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Algorithm Design Techniques: Greedy Algorithms

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Introduction Algorithm Design Techniques –Design of algorithms –Algorithms commonly used to solve problems Greedy, Divide and Conquer, Dynamic Programming, Randomized, Backtracking General approach Examples Time and space complexity (where appropriate)

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Greedy Algorithms Choose the best option during each phase –Dijkstra, Prim, Kruskal Making change –Choose largest bill at each round –Does this always work? Bad examples where greedy does not work?

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Greedy Algorithms Must have –Greedy-choice property: a globally optimal solution can be arrived at by making a locally optimal choice –Optimal substructure: an optimal solution to a problem contains optimal solutions to its subproblems

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Making Change Greedy choice property –Highest denomination coin < n will reside in solution – if not, it will be replaced by two or more smaller coins which will be more coins and not optimal –This is also true for 1, 7, 10 denominations??? Optimal substructure –Solution for (n – highest denomination coin) is optimal

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Scheduling Given jobs j1, j2, j3,..., jn with known running times t1, t2, t3,..., tn – what is the best way to schedule the jobs to minimize average completion time? JobTime j115 j28 j33 j410

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Scheduling j1j2j3j4 15 23 26 36 Average completion time = (15+23+26+36)/4 = 25 j1j2 j3 j4 3 11 21 36 Average completion time = (3+11+21+36)/4 = 17.75

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Scheduling Greedy-choice property: if shortest job does not go first, the y jobs before it will complete 3 time units faster, but j3 will be postponed by time to complete all jobs before it Optimal substructure: if shortest job is removed from optimal solution, remaining solution for n-1 jobs is optimal

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Optimality Proof Total cost of a schedule is N ∑(N-k+1)t ik k=1 t1 + (t1+t2) + (t1+t2+t3)... (t1+t2+...+tn) N N (N+1)∑t ik - ∑k*t ik k=1 k=1 First term independent of ordering, as second term increases, total cost becomes smaller

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Scheduling Suppose there is a job ordering such that x > y and t ix < t iy Swapping jobs (smaller first) increases second term decreasing total cost Show: xt ix + yt iy < yt ix + xt iy xt ix + yt iy = xt ix + yt ix + y(t iy - t ix ) = yt ix + xt ix + y(t iy - t ix ) < yt ix + xt ix + x(t iy - t ix ) = yt ix + xt ix + xt iy - xt ix = yt ix + xt iy

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More Scheduling Multiple processor case –Algorithm?

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More Scheduling Multiple processor case –Algorithm: order jobs shortest first schedule jobs round-robin Minimizing final completion time –When is this useful? –How is this different? –Problem is NP-Complete!

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Huffman Codes 100 ASCII characters Need ceil(log 100) bits to represent each character Large file = lots of bits! Would like to reduce number of bits

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Huffman Codes Idea – encode frequently occurring characters using fewer bits Need to make sure all characters are distinguishable –01 = A 0101 = B –010101 =? AAA, AB, BA No character code should be a prefix of another character code

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Huffman Codes Goal: find a full binary tree of minimum cost where characters are stored in the leaves Cost of tree: sum across all characters of the frequency of the character times its depth in the tree –frequently occurring characters should be highest in the tree

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Huffman Codes t snl eisp CharacterCodeFrequencyTotal Bits a0011030 e011530 i101224 s00000315 t0001416 space111326 newline0000115 total146

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Huffman’s Algorithm How do we produce a code? –Maintain a forest of trees weight of a tree is the sum of the frequencies of the leaves start with C trees to represent each character –weight of each is frequency of that character –Until there is only 1 tree choose the 2 trees with the smallest weights and merge them by creating a new root and making each tree a right or left subtree –Running time – O (ClogC)

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Optimality Proof – Idea 1.The tree must be full if it is not, move leaf with no siblings to its parent 2.Least frequent characters are the deepest nodes if not, a node can be swapped with an ancestor 3.Characters at the same depth can be swapped 4.As trees are merged, optimality holds

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Optimality Proof – Idea Greedy choice property: given x and y -- characters with lowest frequency in alphabet C, there exists an optimal prefix code for C in which the codewords for x and y have the same length and differ only in the last bit –Take an arbitrary optimal prefix code and modify it to make it a tree representing another optimal prefix code such that x and y are sibling leaves of max depth

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Optimality Proof – Idea Optimal substructure: C’ = C – {x, y} U {z} where f[z] = f[x]+f[y] T’ is optimal tree for C’ Replace z in T’ with internal node having x and y as children Result is optimal prefix code for C

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Approximate Bin Packing N items of sizes s 1, s 2,..., s N 0 < s i <= 1 Goal: pack into fewest number of bins of size 1 NP-complete problem, but we can use greedy algorithms to produce solutions not too far from optimal Knapsack problem Examples? –Saving data to external media

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Example – Optimal Packing.8.2.3.7.5.4.1 Input:.2,.5,.4,.7,.1,.3,.8

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On-line vs Off-line On-line –Process one item at a time –Cannot move an item once it is placed Off-line –Look at all items before you place first item

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On-line Algorithms On-line algorithms cannot guarantee optimal solution –Problem: cannot know when input will end –M small items ½-ε – M large items ½+ε –Can fit into M bins with 1 large and 1 small in each bin –If all small come first, place in M separate bins –If input is only M small items, we have used twice as many bins as necessary –There are inputs that force any on-line bin-packing algorithm to use at least 4/3 the optimal number of bins.

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On-line Bin Packing Algorithms Next fit First fit Best fit

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On-line Bin Packing Algorithms Next fit –Algorithm if item first in bin with last item – place there else – place in new bin –(.2,.5) (.4) (.7,.1) (.3) (.8) –Running time? –Let M be the optimal number of bins required to pack a list I of items. Then next fit never uses more than 2M bins. At most, half of the space is wasted (B j + B j+1 > 1)

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On-line Bin Packing Algorithms First fit –Algorithm Scan all bins and place item in first bin large enough to hold it if no bin is large enough, create new bin –(.2,.5,.1) (.4,.3) (.7) (.8) –Running time? –Let M be the optimal number of bins required to pack a list I of items. Then first fit never uses more than ceil(17/10M) bins.

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On-line Bin Packing Algorithms Best fit –Algorithm Scan all bins and place item in bin with tightest fit (will be fullest after item is placed there) if no bin is large enough, create new bin –(.2,.5,.1) (.4) (.7,.3) (.8) –Running time? –Same performance as first fit.

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Off-line Bin Packing Sort items (in decreasing order) first for easier placement of large items Apply first fit or best fit algorithm –First fit – (.8,.2) (.7,.3) (.5,.4,.1) Let M be the optimal number of bins required to pack a list I of items. Then first fit decreasing never uses more than (11/9M)+4 bins.

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