Divide and Conquer

Divide-and-Conquer
Republic of Yemen
THAMAR UNIVERSITY
Faculty of Computer Science&
Information System
1 Eng: Mohammed Hussein

Divide-and-Conquer

Divide-and-Conquer
 The whole problem we want to solve may too big to understand or solve at
once.
 We break it up into smaller pieces, solve the pieces separately, and
combine the separate pieces together.
 Divide-and conquer is a general algorithm design paradigm:
1. Divide: divide the input data S in two or more disjoint subsets S1,
S2, …
2. Recur: solve the subproblems recursively
3. Conquer: combine the solutions for S1, S2, …, into a solution for S
 The base case for the recursion are subproblems of constant size
 Analysis can be done using recurrence equations

Recurrence Equation Analysis
 The conquer step of merge-sort consists of merging two sorted sequences, each with
n/2 elements and implemented by means of a doubly linked list, takes at most bn steps,
for some constant b.
 Likewise, the basis case (n < 2) will take at b most steps.
 Therefore, if we let T(n) denote the running time of merge-sort:
 We can therefore analyze the running time of merge-sort by finding a closed form
solution to the above equation.
 That is, a solution that has T(n) only on the left-hand side.






2if)2/(2
2if
)(
nbnnT
nb
nT

Solving recurrences
 Recurrences are like solving integrals, differential equations, etc.
 Should learn a few tricks.
 Three ways of solving recurrences
1. Guess-and-Test Method(If we know the answer)
2. RecursionTree Method (Very useful !)
3. MasterTheorem (Save our effort)

Guess-and-Test Method
 In the guess-and-test method, we guess a closed form solution and then try to prove it is
true by induction:
 Guess that:T(n) < c n log n.
 Wrong: we cannot make this last line be less than cn log n
ncn
nbncnncn
nbnncn
nbnnnc
nbnnTnT
log
loglog
log)2log(log
log))2/log()2/((2
log)2/(2)(











2iflog)2/(2
2if
)(
nnbnnT
nb
nT
Log(x/y)= log x – log y
Log 2 =1
‫بالتعويض‬

Guess-and-Test Method, Part 2
 Recall the recurrence equation:
 Guess #2 that:T(n) < cn log2 n.
 if c > b.
 So,T(n) is O(n log2 n).
 In general, to use this method, you need to have a good guess and you need to be good at
induction proofs.
ncn
nbncnncnncn
nbnnncn
nbnncn
nbnnnc
nbnnTnT
2
2
22
2
2
log
loglog2log
log))2(log2loglog2(log
log)2log(log
log))2/(log)2/((2
log)2/(2)(












2iflog)2/(2
2if
)(
nnbnnT
nb
nT
(x-y) 2= x2 -2xy +y2
Log 2 =1
nj <= nd if j <= d
Big-Oh of highest degree

Recurrences
 In Mathematic a recurrence is a function that is defined in
terms of one or more base cases, and itself, with smaller
arguments
 Examples:

Master Method (tree)
 The master theorem concerns the recurrence
 a≥1 , b> 1
 n is the size of the problem.
 a is the number of subproblems in the recursion. (‫)الفزوع‬
 n/b is the size of each subproblem. (Here it is assumed that
all subproblems are essentially the same size.)
 f (n) is the cost of the work done outside the recursive calls
)()/()( nfbnaTnT 
merge-sort ‫خوارسمية‬
‫االستدعاء‬ ‫عمليات‬ ‫عند‬‫الذاتي‬‫الدمج‬ ‫عملية‬ ‫تنفيذ‬ ‫مستويات‬ ‫مختلف‬ ‫في‬

The Tree Definition
 The root of this tree is node A.
 Definitions:
 Parent (A)
 Children (E, F)
 Siblings (C, D)
 Root (A)
 Leaf / Leaves
 K, L, F, G, M, I, J…
 The degree of a node is the number of sub-trees of the node.
 The level of a node:
 Initially letting the root be at level one
 the level of the node’s parent plus one. Level H=level(Parent (H)+1));
 The height or depth of a tree is the maximum level of any node in the tree.
 height =Max(level(tree))

Divide and conquer (tree)
 We analyze this in some generality:
 suppose we have a pieces, each of size n/b and merging takes time f(n).
 (In this example a=b=2 and f(n)=O(log n) but it will not always be true that
a=b sometimes the pieces will overlap.)
 For simplicity, let's assume n is a power of b, and that the recursion stops
when n is 1.
 Notice that the size of a node depends only on its level: size(i) = n/(bi).
 What is time taken by a node at level i? : time(i) = f(n/ bi)
 How many levels can we have before we get down to n=1?
 For bottom level, n/bi =1, so n=bi and i=(log n)/(log b).
 How many items at level i?: ai.
)()/()( nfbnaTnT 
time(i) = f(n/ bi)

The Recursion Tree
 Draw the recursion tree for the recurrence relation and look for a pattern:
 At each successive level of recursion the sub problems get halved in size (n/2).
 At the (log n) th level the sub problems get down to size 1, and so the recursion ends.
Level/
depth
T’s size
0 1 n
1 2 n/2
k 2k n/2k
… … …






2if)2/(2
2if
)(
nbnnT
nb
nT
time
bn
bn
bn
…
Total time = bn + bn log n
(last level plus all previous levels)
2k =n  k= log n for each level.
log n
Notice that the size of a node depends only on its level
time=b x size = bn

Steps using recursion tree.
 Height = k=log n
 Levels = log n + 1
 Cost/level = cn
 Bottom node number n = 2log n for
height log n dividing problem by 2
each time.
 Recall that a balanced binary tree of
height k has k + 1 levels.
 Recall: a
log an = n
 so 2
log 2n = 2
log n =n
 Bottom cost = T(1) * n = cn
 Note that T(1) is problem size n=1,
there are n subproblems at bottom.
13 Eng: Mohammed Hussein 2k =n  k= log n for each level.

Merge Sort - Finding O
 Recall that a balanced binary tree of height k has k + 1 levels.
 A problem of size n=16 divided as described by the following recurrence equation
would then be represented by the tree.
 Use the implied coefficient c for arithmetic purposes:
16=2
4

Master Method
 Many divide-and-conquer recurrence equations have the form:
 The MasterTheorem:






dnnfbnaT
dnc
nT
if)()/(
if
)(
.1somefor)()/(provided
)),((is)(then),(is)(if3.
)log(is)(then),log(is)(if2.
)(is)(then),(is)(if1.
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
very small f(n)
moderate f(n)
very large f(n)

Analysis of Merge Sort
Analysis of Merge Sort
 Divide:Trivial.
 Conquer:Recursively sort 2subarrays.
 Combine:Linear-time merge.
 Using Master theorem

Analysis of Binary Search
 Recurrence for the worst case

Master Method, Example 1
 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)()2/(4)( nOnTnT 
Solution: logba =log24=2, so case 1 says T(n) is Θ(n2).

 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)log()2/(2)( nnnTnT 
Solution: logba=log22=1, so case 2 says T(n) is Θ(n log2 n).

 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)log()3/()( nnnTnT 
Solution: logba=0, so case 3 says T(n) is Θ(n log n).

 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)()2/(8)( 2
nnTnT 
Solution: logba=3, so case 1 says T(n) is Θ(n3).

 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)()3/(9)( 3
nnTnT 
Solution: logba=2, so case 3 says T(n) is Θ(n3).

 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)1()2/()(  nTnT
Solution: logba=0, so case 2 says T(n) is Θ(log n).
(binary search)

 The form:
 Example:






dnnfbnaT
dnc
nT
if)()/(
if
)(
log
1loglog
loglog










nfbnaf
nfnTnnf
nnnTnnnf
nnTnOnf
a
kaka
aa
b
bb
bb
)(log)2/(2)( nnTnT 
Solution: logba=1, so case 1 says T(n) is Θ(n).
(heap construction)

Divide and Conquer

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