CS521 Final Term Notes

√2 lg 𝑛 lg 𝑛 1/ lg 𝑛

Asymptotic Notations: · CS521 \ Notes for the Final Exam · Notation · Definition · Limit · Big-O

lim > 0

Small-o

lim

Big-Ω lim = 𝑐 ≥ 0

(non-negative constant)

Small-𝜔

lim = 0

Big-Θ lim = 𝑐

(some constant) · Notes: · Some hierarchies: (> is “o”, = is “Θ”)

lg 𝑛 > (√2) > 2

Divide and Conquer – recurrences:

Break a problem into sub-problems, solve sub problems and merge solutions.

Solving Recurrences:

1. Substitution method:

Substitute 𝑇-terms in the recurrence by the assumption, using constants, and derive upper / lower bounds (or

both). Example:

𝑛

𝑏

𝑏 𝑘=1

𝑘=0

Useful math:

𝑛 𝑛

Harmonic series: ∑𝑛

1 = Θ(lg 𝑛) 𝑛+1

Sorting algorithms:

Insertion Sort:

𝑛 (𝑞 −1) 𝑞−1 when 𝑛 = ∞∶ 𝑎

Each element is pushed as far as possible to the left (sorted) part of the array.

Running time: 𝑂(𝑛2) worst-case (when sorted in reverse), Ω(𝑛) best-case (when already sorted).

Merge-sort:

Divide the sequence into two 𝑛/2 sequences, apply the algorithm recursively and merge solutions.

Quick Sort:

𝑥 – the pivot.

Preferable over merge-sort, as it sorts in-place.

For 𝛼 = 2 we get a balanced tree.

and returns the index of

Heap:

Left filled structure that satisfies partial order (node > its children, no particular relation between children).

Height is lg 𝑛: the cost of insertion, extract-max, delete.

𝑛

Remove-max: remove root and put last node of the heap in its place. Heapify it down, each level switching with the greater child of the two until satisfies heap properties.

Build-heap: for node 𝑛/2 down to 1, heapify down. Amortized analysis – each level 𝑘 but lowest will be visited at most

Heap-sort:

Build a heap and then run ~𝑛 times: Swap the max with the last element, decrease heap size by 1 and heapify the new

root down. Result: the array is sorted.

Selection problems:

The Select algorithm for finding the 𝑖th o.s.:

Find the median for each group and store in 𝐴′.

𝑛

A general recurrence (for constants other than 5):

Calculate the minimal number of elements we know for sure are greater than 𝑥, the medians’ median.

Then we take the complement of that number:

𝑛 For any constant 𝑘 (the default in the selection algorithm is 𝑘 = 5), where we divide into ] ] sets, the 𝑘

number of elements that are definitely greater than 𝑥:

𝑛 𝑛 ] ]⌉ − 2) 𝑛 𝑛 − 2) ⇒

⏟2

˛_2 𝑘

2𝑘

The recursion over the complement is:

𝑛

𝑘 𝑛 ] · ( 𝑛 2𝑘

Then show by substitution that it is bounded / not bounded from above by 𝑐 · 𝑛 (would be bounded for any 𝑘 ≥ 4). Note that we get

Binary Search Trees:

Trees that satisfy: left(x) ≤ key(x) < right(x)

• Left most leaf of 𝑥’s right subtree.