Slides credited from Hsu-Chun Hsiao

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▪

Mini-HW 5 released

▪ Due on 10/26 (Thu) 17:20

▪

Homework 1 due soon

▪

Homework 2

▪ Due on 11/09 (Thur) 17:20 (4 weeks)

▪

TA Recitation (next week)

▪ 10/26 (Thu) at R103

▪ Homework 1 QA

▪

Another course website you can get the videos sooner

▪ http://ada.miulab.tw

▪

Note: if you have questions about the homework, please find TAs

Frequently check the website for the updated information! 2

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▪ Dynamic Programming

▪ DP #1: Rod Cutting

▪ DP #2: Stamp Problem

▪ DP #3: Matrix-Chain Multiplication

▪ DP #4: Sequence Alignment Problem

▪ Longest Common Subsequence (LCS) / Edit Distance

▪ Space Efficient Algorithm

▪ Viterbi Algorithm

▪ DP #5: Weighted Interval Scheduling

▪ DP #6: Knapsack Problem

▪ 0-1 Knapsack

▪ Unbounded Knapsack

▪ Multidimensional Knapsack

▪ Multi-Choice Knapsack

▪ Fractional Knapsack

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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。

6

……

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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▪

http://qstn.co/q/OOZK7sYQ

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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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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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Textbook Chapter 15.4 – Longest common subsequence Textbook Problem 15-5 – Edit distance

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▪

猴子們各自講話，經過語音辨識系統後，哪一支猴子發出最接近英文字”banana”的語音為優勝者

▪

How to evaluate the similarity between two sequences?

12

aeniqadikjaz

svkbrlvpnzanczyqza

banana

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▪

Input: two sequences

▪

Output: longest common subsequence of two sequences

▪ The maximum-length sequence of characters that appear left-to-right (but not necessarily a continuous string) in both sequences

13

X = banana

Y = svkbrlvpnzanczyqza X → ---ba---n-an---a Y → svkbrlvpnzanczyqza X = banana

Y = aeniqadikjaz X → ba-n--an---a- Y → -aeniqadikjaz

4 5

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▪

Input: two sequences

▪

Output: the minimum cost of transformation from X to Y

▪ Quantifier of the dissimilarity of two strings

14

X = banana

Y = svkbrlvpnzanczyqza X → ---ba---n-an---a Y → svkbrlvpnzanczyqza X = banana

Y = aeniqadikjaz X → ba-n--an---a- Y → -aeniqadikjaz

1 deletion, 7 insertions, 1 substitution 12 insertions, 1 substitution

9 13

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▪

Input: two sequences

▪

Output: the minimal cost 𝑀

_𝑚,𝑛

for aligning two sequences

▪ Cost = #insertions × 𝐶_INS + #deletions × 𝐶_DEL + #substitutions × 𝐶_𝑝,𝑞

15

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▪ Subproblems

▪ SA(i, j): sequence alignment between prefix strings 𝑥₁, … , 𝑥_𝑖 and 𝑦₁, … , 𝑦_𝑗

▪ Goal: SA(m, n)

▪ Optimal substructure: suppose OPT is an optimal solution to SA(i, j), there are 3 cases:

▪ Case 1: 𝑥_𝑖 and 𝑦_𝑗 are aligned in OPT (match or substitution)

▪ OPT/{𝑥_𝑖, , 𝑦_𝑗} is an optimal solution of SA(i-1, j-1)

▪ Case 2: 𝑥_𝑖 is aligned with a gap in OPT (deletion)

▪ OPT is an optimal solution of SA(i-1, j)

▪ Case 3: 𝑦_𝑗 is aligned with a gap in OPT (insertion)

▪ OPT is an optimal solution of SA(i, j-1)

16

Sequence Alignment Problem Input: two sequences

Output: the minimal cost 𝑀_𝑚,𝑛 for aligning two sequences

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▪

Suppose OPT is an optimal solution to SA(i, j), there are 3 cases:

▪ Case 1: 𝑥_𝑖 and 𝑦_𝑗 are aligned in OPT (match or substitution)

▪ OPT/{𝑥_𝑖, , 𝑦_𝑗} is an optimal solution of SA(i-1, j-1)

▪ Case 2: 𝑥_𝑖 is aligned with a gap in OPT (deletion)

▪ OPT is an optimal solution of SA(i-1, j)

▪ Case 3: 𝑦_𝑗 is aligned with a gap in OPT (insertion)

▪ OPT is an optimal solution of SA(i, j-1)

▪

Recursively define the value

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▪ Bottom-up method: solve smaller subproblems first

18

X\Y 0 1 2 3 4 5 … n

0 1 : m

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19

X\Y 0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 4 8 12 16 20 24 28 32 36 40 44 48

1 4 7 11 15 19 23 27 31 35 39 43 47 51

2 8 4 8 12 16 20 23 27 31 35 39 43 47

3 12 8 12 8 12 16 20 24 28 32 36 40 44

4 16 12 15 12 15 19 16 20 24 28 32 36 40 5 20 16 19 15 19 22 20 23 27 31 35 39 43 6 24 20 23 19 22 26 22 26 30 34 38 35 39

a e n i q a d i k j a z

b a n a n a

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Seq-Align(X, Y, C_DEL, C_INS, C_p,q) for j = 0 to n

M[0][j] = j * C_INS // |X|=0, cost=|Y|*penalty for i = 1 to m

M[i][0] = i * C_DEL // |Y|=0, cost=|X|*penalty for i = 1 to m

for j = 1 to n

M[i][j] = min(M[i-1][j-1]+C_xi,yi, M[i-1][j]+C_DEL, M[i][j-1]+C_INS) return M[m][n]

▪ Bottom-up method: solve smaller subproblems first Sequence Alignment Problem

Input: two sequences

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21

X\Y 0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 4 8 12 16 20 24 28 32 36 40 44 48

1 4 7 11 15 19 23 27 31 35 39 43 47 51

2 8 4 8 12 16 20 23 27 31 35 39 43 47

3 12 8 12 8 12 16 20 24 28 32 36 40 44

4 16 12 15 12 15 19 16 20 24 28 32 36 40 5 20 16 19 15 19 22 20 23 27 31 35 39 43 6 24 20 23 19 22 26 22 26 30 34 38 35 39

a e n i q a d i k j a z

b a n a n a

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22

Find-Solution(M) if m = 0 or n = 0

return {}

v = min(M[m-1][n-1] + C_xm,yn, M[m-1][n] + C_DEL, M[m][n-1] + C_INS) if v = M[m-1][n] + C_DEL// ↑: deletion

return Find-Solution(m-1, n)

if v = M[m][n-1] + C_INS// ←: insertion return Find-Solution(m, n-1)

return {(m, n)} ∪ Find-Solution(m-1, n-1) // ↖: match/substitution

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Find-Solution(M) if m = 0 or n = 0

return {}

v = min(M[m-1][n-1] + C_xm,yn, M[m-1][n] + C_DEL, M[m][n-1] + C_INS) if v = M[m-1][n] + C_DEL// ↑: deletion

return Find-Solution(m-1, n)

if v = M[m][n-1] + C_INS// ←: insertion return Find-Solution(m, n-1)

return {(m, n)} ∪ Find-Solution(m-1, n-1) // ↖: match/substitution Seq-Align(X, Y, C_DEL, C_INS, C_p,q)

for j = 0 to n

M[0][j] = j * C_INS // |X|=0, cost=|Y|*penalty for i = 1 to m

M[i][0] = i * C_DEL // |Y|=0, cost=|X|*penalty for i = 1 to m

for j = 1 to n

M[i][j] = min(M[i-1][j-1]+C_xi,yi, M[i-1][j]+C_DEL, M[i][j-1]+C_INS) return M[m][n]

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▪

Space complexity

▪

If only keeping the most recent two rows: Space-Seq-Align(X, Y)

24

X\Y 0 1 2 3 … j … n

i - 1 i

The optimal value can be computed, but the solution cannot be reconstructed

X\Y 0 1 2 3 4 5 … n

0 1 : m

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▪

Problem: find the min-cost alignment  find the shortest path

25

Divide-and-Conquer +

Dynamic Programming

a

e p p l

p e

a

X\Y 0 1 2 3

0 0 4 8 12

1 4 7 11 15

2 8 4 8 12

3 12 8 12 8

4 16 12 15 12 5 20 16 19 15 a p e

p p l e a

→ distance = C_INS

↓ distance = C_DEL

↘ distance = C_u,v for edge (u, v) START

END

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𝐹 2,3 = distance of the shortest path

▪

Each edge has a length/cost

▪

𝐹 𝑖, 𝑗 : length of the shortest path from 0,0 to 𝑖, 𝑗 (START  𝑖, 𝑗 )

▪

𝐵 𝑖, 𝑗 : length of the shortest path from 𝑖, 𝑗 to 𝑚, 𝑛 ( 𝑖, 𝑗  END)

▪

𝐹 𝑚, 𝑛 = 𝐵 0,0

26

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

𝐵 2,3 = distance of the shortest path

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▪

Each edge has a length/cost

▪

𝐹 𝑖, 𝑗 : length of the shortest path from 0,0 to 𝑖, 𝑗 (START  𝑖, 𝑗 )

▪

𝐵 𝑖, 𝑗 : length of the shortest path from 𝑖, 𝑗 to 𝑚, 𝑛 ( 𝑖, 𝑗  END)

▪

Forward formulation

▪

Backward formulation

27

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

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▪

Observation 1: the length of the shortest path from 0,0 to 𝑚, 𝑛 that passes through 𝑖, 𝑗 is 𝐹 𝑖, 𝑗 + 𝐵 𝑖, 𝑗

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𝐹 𝑖, 𝑗 : length of the shortest path from 0,0 to 𝑖, 𝑗 𝐵 𝑖, 𝑗 : length of the shortest path from 𝑖, 𝑗 to 𝑚, 𝑛

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

𝐹 𝑖, 𝑗

𝐵 𝑖, 𝑗

 optimal substructure

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▪

Observation 2: for any 𝑣 in {0, … , 𝑛}, there exists a 𝑢 s.t. the shortest path between (0,0) and 𝑚, 𝑛 goes through (𝑢, 𝑣)

29

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5  the shortest path must go across a vertical cut

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▪

Observation 1+2:

30

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5 i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

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▪

Goal: finds optimal solution

31

How to find the value of 𝑢^∗?

▪

Idea: utilize sequence alignment algo.

▪ Call Space-Seq-Align(X,Y[1:v]) to find 𝐹 0, 𝑣 , 𝐹 1, 𝑣 , … , 𝐹 𝑚, 𝑣

▪ Call Back-Space-Seq-Align(X,Y[v+1:n]) to find 𝐵 0, 𝑣 , 𝐵 1, 𝑣 , … , 𝐵 𝑚, 𝑣

▪ Let 𝑢 be the index minimizing 𝐹 𝑢, 𝑣 + 𝐵 𝑢, 𝑣

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▪

Goal: finds optimal solution – DC-Align(X, Y)

32

1. Divide

2. Conquer

3. Combine

▪ Divide the sequence of size n into 2 subsequences

▪ Find 𝑢 to minimize 𝐹 𝑢, 𝑣 + 𝐵 𝑢, 𝑣

▪ Recursive case (𝑛 > 1)

▪ prefix

= DC-Align(X[1:u], Y[1:v])

▪ suffix

= DC-Align(X[u+1:m], Y[v+1:n])

▪ Base case (𝑛 = 1)

▪ Return Seq-Align(X, Y)

▪ Return prefix + suffix

▪ 𝑇 𝑚, 𝑛 = time for running DC-Align(X, Y) with 𝑋 = 𝑚, 𝑌 = 𝑛

Space Complexity:

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▪

Theorem

▪

Proof

▪ There exists positive constants 𝑎, 𝑏 s.t. all

▪ Use induction to prove

33

Inductive hypothesis

when

Practice to check the initial condition

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▪

Given a graph 𝐺 = 𝑉, 𝐸 , each edge 𝑢, 𝑣 ∈ 𝐸 has an associated non- negative probability 𝑝 𝑢, 𝑣 of traversing the edge 𝑢, 𝑣 and producing the corresponding character. Find the most probable path with the

label 𝑠 = 𝜎

₁

, 𝜎

₂

, … , 𝜎

_𝑛

.

34

ㄨㄅㄒㄎㄕ

START

我烏為問

END

爸不

想續小

考看卡

書

試上

白鄉

Find the path from START to END with

highest prob

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35

𝜎

₁

𝜎

₂

… … 𝜎

_𝑛

START END

produce 𝜎₁

produce 𝜎_𝑗

V: vocabulary size

Viterbi has been applied to many AI applications, e.g. speech recognition

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36

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▪

Input: 𝑛 job requests with start times 𝑠

_𝑖

, finish times 𝑓

_𝑖

▪

Output: the maximum number of compatible jobs

▪

The interval scheduling problem can be solved using an “early-finish-time- first” greedy algorithm in 𝑂(𝑛) time

37

“Greedy Algorithm”

Next topic!

time 1

2 3 4 5 6

job index

1 2 3 4 5 6 7 8 9

(38)

▪

Input: 𝑛 job requests with start times 𝑠

_𝑖

, finish times 𝑓

_𝑖

, and values 𝑣

_𝑖

▪

Output: the maximum total value obtainable from compatible jobs

time 38

1

3

3 4

3 1 1

2 3 4 5 6

job index

1 2 3 4 5 6 7 8 9

Assume that the requests are sorted in non-decreasing order (𝑓_𝑖 ≤ 𝑓_𝑗 when 𝑖 < 𝑗) 𝑝(𝑗) = largest index 𝑖 < 𝑗 s.t. jobs 𝑖 and 𝑗 are compatible

e.g. 𝑝 1 = 0, 𝑝 2 = 0, 𝑝 3 = 1, 𝑝 4 = 1, 𝑝 5 = 4, 𝑝 6 = 3

(39)

▪

Subproblems

▪ WIS(i): weighted interval scheduling for the first 𝑖 jobs

▪ Goal: WIS(n)

▪

Optimal substructure: suppose OPT is an optimal solution to WIS(i), there are 2 cases:

▪ Case 1: job 𝑖 in OPT

▪ OPT\{𝑖} is an optimal solution of WIS(p(i))

▪ Case 2: job 𝑖 not in OPT

▪ OPT is an optimal solution of WIS(i-1)

39

Weighted Interval Scheduling Problem

Input: 𝑛 jobs with 𝑠_𝑖, 𝑓_𝑖, 𝑣_𝑖 , 𝑝(𝑗) = largest index 𝑖 < 𝑗 s.t. jobs 𝑖 and 𝑗 are compatible Output: the maximum total value obtainable from compatible

time 1

3

3 4

3 1 1

2 3 4 5 6

job index

2

1 3 4 5 6 7 8 9

4

1

(40)

▪

Optimal substructure: suppose OPT is an optimal solution to WIS(i), there are 2 cases:

▪ Case 1: job 𝑖 in OPT

▪ OPT\{𝑖} is an optimal solution of WIS(p(i))

▪ Case 2: job 𝑖 not in OPT

▪ OPT is an optimal solution of WIS(i-1)

▪

Recursively define the value

40

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41

i 0 1 2 3 4 5 … n

M[i]

WIS(n, s, f, v, p) M[0] = 0

for i = 1 to n

M[i] = max(v[i] + M[p[i]], M[i - 1]) return M[n]

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42

▪ Bottom-up method: solve smaller subproblems first Weighted Interval Scheduling Problem

i 0 1 2 3 4 5 6

M[i] 0 1 3 4 5 6 7

time 1

3

3 4

3 1 1

2 3 4 5 6 job index

1 2 3 4 5 6 7 8 9

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43

WIS(n, s, f, v, p) M[0] = 0

for i = 1 to n

M[i] = max(v[i] + M[p[i]], M[i - 1]) return M[n]

Find-Solution(M, n) if n = 0

return {}

if v[n] + M[p[n]] > M[n-1] // case 1 return {n} ∪ Find-Solution(p[n]) return Find-Solution(n-1) // case 2

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Textbook Exercise 16.2-2

44

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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

45

(46)

▪

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

46

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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

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

47

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)

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

▪

Recursively define the value

48

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

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49

i\w 0 1 2 3 … w … W

0 1 2 i n

(50)

50

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

3 0 4 9 13 20 24

i w_i v_i

1 1 4

2 2 9

3 4 20

𝑊 = 5

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51

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]

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52

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}

return S

(53)

▪

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

53

= 2^𝑚 = 𝑚

(54)

▪

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

54

(55)

▪

Subproblems

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

▪ Goal: U-KP(n, W)

55

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 Can we do better?

(56)

▪ 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)

56

(57)

▪

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

57

只考慮背包還裝的下的情形

(58)

58

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

(59)

59

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

(60)

60

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]

(61)

61

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

C[i] += 1 return C

(62)

▪

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

62

(63)

▪

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)

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

63

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

(64)

▪

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

64

(65)

▪ Step 3: Compute Value of an OPT Solution

▪ Step 4: Construct an OPT Solution by Backtracking

▪ What is the time complexity?

65

(66)

▪

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

66

(67)

▪

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

67

group 1 group 2 group 3

(68)

▪

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

68

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

(69)

▪

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

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

69

(70)

▪

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

70

𝑛_𝑖 + 1

(71)

71

i\w 0 1 2 3 … w … W

0 1 2 i n

(72)

72

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]

(73)

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 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()

(74)

▪

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

74

(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, where we can take any fraction of items

▪

Dynamic programming algorithm should work

▪

Choose maximal

^𝑣^𝑖

𝑤_𝑖

(類似CP值) first

75

“Greedy Algorithm”

Next topic!

Can we do better?

(76)

▪

“Dynamic Programming”: solve many subproblems in polynomial time for which a naïve approach would take exponential time

▪

When to use DP

▪ Whether subproblem solutions can combine into the original solution

▪ When subproblems are overlapping

▪ Whether the problem has optimal substructure

▪ Common for optimization problem

▪

Two ways to avoid recomputation

▪ Top-down with memoization

▪ Bottom-up method

▪

Complexity analysis

▪ Space for tabular filling

▪ Size of the subproblem graph 76

(77)

Course Website: http://ada17.csie.org Email: ada-ta@csie.ntu.edu.tw

77

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

& post to the course website

Slides credited from Hsu-Chun Hsiao

Mini-HW 5 released

Homework 1 due soon

Homework 2

TA Recitation (next week)

Another course website you can get the videos sooner

Note: if you have questions about the homework, please find TAs

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。

http://qstn.co/q/OOZK7sYQ

8

Divide-and-Conquer

Dynamic Programming

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

11

猴子們各自講話，經過語音辨識系統後，哪一支猴子發出最接近英 文字”banana”的語音為優勝者

How to evaluate the similarity between two sequences?

aeniqadikjaz

svkbrlvpnzanczyqza

Input: two sequences

Output: longest common subsequence of two sequences

X = banana

Y = svkbrlvpnzanczyqza X → ---ba---n-an---a Y → svkbrlvpnzanczyqza X = banana

Y = aeniqadikjaz X → ba-n--an---a- Y → -aeniqadikjaz

Input: two sequences

Output: the minimum cost of transformation from X to Y

X = banana

Y = svkbrlvpnzanczyqza X → ---ba---n-an---a Y → svkbrlvpnzanczyqza X = banana

Y = aeniqadikjaz X → ba-n--an---a- Y → -aeniqadikjaz

Input: two sequences

Output: the minimal cost 𝑀

for aligning two sequences

Suppose OPT is an optimal solution to SA(i, j), there are 3 cases:

Recursively define the value

Space complexity

If only keeping the most recent two rows: Space-Seq-Align(X, Y)

Problem: find the min-cost alignment  find the shortest path

a

e p p l

p e

a

Each edge has a length/cost

𝐹 𝑖, 𝑗 : length of the shortest path from 0,0 to 𝑖, 𝑗 (START  𝑖, 𝑗 )

𝐵 𝑖, 𝑗 : length of the shortest path from 𝑖, 𝑗 to 𝑚, 𝑛 ( 𝑖, 𝑗  END)

𝐹 𝑚, 𝑛 = 𝐵 0,0

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

Each edge has a length/cost

𝐹 𝑖, 𝑗 : length of the shortest path from 0,0 to 𝑖, 𝑗 (START  𝑖, 𝑗 )

𝐵 𝑖, 𝑗 : length of the shortest path from 𝑖, 𝑗 to 𝑚, 𝑛 ( 𝑖, 𝑗  END)

Forward formulation

Backward formulation

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

Observation 1: the length of the shortest path from 0,0 to 𝑚, 𝑛 that passes through 𝑖, 𝑗 is 𝐹 𝑖, 𝑗 + 𝐵 𝑖, 𝑗

i = 0

4 1 2 3

j = 0 1 2 3 4 5 6 7

5

 optimal substructure

Observation 2: for any 𝑣 in {0, … , 𝑛}, there exists a 𝑢 s.t. the shortest path between (0,0) and 𝑚, 𝑛 goes through (𝑢, 𝑣)

i = 0

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。

猴子們各自講話，經過語音辨識系統後，哪一支猴子發出最接近英文字”banana”的語音為優勝者

ㄨㄅㄒㄎㄕ