The Maximum Subarray Problem
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The document discusses the maximum subarray problem and different solutions to solve it. It defines the problem as finding a contiguous subsequence within a given array that has the largest sum. It presents a brute force solution with O(n2) time complexity and a more efficient divide and conquer solution with O(nlogn) time complexity. The divide and conquer approach recursively finds maximum subarrays in the left and right halves of the array and the maximum crossing subarray to return the overall maximum.
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![掘恰温馨沿鉛艶
Here, the subarray A[811] is the Maximum
Subarray, with sum 43, has the greatest sum of
any contiguous subarray of array A.](https://image.slidesharecdn.com/d2216e5c-75da-44c5-ab56-829b21931b58-150413003852-conversion-gate01/85/The-Maximum-Subarray-Problem-3-320.jpg)
![Divide and Conquer Solution
First we Divide the array A [low high] into two
subarrays A [low mid] (left) and A [mid +1
high] (right).
We recursively find the maximum subarray of A
[low mid] (left), and the maximum of A [mid
+1 high] (right).
We also find maximum crossing subarray that
crosses the midpoint.
Finally we take a subarray with the largest sum
out of three.](https://image.slidesharecdn.com/d2216e5c-75da-44c5-ab56-829b21931b58-150413003852-conversion-gate01/85/The-Maximum-Subarray-Problem-5-320.jpg)
![Pseudo Code
Max-subarray(A, Left, Right)
if (Right == Left)
return (left, right, A[left])
else mid= [(left+right)/2]
L1=Find-Maximum-Subarray(A,left,mid)
R1=Find-Maximum-Subarray(A,mid+1,right)
M1=Find-Max-Crossing-Subarray(A,left,mid,right)
If sum(L1) > sum(R1) and sum(L1) > sum(M1)
Return L1
elseif sum(R1) > sum(L1) and sum(R1) > sum(M1)
Return R1
Else return M1](https://image.slidesharecdn.com/d2216e5c-75da-44c5-ab56-829b21931b58-150413003852-conversion-gate01/85/The-Maximum-Subarray-Problem-8-320.jpg)
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Lower bounds help establish the minimum efficiency of algorithms. There are four main methods to determine lower bounds: trivial counting arguments, information theory, adversary arguments, and problem reduction. Decision trees can also help analyze comparison-based algorithms and establish lower bounds.Unit-3 greedy method, Prim's algorithm, Kruskal's algorithm.pdf
Unit-3 greedy method, Prim's algorithm, Kruskal's algorithm.pdfyashodamb
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The Greedy Method: Introduction, Huffman Trees and codes, Minimum
Coin Change problem, Knapsack problem, Job sequencing with deadlines,
Minimum Cost Spanning Trees, Single Source Shortest paths. Solution 3.
Solution 3.sansaristic
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The document contains exercises, hints, and solutions for analyzing algorithms from a textbook. It includes problems related to brute force algorithms, sorting algorithms like selection sort and bubble sort, and evaluating polynomials. The solutions analyze the time complexity of different algorithms, such as proving that a brute force polynomial evaluation algorithm is O(n^2) while a modified version is linear time. It also discusses whether sorting algorithms like selection sort and bubble sort preserve the original order of equal elements (i.e. whether they are stable).Divide and Conquer / Greedy Techniques
Divide and Conquer / Greedy TechniquesNirmalavenkatachalam
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Divide and conquer is an algorithm design paradigm where a problem is broken into smaller subproblems, those subproblems are solved independently, and then their results are combined to solve the original problem. Some examples of algorithms that use this approach are merge sort, quicksort, and matrix multiplication algorithms like Strassen's algorithm. The greedy method works in stages, making locally optimal choices at each step in the hope of finding a global optimum. It is used for problems like job sequencing with deadlines and the knapsack problem. Minimum cost spanning trees find subgraphs of connected graphs that include all vertices using a minimum number of edges.Programming Exam Help 
Programming Exam Help Programming Exam Help
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I am Joanna R. I am a Programming Exam Expert at programmingexamhelp.com. I hold a Bachelor of Information Technology from, California Institute of Technology, United States. I have been helping students with their exams for the past 11 years. You can hire me to take your exam in Programming.
Visit programmingexamhelp.com or email support@programmingexamhelp.com. You can also call on +1 678 648 4277 for any assistance with the Programming Exam.Algorithm Homework Help
Algorithm Homework HelpProgramming Homework Help
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This document provides information about homework help services and programming homework help. It lists contact details including a phone number and email address to get help with homework queries. It also provides a website URL for programming homework help. The document then presents sample programming homework problems and solutions.Learn Dynamic Programming Roadmap at Tutort Academy
Learn Dynamic Programming Roadmap at Tutort AcademyTutort Academy
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Dynamic Programming is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. Undecidable Problems and Approximation Algorithms
Undecidable Problems and Approximation AlgorithmsMuthu Vinayagam
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The document discusses algorithm limitations and approximation algorithms. It begins by explaining that some problems have no algorithms or cannot be solved in polynomial time. It then discusses different algorithm bounds and how to derive lower bounds through techniques like decision trees. The document also covers NP-complete problems, approximation algorithms for problems like traveling salesman, and techniques like branch and bound. It provides examples of approximation algorithms that provide near-optimal solutions when an optimal solution is impossible or inefficient to find.test pre
test prefarazch
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The document discusses greedy algorithms and their properties. It describes how greedy algorithms work by making locally optimal choices at each step in the hope of reaching a globally optimal solution. Two examples are given: the activity selection problem and finding minimum spanning trees. Prim's algorithm for finding minimum spanning trees is described in detail, showing how it works by always selecting the lightest edge between the growing tree and remaining vertices.The Maximum Subarray Problem
- 1. The Maximum Subarray Problem Defining problem , its brute force solution, divide and conquer solution Presented by: Kamran Ashraf
- 2. Formal Problem Definition Given a sequence of numbers <a1,a2,..an> we work to find a subsequence of A that is contiguous and whose values have the maximum sum. Maximum Subarray problem is only challenging when having both positive and negative numbers.
- 3. 掘恰温馨沿鉛艶 Here, the subarray A[811] is the Maximum Subarray, with sum 43, has the greatest sum of any contiguous subarray of array A.
- 4. Brute Force Solution To solve the Maximum Subarray Problem we can use brute force solution however in brute force we will have to compare all possible continuous subarray which will increase running time. Brute force will result in 儔(n2), and the best we can hope for is to evaluate each pair in constant time which will result in 畚(n2). Divide and Conquer is a more efficient method for solving large number of problems resulting in 儔(nlogn).
- 5. Divide and Conquer Solution First we Divide the array A [low high] into two subarrays A [low mid] (left) and A [mid +1 high] (right). We recursively find the maximum subarray of A [low mid] (left), and the maximum of A [mid +1 high] (right). We also find maximum crossing subarray that crosses the midpoint. Finally we take a subarray with the largest sum out of three.
- 6. 掘恰温馨沿鉛艶
- 7. Time Analysis Find-Max-Cross-Subarray takes: 儔 (n) time Two recursive calls on input size n/2 takes: 2T(n/2) time Hence: T(n) = 2T(n/2) + 儔 (n) T(n) = 儔 (n log n)
- 8. Pseudo Code Max-subarray(A, Left, Right) if (Right == Left) return (left, right, A[left]) else mid= [(left+right)/2] L1=Find-Maximum-Subarray(A,left,mid) R1=Find-Maximum-Subarray(A,mid+1,right) M1=Find-Max-Crossing-Subarray(A,left,mid,right) If sum(L1) > sum(R1) and sum(L1) > sum(M1) Return L1 elseif sum(R1) > sum(L1) and sum(R1) > sum(M1) Return R1 Else return M1