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Transform and Conquer

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balanced search trees, heaps and heapsort, polynomial evaluation by Horner s rule, Fast Fourier Transform a different problem altogether (problem reduction) – PowerPoint PPT presentation

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Title: Transform and Conquer


1
Transform and Conquer
  • Solve problem by transforming into
  • a more convenient instance of the same problem
    (instance simplification)
  • Presorting, Gaussian elimination, matrix
    inversion, determinant computation
  • a different representation of the same instance
    (representation change)
  • balanced search trees, heaps and heapsort,
    polynomial evaluation by Horners rule, Fast
    Fourier Transform
  • a different problem altogether (problem
    reduction)
  • reductions to graph problems, linear programming

2
Instance simplification - Presorting
  • Solve instance of problem by transforming into
    another simpler/easier instance of the same
    problem
  • Presorting
  • Many problems involving lists are easier when
    list is sorted.
  • element uniqueness
  • computing the mode
  • finding repeated elements
  • searching
  • computing the median (selection problem)

3
Selection Problem
  • Find the k-th smallest element in A1,An.
  • minimum k 1
  • maximum k n
  • median k n/2
  • Presorting-based algorithm
  • sort list
  • return Ak
  • Partition-based algorithm (decrease conquer)
  • pivot/split at As using partitioning algorithm
  • if sk return As
  • else if sltk repeat with sublist As1,An.
  • else if sgtk repeat with sublist A1,As-1.

4
Notes on Selection Problem
  • Presorting-based algorithm O(nlgn) T(1)
    O(nlgn)
  • Partition-based algorithm (decrease conquer)
  • worst case T(n) T(n-1) (n1) -gt T(n2)
  • best case T(n)
  • average case T(n) T(n/2) (n1) -gt T(n)
  • Bonus also identifies the k smallest elements
  • Special cases max, min better, simpler linear
    algorithm (brute force)
  • Conclusion Presorting does not help in this
    case.

5
Finding repeated elements
  • Presorting-based algorithm
  • use mergesort (optimal) T(nlgn)
  • scan array to find repeated adjacent elements
    T(n)
  • in total it makes T(nlgn)
  • Brute force algorithm T(n2)
  • Conclusion Presorting yields significant
    improvement

6
Checking element uniqueness
  • Brute force algorithm T(n2)
  • Algorithm PresortedElementUniqueness
  • Sort the array A
  • for i ? 0 to n-2 do
  • if AiAi1 return false
  • else return true
  • Conclusion Presorting again improves
  • Similar improvement for mode

7
Checking mode
  • Mode is the most often met element
  • Brute force scan the list and compute the
    frequencies. Then find the largest frequency
  • Algorithm PresortedMode
  • Sort the array A
  • i ? 1 modefrequency ? 0
  • while i ? n-1 do
  • runlength?1 runvalue?Ai
  • while irunlengthn-1 and Airunlengthrunvalu
    e
  • runlength ? runlength1
  • if runlengthgt modefrequency
  • modefrequency?runlength, modevalue?runvalue
  • i?irunlength
  • return modevalue
  • Conclusion Presorting again improves

8
Gaussian Elimination
  • Given a system of two linear equations with two
    unknowns
  • a11x a12y b1
  • a21x a22y b2
  • It has a unique solution unless the coefficients
    are proportional
  • Express one variable as function of the other and
    substitute to solve one equation.
  • What if the system has n equations and n unknowns
    ?

9
Gaussian Elimination (2)
  • Transform Axb to Axb, where A upper
    triangular
  • Then, solution is possible with backward
    substitution
  • Elementary operations
  • Exchange equations
  • Replace an equation with a nonzero multiple
  • Replace an equation with a sum or difference of
    this equation and some multiple of another
    equation
  • Example
  • 2x1 - x2 x3 1
  • 4x1 x2 - x3 5
  • x1 x2 x3 0

10
Gaussian Elimination (3)
  • Algorithm GaussElimination
  • for i ? 1 to n do Ai,n1?bi
  • for i? 1 to n-1 do
  • for j?i1 to n do
  • for k?i to n1 do
  • Aj,k?Aj,k-Ai,kAj,i/Ai,i
  • Potential problems if Ai,i is zero or very small

11
Gaussian Elimination (partial pivoting)
  • Algorithm GaussElimination2
  • for i ? 1 to n do Ai,n1?bi
  • for i? 1 to n-1 do
  • pivotrow?i
  • for j?i1 to n do
  • if Aj,igtApivot,i pivotrow?j
  • for k?i to n1 do
  • swap(Ai,k,Apivotrow,k)
  • for j?i1 to n do
  • temp?Aj,i/Ai,i
  • for k?I to n1 do
  • Aj,k?Aj,k-AI,ktemp
  • Efficiency

12
LU decomposition
  • Byproduct of Gaussian Elimination
  • Example ALU
  • LUxb. Denote yUx ? Lyb
  • Solve Lyb, then solve Uxy
  • Solve with as many times with different bs.
  • No extra space

1 0 0
2 1 0
1/2 1/2 1
2 -1 1
0 3 -3
0 0 2
13
Computing a matrix inverse
  • AA-1I
  • A singular matrix does not have an inverse
  • A matrix is singular if and only if one of the
    rows is a linear combination of the other rows.
  • Apply Gaussian elimination. If it yields an
    upper-triangular with no zeros on the diagonal,
    then the matrix is not singular
  • Axjej

14
Computing the determinant
  • A well-known recursive formula
  • What if n is large ? Efficiency ?
  • Apply Gaussian elimination.
  • The determinant of an upper-triangular matrix is
    the product of elements on its diagonal.
  • Efficiency ?
  • Cramers rule

15
Taxonomy of Searching Algorithms
  • Elementary searching algorithms
  • sequential search
  • binary search
  • binary tree search
  • Balanced tree searching
  • AVL trees
  • red-black trees
  • multiway balanced trees (2-3 trees, 2-3-4 trees,
    B trees)
  • Hashing
  • separate chaining
  • open addressing

16
Balanced trees AVL trees
  • For every node, difference in height between left
    and right subtree is at most 1
  • AVL property is maintained through rotations,
    each time the tree becomes unbalanced
  • lg n h 1.4404 lg (n 2) - 1.3277
    average 1.01 lg n
    0.1 for large n
  • Disadvantage needs extra storage for maintaining
    node balance
  • A similar idea red-black trees (height of
    subtrees is allowed to differ by up to a factor
    of 2)

17
AVL tree rotations
  • Small examples
  • 1, 2, 3
  • 3, 2, 1
  • 1, 3, 2
  • 3, 1, 2
  • Larger example 4, 5, 7, 2, 1, 3, 6
  • See figures 6.4, 6.5 for general cases of
    rotations

18
General case single R-rotation
19
Double LR-rotation
20
Balance factor
  • Algorithm maintains balance factor for each node.
    For example

21
Heapsort
  • Definition
  • A heap is a binary tree with the following
    conditions
  • it is essentially complete
  • The key at each node is keys at its children

22
Definition implies
  • Given n, there exists a unique binary tree with n
    nodes that is essentially complete, with h lg n
  • The root has the largest key
  • The subtree rooted at any node of a heap is also
    a heap

23
Heapsort Algorithm
  • Build heap
  • Remove root exchange with last (rightmost) leaf
  • Fix up heap (excluding last leaf)
  • Repeat 2, 3 until heap contains just one node.

24
Heap construction
  • Insert elements in the order given breadth-first
    in a binary tree
  • Starting with the last (rightmost) parental node,
    fix the heap rooted at it, if it does not satisfy
    the heap condition
  • exchange it with its largest child
  • fix the subtree rooted at it (now in the childs
    position)
  • Example 2 3 6 7 5 9

25
Root deletion
  • The root of a heap can be deleted and the heap
    fixed up as follows
  • exchange the root with the last leaf
  • compare the new root (formerly the leaf) with
    each of its children and, if one of them is
    larger than the root, exchange it with the larger
    of the two.
  • continue the comparison/exchange with the
    children of the new root until it reaches a level
    of the tree where it is larger than both its
    children

26
Representation
  • Use an array to store breadth-first traversal of
    heap tree
  • Example
  • Left child of node j is at 2j
  • Right child of node j is at 2j1
  • Parent of node j is at j /2
  • Parental nodes are represented in the first n
    /2 locations

9
5
3
1
4
2
27
Bottom-up heap construction algorithm
28
Analysis of Heapsort
  • See algorithm HeapBottomUp in section 6.4
  • Fix heap with problem at height j 2j
    comparisons
  • For subtree rooted at level i it does 2(h-i)
    comparisons
  • Total for heap construction phase

h-1
S
2(h-i) 2i 2 ( n lg (n 1)) T(n)
i0
nodes at level i
29
Analysis of Heapsort (continued)
  • Recall algorithm
  • Build heap
  • Remove root exchange with last (rightmost) leaf
  • Fix up heap (excluding last leaf)
  • Repeat 2, 3 until heap contains just one node.

T(n)
T(log n)
n 1 times
Total T(n) T( n log n) T(n log n)
  • Note this is the worst case. Average case also
    T(n log n).

30
Priority queues
  • A priority queue is the ADT of an ordered set
    with the operations
  • find element with highest priority
  • delete element with highest priority
  • insert element with assigned priority
  • Heaps are very good for implementing priority
    queues

31
Insertion of a new element
  • Insert element at last position in heap.
  • Compare with its parent and if it violates heap
    condition exchange them
  • Continue comparing the new element with nodes up
    the tree until the heap condition is satisfied
  • Example
  • Efficiency

32
Bottom-up vs. Top-down heap construction
  • Top down Heaps can be constructed by
    successively inserting elements into an
    (initially) empty heap
  • Bottom-up Put everything in and then fix it
  • Which one is better?

33
Horners rule
  • Horner published in early 19th century
  • According to Knuth, the method was used by Newton
  • Evaluate a polynomial at a point x
  • p(x) anxn an-1xn-1 a1x a0
  • p(x) ( (anx an-1) x )x a0
  • Example evaluate p(x)2x4-x33x2x-5 at x3
  • p(x) x (x (x (2x-1) 3) 1) - 5
  • Visualization by a table

34
Horners rule 2
  • Algorithm Horner(P0..n,x)
  • // Evaluate polynomial at a given point
  • // Input an array P0..n of coefficients and a
    number x
  • //Output the value of polynomial at point x
  • p ? Pn
  • for i ? n-1 down to 0 do
  • p ? xp Pi
  • return p
  • Efficiency ?
  • Byproduct coefficients of the quotient of the
    division of p(x) by (x-x0)

35
Binary exponentiation
  • Horner is not efficient to compute p(x)xn at xa
  • Degenerate to brute force
  • Let the binary representation nblbl-1 bi b1b0
  • p(x) blxl bl-1xl-1 b1x b0 and x2
  • Algorithm LeftRightBinaryExponentiation
  • product ? a
  • for i ? l-1 down to 0 do
  • product ? product product
  • if bi?1 then product ? producta
  • return product
  • Example compute a13, n131101
  • Efficiency

36
Binary exponentiation (2)
  • Compute an
  • Consider n bl2l bl-12l-1 b12 b0
  • and multiply independent powers terms
  • Algorithm RightLeftBinaryExponentiation
  • term ? a
  • if b01 then product ? a
  • else product ? 1
  • for i? 1 to l do
  • term ? term term
  • if bi 1 then product ? product term
  • return product
  • Example compute a13, n131101
  • Efficiency

37
Least common multiple
  • lcm(24,60)120, lcm(11,5)55
  • Example 24 2 x 2 x 2 x 3
  • 60 2 x 2 x 3 x 5
  • lcm(24,60) (2x2x3) x 2 x 5
  • Efficiency (a list of primes is required)
  • lcm(m,n) mn / gcd(m,n)

38
Counting paths in a graph
  • The number of different paths of length kgt0 from
    node i to node j equals the (i,j) element of the
    Ak, where A the adjacency matrix
  • Example
  • Efficiency

39
Reduction to graph problems
  • Applies for a variety of games and puzzles
  • Build the state-space graph
  • Example peasant, wolf, goat, cabbage
  • Traverse the graph by applying what?
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