Transform & Conquer Replacing One Problem With Another Saadiq Moolla.

Transform & Conquer

Replacing One Problem With

Another

Saadiq Moolla

Introduction

• Two Stage Solution−Transform into Another Problem−Solve New Problem

• Three Variations−Instance Simplification−Representation Change−Problem Reduction

Initial Problem

New Representati

on

Solution

Instance Simplification

• Reducing the Problem to a Simpler

One

• Techniques:−Presorting−Gaussian Elimination−Search Trees

Presorting

• Example: Given a random list of

numbers, determine if there are any

duplicates.

• Brute Force: Compare Every Pair

• Transform and Conquer:−Sort the List−Compare A [i] with A [i + 1] i

A

Presorting (Analysis)

• Old Idea, Many Different Ways

• Efficiency Dependant on Algorithm

• Compare Benefits vs Time Required

• Useful if Operation Repeated

Gaussian Eliminationa11x + a12x + … + a1nx = b1

a21x + a22x + … + a2nx = b2…

an1x + an2x + … + annx = b1

a’11x + a’12x + … + a’1nx = b’1

a’22x + … + a’2nx = b2…a’nnx = b1

Binary Search Trees8

415

7 91

5

Binary Search Trees (Analysis)• Time Efficiency

• Slow

• Methods to Balance

• Special Trees:−AVL Trees−B-trees

Heaps

• Type of Binary Tree

• Requirements:−Essentially Complete−Parental Dominance

9

5 7

4 2 1

Properties of Heaps

• Height is log2n

• Root is largest element

• Node + Descendents = Heap

• Stored in Array:−Children of A [i] are A [2i] and A [2i +

1]

Heap Insertion9

5 7

4 2 1 8

9

5

74 2 1

8

Heaps (Analysis)

• Sorting

• Priority Queue

• O (n log n)

Representation Change

• Change One Problem Into Another

• Steps:−Identify−Transform−Solve

• Mathematical Modeling

Problem Reduction

• Reduce to a Known Problem

• Use Known Algorithms:−Dijkstra’s algorithm−Horner’s Rule−Karp – Rabin algorithm−etc.

Horner’s Rule

• Used to Evaluate Polynomials

• 5x^2 – 3x + 8 » (5x + 3)x + 8

• 7x^3 + x^2 – 9x – 2 » ((7x + 1)x –

9)x – 2

• Linear

Horner’s Algorithmalgorithm Horner (x, P [])

// Evaluates a polynomial with coefficients P [] at xfor i ← n – 1 downto 0 do

v ← x * v + P [i]return v

Fast Exponentiation

• Evaluate x^n

• 2^10 = 2^10102 = 2^8 * 2^2

Fast Exponentiation Algorithm

Algorithm FastExp (x, n)// Returns x^nterm ← xproduct ← 1while n > 0 do

if n mod 2 = 1 thenproduct ← product * term

term ← term * termn ← └ n/2 ┘

return product