COMPSCI 102 Introduction to Discrete Mathematics Counting III

COMPSCI 102 Introduction to Discrete Mathematics

Counting III Lecture 8 (September 21, 2007) 1 X+ 2 X+ 3 X

How many ways to rearrange the letters in the word “SYSTEMS”?

SYSTEMS 7 places to put the Y, 6 places to put the T, 5 places to put the E, 4 places to put the M, and the S’s are forced 7 X 6 X 5 X 4 = 840

SYSTEMS Let’s pretend that the S’s are distinct: S 1 YS 2 TEMS 3 There are 7! permutations of S 1 YS 2 TEMS 3 But when we stop pretending we see that we have counted each arrangement of SYSTEMS 3! times, once for each of 3! rearrangements of S 1 S 2 S 3 7! = 840 3!

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

Remember: The number of ways to arrange n symbols with r 1 of type 1, r 2 of type 2, …, rk of type k is: n! r 1!r 2! … rk!

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 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 4 /’s 24 4

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

How many integer solutions to the following equations? x 1 + x 2 + x 3 + x 4 + x 5 = 20 x 1, x 2, x 3, x 4, x 5 ≥ 0 Think of xk are being the number of gold bars that are allotted to pirate k 24 4

How many integer solutions to the following equations? x 1 + x 2 + x 3 + … + xn = k x 1, x 2, x 3, …, xn ≥ 0 n+k-1 n-1 = n+k-1 k

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 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 n types of elements is: n+k-1 = n-1 k

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 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

There is a correspondence between paths in a choice tree and the cross terms of the product of polynomials!

Choice Tree for Terms of (1+X)3 1 1 X X 1 X 1 X X 2 X 3 Combine like terms to get 1 + 3 X 2 + X 3

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 Binomial Coefficients binomial expression

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 + S + T + Y)5? 5!

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 BA 3 N 2 in the expansion of (B + A + N)6? The number of ways to rearrange the letters in the word BANANA

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!

There is much, much more to be said about how polynomials encode counting questions!

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 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 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 subsets with odd size

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 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 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 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 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 that many fundamental mathematical identities emerge…

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 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 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 =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 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 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 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 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 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 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

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

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 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 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 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 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 = 1 2 kth avenue 3 2 n n 4

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 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 <k, 0, 0, 0, …> <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 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>; 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, …> 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 X

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, . . . > will be represented by the formal power series: PV = i=0 a i. X i

V ! = < a 0, a 1, a 2, . . . > PV = a i. X i i=0 <0> is represented by 0 <k> 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 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 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! + RIGHT(V!); PV = (1+ X)n V! = nth row of Pascal’s triangle

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