Y.-W. Chang 1Unit 5

Unit 5: Greedy Algorithms․Course contents:

⎯ Elements of the greedy strategy⎯ Activity selection⎯ Knapsack problem⎯ Huffman codes⎯ Task scheduling

․Reading:⎯ Chapter 16

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Greedy Algorithm: Vertex Cover․A vertex cover of an undirected graph G=(V, E) is a subset V' ⊆ V

such that if (u, v) ∈ E, then u ∈ V' or v ∈ V', or both.⎯ The set of vertices covers all the edges.

․The size of a vertex cover is the number of vertices in the cover.․The vertex-cover problem is to find a vertex cover of minimum

size in a graph.․Greedy heuristic: cover as many edges as possible (vertex with

the maximum degree) at each stage and then delete the covered edges.

․The greedy heuristic cannot always find an optimal solution!⎯ The vertex-cover problem is NP-complete.

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A Greedy Algorithm․A greedy algorithm always makes the choice that looks best at

the moment.․An Activity-Selection Problem: Given a set S = {1, 2, …, n} of n

proposed activities, with a start time si and a finish time fi for each activity i, select a maximum-size set of mutually compatible activities.⎯ If selected, activity i takes place during the half-open time

interval [si, fi).⎯ Activities i and j are compatible if [si, fi) and [sj, fj) do not

overlap (i.e., si ≥ fj or sj ≥ fi).

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Activity Selection

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The Activity-Selection AlgorithmGreedy-Activity-Selector(s,f)/* Assume f1 ≤ f2 ≤ … ≤ fn. */ 1. n ← length[s];2. A ← {1};3. j ← 1;4. for i ← 2 to n5. if si ≥ fj6. A ← A ∪ {i};7. j ← i;8. return A.

․Theorem: Algorithm Greedy-Activity-Selector produces solutions of maximum size for the activity-selection problem.⎯ (Greedy-choice property) Suppose A ⊆ S is an optimal solution.

Show that if the first activity in A activity k ≠ 1, then B = A - {k} ∪ {1} is an optimal solution.

⎯ (Optimal substructure) Show that if A is an optimal solution to S, then A' = A - {1} is an optimal solution to S' = {i ∈ S: si ≥ f1}.

⎯ Prove by induction on the number of choices made.

․Time complexity excluding sorting: O(n)

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Elements of the Greedy Strategy․When to apply greedy algorithms?

⎯ Greedy-choice property: A global optimal solution can be arrived at by making a locally optimal (greedy) choice.

Dynamic programming needs to check the solutions to subproblems.

⎯ Optimal substructure: An optimal solution to the problem contains within its optimal solutions to subproblems.

E.g., if A is an optimal solution to S, then A' = A - {1} is an optimal solution to S' = {i ∈ S: si ≥ f1}.

․Greedy algorithms (heuristics) do not always produce optimal solutions.

․Greedy algorithms v.s. dynamic programming (DP)⎯ Common: optimal substructure⎯ Difference: greedy-choice property⎯ DP can be used if greedy solutions are not optimal.

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Knapsack Problem․ Knapsack Problem: Given n items, with ith item worth vi dollars and

weighing wi pounds, a thief wants to take as valuable a load as possible, but can carry at most W pounds in his knapsack.

․ The 0-1 knapsack problem: Each item is either taken or not taken (0-1 decision).

․ The fractional knapsack problem: Allow to take fraction of items.․ Exp: = (60, 100, 120), = (10, 20, 30), W = 50

․ Greedy solution by taking items in order of greatest value per pound is optimal for the fractional version, but not for the 0-1 version.

․ The 0-1 knapsack problem is NP-complete, but can be solved in O(nW) time by DP. (A polynomial-time DP??)

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Coding

․Is used for data compression, instruction-set encoding, etc.

․Binary character code: character is represented by a unique binary string⎯ Fixed-length code (block code): a: 000, b:

001, …, f: 101 ⇒ ace ↔ 000 010 100.⎯ Variable-length code: frequent characters ⇒ short

codeword; infrequent characters ⇒ long codeword

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Binary Tree v.s. Prefix Code․Prefix code: No code is a prefix of some other code.

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Optimal Prefix Code Design․Coding Cost of T: B(T)=∑ c ∈ C f(c)dT(c)

⎯ c : each character in the alphabet C⎯ f(c): frequency of c⎯ dT(c): depth of c's leaf (length of the codeword of c)

․Code design: Given f(c1), f(c2), …, f(cn), construct a binary tree with n leaves such that B(T) is minimized.⎯ Idea: more frequently used characters use shorter depth.

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Huffman's Procedure․Pair two nodes with the least costs at each step.

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Huffman's Algorithm

․Time complexity: O(n lg n).⎯ Extract-Min(Q) needs O(lg n) by a heap operation.⎯ Requires initially O(n lg n) time to build a binary heap.

Huffman(C)1. n ← |C|;2. Q ← C;3. for i ← 1 to n -14. z ← Allocate-Node();5. x ← left[z] ← Extract-Min(Q);6. y ← right[z] ← Extract-Min(Q);7. f[z] ← f[x] + f[y];8. Insert(Q, z);9. return Extract-Min(Q)

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Huffman’s Algorithm: Greedy Choice

․Greedy choice: Two characters x and y with the lowest frequencies must have the same length and differ only in the last bit.

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Huffman's Algorithm: Optimal Substructure․Optimal substructure: Let T be a full binary tree for an

optimal prefix code over C. Let z be the parent of two leaf characters x and y. If f [z] = f [x] + f [y], tree T' = T -{x, y} represents an optimal prefix code for C'= C - {x, y} ∪ {z}.

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Task Scheduling․The task scheduling problem: Schedule unit-time tasks

with deadlines and penalties s.t. the total penalty for missed deadlines is minimized.⎯ S = {1, 2, …, n} of n unit-time tasks.⎯ Deadlines d1, d2, …, dn for tasks, 1 ≤ di ≤ n.⎯ Penalties w1, w2, …, wn : wi is incurred if task i misses

deadline.․Set A of tasks is independent if ∃ a schedule with no late

tasks. ․Nt(A): number of tasks in A with deadlines t or earlier,

t = 1, 2, …, n.․Three equivalent statements for any set of tasks A

1. A is independent.2. Nt(A) ≤ t, t = 1, 2, …, n.3. If the tasks in A are scheduled in order of nondecreasing

deadlines, then no task is late.

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Greedy Algorithm: Task Scheduling․The optimal greedy scheduling algorithm:

1. Sort penalties in non-increasing order.2. Find tasks of independent sets: no late task in the sets.3. Schedule tasks in a maximum independent set in order

of nondecreasing deadlines.4. Schedule other tasks (missing deadlines) at the end

arbitrarily.

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Two-layer Channel Routing․ Local density at column i: total # of nets that cross column i.․ Channel density: maximum local density.․ # of horizontal tracks required ≥ channel density.

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Standard Cell Layout

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Channel Routing Problem

․ Assignments of horizontal segments of nets to tracks. ․ Assignments of vertical segments to connect.

⎯ horizontal segments of the same net in different tracks, and⎯ the terminals of the net to horizontal segments of the net.

․ Horizontal and vertical constraints must not be violated.⎯ Horizontal constraints between two nets: The horizontal span of two

nets overlaps each other.⎯ Vertical constraints between two nets: There exists a column such that

the terminal on top of the column belongs to one net and the terminal on bottom of the column belongs to the other net.

․ Objective: Channel height is minimized (i.e., channel area is minimized).

Horizontal constraint

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2-L Channel Routing: Basic Left-Edge Algorithm

․No vertical constraint.․One layer for horizontal nets, one for vertical nets.․Doglegs are not allowed.․Treat each net as an interval.․Greedy Algorithm

⎯ Intervals are sorted according to their left-end x-coordinates.⎯ Intervals (nets) are routed one-by-one according to the order.⎯ For a net, tracks are scanned from top to bottom, and the first

track that can accommodate the net is assigned to the net.

․Optimality: produces a routing solution with the minimum # of tracks if no vertical constraint.

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Basic Left-Edge AlgorithmBasic_Left-Edge(U, track[j]) /* U: set of unassigned intervals (nets) I1, …, In; *//* Ij=[sj, ej]: interval j with left-end x-coordinate sj and right-end ej; *//* track[j]: track to which net j is assigned. */

1. U ← { I1, I2, …, In};2. t ← 0 ;3. while (U ) 4. t ← t + 1; 5. watermark ← 0; 6. while (there is an Ij U s.t. sj > watermark)7. Pick the interval Ij U with sj > watermark, nearest watermark;8. track[j] ← t; 9. watermark ← ej; 10. U ← U - {Ij};

≠ ∅

∈∈

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Basic Left-Edge Example

․ U = {I1, I2, …, I6}.⎯ I1 = [1, 3], I2 = [2, 6], I3 = [4, 8], I4 = [5, 10], I5= [7, 11], I6 = [9, 12].

․ t =1:⎯ Route I1: watermark = 3;⎯ Route I3: watermark = 8;⎯ Route I6: watermark = 12;

․ t = 2:⎯ Route I2: watermark = 6;⎯ Route I5: watermark = 11;

․ t = 3: Route I4

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Over-the-Cell Routing․Routing over the cell rows is possible due to the limited use

of the 2nd (M2) metal layers within the cells.․Divide the over-the-cell routing problem into 3 steps: (1)

routing over the cell, (2) choosing the net segments, and (3) routing within the channel.

․Reference: Cong & Liu, “Over-the-cell channel routing,”IEEE TCAD, Apr. 1990.

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Over-the-Cell Channel Routing

․Cong & Liu, “Over-the-cell channel routing,” IEEE TCAD, Apr. 1990.

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Supowit's Algorithm․Supowit, “Finding a maximum plannar subset of a set of nets

in a channel,” IEEE TCAD, 1987.․Problem: Given a set of chords, find a maximum plannar

subset of chords.⎯ Label the vertices on the circle 0 to 2n-1.⎯ Compute MIS(i, j): size of maximum independent set between

vertices i and j, i < j.⎯ Answer = MIS(0, 2n-1).

Vertices on the circle

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Dynamic Programming in Supowit's Algorithm

․Apply dynamic programming to compute MIS(i, j ).