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Amortized analysis

Yan Gu

CS218: Design and Analysis of algorithms

Exam structure (55 points in total)

• Basic knowledge (multiple choices), 10 pts
• DP basics, 9 pts
• Greedy proof, 5 pts
• Augmented-tree design, 5 pts
• Graph basics, 9 pts
• Analysis basics, 6 pts
• Algorithm design, 16 pts
• 60 points in total
• Plus bonus points
• Basic knowledge (multiple choices), 10 pts
• DP basics, 9 pts
• Greedy proof, 5 pts
• Augmented-tree design, 5 pts
• Graph basics, 9 pts
• Analysis basics, 6 pts
• Algorithm design, 16 pts
• 60 points in total
• Plus bonus points

CEP prob 2 CEP prob 1 Midterm Final

• Multiple choices, 5 pts
• DP basics, 9 pts
• Greedy proof, 5 pts
• Augmented-tree design, 5 pts
• Algorithm design, 8 pts
• 32 points in total
  • Multiple choices, 5 pts
  • DP basics, 9 pts
  • Greedy proof, 5 pts
  • Graph basics, 9 pts
  • Analysis basics, 6 pts
  • Algorithm design, 8+8 pts
  • 50 points in total

Amortized analysis

Yan Gu

CS218: Design and Analysis of algorithms

Hashing and hash table

Hash function

• Maps arbitrary data to fixed-size values
  • Usually integers
• The same data are always mapped to the same value
• Different data are unlikely to be mapped to the same value
  • Collision: two keys are hashed to the same hash value

Hash table and hash functions

• E.g., Strings - > integers
• A[“John Smith”] = false
  • A[2] = false (John Smith)
• A[“Lisa Smith”] = true
  • A[1] = true (Lisa Smith)
• X = A[“John Smith”]
  • X = A[2]
• A[“Sandra Dee”] = false
  • A[2]?

Simple uniform hashing strategy

• For each element with key 𝑥 , find a random position 𝒉𝟏 𝒙 ;
  • if there’s a collision, try another (i.e., 𝒉𝟐 𝒙 )
• Say insert key to be 5
• Then insert key to be 88
• What’s the expected number of retries?

Ways to deal with collisions

• When an element is mapped to a index 𝒊 , but finds out that position 𝒊 has been taken by another element?
• Open addressing / closed hashing
  • Find another empty position (e.g., linear probing: try the next position)
  • It can also be using other ways to find the next empty position, not necessarily try the next position (e.g., probe quadratically)
• Closed addressing / open hashing
  • Throw the element still to position 𝑖
  • All elements in position 𝑖 will be further organized as another data structure, e.g., a linked list

Open addressing vs. closed addressing Open Addressing Closed Addressing Source: https://programming.guide/h ash-tables-open-vs-closed- addressing.html

In the above analysis, it’s very important that the hash table load factor is a below 1/

• That’s why we know that an insertion succeeds with probability ½
• So the cost is 1/2+1/4+1/8+… = O(1)
• Actually, any constant works
• But, how can we guarantee that?

Resizing hash table

• When the load factor of the hash table is more than ½
• i.e., when we have more than n/2 elements in the hash table of size n
• Resize the table 3 5 8 7 2 Array of size 2n Array of size n, half full 3 8 7 5 2

What is the cost of an insertion?

• Worst case cost: 𝑶(𝒏)?
  • 𝑛 is the current table size
• However, this happens very rarely - at least every 𝑶 𝒏 insertions!
• All the rest of the insertions cost 𝑂( 1 )!
• Every 𝑡 insertions, we have an insertion of cost 𝑂(𝑡)
• Somehow “on average”, the cost is still a constant? Initial total size Current #slots filled #insertions before resizing Resizing cost Phase 1 k 0 k/2 k/ Phase 2 2k k/2 k/2 k Phase 3 4k k k 2k Phase 4 8k 2k 2k 4k …… Happens after k/2 insertions Happens after k/2 insertions Happens after k insertions Happens after 2k insertions (Assume unit cost per insertion and per rehash)

Amortized Analysis