Algorithm Presentation Group: Tarjan Union Find Algorithm Group: - - PowerPoint PPT Presentation

Algorithm Presentation

▪ Group: Tarjan

Union Find Algorithm

Group: Tarjan Dio, Zack, Chips

• 2016. 12

Contents

1

Basic Idea

2

Union By Rank

3

Path Compression

4

Analysis of Path Compression

5

Further Analysis

Contents

1 2 3 4 5

Basic Idea Union By Rank Path Compression Analysis of Path Compression Further Analysis

▪ Used to manage a bunch of disjoint sets. ▪ Find (v): return a canonical element of the set containing v. ▪ Union (u, v): merges the sets containing the canonical elements u and v. ▪ Make_Set (v): add some new element into the bunch. ▪ Example: Find (6) : return B

Bas asic idea ea

1, 2, 4 3 5, 7, 9 6, 8

A B C D

1, 2, 4 3 5, 6, 7, 8, 9

A B C Union (5, 6) =>

Contents

1 2 3 4 5

Basic Idea Union By Rank Path Compression Analysis of Path Compression Further Analysis

▪ Represent each set as a tree with data elements at the vertices. ▪ Each element u has a pointer to its parent. ▪ The root serves as the canonical element of the set. ▪ Observation: 𝑠𝑏𝑜𝑙(𝑦) = ℎ𝑓𝑗𝑕ℎ𝑢 𝑝𝑔 𝑡𝑣𝑐𝑢𝑠𝑓𝑓 𝑠𝑝𝑝𝑢𝑓𝑒 𝑐𝑧 𝑦. ▪ Lemma: 𝑡𝑗𝑨𝑓(𝑡𝑣𝑐𝑢𝑠𝑓𝑓 𝑝𝑔 𝑦) ≥ 2𝑠𝑏𝑜𝑙(𝑦) ▪ Corollary 1: 𝑠𝑏𝑜𝑙 𝑦 ≤ log 𝑛 ▪ Corollary 2: (𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑓𝑚𝑓𝑛𝑓𝑜𝑢 𝑠𝑏𝑜𝑙𝑓𝑒 𝑙) ≤ 𝑜/2𝑙

Rep epresent as as a a tree ee

Union by rank ank

Union (5, 6) => Union (4, 5) =>

Contents

1 2 3 4 5

Basic Idea Union By Rank Path Compression Analysis of Path Compression Further Analysis

▪ After find, change the parent pointers of all vertices along the path directly to v. ▪ Example:

Path Compression

▪ Observation: 𝑠𝑏𝑜𝑙 𝑦 ≠ ℎ𝑓𝑗𝑕ℎ𝑢 𝑦 𝑗𝑡 𝑞𝑝𝑡𝑡𝑗𝑐𝑚𝒇 ▪ Corollary 1, 2 still hold Corollary 1: 𝑠𝑏𝑜𝑙 𝑦 ≤ log 𝑛

Corollary 2: 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑓𝑚𝑓𝑛𝑓𝑜𝑢 𝑠𝑏𝑜𝑙𝑓𝑒 𝑙 ≤ 𝑜/2𝑙

5

3 3 2 1

8 6 5

3 3 2

8 0 6

1 𝐺𝑗𝑜𝑒(8) →

Contents

1 2 3 4 5

Basic Idea Union By Rank Path Compression Analysis of Path Compression Further Analysis

▪ Define 𝑕𝑏𝑞 (𝑦) = 𝑠𝑏𝑜𝑙 (𝑞𝑏𝑠𝑓𝑜𝑢) – 𝑠𝑏𝑜𝑙(𝑦). (In the picture, blue number is rank, red number is gap). ▪ A vertex is happy if 𝑕𝑏𝑞(𝑦) > 𝑠𝑏𝑜𝑙(𝑦). (i.e. rank (parent) > 2rank (x)) and sad otherwise. ▪ How many vertices can be happy in the path from x to root? ▪ Assume x1, x2, x3, …, xk is the sequence of happy vertices

• n the path. 𝑠𝑏𝑜𝑙 𝑦𝑗 ≥ 2𝑗 − 1

𝑀𝑝𝑕(𝑛) ≥ 𝑠𝑏𝑜𝑙(𝑠𝑝𝑝𝑢) >= 2𝑙 − 1 𝑙 ≤ 𝑚𝑝𝑕𝑚𝑝𝑕(𝑛);

Analysis of Path Compression

u

11 8

v

19

z

5 6

y

3 2

x

3 Sad Happy Happy Sad

▪ Observation: n: every time x is part of a find operation, its gap increase by 1. ▪ Observation n 1: along path, ℎ𝑏𝑞𝑞𝑧 𝑤𝑓𝑠𝑢𝑗𝑑𝑓𝑡 ≤ 𝑚𝑝𝑕𝑚𝑝𝑕 𝑛 ▪ Observation n 2: A sad vertex will be happy after rank(x) find ops.

Analysis of Path Compression

u

root

2

y

50 51 28

x

30 20 1

root

51

y

50

x

30

u

2 𝐺𝑗𝑜𝑒(𝑣) → 1 21 49

▪ Suppose there are m find operations.

▪ 𝑗=1

𝑛 𝑢𝑗𝑛𝑓 𝑔𝑝𝑠 𝑗𝑢ℎ 𝑔𝑗𝑜𝑒 𝑝𝑞 = 𝑗=1 𝑛 (ℎ𝑏𝑞𝑞𝑧 𝑗𝑜 𝑗𝑢ℎ find + sad in 𝑗𝑢ℎ 𝑔𝑗𝑜𝑒)

▪ ≤ 𝑛𝑚𝑝𝑕𝑚𝑝𝑕 𝑛 + 𝑗=1

𝑛 𝑡𝑏𝑒 𝑗𝑜 𝑗𝑢ℎ 𝑔𝑗𝑜𝑒

▪ = 𝑛𝑚𝑝𝑕𝑚𝑝𝑕 𝑛 + 𝑦∈𝑛 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑗 𝑡. 𝑢. 𝑦 𝑗𝑡 𝑡𝑏𝑒 𝑗𝑜 𝑗𝑢ℎ 𝑔𝑗𝑜𝑒

▪ Let 𝑇𝑗,𝑦 = 1, 𝑗𝑔 𝑦 𝑏𝑞𝑞𝑓𝑏𝑠𝑡 𝑏𝑡 𝑏 𝑡𝑏𝑒 𝑤𝑓𝑠𝑢𝑓𝑦 𝑗𝑜 𝑗𝑢ℎ 𝑔𝑗𝑜𝑒 𝑝𝑞, 0 𝑓𝑚𝑡𝑓. ▪ Let 𝐼𝑗,𝑦 = 1, 𝑗𝑔 𝑦 𝑏𝑞𝑞𝑓𝑏𝑠𝑡 𝑏𝑡 𝑏 ℎ𝑏𝑞𝑞𝑧 𝑤𝑓𝑠𝑢𝑓𝑦 𝑗𝑜 𝑗𝑢ℎ 𝑔𝑗𝑜𝑒 𝑝𝑞, 0 𝑓𝑚𝑡𝑓.

▪ 𝑈𝑗𝑛𝑓𝑡 𝑔𝑝𝑠 𝑏𝑚𝑚 𝑔𝑗𝑜𝑒𝑡 = 𝑗,𝑦 𝑇𝑦,𝑗 + 𝑗,𝑦 𝐼𝑦,𝑗 ▪ = 𝑗 𝑦 𝐼𝑦,𝑗 + 𝑗 𝑦 𝑇𝑦,𝑗 ≤ 𝑛𝑚𝑝𝑕𝑚𝑝𝑕 𝑛 + 𝑦 𝑠𝑏𝑜𝑙(𝑦) ▪ And 𝑦 𝑠𝑏𝑜𝑙(𝑦) ≤ 𝑙 𝑙 × 𝑜𝑣𝑛 𝑝𝑔 𝑦 𝑡. 𝑢. 𝑠𝑏𝑜𝑙 𝑦 = 𝑙 = 𝑙 𝑙 × 𝑜/2𝑙 ≤ 2𝑜

▪ So, 𝑈𝑗𝑛𝑓𝑡 𝑔𝑝𝑠 𝑏𝑚𝑚 𝑔𝑗𝑜𝑒𝑡 ≤ 𝑛𝑚𝑝𝑕𝑚𝑝𝑕 𝑛 + 2𝑜

Amortized Analysis

Contents

1 2 3 4 5

Basic Idea Union By Rank Path Compression Analysis of Path Compression Further Analysis

▪ We have proved that the time complexity of m FIND() operations is: 𝑃 #ℎ𝑏𝑞𝑞𝑧 𝑤𝑓𝑠𝑢𝑓𝑦 𝑤𝑗𝑡𝑗𝑢𝑡 + 𝑃 #𝑡𝑏𝑒 𝑤𝑓𝑠𝑢𝑓𝑦 𝑤𝑗𝑡𝑗𝑢𝑡 = 𝑃 𝑛𝑚𝑝𝑕𝑚𝑝𝑕𝑜 + 𝑃 𝑜 ▪ Let’s have a try

Fu Further Analysis

16 Maybe we can improve that part. And don’t change this part.

▪ Note that 𝑃 #𝑡𝑏𝑒 𝑤𝑓𝑠𝑢𝑓𝑦 𝑤𝑗𝑡𝑗𝑢𝑡 =

𝑤

𝑠𝑏𝑜𝑙(𝑤) = 𝑜

𝑙 𝑚𝑝𝑕𝑜 𝑙

2𝑙 = 𝑃(𝑜) ▪ This 𝑙 means 𝑛𝑗𝑜 𝑠𝑏𝑜𝑙 𝑤′𝑡 𝑞𝑏𝑠𝑓𝑜𝑢 − 𝑠𝑏𝑜𝑙(𝑤) if we want 𝑤 to be happy ▪ Obs. 1. As long as it’s not expensive than exponential function, the result is still 𝑃(𝑜). ▪ Obs. 2. We change the definition of “happy”, we may improve the complexity (of the “happy part”)

A tiny work

17

▪ In the previous work, we said a vertex 𝑌 is happy if 𝑠𝑏𝑜𝑙(𝑌′𝑡 𝑞𝑏𝑠𝑓𝑜𝑢) > 𝑠𝑏𝑜𝑙(𝑌) ▪ If we make a more strict constrain, we can reduce the number of “happy vertices”. ▪ Let's replace 𝑠𝑏𝑜𝑙(𝑌) by 𝑠𝑏𝑜𝑙(𝑌)2.

A tiny work

18

▪ De

• Def. We say a vertex 𝑤 is second-order happy iff 𝑠𝑏𝑜𝑙 𝑤′𝑡 𝑞𝑏𝑠𝑓𝑜𝑢 >

𝑠𝑏𝑜𝑙(𝑤)2 ▪ Now, the longest chain of second-order happy vertices must have the following ranks: 0,1,2,4,16,256,65536, … ▪ The log of ranks are like this: 0,1,2,4,8,16, … ▪ We can say that the number of happy vertices is 𝑃 𝑚𝑝𝑕𝑚𝑝𝑕 𝑠𝑏𝑜𝑙 𝑠𝑝𝑝𝑢 = 𝑃(𝑚𝑝𝑕𝑚𝑝𝑕𝑚𝑝𝑕𝑜)

A tiny work

▪ Let’s look at the “sad part”: 𝑃 #𝑡𝑏𝑒 𝑤𝑓𝑠𝑢𝑓𝑦 𝑤𝑗𝑡𝑗𝑢𝑡 =

𝑤

𝑠𝑏𝑜𝑙(𝑤)2 − 𝑠𝑏𝑜𝑙(𝑤) <

𝑤

𝑠𝑏𝑜𝑙(𝑤)2 = 𝑜

𝑙=0 𝑚𝑝𝑕𝑜 𝑙2

2𝑙 ▪ Just use some simple math in your high school, we can show that

𝑙=0

𝑙2 2𝑙 < 6 ▪ Now the complexity of m FIND() operations is 𝑃 𝑛𝑚𝑝𝑕𝑚𝑝𝑕𝑚𝑝𝑕𝑜 + 𝑃(𝑜)

A tiny work

Can an we e do bet etter?

𝑙=0 𝑚𝑝𝑕𝑜 𝑙2

2𝑙 ▪ 𝑙2 is too smaller than 2𝑙, we still have much room for improving the complexity ▪ We say second-order happy is defined as 𝑠𝑏𝑜𝑙 𝑞𝑏𝑠𝑓𝑜𝑢 > 𝑠𝑏𝑜𝑙(𝑤)2 ▪ That is to say 𝑚𝑝𝑕𝑚𝑝𝑕 𝑠𝑏𝑜𝑙 𝑞𝑏𝑠𝑓𝑜𝑢 > 𝑚𝑝𝑕𝑚𝑝𝑕 𝑠𝑏𝑜𝑙 𝑤 + 1 ▪ It’s easy to define “k-th order” happy

Mo More power!

▪ De

• Def. We say a vertex 𝑤 is k-th order happy iff

𝑚𝑝𝑕 𝑙 𝑠𝑏𝑜𝑙 𝑤′𝑡 𝑞𝑏𝑠𝑓𝑜𝑢 > 𝑚𝑝𝑕(𝑙) 𝑠𝑏𝑜𝑙 𝑤 + 1 ▪ 𝑚𝑝𝑕 𝑙 𝑦 means k iterations of log: 𝑚𝑝𝑕(𝑙)𝑦 = 𝑚𝑝𝑕𝑚𝑝𝑕 … 𝑚𝑝𝑕𝑦 ▪ Thus the number of k-th order happy vertices is 𝑃 𝑚𝑝𝑕 𝑙 𝑠𝑏𝑜𝑙 𝑠𝑝𝑝𝑢 = 𝑃(𝑚𝑝𝑕 𝑙+1 𝑜) ▪ Note that we may face the problem that we must compute log(0), we can ignore those vertices with small ranks (the number of which is independent of

• f n, i.e. 𝑃(1)) when we consider the number of transferred vertices in FIND()

Mo More power!

𝑙 𝑢𝑗𝑛𝑓𝑡

▪ Now, if we want a vertex happy, the maximal transfer operations must be 22…2𝑚𝑝𝑕 𝑙 𝑠𝑏𝑜𝑙 𝑤 +1 − 𝑠𝑏𝑜𝑙(𝑤) ▪ We can define a function to replace the complex expression ▪ Def. 𝐺

𝑙 𝑦 = 2𝑦

(𝑙 = 1) 2𝐺𝑙−1(𝑚𝑝𝑕𝑦) (𝑙 > 1) ▪ So the “sad part” becomes

𝑤

𝐺

𝑙 𝑦 − 𝑠𝑏𝑜𝑙 𝑤 < 𝑜 𝑙=𝑙0 𝑚𝑝𝑕𝑜 𝐺 𝑙(𝑦)

2𝑙

Mo More power!

▪ If the series 𝑙=𝑙0

∞ 𝐺𝑙(𝑦) 2𝑙 converges, the partial sum must be bounded

▪ That case, we need lim

𝑜→∞

𝐺

𝑙(𝑜)

2𝑜 = 0

Mo More power!

▪ In fact, if lim

𝑦→+∞ 𝐺𝑙−1(𝑦) 2𝑦

= 0, i.e. lim

𝑦→+∞ 𝐺𝑙−1(𝑚𝑝𝑕𝑦) 𝑦

= 0, we can see lim

𝑦→+∞

𝐺

𝑙(𝑦)

2𝑦 = lim

𝑦→+∞

2𝐺

𝑙−1(𝑚𝑝𝑕𝑦)

2𝑦 = lim

𝑦→+∞2𝑦(𝐺𝑙−1 𝑚𝑝𝑕𝑦 𝑦 −1) = lim 𝑦→+∞2−𝑦 = 0

▪ Besides, when k=1, lim

𝑦→+∞ 𝐺𝑙(𝑦) 2𝑦

= 0 ▪ So, lim

𝑜→∞ 𝐺𝑙(𝑜) 2𝑜

= 0 is true, and we also have lim

𝑦→+∞ 𝐺𝑙(𝑚𝑝𝑕𝑦) 𝑦

= 0 ▪ Now we use Cauchy test lim

𝑜→∞

𝑜 𝐺

𝑙(𝑜)

2𝑜 = lim

𝑜→∞2𝐺𝑙−1 𝑚𝑝𝑕𝑜 /𝑜−1 = 1

2 < 1

We e nee eed some e math

▪ Now we have 𝑙=𝑙0

𝑚𝑝𝑕𝑜 𝐺𝑙(𝑦) 2𝑙 < 𝑙=𝑙0 ∞ 𝐺𝑙(𝑦) 2𝑙

= 𝐷𝑝𝑜𝑡𝑢𝑏𝑜𝑢 ▪ So for ∀𝑙 ∈ 𝑂 the complexity is 𝑃 𝑛𝑚𝑝𝑕 𝑙 𝑜 + 𝑃(𝑜)

Mo More power!

Can an we e do bet etter?

▪ In fact, the complexity is 𝑃 𝑛𝛽 𝑜 + 𝑃(𝑜), where 𝛽(𝑜) is the inverse function of Ackermann Function, which is smaller than 𝑚𝑝𝑕(𝑙)𝑜 for any k

Yes es, , we e can an

单击此处编辑母版标题样式

Thanks! Thanks!