Algorithm Design and Analysis Greedy Algorithm (1)
http://ada.miulab.tw
Outline
• Greedy Algorithms
• Greedy #1: Activity-Selection / Interval Scheduling
• Greedy #2: Coin Changing
• Greedy #3: Fractional Knapsack Problem
• Greedy #4: Breakpoint Selection
• Greedy #5: Huffman Codes
• Greedy #6: Task-Scheduling
• Greedy #7: Scheduling to Minimize Lateness
Algorithm Design Strategy
• Do not focus on “specific algorithms”
• But “some strategies” to “design” algorithms
• First Skill: Divide-and-Conquer (各個擊破/分治)
• Second Skill: Dynamic Programming (動態規劃)
• Third Skill: Greedy (貪婪法則)
Greedy Algorithms
Textbook Chapter 16 – Greedy Algorithms
Textbook Chapter 16.2 – Elements of the greedy strategy
What is Greedy Algorithms?
• always makes the choice that looks best at the moment
• makes a locally optimal choice in the hope that this choice will lead to a globally optimal solution
• not always yield optimal solution; may end up at local optimal
Greedy: move towards max gradient and hope it is global maximum local maximal
global maximal
local maximal
Algorithm Design Paradigms
• Dynamic Programming
• has optimal substructure
• make an informed choice after getting optimal solutions to subproblems
• dependent or overlapping subproblems
• Greedy Algorithms
• has optimal substructure
• make a greedy choice before solving the subproblem
• no overlapping subproblems
✓ Each round selects only one subproblem
✓ The subproblem size decreases
Optimal Solution
Possible Case 1 Possible
Case 2
Possible Case k
max /min
Subproblem Solution Subproblem
Solution
Subproblem Solution
+
+
+
= Optimal
Solution
Greedy Choice
Subproblem Solution
= +
Greedy Procedure
1. Cast the optimization problem as one in which we make a choice and remain one subproblem to solve
2. Demonstrate the optimal substructure
✓ Combining an optimal solution to the subproblem via greedy can arrive an optimal solution to the original problem
3. Prove that there is always an optimal solution to the original problem that
makes the greedy choice
Greedy Algorithms
To yield an optimal solution, the problem should exhibit
1. Optimal Substructure : an optimal solution to the problem contains within its optimal solutions to subproblems
2. Greedy-Choice Property : making locally optimal (greedy) choices leads to
a globally optimal solution
Proof of Correctness Skills
• Optimal Substructure : an optimal solution to the problem contains within it optimal solutions to subproblems
• Greedy-Choice Property : making locally optimal (greedy) choices leads to a globally optimal solution
• Show that it exists an optimal solution that “contains” the greedy choice using exchange argument
• For any optimal solution OPT, the greedy choice 𝑔 has two cases
• 𝑔 is in OPT: done
• 𝑔 not in OPT: modify OPT into OPT’ s.t. OPT’ contains 𝑔 and is at least as good as OPT
OPT OPT’
𝑔
✓ If OPT’ is better than OPT, the property is proved by contradiction
✓ If OPT’ is as good as OPT, then we showed that there exists an optimal solution containing 𝑔 by construction
Activity-Selection / Interval Scheduling
Textbook Chapter 16.1 – An activity-selection problem
Activity-Selection/ Interval Scheduling
• Input: 𝑛 activities with start times 𝑠
𝑖and finish times 𝑓
𝑖(the activities are sorted in monotonically increasing order of finish time 𝑓
1≤ 𝑓
2≤ ⋯ ≤ 𝑓
𝑛)
• Output: the maximum number of compatible activities
• Without loss of generality: 𝑠
1< 𝑠
2< ⋯ < 𝑠
𝑛and 𝑓
1< 𝑓
2< ⋯ < 𝑓
𝑛• 大的包小的則不考慮大的 → 用小的取代大的一定不會變差
time 1
2 3 4 5 6
activity index
Weighted Interval Scheduling
• Subproblems
• WIS(i): weighted interval scheduling for the first 𝑖 jobs
• Goal: WIS(n)
• Dynamic programming algorithm
i 0 1 2 3 4 5 … n
M[i]
Set 𝑣𝑖 = 1 for all 𝑖 to formulate it into the activity-selection problem
Weighted Interval Scheduling Problem
Input: 𝑛 jobs with 𝑠𝑖, 𝑓𝑖, 𝑣𝑖 , 𝑝(𝑗) = largest index 𝑖 < 𝑗 s.t. jobs 𝑖 and 𝑗 are compatible Output: the maximum total value obtainable from compatible
Activity-Selection Problem
• Dynamic programming
• Optimal substructure is already proved
• Greedy algorithm
select the 𝑖-th activity
Why does the 𝑖-th activity must appear
in an OPT?
Activity-Selection Problem
Input: 𝑛 activities with 𝑠𝑖, 𝑓𝑖 , 𝑝(𝑗) = largest index 𝑖 < 𝑗 s.t. 𝑖 and 𝑗 are compatible Output: the maximum number of activities
Greedy-Choice Property
• Goal:
• Proof
• Assume there is an OPT solution for the first 𝑖 − 1 activities (𝑀𝑖−1)
• 𝐴𝑗 is the last activity in the OPT solution →
• Replacing 𝐴𝑗 with 𝐴𝑖 does not make the OPT worse
time 1
2 :
i i - 1
:
activity index
1 2 3 4 5 6 7 8 9
Pseudo Code
Act-Select(n, s, f, v, p) M[0] = 0
for i = 1 to n if p[i] >= 0
M[i] = 1 + M[p[i]]
return M[n]
Find-Solution(M, n) if n = 0
return {}
return {n} ∪ Find-Solution(p[n])
Activity-Selection Problem
Input: 𝑛 activities with 𝑠𝑖, 𝑓𝑖 , 𝑝(𝑗) = largest index 𝑖 < 𝑗 s.t. 𝑖 and 𝑗 are compatible Output: the maximum number of activities
Coin Changing
Textbook Exercise 16.1
Coin Changing Problem
• Input: 𝑛 dollars and unlimited coins with values 𝑣
𝑖(1, 5, 10, 50)
• Output: the minimum number of coins with the total value 𝑛
• Cashier’s algorithm: at each iteration, add the coin with the largest value no more than the current total
Does this algorithm return the OPT?
Step 1: Cast Optimization Problem
• Subproblems
• C(i): minimal number of coins for the total value 𝑖
• Goal: C(n)
Coin Changing Problem
Input: 𝑛 dollars and unlimited coins with values 𝑣𝑖 (1, 5, 10, 50) Output: the minimum number of coins with the total value 𝑛
Step 2: Prove Optimal Substructure
• Suppose OPT is an optimal solution to C(i) , there are 4 cases:
• Case 1: coin 1 in OPT
• OPT\coin1 is an optimal solution of C(i – v1)
• Case 2: coin 2 in OPT
• OPT\coin2 is an optimal solution of C(i – v2)
• Case 3: coin 3 in OPT
• OPT\coin3 is an optimal solution of C(i – v3)
• Case 4: coin 4 in OPT
• OPT\coin4 is an optimal solution of C(i – v4)
Coin Changing Problem
Input: 𝑛 dollars and unlimited coins with values 𝑣𝑖 (1, 5, 10, 50) Output: the minimum number of coins with the total value 𝑛
Step 3: Prove Greedy-Choice Property
• Greedy choice: select the coin with the largest value no more than the current total
• Proof via contradiction (use the case 10 ≤ 𝑖 < 50 for demo)
• Assume that there is no OPT including this greedy choice (choose 10)
→ all OPT use 1, 5, 50 to pay 𝑖
• 50 cannot be used
• #coins with value 5 < 2 → otherwise we can use a 10 to have a better output
• #coins with value 1 < 5 → otherwise we can use a 5 to have a better output
• We cannot pay 𝑖 with the constraints (at most 5 + 4 = 9) Coin Changing Problem
Input: 𝑛 dollars and unlimited coins with values 𝑣𝑖 (1, 5, 10, 50) Output: the minimum number of coins with the total value 𝑛
To Be Continued…
Question?
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