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# Algorithm Design and Complexity - Course 5

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### Algorithm Design and Complexity - Course 5

1. 1. Algorithm Design and Complexity Course 5
2. 2. Overview      Greedy Algorithms Activity Selection Huffman Trees Greedy vs Dynamic Programming Knapsack Problem    Greedy DP Generic problem
3. 3. Greedy Algorithms   Efficient method to solve some optimization problems The solutions to an optimization problem must satisfy a global optimum     Advantages:    More difficult to verify Simplification: choose the solution that looks best at each step This is called a locally optimal solution Simpler to build the solution Less time / Better complexity Disadvantage:   The locally optimal solution does not always lead to the globally optimal solution May not correctly solve the problem (but may provide good approximations)
4. 4. Greedy Algorithms (2)   At each step, we choose the best solution according to the local optimum (greedy) choice We abandon all the other possible solutions   We‟ll look at two problems that have a greedy solution that leads to the global optimum as well    The solving paths that are not considered by the greedy choice are discarded! Activity selection Huffman trees Greedy is an algorithm design technique (pattern)!
5. 5. General Greedy Scheme SolveGreedy(Local_choice, Problem) partial_sols = InitialSolution(problem); // determine the starting point final_sols = Φ; WHILE (partial_sols ≠ Φ) FOREACH (s IN partial_sols) IF (s is a solution for Problem) { final_sols = final_sols U {s}; partial_sols = partial_sols {s}; } ELSE // can you optimize current solving path locally ? IF(CanOptimize(s, Local_choice, Problem)) // YES partial_sols = partial_sols {s} U OptimizeLocally(s, Local_choice, Problem) ELSE partial_sols = partial_sols {s}; // NO RETURN final_sols;  Most times we follow only a single solving path!
6. 6. Activity Selection Problem  Given a set of n activities that require exclusive use of a common resource for a given period of time, determine the largest subset of non-overlapping activities    These activities are called mutually compatible There might be more than a single solution We want to identify one of these best solutions   Similar to DP, not suitable for finding all possible solutions Notations:    S = {a1, … , an} are the activities Each activity has a start time, si, and a finish time, fi Each activity requires the common resource for the interval [si, fi)
7. 7. Activity Selection Problem (2)  E.g.    Activity = classes Activity = processes Resource = classroom Resource = CPU There exist some other activity selection problems that are more difficult:   Maximize the usage time of the resource Maximize income if each activity pays for the usage of the resource
8. 8. Example – from CLRS  We can devise a greedy solution if we consider the activities sorted by their finish times i 1 2 3 4 5 6 7 8 9 s[i] 1 2 4 1 5 8 9 11 13 f[i] 3 5 7 8 9 10 11 14 16  Solution: {a1, a3, a6, a8}  Not unique: {a2, a5, a7, a9}
9. 9. Define the Sub-Problems   First, define the similar sub-problems Let‟s consider the subset of activities that:    Start after ai finishes (start after fi) Finish before aj starts (finish before sj) They are compatible with all activities that:   Finish before fi Start after sj  Si,j = {all ak in S | fi <= sk < fk < sj}  We also add two invented activities:   a0 = [-INF, 0) an+1 = [INF, INF + 1)
10. 10. Define the Sub-Problems (2)  S0,n+1 = S = the entire set of activities  When the activities are sorted by their finish time   f0 <= f1 <= f2 <= … <= fn <= fn+1 Si,j = Φ if i > j    fi <= sk < fk < sj < fj => fi < fj Therefore, the sub-problems are Si,j with 0 <= i < j <= n+1
11. 11. Optimal Substructure   Suppose an optimal solution to Si,j includes the activity ak Then, we need to solve two sub-problems:    Therefore, the solution to Si,j is made of:     Si,k: all activities that start after ai and finish before ak Sk,j: all activities that start after ak and finish before aj The solution to Si,k ak The solution to Sk,j Because ak is compatible with both Si,k and Sk,j  |Solution to Si,j| = |Solution to Si,k| + 1 + |Solution to Sk,j|
12. 12. Optimal Substructure (2)  If an optimal solution to Si,j includes ak, then the subsolutions for Si,k and Sk,j must also be optimal Ai,j = optimal solution for Si,j  Ai,j = Ai,k U {ak} U Ak,j     If Si,j is not empty We know ak c[i, j] = |Ai,j| = maximum size of the subset of mutually compatible activities in Ai,j  c[i, j] = 0 if i >= j
13. 13. Recursive Formulation  As we do not know the value of k, we must try all the possible choices in order to find it  Now, we can solve this problem using DP     O(n2) sub-problems O(n) choices at each step O(n3) complexity for the DP solution We can find a better one by using a greedy strategy!
14. 14. Greedy Choice      Theorem If Si,j is not empty and am is the activity with the earliest finish time in Si,j Then, am is used by at least one of the maximum size subset of mutually independent activities in Si,j Si,m = Φ , therefore only Sm,j needs to be solved For any other solution to Si,j , we can replace the activity that finishes earliest in this solution (let‟s call it ak) with am, and these activities are still mutually independent, as am finishes earlier than ak
15. 15. Greedy Choice (2)   The previous theorem offers the greedy choice The number of sub-problems considered in the optimal solution at each step:    The number of choices to be considered at each step:    DP: 2 Greedy: 1 DP: j-i-1 Greedy: 1 As we have a single choice and a single subproblem to solve, we can solve the problem topdown
16. 16. Greedy Solution  In order to solve Si,j  Just choose the activity with the earliest finish time in Si,j    am Then, solve Sm,j In order to solve S = S0,n+1      First choice am1 (is always a1 – why?) for S0,n+1 Then need to solve Sm1,n+1 Second choice am2 for Sm1,n+1 Then need to solve Sm2,n+1 …
17. 17. Recursive Algorithm  Because the greedy algorithm considers the activities sorted by their finish time, we first need to sort by the finish time!  O(n logn) RecursiveActivitySelection(s, f, i, n) m = i +1 WHILE (m <= n AND s[m] < f[i]) m++ // find the activity with the earliest // start time that starts after activity i finishes IF (m <= n) THEN RETURN {am} U RecursiveActivitySelection(s, f, m, n) RETURN Φ Initial call: RecursiveActivitySelection(s, f, 0, n) Complexity: (n) – go through each activity once
18. 18. Iterative Algorithm  Can turn the recursive algorithm into an iterative one IterativeActivitySelection(s, f, n) A = {a1} i=1 FOR (m = 2..n) IF (s[m] < f[i]) CONTINUE ELSE A = A U {am} i=m RETURN A Complexity: (n) – go once through each activity
19. 19. Huffman Trees  Efficient method of compressing files   Especially text files Builds a Huffman tree in a greedy fashion   Specific for the encoded text/file It is used for compressing the file  The compressed file and the Huffman tree are used to recreate the original file  Example text: “ana are mere”
20. 20. Huffman Trees (2)    K – set of keys that are encoded (the characters in the original text file) In the original text, all the keys are represented on the same number of bits Objective: we want to find an alternative representation for each key such that:    The keys that are most frequent are represented on a smaller number of bits than the ones that are less frequent We are able to distinguish easily in this new representation what are the keys that were in the original file Example: text files   Original representation: char – 8 bits or ASCII – 7 bits New representation: 1 bit for the most frequent character in the encoded text and so on…
21. 21. Huffman Trees (3)  Huffman encoding tree:     An ordered binary tree Only the leaves contain the keys from the set K All internal nodes must have exactly 2 children The edges are coded:      0 – left edge 1 – right edge The code in the new representation for each key is the set of codes from the root to the leaf containing that key Start from the frequency of appearance of each key in the original file: p(k) for each k in K Example: “ana are mere”  p(a) = p(e) = 0.25; p(n) = p(m) = 0.083;p(r) = p( ) = 0.166
22. 22. The Huffman Tree     T – encoding tree for the set of keys K code_length(k) – the length of the code for key k in tree T level(k, T) – the level in tree T for the leaf corresponding to key k The cost of an encoding tree T for a set of keys K that have the frequencies p: Cost (T ) code _ length(k ) * p(k ) k K  level (k , T ) * p(k ) k K Huffman Tree = An encoding tree of minimum cost for a set of keys K with frequencies p   The codes in this tree are called Huffman codes Optimization problem!
23. 23. Building the Huffman Tree We can devise a greedy algorithm for building a Huffman tree for any set of keys K  Steps: 1. For each key k in K build a simple tree with a single node that contains k and has the weight w = p(k). Let the forest of trees be called Forest. 2. Choose any two trees from Forest that have the minimum weights. Let them be t1 and t2. 3. Remove t1 and t2 from Forest and add a new tree:  a) b) c) That has a new root r that does not contain any key (as it is not a leaf) The two descendents of r are t1 and t2 respectively. The weight of the new tree is w(r) = w(t1) + w(t2) Repeat steps 2 and 3 until Forest contains a single tree 4.  => the Huffman tree
24. 24. Example  Input: “ana are mere”  p(„a‟) = p(„e‟) = 0.25; p(„n‟) = p(„m‟) = 0.083; p(„r‟) = p(„ „) = 0.166  Initially: W(a)= 0.25 W(e)= 0.25 W(r)= 0.16 W( )= 0.16 W(m)= 0.08 W(n)= 0.08
25. 25. Example – Building the Huffman Tree   Intermediate steps: on whiteboard Solution:    Encoding: „a‟ : 00 , „e‟ : 11 , „r‟ : 010 , „ ‟ : 011 , „m‟ : 100 , „n‟ : 101 Cost of the tree: Cost(Tree) = 2 * 0.25 + 2 * 0.25 + 3 * 0.083 + 3 * 0.083 + 3 * 0.166 + 3 * 0.166 = 2.2 bits Huffman Tree 0 W(a+r+ )=0.57 0 1 W(r+ )=0.32 W(a) 1 0 W(r) W( ) 1 W(m+n+e)=0.41 0 1 W(m+n)=0.16 W(e) 0 1 W(m) W(n)
26. 26. Algorithm for building the Huffman Tree   On the whiteboard Straightforward from the pseudocode
27. 27. Decoding the File  Encoded text:   0010100011000101000111001101011 a n a „‟ a r e „‟ m e r e  We also need the Huffman tree  Starting from the first bit, we walk the tree from the root to the first leaf we encounter   When at a leaf, append the key corresponding to that leaf to the decoded text Go to the root again and repeat until we reach the end of the encoded text
28. 28. Greedy Algorithms – Conclusions   Greedy algorithms that build the globally optimal solution can be devised for some problems that have an optimal substructure Steps for devising a greedy algorithm:  Determine the optimal substructure  Develop a recursive solution  Prove that at any stage of recursion, one of the optimal choices is the greedy choice. Therefore, it‟s always safe to make the greedy choice  Show that all but one of the sub-problems resulting from the greedy choice are empty  Develop a recursive greedy algorithm  Convert it to an iterative algorithm
29. 29. Greedy Algorithms – Conclusions (2)  Properties for optimization problems that accept correct greedy solutions:    Optimal substructure Greedy choice property Preprocessing is essential for efficient greedy algorithms:  E.g. sort some data prior to process it with the greedy algorithm
30. 30. Greedy vs. DP  Similarities:     Optimization problems Optimal substructure (including division into subproblems) Make a choice at each step Differences:     Greedy: 1 choice, 1 sub-problem to be solved Greedy is top-down, DP is bottom-up Greedy has the greedy choice property Greedy does not use memoization as the other subproblems are not important (they are discarded if they are not used by the greedy choice)
31. 31. Knapsack Problem  Given a set on n items:    Which are the items that should be carried in order to maximize the total value that can be carried in a knapsack of total weight W?   Values v[i] Weights w[i] Optimization problem Similar to the change-making problem  Given a set of divisions (coins and banknotes for a currency), find the minimum number of coins and banknotes needed to change a given amount of money
32. 32. Knapsack Problem (2)  Can be solved efficiently if:  Are allowed to carry fractions of the items    Fractional knapsack problem Greedy solution: sort the items according to the ratio v[i]/w[i] and choose the items in the order of the highest ratio until the knapsack is full We are not allowed to carry fractions of the items    Integer (0/1) knapsack problem But the values for weights and values are relatively small integers DP solution: on whiteboard
33. 33. Knapsack Problem (3)  However, in the general case:    Real values for weights Very high values for weights The problem can only be solved using a backtracking approach   The problem is NP-complete The class of the most difficult problems that can be solved on a computer (at this moment, it‟s considered that these problems cannot be solved in polynomial time)
34. 34. References  CLRS – Chapter 16  MIT OCW – Introduction to Algorithms – video lecture 16  http://www.math.fau.edu/locke/Greedy.htm