level n 4 3 2 1 1 1 6 0 0 1 5 1

Level n 4 3 2 1 1 1 6 0 0 1 5 1 1 4 1 3 2 6 1 3 1 1 4 2 kth avenue 3 1 4 1 1 10 10 5 15 20 15 6

Level n 4 3 2 1 0 1 kth avenue 1 1 1 3

Level n 4 3 2 1 0 1 kth avenue 1 1 1 3 2 1 3 1 1 4 1 6 1 1 5 10 + 10 5 6 15 20 15 6 1 n k 4 n-1 = + k-1 k

Level n 4 3 2 1 0 0 n k=0 n k 2 =

Level n 4 3 2 1 0 0 n k=0 n k 2 = 1 2 kth avenue 3 2 n n 4

Level n 4 3 2 1 0 0 n i=k i k 1 2

Level n 4 3 2 1 0 0 n i=k i k 1 2 kth avenue 3 n+1 = k+1 4

Vector Programs Let’s define a (parallel) programming language called VECTOR that operates on possibly

Vector Programs Let’s define a (parallel) programming language called VECTOR that operates on possibly infinite vectors of numbers. Each variable V! can be thought of as: < * , * , *, *, … >

Vector Programs Let k stand for a scalar constant <k> will stand for the

Vector Programs Let k stand for a scalar constant will stand for the vector <0> = <0, 0, …> <1> = <1, 0, 0, 0, …> V! + T! means to add the vectors position-wise <4, 2, 3, …> + <5, 1, 1, …. > = <9, 3, 4, …>

Vector Programs RIGHT(V!) means to shift every number in V! one position to the

Vector Programs RIGHT(V!) means to shift every number in V! one position to the right and to place a 0 in position 0 RIGHT( <1, 2, 3, …> ) = <0, 1, 2, 3, …>

Vector Programs Example: Store: V! : = <6>; V! : = RIGHT(V!) + <42>;

Vector Programs Example: Store: V! : = <6>; V! : = RIGHT(V!) + <42>; V! : = RIGHT(V!) + <13>; V! = <6, 0, 0, 0, …> V! = <42, 6, 0, 0, …> V! = <2, 42, 6, 0, …> V!= <13, 2, 42, 6, …> V! = < 13, 2, 42, 6, 0, 0, 0, … >

Vector Programs Example: Store: V! : = <1>; V! = <1, 0, 0, 0,

Vector Programs Example: Store: V! : = <1>; V! = <1, 0, 0, 0, …> V! = <1, 1, 0, 0, …> V! = <1, 2, 1, 0, …> V!= <1, 3, 3, 1, …> Loop n times V! : = V! + RIGHT(V!); V! = nth row of Pascal’s triangle

1 X + 2 X + Vector programs can be implemented by polynomials! 3

1 X + 2 X + Vector programs can be implemented by polynomials! 3 X

Programs Polynomials The vector V! = < a 0, a 1, a 2, .

Programs Polynomials The vector V! = < a 0, a 1, a 2, . . . > will be represented by the polynomial: PV = i=0 a i. X i

Formal Power Series The vector V! = < a 0, a 1, a 2,

Formal Power Series The vector V! = < a 0, a 1, a 2, . . . > will be represented by the formal power series: PV = i=0 a i. X i

V ! = < a 0, a 1, a 2, . . . >

V ! = < a 0, a 1, a 2, . . . > PV = a i. X i i=0 <0> is represented by 0 is represented by k V! + T! is represented by RIGHT(V!) is represented by (PV + PT) (PV X)

Vector Programs Example: V! : = <1>; PV : = 1; Loop n times

Vector Programs Example: V! : = <1>; PV : = 1; Loop n times V! : = V! + RIGHT(V!); PV : = PV + PV X; V! = nth row of Pascal’s triangle

Vector Programs Example: V! : = <1>; PV : = 1; Loop n times

Vector Programs Example: V! : = <1>; PV : = 1; Loop n times V! : = V! + RIGHT(V!); PV : = PV(1+X); V! = nth row of Pascal’s triangle

Vector Programs Example: V! : = <1>; Loop n times V! : = V!

Vector Programs Example: V! : = <1>; Loop n times V! : = V! + RIGHT(V!); PV = (1+ X)n V! = nth row of Pascal’s triangle

 • Polynomials count • Binomial formula • Combinatorial proofs of binomial identities •

• Polynomials count • Binomial formula • Combinatorial proofs of binomial identities • Vector programs Here’s What You Need to Know…