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Transcript
```CSC 331: Algorithm Analysis
Divide-and-Conquer Algorithms
The divide-and-conquer strategy solves a problem by:
Breaking it into subproblems that are themselves smaller instances of the
same type of problem.
Recursively solving these subproblems.
CSC 331: Algorithm Analysis
Recurrence Relations
Divide-and-conquer algorithms often follow a generic pattern:
They tackle a problem of size n by recursively solving a subproblems of size n/b
and then combining these answers in O(nd) time, for some a, b, d > 0.
Their running time can therefore be captured by the equation T(n) = aT(⎡n/b⎤) +
O(nd).
If T(n) = aT(⎡n/b⎤) + O(nd) for some constants a > 0, b > 1, and d ≥ 0, then
The Master Theorem tells us the running times of most of the
divide-and-conquer procedures we are likely to use.
CSC 331: Algorithm Analysis
Binary Search
Given a sorted array of numbers, return the index of the number k or -1 if it’s not
there.
We could go through each number and check to see if that number is k. If so, return
the index.
What’s the running time?
Instead, we want to USE the fact that the array is sorted.
What is a better way? (Think high/low game).
Input: Look for the 13 in:
2
3
5
6
7
13
19
100
Look at middle number, 7. 13>7, so recurse on second half:
13
19
100
Look at middle number, 19. 13<19, so recurse on first half:
13
Look at middle number, 13. l found it, with only 3 recursions.
Search in part of Array A for element k, and return its
index. Assumes A is sorted.
function binarySearch(A[left..right], k)
// Base case: 0 elements left, I didn’t find it.
mid<- (right-left/2) + left
if k = A[mid] return mid
if k < A[mid] return binarySearch(A[left..mid-1])
if k > A[mid] return binarySearch(A[mid+1..right])
What is the recurrence relation?
What is the running time?
function binarySearch(A[left..right], k)
// Base case: 0 elements left, I didn’t find it.
mid<- (right-left)/2 + left
if k = A[mid] return mid
if k < A[mid] return A[left..mid-1]
if k > A[mid] return A[mid+1..right]
CSC 331: Algorithm Analysis
Mergesort
The problem of sorting a list of numbers lends itself immediately to a divide-andconquer strategy.
Split the list into two halves.
Recursively sort each half.
Merge the two sorted sublists.
Input:
10
10
10
10
2
2
2
5
5
7
13
3
5
2
3
5
1
7
3
7
3
7
13
6
1
6
13
13
1
1
6
6
10
2
2
5
10
2
3
3
1
3
5
2
7
5
7
10
3
13
13
1
5
6
1
7
6
1
6
7
10
13
6
13
function mergesort(a[1...n])
if n > 1:
return merge(mergesort(a[1...⎣n/2⎦]),
mergesort(a[⎣n/2⎦+ 1...n]))
else:
return a
function merge(x[1...k], y[1...m])
if k = 0: return y[1...m]
if m = 0: return x[1...k]
if x[1] ≤ y[1]:
return x[1] ✪ merge(x[2...k], y[1...m])
else:
return y[1] ✪ merge(x[1...k], y[2...m])
where ✪ is concatenation.
function merge(x[1...k], y[1...m])
if k = 0: return y[1...m]
if m = 0: return x[1...k]
if x[1] ≤ y[1]:
return x[1] ✪ merge(x[2...k], y[1...m])
else:
return y[1] ✪ merge(x[1...k], y[2...m])
This merge procedure does a constant amount of work per recursive call (provided
the required array space is allocated in advance), for a total running time of O(k + m).
Thus merge’s are linear, and the overall time take by
mergesort is
T(n) = 2T(n/2) + O(n).
a = 2, b = 2, d = 1 ➠ O(n log n)
CSC 331: Algorithm Analysis
Medians
What is the median of [45, 1, 10, 30, 25]?
25
How do you compute the median of n numbers?
Sorting takes O(n log n), can we do better?
Sorting is doing far more work than we really need -- we just want the
middle element and don’t care about the relative ordering of the rest of
them.
When looking for a recursive solution, it is paradoxically often
easier to work with a more general version of the problem.
Selection
Input: A list of numbers S; an integer k
Output: The kth smallest element of S
For instance, if k = 1, the minimum of S is sought.
For median, k = ⎣⎜S⎟/ 2⎦.
For any number v, split list S into three categories:
 elements smaller than v (SL)
 elements equal to v (SV)
 elements greater than v (SR)
The three sublists can be computed in linear time O(n).
For instance, if the array
S:
2
36
5
21
8
13 11 20
5
is split on v = 13, the three subarrays generated are:
SL:
2
5
8
Sv: 13
SR: 36 21 20
11
5
4
1
4
1
Now we can instantly narrow down the search to one of these
sublists.
SL:
2
SV:
13
SR:
36
5
8
21
20
11
5
4
1
If we wanted, say, the ninth-smallest element of S, we know it must be the
first-smallest element of SR since | SL | + | SV | = 8.
selection(S,9) = selection(SR,1)
if k ≤ | SL |
selection(S,k) = selection(SL, k)
if | SL | < k ≤ | SL | + | SV |
selection(S,k) = v
if k > | SL | + | SV |
selection(S,k) =
selection(SR, k - | SL | - | SV |)
If we could always pick v so that ⎜SL⎟, ⎟ SR⎟≈ 1/2(S)
T(n) = T(n/2) + O(n).
a = 1, b = 2, d = 1 ➠ O(n)
But this requires picking v to be the median, which is our
ultimate goal!
We will pick v randomly from S.
Naturally, the running time of our algorithm depends on the random
choice of v.
The worst-case scenario would be if we kept picking v to be the
largest (or smallest) element of the array.
n + (n - 1) + (n - 2) + ... + n/2 = Θ(n2)
The best-case scenario would be if we picked v to be the median
element of the array.
Θ(n)
Where does the average running time lie?
v is good if it lies within the 25th to 75th percentile of the array.
A good v ensures that the sublists SL and SR have size at most threefourths that of S.
Half of the elements of any list must fall between the 25th and 75th
percentile.
Lemma: On average a fair coin needs to be tossed two
times before a “heads” is seen.
After two split operations on average, the array will shrink to at most three-fourths
of its size.
Letting T(n) be the expected running time on an array of size n, we get
T(n) ≤ T(3n/4) + O(n).
a = 1, b = 4/3, d = 1 ➠ O(n)
So, the overall ideas:
Often the obvious way to solve a problem is not the
most efficient.
Try to find a way to divide the problem into pieces.
IF you know the answer (recursively) to the piece,