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CS 311 – Design and Algorithms Analysis PSU
Divide and Conquer Strategy
2CS 311 – Design and Algorithms Analysis PSU
General Concept of Divide & Conquer
Given a function to compute on n inputs, the divide-and-conquer strategy consists of: splitting the inputs into k distinct subsets, 1kn,
yielding k subproblems. solving these subproblems combining the subsolutions into solution of the whole. if the subproblems are relatively large, then
divide_Conquer is applied again. if the subproblems are small, the are solved without
splitting.
3CS 311 – Design and Algorithms Analysis PSU
The Divide and Conquer Algorithm
Divide_Conquer(problem P){
if Small(P) return S(P);else { divide P into smaller instances P1, P2, …, Pk, k1; Apply Divide_Conquer to each of these subproblems; return Combine(Divide_Conque(P1), Divide_Conque(P2),…, Divide_Conque(Pk));}
}
4CS 311 – Design and Algorithms Analysis PSU
Divde_Conquer recurrence relation
The computing time of Divde_Conquer is
T(n) is the time for Divde_Conquer on any input size n. g(n) is the time to compute the answer directly (for
small inputs) f(n) is the time for dividing P and combining the
solutions.
)()(...)()(
)()(
21 nfnTnTnT
ngnT
k
n small
otherwise
5CS 311 – Design and Algorithms Analysis PSU
Three Steps of The Divide and Conquer Approach
The most well known algorithm design strategy:1. Divide the problem into two or more smaller
subproblems.
2. Conquer the subproblems by solving them recursively.
3. Combine the solutions to the subproblems into the solutions for the original problem.
6CS 311 – Design and Algorithms Analysis PSU
A Typical Divide and Conquer Case
subproblem 2 of size n/2
subproblem 1 of size n/2
a solution to subproblem 1
a solution tothe original problem
a solution to subproblem 2
a problem of size n
7CS 311 – Design and Algorithms Analysis PSU
An Example: Calculating a0 + a1 + … + an-1
Efficiency: (for n = 2k)
A(n) = 2A(n/2) + 1, n >1
A(1) = 0;
ALGORITHM RecursiveSum(A[0..n-1])//Input: An array A[0..n-1] of orderable elements//Output: the summation of the array elementsif n > 1
return ( RecursiveSum(A[0.. n/2 – 1]) + RecursiveSum(A[n/2 .. n– 1]) )
8CS 311 – Design and Algorithms Analysis PSU
General Divide and Conquer recurrence
The Master Theorem
T(n) = aT(n/b) + f (n), where f (n) ∈ Θ(nk) 1. a < bk T(n) ∈ Θ(nk) 2. a = bk T(n) ∈ Θ(nk lg n ) 3. a > bk T(n) ∈ Θ(nlog b a)
the time spent on solving a subproblem of size n/b.
the time spent on dividing the problem
into smaller ones and combining their solutions.
9CS 311 – Design and Algorithms Analysis PSU
Divide and Conquer Other Examples
Sorting: mergesort and quicksort
Binary search
Matrix multiplication-Strassen’s algorithm
10CS 311 – Design and Algorithms Analysis PSU
The Divide, Conquer and Combine Steps in Quicksort
Divide: Partition array A[l..r] into 2 subarrays, A[l..s-1] and A[s+1..r] such that each element of the first array is ≤A[s] and each element of the second array is ≥ A[s]. (computing the index of s is part of partition.) Implication: A[s] will be in its final position in the
sorted array. Conquer: Sort the two subarrays A[l..s-1] and
A[s+1..r] by recursive calls to quicksort Combine: No work is needed, because A[s] is already
in its correct place after the partition is done, and the two subarrays have been sorted.
11CS 311 – Design and Algorithms Analysis PSU
Quicksort Select a pivot whose value we are going to divide
the sublist. (e.g., p = A[l]) Rearrange the list so that it starts with the pivot
followed by a ≤ sublist (a sublist whose elements are all smaller than or equal to the pivot) and a ≥ sublist (a sublist whose elements are all greater than or equal to the pivot ) (See algorithm Partition in section 4.2)
Exchange the pivot with the last element in the first sublist(i.e., ≤ sublist) – the pivot is now in its final position
Sort the two sublists recursively using quicksort.
p
A[i]≤p A[i] ≥ p
12CS 311 – Design and Algorithms Analysis PSU
The Quicksort Algorithm
ALGORITHM Quicksort(A[l..r])//Sorts a subarray by quicksort//Input: A subarray A[l..r] of A[0..n-1],defined by its left
and right indices l and r//Output: The subarray A[l..r] sorted in nondecreasing
orderif l < r
s Partition (A[l..r]) // s is a split positionQuicksort(A[l..s-1])Quicksort(A[s+1..r]
13CS 311 – Design and Algorithms Analysis PSU
The Partition Algorithm
ALGORITHM Partition (A[l ..r])//Partitions a subarray by using its first element as a pivot//Input: A subarray A[l..r] of A[0..n-1], defined by its left and right
indices l and r (l < r)//Output: A partition of A[l..r], with the split position returned as this
function’s valueP A[l]i l; j r + 1;Repeat
repeat i i + 1 until A[i]>=p //left-right scanrepeat j j – 1 until A[j] <= p//right-left scanif (i < j) //need to continue with the scan
swap(A[i], a[j])until i >= j //no need to scanswap(A[l], A[j])return j
14CS 311 – Design and Algorithms Analysis PSU
Quicksort Example
15 22 13 27 12 10 20 25
15CS 311 – Design and Algorithms Analysis PSU
Efficiency of Quicksort
Based on whether the partitioning is balanced. Best case: split in the middle — Θ( n log n)
T(n) = 2T(n/2) + Θ(n) //2 subproblems of size n/2 each
Worst case: sorted array! — Θ( n2) T(n) = T(n-1) + n+1 //2 subproblems of size 0 and n-1 respectively
Average case: random arrays — Θ( n log n)
16CS 311 – Design and Algorithms Analysis PSU
Binary Search – an Iterative Algorithm
ALGORITHM BinarySearch(A[0..n-1], K)//Implements nonrecursive binary search//Input: An array A[0..n-1] sorted in ascending order and // a search key K//Output: An index of the array’s element that is equal to K// or –1 if there is no such elementl 0, r n – 1while l r do //l and r crosses over can’t find K.
m (l + r) / 2if K = A[m] return m //the key is foundelse
if K < A[m] r m – 1 //the key is on the left half of the arrayelse l m + 1 // the key is on the right half of the
arrayreturn -1
17CS 311 – Design and Algorithms Analysis PSU
Strassen’s Matrix Multiplication
Brute-force algorithmc00 c01 a00 a01 b00 b01
= *
c10 c11 a10 a11 b10 b11
a00 * b00 + a01 * b10 a00 * b01 + a01 * b11
=
a10 * b00 + a11 * b10 a10 * b01 + a11 * b11
8 multiplications
4 additions
Efficiency class: (n3)
18CS 311 – Design and Algorithms Analysis PSU
Strassen’s Matrix Multiplication
Strassen’s Algorithm (two 2x2 matrices)c00 c01 a00 a01 b00 b01
= *c10 c11 a10 a11 b10 b11
m1 + m4 - m5 + m7 m3 + m5
= m2 + m4 m1 + m3 - m2 + m6
m1 = (a00 + a11) * (b00 + b11) m2 = (a10 + a11) * b00 m3 = a00 * (b01 - b11) m4 = a11 * (b10 - b00) m5 = (a00 + a01) * b11 m6 = (a10 - a00) * (b00 + b01) m7 = (a01 - a11) * (b10 + b11)
7 multiplications
18 additions
19CS 311 – Design and Algorithms Analysis PSU
Strassen’s Matrix Multiplication
Two nxn matrices, where n is a power of two (What about the situations where n is not a power of two?)
C00 C01 A00 A01 B00 B01
= *
C10 C11 A10 A11 B10 B11
M1 + M4 - M5 + M7 M3 + M5
=
M2 + M4 M1 + M3 - M2 + M6
20CS 311 – Design and Algorithms Analysis PSU
Submatrices:
M1 = (A00 + A11) * (B00 + B11)
M2 = (A10 + A11) * B00
M3 = A00 * (B01 - B11)
M4 = A11 * (B10 - B00)
M5 = (A00 + A01) * B11
M6 = (A10 - A00) * (B00 + B01)
M7 = (A01 - A11) * (B10 + B11)
21CS 311 – Design and Algorithms Analysis PSU
Efficiency of Strassen’s Algorithm
If n is not a power of 2, matrices can be padded with rows and columns with zeros
Number of multiplications
Number of additions
Other algorithms have improved this result, but are even more complex
22CS 311 – Design and Algorithms Analysis PSU
Important Recurrence Types
Decrease-by-one recurrences A decrease-by-one algorithm solves a problem by
exploiting a relationship between a given instance of size n and a smaller size n – 1.
Example: n! The recurrence equation for investigating the time
efficiency of such algorithms typically has the formT(n) = T(n-1) + f(n)
Decrease-by-a-constant-factor recurrences A decrease-by-a-constant algorithm solves a problem by
dividing its given instance of size n into several smaller instances of size n/b, solving each of them recursively, and then, if necessary, combining the solutions to the smaller instances into a solution to the given instance.
Example: binary search. The recurrence equation for investigating the time
efficiency of such algorithms typically has the form T(n) = aT(n/b) + f (n)
23CS 311 – Design and Algorithms Analysis PSU
Decrease-by-one Recurrences
One (constant) operation reduces problem size by one.
T(n) = T(n-1) + c T(1) = dSolution:
A pass through input reduces problem size by one.
T(n) = T(n-1) + c n T(1) = dSolution:
T(n) = (n-1)c + d linear
T(n) = [n(n+1)/2 – 1] c + d quadratic
24CS 311 – Design and Algorithms Analysis PSU
Decrease-by-a-constant-factor recurrences – The Master Theorem
T(n) = aT(n/b) + f (n), where f (n) ∈ Θ(nk) , k>=0
1. a < bk T(n) ∈ Θ(nk) 2. a = bk T(n) ∈ Θ(nk log n ) 3. a > bk T(n) ∈ Θ(nlog b a)
4. Examples:1. T(n) = T(n/2) + 12. T(n) = 2T(n/2) + n3. T(n) = 3 T(n/2) + n
Θ(nlog2
3)
Θ(logn)
Θ(nlogn)