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Algorithms and Data Structures Lecture IIIPowerPoint Presentation

Algorithms and Data Structures Lecture III

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This Lecture

- Divide-and-conquer technique for algorithm design. Example problems:
- Tiling
- Searching (binary search)
- Sorting (merge sort).

Tiling: Trivial Case (n = 1)

- Trivial case (n = 1): tiling a 2x2 board with a hole:

- Idea – try somehow to reduce the size of the original problem, so that we eventually get to the 2x2 boards which we know how to solve…

Tiling: Dividing the Problem

- To get smaller square boards let’s divide the original board into for boards

- Great! We have one problem of the size 2n-1x2n-1!
- But: The other three problems are not similar to the original problems – they do not have holes!

Tiling: Dividing the Problem

- Idea: insert one tromino at the center to get three holes in each of the three smaller boards

- Now we have four boards with holes of the size 2n-1x2n-1.
- Keep doing this division, until we get the 2x2 boards with holes – we know how to tile those

Tiling: Algorithm

INPUT: n – the board size (2nx2n board), L – location of the hole.

OUTPUT: tiling of the board

Tile(n, L)

if n = 1 then

Trivial case

Tile with one tromino

return

Divide the board into four equal-sized boards

Place one tromino at the center to cut out 3 additional holes

Let L1, L2, L3, L4 denote the positions of the 4 holes

Tile(n-1, L1)

Tile(n-1, L2)

Tile(n-1, L3)

Tile(n-1, L4)

Divide and Conquer

- Divide-and-conquer method for algorithm design:
- If the problem size is small enough to solve it in a straightforward manner, solve it. Else:
- Divide: Divide the problem into two or more disjoint subproblems
- Conquer: Use divide-and-conquer recursively to solve the subproblems
- Combine: Take the solutions to the subproblems and combine these solutions into a solution for the original problem

- If the problem size is small enough to solve it in a straightforward manner, solve it. Else:

Tiling: Divide-and-Conquer

- Tiling is a divide-and-conquer algorithm:
- Just do it trivially if the board is 2x2, else:
- Divide the board into four smaller boards (introduce holes at the corners of the three smaller boards to make them look like original problems)
- Conquer using the same algorithm recursively
- Combine by placing a single tromino in the center to cover the three introduced holes

Binary Search

- Find a number in a sorted array:
- Just do it trivially if the array is of one element
- Else divide into two equal halves and solve each half
- Combine the results

INPUT: A[1..n] – a sorted (non-decreasing) array of integers, s – an integer.

OUTPUT: an index j such that A[j] = s. NIL, if "j (1£j£n): A[j] ¹ s

Binary-search(A, p, r, s):

if p = r then

if A[p] = s then return p

else return NIL

q¬ë(p+r)/2û

ret ¬ Binary-search(A, p, q, s)

if ret = NIL then

return Binary-search(A, q+1, r, s)

else return ret

Recurrences

- Running times of algorithms with Recursive calls can be described using recurrences
- A recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs
- Example: Binary Search

Binary Search (improved)

INPUT: A[1..n] – a sorted (non-decreasing) array of integers, s – an integer.

OUTPUT: an index j such that A[j] = s. NIL, if "j (1£j£n): A[j] ¹ s

Binary-search(A, p, r, s):

if p = r then

if A[p] = s then return p

else return NIL

q¬ë(p+r)/2û

if A[q] £ s then return Binary-search(A, p, q, s)

else return Binary-search(A, q+1, r, s)

- T(n) = Q(n) – not better than brute force!
- Clever way to conquer:
- Solve only one half!

Running Time of BS

- T(n) = Q(lg n) !

Merge Sort Algorithm

- Divide: If S has at least two elements (nothing needs to be done if S has zero or one elements), remove all the elements from S and put them into two sequences, S1 and S2 , each containing about half of the elements of S. (i.e. S1 contains the firstén/2ù elements and S2 contains the remaining ën/2û elements).
- Conquer: Sort sequences S1 and S2 using Merge Sort.
- Combine: Put back the elements into S by merging the sorted sequences S1 and S2 into one sorted sequence

Merge Sort: Algorithm

Merge-Sort(A, p, r)

if p < r then

q¬ ë(p+r)/2û

Merge-Sort(A, p, q)

Merge-Sort(A, q+1, r)

Merge(A, p, q, r)

Merge(A, p, q, r)

Take the smallest of the two topmost elements of sequences A[p..q] and A[q+1..r] and put into the resulting sequence. Repeat this, until both sequences are empty. Copy the resulting sequence into A[p..r].

Merge Sort Summarized

- To sort n numbers
- if n=1 done!
- recursively sort 2 lists of numbers ën/2û and én/2ù elements
- merge 2 sorted lists in Q(n) time

- Strategy
- break problem into similar (smaller) subproblems
- recursively solve subproblems
- combine solutions to answer

Running time of MergeSort

- Again the running time can be expressed as a recurrence:

Repeated Substitution Method

- Let’s find the running time of merge sort (let’s assume that n=2b, for some b).

Example: Finding Min and Max

- Given an unsorted array, find a minimum and a maximum element in the array

INPUT: A[l..r] – an unsorted array of integers, l £ r.

OUTPUT: (min, max) such that "j (l£j£r): A[j] ³min and A[j] £max

MinMax(A, l, r):

if l = r then return (A[l], A[r]) Trivial case

q¬ ë(l+r)/2û Divide

(minl, maxl)¬ MinMax(A, l, q)

(minr, maxr)¬ MinMax(A, q+1, r)

if minl < minr then min = minl else min = minr

if maxl > maxr then max = maxl else max = maxr

return (min, max)

Conquer

Combine

Next Week

- Analyzing the running time of recursive algorithms (such as divide-and-conquer)

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