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Divide and Conquer

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Recall Complexity Analysis – Comparison of algorithm – Big O Simplification From source code – Recursive

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Today Topic Divide and Conquer

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DIVIDE AND CONQUER Algorithm Design Technique

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Divide and Conquer

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Induction

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Induction Example

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The Inductive Step The inductive step – Assume that it is true for (n-1) – Check the case (n) 1 + 2 + 3 +… + n = (1 + 2 + 3 + … + (n-1) ) + n = (n-1)(n) / 2 + n = n((n-1)/2 + 1) = n(n/2-1/2 + 1) = n(n/2 + 1/2) = n(n+1)/2

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The Inductive Step The inductive step – Assume that it is true for (n-1) – Check the case (n) 1 + 2 + 3 +… + n = (1 + 2 + 3 + … + (n-1) ) + n = (n-1)(n) / 2 + n = n((n-1)/2 + 1) = n(n/2-1/2 + 1) = n(n/2 + 1/2) = n(n+1)/2 By using (n-1)

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The Induction By using the result of the smaller case We can proof the larger case

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Key of Divide and Conquer

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Steps in D&C

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Code Example ResultType DandC(Problem p) { if (p is trivial) { solve p directly return the result } else { divide p into p 1,p 2,...,p n for (i = 1 to n) r i = DandC(p i ) combine r 1,r 2,...,r n into r return r }

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Code Example ResultType DandC(Problem p) { if (p is trivial) { solve p directly t s return the result } else { divide p into p 1,p 2,...,p n t d for (i = 1 to n) t r r i = DandC(p i ) combine r 1,r 2,...,r n into r t c return r }

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Examples Let’s see some examples

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MERGE SORT

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The Sorting Problem Given a sequence of numbers – A = [a 1,a 2,a 3,…,a n ] Output – The sequence A that is sorted from min to max

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The Question What if we know the solution of the smaller problem? – What is the smaller problem? Try the same problem sorting What if we know the result of the sorting of some elements?

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The Question How to divide? – Let’s try sorting of the smaller array Divide exactly at the half of the array How to conquer? – Merge the result directly

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Idea Simplest dividing – Divide array into two array of half size Laborious conquer – Merge the result

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Divide

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Merge

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Analysis T(n) = 2T(n/2) + O(n) Master method – T(n) = O(n lg n)

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QUICK SORT

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The Sorting Problem (again) Given a sequence of numbers – A = [a 1,a 2,a 3,…,a n ] Output – The sequence A that is sorted from min to max

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Problem of Merge Sort Need the use of external memory for merging Are there any other way such that conquering is not that complex?

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The Question How to divide? – Try doing the same thing as the merge sort – Add that every element in the first half is less than the second half Can we do that? How to conquer? – The sorted result should be easier to merge?

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Idea Laborious dividing – Divide array into two arrays Add the requirement of value of the array Simplest conquer – Simply connect the result

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Is dividing scheme possible? Can we manage to have two subproblems of equal size – That satisfy our need? Any trade off?

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The Median

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Simplified Division

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partitioning First k element Other n- k element

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Solve by Recursion sorting

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Do nothing!

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Analysis T(n) = T(k) + T(n – k) + Θ(n) There can be several cases – Up to which K that we chosen

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Analysis : Worst Case K always is 1 st What should be our T(N) ? What is the time complexity

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Analysis : Worst Case K always is 1 st T(n) = T(1) + T(n – 1) + Θ(n) = T(n – 1) + Θ(n) = Σ Θ(i) = Θ(n 2 ) Not good

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Analysis : Best Case K always is the median What should be our T(N) ? What is the time complexity

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Analysis : Best Case K always is the median T(n) = 2T(n/2) + Θ(n) = Θ(n log n) The same as the merge sort (without the need of external memory)

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Fixing the worst case When will the worst case happen? – When we always selects the smallest element as a pivot – Depends on pivot selection strategy If we select the first element as a pivot – What if the data is sorted?

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Fixing the worst case Select wrong pivot leads to worst case. There will exist some input such that for any strategy of “deterministic” pivot selection leads to worst case.

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Fixing the worst case Use “non-deterministic” pivot selection – i.e., randomized selection Pick a random element as a pivot It is unlikely that every selection leads to worst case We can hope that, on average, – it is O(n log n)

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MODULO EXPONENTIAL

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The Problem Given x, n and k Output: the value of x n mod k

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Naïve method res = x mod k; for (i = 2 to n) do { res = res * x; res = res mod k; } Θ(n) Using basic facts: (a * b) mod k = ((a mod k) * (b mod k)) mod k

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The Question What if we knows the solution to the smaller problem? – Smaller problem smaller N – x n ?? from x (n-e) ? Let’s try x (n/2)

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The Question How to divide? – x n = x n/2 * x n/2 (n is even) – x n = x n/2 * x n/2 * x (n is odd) How to conquer? – Compute x n/2 mod k Square and times with x, if needed, mod k afterward

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Analysis T(n) = T(n/2) + Θ(1) = O(log n)

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Example 2 92 mod 10 2 92 = 2 46 × 2 46 2 46 = 2 23 × 2 23 2 23 = 2 11 × 2 11 × 2 2 11 = 2 5 × 2 5 × 2 2 5 = 2 2 × 2 2 × 2 2 2 = 2 1 × 2 1 = 4×4 mod 10 = 6 = 8×8 mod 10 = 4 = 8×8×2 mod 10 = 8 = 2×2×2 mod 10 = 8 = 4×4×2 mod 10 = 2 = 2×2 mod 10 = 4

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MAXIMUM CONTIGUOUS SUM OF SUBSEQUENCE (MCS)

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The MCS Problem Given a sequence of numbers – A = [a 1,a 2,a 3,…,a n ] There are n members Find a subsequence s = [a i,a i+1,a i+2,…,a k ] – Such that the sum of the element in s is maximum

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Example Sum = 11

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Naïve approach Try all possible sequences – How many sequence? – How much time does it use to compute the sum?

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Naïve approach

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The DC Approach What if we know the solution of the smaller problem – What if we know MSS of the first half and the second half?

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The Question How to divide? – By half of the member How to conquer? – Does the result of the subproblems can be used to compute the solution? – Let’s see

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Combining the result from sub MSS

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4-35-226-2

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4-35-226-2 Combining the result from sub MSS Sum = 6Sum = 8 Shall this be our answer?

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Combining the result from sub MSS Do we consider all possibilities? 4-35-2 26-2

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Combining the result from sub MSS Considered all pairs in each half But not the pair including member from both half There are (n/2) 2 additional pairs 4-35-2 26-2

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Compute max of cross over Consider this The max sequence from the green part is always the same, regardless of the pink part The sequence always start at the left border 4-35-2 26-2

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Compute max of cross over Similarly The max sequence from the pink part is always the same, regardless of the green part The sequence always start at the right border 4-35-2 26-2

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4-35-2 26-2 Compute max of cross over Just find the marginal max from each part Sum = -1

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Compute max of cross over Just find the marginal max from each part 4-35-2 26-2 Sum = 1

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Compute max of cross over Just find the marginal max from each part 4-35-2 26-2 Sum = 7Sum = 7

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Compute max of cross over Just find the marginal max from each part Less than previous 4-35-2 26-2 Sum = 5Sum = 5

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Compute max of cross over Just find the marginal max from each part 4-35-2 26-2 Sum = -2

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Compute max of cross over Just find the marginal max from each part 4-35-2 26-2 Sum = 3Sum = 3

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4-35-2 26-2 Compute max of cross over Just find the marginal max from each part Sum = 0Sum = 0

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4-35-2 26-2 Compute max of cross over Just find the marginal max from each part Sum = 4Sum = 4 This is max

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4-35-2 26-2 Compute max of cross over Just find the marginal max from each part Sum = 7Sum = 7Sum = 4Sum = 4 Total = 11 takes Θ(n/2)

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Combine Max from the three parts – The left half – The right half – The cross over part Use the one that is maximum

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Analysis T(n) = 2 T(n/2) + Θ(n) = Θ(n log n) a = 2, b = 2, c = log 2 2=1, f(n) = O(n) n c =n 2 = Θ(n) – f(n) = Θ(n c )this is case 1 of the master’s method Hence, T(n) = Θ(n log n) Left and right parts Cross over part

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CLOSEST PAIR

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The Problem Given – N points in 2D (x 1,y 1 ), (x 2,y 2 ), …,(x n,y n ) Output – A pair of points from the given set (x a,y a ), (x b,y b ) 1 <= a,b <= n Such that the distance between the points is minimal

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Input Example

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Output Example

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The Naïve Approach

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DC approach What if we know the solution of the smaller problem – What if we know the Closest Pair of half of the points? Which half?

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Divide by X axis Find closest pair of the left side Find closest pair of the right side

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Conquer Like the MSS problem – Solutions of the subproblems do not cover every possible pair of points – Missing the pairs that “span” over the boundary

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Divide by X axis Find closest pair of the left side Find closest pair of the right side

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Conquer

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Find Closest Spanning Pair Should we consider this one? Why?

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Find Closest Spanning Pair

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Possible Spanning Pair Consider pairs only in this strip One point from the left side Another point from the right side Consider pairs only in this strip One point from the left side Another point from the right side Any point outside the strip, if paired, its distance will be more than b

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Point in Strips

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Point in Strips is O(N)

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The Solution

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Spanning Pair to be considered

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Question is still remains How many pair to be checked? A point to check

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Implementation Detail

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Naïve approach

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Better Approach

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Analysis T(n) = 2 T(n/2) + 4O(n) + O(n) = Θ(n log n) a = 2, b = 2, c = log 2 2=1, f(n) = O(n) n c =n 2 = Θ(n) – f(n) = Θ(n c )this is case 1 of the master’s method Hence, T(n) = Θ(n log n) Left and right parts Point in strip Divide the list

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MATRIX MULTIPLICATION

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The Problem Given two square matrix – A n x n and B n x n – A R n x n and B R n x n Produce – C = AB

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Multiplying Matrix c i,j = Σ(a i,k *b k,j )

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Naïve Method for (i = 1; i <= n;i++) { for (j = 1; j <= n;j++) { sum = 0; for (k = 1;k <= n;k++) { sum += a[i][k] * b[k][j]; } c[i][j] = sum; } O(N 3 )

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Simple Divide and Conquer Divide Matrix into Block Multiply the block

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Simple Divide and Conquer What is the complexity? T(n) = 8T(n/2) + O(n 2 ) Master method gives O(n 3 ) – Still the same

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Strassen’s Algorithm We define Note that each M can be computed by one single multiplication of n/2 * n/2 matrices The number of addition is 10(n/2) 2

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Strassen’s Algorithm Compute the result 8 more additions of n/2 * n/2 matrices Hence, total addition is 18 (n/2) 2

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Analysis T(n) = 7T(n/2) + O(n 2 ) – Using Master’s method – a = 7, b = 2, f(n) = O(n 2 ) c = log 2 7 ≈ 2.807 – Hence, f(n) = O(n 2.807 ) this is case 1 of the master method So, T(n) = O(n 2.807 )

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