Data Structures and Algorithms I Day 1, 8/30/11 CMP 338 Data Structures and Algorithms I Day 1, 8/30/11

The Course Instructor: Bowen Alpern Texts: Email: alpern@acm.org Office hour: GI 137-I Tuesday 4-5pm (and by appointment) (poor speller, nominal aphasia, limited teaching experience) Texts: Algorithms, fourth edition, Sedgewick and Wayne Booksite: algs4.cs.princeton.edu Format: T, Th 6-7:40pm, room 333 Gillet Lecture (100 min) Grade policy (tentative) Program assignments 40% Exams 40% (4 in class, 5% each, final 20%) Homework 20%

CMP 338 “Algorithms + Data Structures = Programs” - Niklaus Wirth Algorithm – a method for solving a problem Data Structure – a method for storing information Why study algorithms (and data structures)?

I'm thinking of a number ...

I'm thinking of a number ... 0 ≤ x < 1024 = 210

I'm thinking of a number ... 0 ≤ x < 1024 = 210 Queries: is x ≤ y ? (for any integer y)

I'm thinking of a number ... 0 ≤ x < 1024 = 210 Queries: is x ≤ y ? (for any integer y) Algorithm: lo = 0; hi =1023 while (lo < hi) if (x ≤ lo + (lo+hi)/2) then hi = lo + (lo+hi)/2 else lo = lo + (lo+hi)/2 + 1 return lo // or hi

Analysis If 0 ≤ x < N = 2n (n = log2 N ≡ lg N) x has n bits Each query resolves 1 bit Why? Each query consists of a single comparison Therefore: Guessing one of N consecutive integers can be done with ~ lg N comparisons

So What?

So What? Same algorithm tests membership in a sorted list

So What? Same algorithm tests membership in a sorted list boolean member(Item i, List<Item> l) lo = 0; hi = l.length -1 while (lo < hi) int m = lo + (lo+hi)/2 if (0 ≤ i.compareTo(l.get(m))) then hi = m else lo = m + 1 return i.equals(l.get(lo))

Search Binary search Work to look up 1 of N keys ~ lg N comparisons Can we do better? Kabbalah Assign a numeric value to each letter of an alphabet Tries Work to look up key ~ ||key|| Hashing Work to look up key ~ c (average case)

Polynomials Polynomials over x with integer coefficients Operations: +, -, ●, /, %, evaluate, … First implementation: array of ints p[i] – coefficient of xi Good for dense polynomials Second implementation: linked list of nodes int coefficient Nomial link Good for sparse polynomials

Matrices Implementation: 2-D array of double Multiply-add: for (int i=0; i<M; i++) for (int j=0; j<N; j++) for (int k=0; k<L, k++) c[i][j] += a[i][k]*b[k][j] Analysis: ~ MNL double multiply-adds ~ N3 for square matrices Data structures for sparse matrices

Selection Select the kth smallest entry in a collection Application: sort the √N smallest entries Select the √Nth smallest Partition on k = √N Sort the partition with smallest entries Analysis: Select ??? Partition ~ N (next slide) Sort ~ c √N log N << N Can this be done in linear time?

Partitioning int partition(Item[] a, Item p, int lo, int hi) int i = lo-1, j = hi while (true) while (less(a[++i],p) if (i == hi) break; while (less(p, a[--j]) if (j == lo) break; swap(a, i, j) swap(a, lo, j) return j; // a[lo..j] < p & p < a[j+1..hi]

Analysis Partitioning N Items ~ 2N compares Average case: Two partitions are approximately equal in size Worst case: One partition is empty (or has only 1 Item)

Selection Algorithm Item select(Item[] a, int k, int lo, int hi) if (k == 0 && lo == hi) return a[lo] Item p = choosePivot(a, lo, hi) // a[lo+(lo+hi)/2] int j =partition(a, p, lo, hi) if (k <= j) return select(a, k, lo, j) else return select(a, k-(j+1), j+1, hi)

Selection Analysis Average case: T(N) < 2N + T(N/2 + c√N) ~ c' N Worst case: T(N) < 2N + T(N-1) ~ c N2 ~ c k N How can worst case performance be improved? Answer: choose a better pivot!

Choosing a Good Pivot Item pivot(Item[] a, int lo, int hi) Item[] b = new Item[N/5] for (int j=0; j<N/5; j++) b[j] = median5(a, j*5) Item p = select(b, N/10, 0, (N/5)-1) return p Item median5(Item[a], int lo) sort(a, lo, lo+4) return a[lo+2]

Analysis Computing b ~ c N Pivot : Tp(N) < c' N + Ts(N/5) Select: Ts(N) < c'' N + Ts(N/5) + Ts(7N/10) ~ c''' N Why Ts(7N/10)? Largest partition size <= 70% N Pivot is less than half of b Each element of b is less than 2 elements of a Pivot is less than 30% of a Similarly, pivot is greater than 30% of a

Homework 1) Create a Matrix library Constructor create a random N x M matrix void mutiplyAdd does c += a*b for conformant matrices a, b, and c 2) Code partition() Get the details right 3) Code select() 4) Code pivot()