Comp Sci 102 Discrete Math for Computer Science

Comp. Sci 102 Discrete Math for Computer Science 1 1 1 1 2 3 4 5 6 1 3 6 10 15 1 1 4 10 20 1 5 15 March 27, 2012 Prof. Rodger 1 6 1 Lecture adapted from Bruce Maggs/Lecture developed at Carnegie Mellon, primarily by Prof. Steven Rudich.

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Counting III 1 X+ 2 X+ 3 X

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

Example 4 0 4 1 4 2 24 4 3 4 4 = 16 Number of subsets of size 0, size 1, size 2, size 3, and size 4

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 Example jth street 4 3 2 1 0 0 1 2 kth avenue 3 4 2 cd street, 3 rd ave j+k k = 5 3 6 points on that row, this is the 4 th binomial coefficient of 6 items

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 2 1 0 0 1 2 kth avenue 3 4 Example: Level 4, 3 rd avenue 4 3 There are n k 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 n k=0 3 n k 2 1 2 = 0 0 2 n n 1 2 kth avenue 3 4 Example: Show for n=3

Show for n=3 2 3 3 + 1 0 12 20 + 32 2 2 3 3 3 + + 2 + 32 + 2 = 12 6 3 =

Level n 4 3 2 1 0 0 1 2 kth avenue 3 4 Show for k=2, n=5 n i=k i k n+1 = k+1

Show for k=2, n=5 3 2 + 2 2 1 + 3 4 5 + + 2 2 + 6 3 = 10 = 20

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…