Slides credited from Hsueh-I Lu, Hsu-Chun Hsiao, & Michael Tsai

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▪

Mini-HW 4 released

▪ Due on 10/18 (Thu) 14:20

▪

Homework 1 due a week later

▪ A4 hardcopy submitted before the class starts

▪ Softcopy submitted to NTU COOL before the deadline

▪

Homework 2 released

▪ Due on 11/06 (Tue) 18:00 (3.5 weeks)

▪ Submitted to NTU COOL only

Frequently check the website for the updated information!

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▪ Dynamic Programming

▪ DP #1: Rod Cutting

▪ DP #2: Stamp Problem

▪ DP #3: Knapsack Problem

▪ 0/1 Knapsack

▪ Unbounded Knapsack

▪ Multidimensional Knapsack

▪ Fractional Knapsack

▪ DP #4: Matrix-Chain Multiplication

▪ DP #5: Sequence Alignment Problem

▪ Longest Common Subsequence (LCS) / Edit Distance

▪ Viterbi Algorithm

▪ Space Efficient Algorithm

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▪

有100個死囚，隔天執行死刑，典獄長開恩給他們一個存活的機會。

▪

當隔天執行死刑時，每人頭上戴一頂帽子(黑或白)排成一隊伍，在

死刑執行前，由隊伍中最後的囚犯開始，每個人可以猜測自己頭上的帽子顏色(只允許說黑或白)，猜對則免除死刑，猜錯則執行死刑。

▪

若這些囚犯可以前一天晚上先聚集討論方案，是否有好的方法可以

使總共存活的囚犯數量期望值最高？

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▪

囚犯排成一排，每個人可以看到前面所有人的帽子，但看不到自己及後面囚犯的。

▪

由最後一個囚犯開始猜測，依序往前。

▪

每個囚犯皆可聽到之前所有囚犯的猜測內容。

……

Example: 奇數者猜測內容為前面一位的帽子顏色 → 存活期望值為75人有沒有更多人可以存活的好策略?

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8

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▪

Divide-and-Conquer

▪ partition the problem into independent or disjoint subproblems

▪ repeatedly solving the common subsubproblems

→ more work than necessary

▪

Dynamic Programming

▪ partition the problem into dependent or overlapping subproblems

▪ avoid recomputation

✓Top-down with memoization

✓Bottom-up method

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▪

DP procedure

1. Characterize the structure of an optimal solution

2. Recursively define the value of an optimal solution

3. Compute the value of an optimal solution, typically in a bottom-upfashion

4. Construct an optimal solution from computed information

▪

Two key properties of DP for optimization

▪ Overlapping subproblems

▪ Optimal substructure – an optimal solution can be constructed from optimal solutions to subproblems

✓Reduce search space (ignore non-optimal solutions)

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Textbook Chapter 15.1 – Rod Cutting

11

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▪

Input: a rod of length 𝑛 and a table of prices 𝑝

_𝑖

for 𝑖 = 1, … , 𝑛

▪

Output: the maximum revenue 𝑟

_𝑛

obtainable by cutting up the rod and selling the pieces

length 𝑖 (m) 1 2 3 4 5

price 𝑝_𝑖 1 5 8 9 10

4m

2m

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▪

A rod with the length = 4

length 𝑖 (m) 1 2 3 4 5

price 𝑝_𝑖 1 5 8 9 10

4m

3m 1m

2m 2m

1m 3m

2m 1m 1m

1m 2m 1m

1m 1m 1m

1m

→ 9

→ 8 + 1 = 9

→ 5 + 5 = 10

→ 1 + 8 = 9

→ 5 + 1 + 1 = 7

→ 1 + 5 + 1 = 7

→ 1 + 1 + 5 = 7

→ 1 + 1 + 1 + 1 = 4

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▪

A rod with the length = 𝑛

▪ For each integer position, we can choose “cut” or “not cut”

▪ There are 𝑛 – 1 positions for consideration

▪

The total number of cutting results is 2

^𝑛−1

= Θ 2

^𝑛−1

length 𝑖 (m) 1 2 3 4 5

price 𝑝_𝑖 1 5 8 9 10

n

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▪

We use a recursive function to solve the subproblems

▪

If we know the answer to the subproblem, can we get the answer to the original problem?

▪

Optimal substructure – an optimal solution can be constructed from optimal solutions to subproblems

𝑟_𝑛−𝑖 𝑟_𝑖

no cut

cut at the i-th position (from left to right)

𝑟_𝑛: the maximum revenue obtainable for a rod of length 𝑛

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▪

Version 1

▪

Version 2

▪ try to reduce the number of subproblems → focus on the left-most cut no cut

cut at the i-th position (from left to right)

left-most value maximum value obtainable 𝑟_𝑛−𝑖

𝑝_𝑖

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▪

Focus on the left-most cut

▪ assume that we always cut from left to right → the first cut

optimal solution to subproblems

𝑟_𝑛−𝑖 𝑝_𝑖

𝑟_𝑛−1 𝑝₁

𝑟_𝑛−2 𝑝₂

: : optimal solution

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▪

𝑇 𝑛 = time for running

Cut-Rod(p, n) Cut-Rod(p, n)

// base case if n == 0

return 0

// recursive case q = -∞

for i = 1 to n

q = max(q, p[i] + Cut-Rod(p, n - i)) return q

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▪

Rod cutting problem

Cut-Rod(p, n) // base case if n == 0

return 0

// recursive case q = -∞

for i = 1 to n

q = max(q, p[i] + Cut-Rod(p, n - i)) return q

CR(4)

CR(3) CR(0)

CR(2) CR(1) CR(1) CR(0)

CR(1) CR(0) CR(0)

CR(0)

Calling overlapping subproblems result in poor efficiency

CR(2) CR(1)

CR(0) CR(0)

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▪

Idea: use space for better time efficiency

▪

Rod cutting problem has overlapping subproblems and optimal substructures

→ can be solved by DP

▪

When the number of subproblems is polynomial, the time complexity is polynomial using DP

▪

DP algorithm

▪ Top-down: solve overlapping subproblems recursively with memoization

▪ Bottom-up: build up solutions to larger and larger subproblems

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▪

Top-Down with Memoization

▪ Solve recursively and memo the subsolutions (跳著填表)

▪ Suitable that not all

subproblems should be solved

▪

Bottom-Up with Tabulation

▪ Fill the table from small to large

▪ Suitable that each small problem should be solved

f(0) f(1) f(2) … f(n) f(0) f(1) f(2) … f(n)

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Memoized-Cut-Rod(p, n)

// initialize memo (an array r[] to keep max revenue) r[0] = 0

for i = 1 to n

r[i] = -∞ // r[i] = max revenue for rod with length=i return Memorized-Cut-Rod-Aux(p, n, r)

Memoized-Cut-Rod-Aux(p, n, r) if r[n] >= 0

return r[n] // return the saved solution q = -∞

for i = 1 to n

q = max(q, p[i] + Memoized-Cut-Rod-Aux(p, n-i, r)) r[n] = q // update memo

return q

▪

𝑇 𝑛 = time for running

Memoized-Cut-Rod(p, n)

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Bottom-Up-Cut-Rod(p, n) r[0] = 0

for j = 1 to n // compute r[1], r[2], ... in order q = -∞

for i = 1 to j

q = max(q, p[i] + r[j - i]) r[j] = q

return r[n]

▪

𝑇 𝑛 = time for running

Bottom-Up-Cut-Rod(p, n)

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▪

Input: a rod of length 𝑛 and a table of prices 𝑝

_𝑖

for 𝑖 = 1, … , 𝑛

▪

Output: the maximum revenue 𝑟

_𝑛

obtainable and the list of cut pieces

length 𝑖 (m) 1 2 3 4 5

price 𝑝_𝑖 1 5 8 9 10

4m

2m

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▪

Add an array to keep the cutting positions cut

Extended-Bottom-Up-Cut-Rod(p, n) r[0] = 0

for j = 1 to n //compute r[1], r[2], ... in order q = -∞

for i = 1 to j

if q < p[i] + r[j - i]

q = p[i] + r[j - i]

cut[j] = i // the best first cut for len j rod r[i] = q

return r[n], cut

Print-Cut-Rod-Solution(p, n)

(r, cut) = Extended-Bottom-up-Cut-Rod(p, n) while n > 0

print cut[n]

n = n – cut[n] // remove the first piece

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f(0) f(1) f(2) … f(n)

▪

Top-Down with Memoization

▪ Better when some subproblems not be solved at all

▪ Solve only the required parts of subproblems

▪

Bottom-Up with Tabulation

▪ Better when all subproblems must be solved at least once

▪ Typically outperform top-down method by a constant factor

▪ No overhead for recursive calls

▪ Less overhead for maintaining the table

f(0) f(1) f(2) … f(n)

F(5) F(4) F(3) F(2) F(1) F(0)

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▪ Approach 1: approximate via (#subproblems) * (#choices for each subproblem)

▪

For rod cutting

▪ #subproblems = n

▪ #choices for each subproblem = O(n)

▪ → T(n) is about O(n²)

▪ Approach 2: approximate via subproblem graphs

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▪

The size of the subproblem graph allows us to estimate the time complexity of the DP algorithm

▪

A graph illustrates the set of subproblems involved and how subproblems depend on another 𝐺 = 𝑉, 𝐸 (E: edge, V: vertex)

▪ 𝑉 : #subproblems

▪ A subproblem is run only once

▪ |𝐸|: sum of #subsubproblems are needed for each subproblem

▪ Time complexity: linear to 𝑂( 𝐸 + 𝑉 )

F(5) F(4)

F(3)

F(2)

F(1)

F(0)

Bottom-up: Reverse Topological Sort Top-down: Depth First Search

Graph Algorithm (taught later)

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1.

Characterize the structure of an optimal solution

✓

Overlapping subproblems: revisit same subproblems

✓

Optimal substructure: an optimal solution to the problem contains within it optimal solutions to subproblems

2.

Recursively define the value of an optimal solution

✓

Express the solution of the original problem in terms of optimal solutions for subproblems

3.

Compute the value of an optimal solution

✓

typically in a bottom-up fashion

4.

Construct an optimal solution from computed information

✓

Step 3 and 4 may be combined

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1. Characterize the structure of an optimal solution

2. Recursively define the value of an optimal solution

3. Compute the value of an optimal solution

4. Construct an optimal solution from computed information

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▪

Step 1-Q1: What can be the subproblems?

▪

Step 1-Q2: Does it exhibit optimal structure? (an optimal solution can be represented by the optimal solutions to subproblems)

▪ Yes. → continue

▪ No. → go to Step 1-Q1 or there is no DP solution for this problem Rod Cutting Problem

Input: a rod of length 𝑛 and a table of prices 𝑝_𝑖 for 𝑖 = 1, … , 𝑛 Output: the maximum revenue 𝑟_𝑛 obtainable

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▪

Step 1-Q1: What can be the subproblems?

▪ Subproblems: Cut-Rod(0), Cut-Rod(1), …, Cut-Rod(n-1)

▪ Cut-Rod(i): rod cutting problem with length-i rod

▪ Goal: Cut-Rod(n)

▪ Suppose we know the optimal solution to Cut-Rod(i), there are i cases:

▪ Case 1: the first segment in the solution has length 1

:

▪ Case i: the first segment in the solution has length i Rod Cutting Problem

從solution中拿掉一段長度為1的鐵條, 剩下的部分是Cut-Rod(i-1)的最佳解從solution中拿掉一段長度為2的鐵條, 剩下的部分是Cut-Rod(i-2)的最佳解

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▪

Step 1-Q2: Does it exhibit optimal structure? (an optimal solution can be represented by the optimal solutions to subproblems)

▪

Yes. Prove by contradiction.

Rod Cutting Problem

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▪ Suppose we know the optimal solution to Cut-Rod(i), there are i cases:

:

▪ Case i: the first segment in the solution has length i

▪ Recursively define the value Rod Cutting Problem

從solution中拿掉一段長度為1的鐵條, 剩下的部分是Cut-Rod(i-1)的最佳解從solution中拿掉一段長度為2的鐵條, 剩下的部分是Cut-Rod(i-2)的最佳解

從solution中拿掉一段長度為i的鐵條, 剩下的部分是Cut-Rod(0)的最佳解

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▪ Bottom-up method: solve smaller subproblems first

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Rod Cutting Problem

i 0 1 2 3 4 5 … n

r[i]

Bottom-Up-Cut-Rod(p, n) r[0] = 0

for i = 1 to j

q = max(q, p[i] + r[j - i]) r[j] = q

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▪ Bottom-up method: solve smaller subproblems first Rod Cutting Problem

i 0 1 2 3 4 5 … n

r[i] 0

cut[i] 0 1

1

2 5

3 8

2 10

length 𝑖 1 2 3 4 5

price 𝑝_𝑖 1 5 8 9 10

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Cut-Rod(p, n) r[0] = 0

for i = 1 to j

if q < p[i] + r[j - i]

q = p[i] + r[j - i]

cut[j] = i // the best first cut for len j rod r[i] = q

return r[n], cut

Print-Cut-Rod-Solution(p, n) (r, cut) = Cut-Rod(p, n) while n > 0

print cut[n]

n = n – cut[n] // remove the first piece

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▪ Input: the postage 𝑛 and the stamps with values 𝑣

₁

, 𝑣

₂

, … , 𝑣

_𝑘

▪ Output: the minimum number of stamps to cover the postage

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▪

The optimal solution 𝑆

_𝑛

can be recursively defined as

Stamp(v, n) r_min = ∞

if n == 0 // base case return 0

for i = 1 to k // recursive case r[i] = Stamp(v, n - v[i])

if r[i] < r_min r_min = r[i]

return r_min + 1

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▪

Subproblems

▪ S(i): the min #stamps with postage i

▪ Goal: S(n)

▪

Optimal substructure: suppose we know the optimal solution to S(i), there are k cases:

▪ Case 1: there is a stamp with v₁ in OPT

▪ Case 2: there is a stamp with v₂ in OPT

:

▪ Case k: there is a stamp with v_k in OPT

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Stamp Problem

Input: the postage 𝑛 and the stamps with values 𝑣₁, 𝑣₂, … , 𝑣_𝑘 Output: the minimum number of stamps to cover the postage

從solution中拿掉一張郵資為v₁的郵票, 剩下的部分是S(i-v[1])的最佳解從solution中拿掉一張郵資為v₂的郵票, 剩下的部分是S(i-v[2])的最佳解

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▪ Suppose we know the optimal solution to S(i), there are k cases:

▪ Case 1: there is a stamp with v₁ in OPT

▪ Case 2: there is a stamp with v₂ in OPT

:

▪ Case k: there is a stamp with v_k in OPT

▪

Recursively define the value

Stamp Problem

從solution中拿掉一張郵資為v₁的郵票, 剩下的部分是S(i-v[1])的最佳解從solution中拿掉一張郵資為v₂的郵票, 剩下的部分是S(i-v[2])的最佳解

從solution中拿掉一張郵資為v_k的郵票, 剩下的部分是S(i-v[k])的最佳解

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43

Stamp Problem

i 0 1 2 3 4 5 … n

S[i]

Stamp(v, n) S[0] = 0

for i = 1 to n // compute r[1], r[2], ... in order r_min = ∞

for j = 1 to k

if S[i - v[j]] < r_min r_min = 1 + S[i – v[j]]

S[i] = r_min

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Stamp(v, n) S[0] = 0

for i = 1 to n r_min = ∞

for j = 1 to k

if S[i - v[j]] < r_min r_min = 1 + S[i – v[j]]

B[i] = j // backtracking for stamp with v[j]

S[i] = r_min return S[n], B

Print-Stamp-Selection(v, n) (S, B) = Stamp(v, n)

while n > 0 print B[n]

n = n – v[B[n]]

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Textbook Exercise 16.2-2

45

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▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊

▪

Variants of knapsack problem

▪ 0-1 Knapsack Problem: 每項物品只能拿一個

▪ Unbounded Knapsack Problem: 每項物品可以拿多個

▪ Multidimensional Knapsack Problem: 背包空間有限

▪ Multiple-Choice Knapsack Problem: 每一類物品最多拿一個

▪ Fractional Knapsack Problem: 物品可以只拿部分

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▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊

▪

Variants of knapsack problem

▪ 0-1 Knapsack Problem: 每項物品只能拿一個

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▪

Subproblems

▪ ZO-KP(i, w): 0-1 knapsack problem within 𝑤 capacity for the first 𝑖 items

▪ Goal: ZO-KP(n, W)

▪

Optimal substructure: suppose OPT is an optimal solution to ZO-KP(i, w), there are 2 cases:

▪ Case 1: item 𝑖 in OPT

▪ OPT\{𝑖} is an optimal solution of ZO-KP(i - 1, w - w_i)

▪ Case 2: item 𝑖 not in OPT 0-1 Knapsack Problem

Input: 𝑛 items where 𝑖-th item has value 𝑣_𝑖 and weighs 𝑤_𝑖

Output: the max value within 𝑊 capacity, where each item is chosen at most once ZO-KP(i) ZO-KP(i, w)

consider the available capacity

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▪

Optimal substructure: suppose OPT is an optimal solution to ZO-KP(i, w), there are 2 cases:

▪ OPT\{𝑖} is an optimal solution of ZO-KP(i - 1, w - w_i)

▪ Case 2: item 𝑖 not in OPT

▪ OPT is an optimal solution of ZO-KP(i - 1, w)

▪

Recursively define the value

0-1 Knapsack Problem

Output: the max value within 𝑊 capacity, where each item is chosen at most once

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i\w 0 1 2 3 … w … W

0 1 2 i

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i\w 0 1 2 3 4 5

0 0 0 0 0 0 0

1 0 4 4 4 4 4

2 0 4 9 13 13 13

i w_i v_i

1 1 4

2 2 9

3 4 20

𝑊 = 5

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ZO-KP(n, v, W) for w = 0 to W

M[0, w] = 0 for i = 1 to n

for w = 0 to W if(w_i > w)

M[i, w] = M[i-1, w]

else

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ZO-KP(n, v, W) for w = 0 to W

M[0, w] = 0 for i = 1 to n

for w = 0 to W if(w_i > w)

M[i, w] = M[i-1, w]

else

M[i, w] = max(v_i + M[i-1, w-w_i], M[i-1, w]) return M[n, W]

Find-Solution(M, n, W) S = {}

w = W

for i = n to 1

if M[i, w] > M[i – 1, w] // case 1 w = w – w_i

S = S ∪ {i}

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▪

Polynomial: polynomial in the length of the input (#bits for the input)

▪

Pseudo-polynomial: polynomial in the numeric value

▪

The time complexity of 0-1 knapsack problem is Θ 𝑛𝑊

▪ 𝑛: number of objects

▪ 𝑊: knapsack’s capacity (non-negative integer)

▪ polynomial in the numeric value

= pseudo-polynomial in input size

= exponential in the length of the input

▪

Note: the size of the representation of 𝑊 is log

₂

𝑊

= 2^𝑚 = 𝑚

(55)

▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊

▪

Variants of knapsack problem

▪ Unbounded Knapsack Problem: 每項物品可以拿多個

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▪

Subproblems

▪ U-KP(i, w): unbounded knapsack problem with 𝑤 capacity for the first 𝑖 items

▪ Goal: U-KP(n, W)

Unbounded Knapsack Problem

Input: 𝑛 items where 𝑖-th item has value 𝑣_𝑖 and weighs 𝑤_𝑖, each has unlimited supplies Output: the max value within 𝑊 capacity

0-1 Knapsack Problem Unbounded Knapsack Problem

each item can be chosen at most once each item can be chosen multiple times a sequence of binary choices: whether

to choose item 𝑖

a sequence of 𝑖 choices: which one (from 1 to 𝑖) to choose

Time complexity = Θ 𝑛𝑊 Time complexity = Θ 𝑛²𝑊

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▪ Subproblems

▪ U-KP(w): unbounded knapsack problem with 𝑤 capacity

▪ Goal: U-KP(W)

▪ Optimal substructure: suppose OPT is an optimal solution to U-KP(w), there are 𝑛 cases:

▪ Case 1: item 1 in OPT

▪ Removing an item 1 from OPT is an optimal solution of U-KP(w – w₁)

▪ Case 2: item 2 in OPT

▪ Removing an item 2 from OPT is an optimal solution of U-KP(w – w₂) :

▪ Case 𝑛: item 𝑛 in OPT

▪ Removing an item 𝑛 from OPT is an optimal solution of U-KP(w - w_n)

(58)

▪

Optimal substructure: suppose OPT is an optimal solution to U-KP(w), there are 𝑛 cases:

▪ Case 𝑖: item 𝑖 in OPT

▪ Removing an item i from OPT is an optimal solution of U-KP(w – w₁)

▪

Recursively define the value

只考慮背包還裝的下的情形

(59)

▪ Bottom-up method: solve smaller subproblems first Unbounded Knapsack Problem

w 0 1 2 3 4 5 … W

M[w]

i w_i v_i

1 1 4

2 2 9

3 4 20

𝑊 = 5

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▪ Bottom-up method: solve smaller subproblems first Unbounded Knapsack Problem

w 0 1 2 3 4 5

M[w] 0

i w_i v_i

1 1 4

2 2 9

3 4 17

𝑊 = 5

4 9 13 18 22

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U-KP(v, W)

for w = 0 to W M[w] = 0 for w = 0 to W

for i = 1 to n if(w_i <= w)

tmp = v_i + M[w - w_i] M[w] = max(M[w], tmp)

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U-KP(v, W)

for w = 0 to W M[w] = 0

for w = 0 to W for i = 1 to n

if(w_i <= w)

tmp = v_i + M[w - w_i] M[w] = max(M[w], tmp) return M[W]

Find-Solution(M, n, W) for i = 1 to n

C[i] = 0 // C[i] = # of item i in solution w = W

for i = i to n while w > 0

if(w_i <= w && M[w] == (v_i + M[w - w_i])) w = w - w_i

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▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊

▪

Variants of knapsack problem

▪ Multidimensional Knapsack Problem: 背包空間有限

(64)

▪

Subproblems

▪ M-KP(i, w, d): multidimensional knapsack problem with 𝑤 capacity and 𝑑 size for the first 𝑖 items

▪ Goal: M-KP(n, W, D)

▪

Optimal substructure: suppose OPT is an optimal solution to M-KP(i, w, d) , there are 2 cases:

▪ OPT\{𝑖} is an optimal solution of M-KP(i - 1, w - w_i, d – d_i)

Multidimensional Knapsack Problem

Input: 𝑛 items where 𝑖-th item has value 𝑣_𝑖, weighs 𝑤_𝑖, and size 𝑑_𝑖

Output: the max value within 𝑊 capacity and with the size of 𝑫, where each item is chosen at most once

(65)

▪

Optimal substructure: suppose OPT is an optimal solution to M-KP(i, w, d), there are 2 cases:

▪ OPT\{𝑖} is an optimal solution of M-KP(i - 1, w - w_i, d – d_i)

▪ OPT is an optimal solution of M-KP(i - 1, w, d)

▪

Recursively define the value

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▪ Step 3: Compute Value of an OPT Solution

▪ Step 4: Construct an OPT Solution by Backtracking

▪ What is the time complexity?

(67)

▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊

▪

Variants of knapsack problem

▪ Multiple-Choice Knapsack Problem: 每一類物品最多拿一個

(68)

▪

Input: 𝑛 items

▪ 𝑣_𝑖,𝑗: value of 𝑗-th item in the group 𝑖

▪ 𝑤_𝑖,𝑗: weight of 𝑗-th item in the group 𝑖

▪ 𝑛_𝑖: number of items in group 𝑖

▪ 𝑛: total number of items (σ 𝑛_𝑖)

▪ 𝐺: total number of groups

▪

Output: the maximum value for the knapsack with capacity of 𝑊,

where the item from each group can be selected at most once

(69)

▪

Subproblems

▪ MC-KP(w): 𝑤 capacity

▪ MC-KP(i, w): 𝑤 capacity for the first 𝑖 groups

▪ MC-KP(i, j, w): 𝑤 capacity for the first 𝑗 items from first 𝑖 groups Multiple-Choice Knapsack Problem

Input: 𝑛 items with value 𝑣_𝑖,𝑗 and weighs 𝑤_𝑖,𝑗 (𝑛_𝑖: #items in group 𝑖, 𝐺: #groups) Output: the max value within 𝑊 capacity, where each group is chosen at most once

Which one is more suitable for this problem?

the constraint is for groups

(70)

▪

Subproblems

▪ MC-KP(i, w): multi-choice knapsack problem with 𝑤 capacity for the first 𝑖 groups

▪ Goal: MC-KP(G, W)

▪

Optimal substructure: suppose OPT is an optimal solution to MC-KP(i, w), for the group 𝑖, there are 𝑛

_𝑖

+ 1 cases:

▪ Case 1: no item from 𝑖-th group in OPT

▪ OPT is an optimal solution of MC-KP(i - 1, w)

:

▪ Case 𝑗 + 1: 𝑗-th item from 𝑖-th group (item_i,j) in OPT Multiple-Choice Knapsack Problem

(71)

▪

Optimal substructure: suppose OPT is an optimal solution to MC-KP(i, w), for the group 𝑖, there are 𝑛

_𝑖

+ 1 cases:

▪ Case 1: no item from 𝑖-th group in OPT

▪ OPT is an optimal solution of MC-KP(i - 1, w)

▪ Case 𝑗 + 1: 𝑗-th item from 𝑖-th group (item_i,j) in OPT

▪ OPT\item_i,j is an optimal solution of MC-KP(i - 1, w – w_i,j)

▪

Recursively define the value

71

Multiple-Choice Knapsack Problem

(72)

i\w 0 1 2 3 … w … W

0 1 2 i

(73)

MC-KP(n, v, W) for w = 0 to W

M[0, w] = 0

for i = 1 to G // consider groups 1 to i for w = 0 to W // consider capacity = w

M[i, w] = M[i - 1, w]

for j = 1 to n_i // check j-th item in group i if(v_i,j + M[i - 1, w - w_i,j] > M[i, w])

M[i, w] = v_i,j + M[i - 1, w - w_i,j] return M[G, W]

(74)

MC-KP(n, v, W) for w = 0 to W

M[0, w] = 0

for i = 1 to G // consider groups 1 to i for w = 0 to W // consider capacity = w

M[i, w] = M[i - 1, w]

for j = 1 to n_i // check items in group i if(v_i,j + M[i - 1, w - w_i,j] > M[i, w])

M[i, w] = v_i,j + M[i - 1, w - w_i,j] B[i, w] = j

return M[G, W], B[G, W]

Practice to write the pseudo code for Find-Solution()

(75)

▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊

▪

Variants of knapsack problem

▪ Fractional Knapsack Problem: 物品可以只拿部分

(76)

▪

Input: 𝑛 items where 𝑖-th item has value 𝑣

_𝑖

and weighs 𝑤

_𝑖

(𝑣

_𝑖

and 𝑤

_𝑖

are positive integers)

▪

Output: the maximum value for the knapsack with capacity of 𝑊, where we can take any fraction of items

▪

Dynamic programming algorithm should work

▪

Choose maximal

^𝑣^𝑖

𝑤_𝑖

(類似CP值) first

“Greedy Algorithm”

Next topic!

Can we do better?

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77

(78)

Course Website: http://ada.miulab.tw Email: ada-ta@csie.ntu.edu.tw

78

Important announcement will be sent to @ntu.edu.tw mailbox

& post to the course website

Slides credited from Hsueh-I Lu, Hsu-Chun Hsiao, & Michael Tsai

Mini-HW 4 released

Homework 1 due a week later

Homework 2 released

有100個死囚，隔天執行死刑，典獄長開恩給他們一個存活的機會。

當隔天執行死刑時，每人頭上戴一頂帽子(黑或白)排成一隊伍，在

死刑執行前，由隊伍中最後的囚犯開始，每個人可以猜測自己頭上 的帽子顏色(只允許說黑或白)，猜對則免除死刑，猜錯則執行死刑。

若這些囚犯可以前一天晚上先聚集討論方案，是否有好的方法可以

使總共存活的囚犯數量期望值最高？

囚犯排成一排，每個人可以看到前面所有人的帽子，但看不到自己 及後面囚犯的。

由最後一個囚犯開始猜測，依序往前。

每個囚犯皆可聽到之前所有囚犯的猜測內容。

8

Divide-and-Conquer

Dynamic Programming

DP procedure

Two key properties of DP for optimization

11

Input: a rod of length 𝑛 and a table of prices 𝑝

for 𝑖 = 1, … , 𝑛

Output: the maximum revenue 𝑟

obtainable by cutting up the rod and selling the pieces

A rod with the length = 4

A rod with the length = 𝑛

The total number of cutting results is 2

= Θ 2

We use a recursive function to solve the subproblems

If we know the answer to the subproblem, can we get the answer to the original problem?

Optimal substructure – an optimal solution can be constructed from optimal solutions to subproblems

Version 1

Version 2

Focus on the left-most cut

𝑇 𝑛 = time for running

Rod cutting problem

Idea: use space for better time efficiency

Rod cutting problem has overlapping subproblems and optimal substructures

→ can be solved by DP

When the number of subproblems is polynomial, the time complexity is polynomial using DP

DP algorithm

Top-Down with Memoization

Bottom-Up with Tabulation

𝑇 𝑛 = time for running

𝑇 𝑛 = time for running

Input: a rod of length 𝑛 and a table of prices 𝑝

for 𝑖 = 1, … , 𝑛

Output: the maximum revenue 𝑟

obtainable and the list of cut pieces

Add an array to keep the cutting positions cut

Top-Down with Memoization

Bottom-Up with Tabulation

▪ Approach 1: approximate via (#subproblems) * (#choices for each subproblem)

For rod cutting

▪ Approach 2: approximate via subproblem graphs

The size of the subproblem graph allows us to estimate the time complexity of the DP algorithm

A graph illustrates the set of subproblems involved and how subproblems depend on another 𝐺 = 𝑉, 𝐸 (E: edge, V: vertex)

Characterize the structure of an optimal solution

Overlapping subproblems: revisit same subproblems

Optimal substructure: an optimal solution to the problem contains within it optimal solutions to subproblems

Recursively define the value of an optimal solution

Express the solution of the original problem in terms of optimal solutions for subproblems

Compute the value of an optimal solution

typically in a bottom-up fashion

Construct an optimal solution from computed information

Step 3 and 4 may be combined

1. Characterize the structure of an optimal solution

2. Recursively define the value of an optimal solution

3. Compute the value of an optimal solution

4. Construct an optimal solution from computed information

Step 1-Q1: What can be the subproblems?

Step 1-Q2: Does it exhibit optimal structure? (an optimal solution can be represented by the optimal solutions to subproblems)

Step 1-Q1: What can be the subproblems?

Step 1-Q2: Does it exhibit optimal structure? (an optimal solution can be represented by the optimal solutions to subproblems)

Yes. Prove by contradiction.

38

▪ Input: the postage 𝑛 and the stamps with values 𝑣

, 𝑣

, … , 𝑣

▪ Output: the minimum number of stamps to cover the postage

The optimal solution 𝑆

can be recursively defined as

死刑執行前，由隊伍中最後的囚犯開始，每個人可以猜測自己頭上的帽子顏色(只允許說黑或白)，猜對則免除死刑，猜錯則執行死刑。

囚犯排成一排，每個人可以看到前面所有人的帽子，但看不到自己及後面囚犯的。