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101 lines
2.3 KiB
Markdown
101 lines
2.3 KiB
Markdown
---
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title: "heap"
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tags:
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- cosc201
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- datastructure
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---
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A tree where:
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1. every elements should be greater than ites children
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2. the structure should be filled from top to bottom and left to right
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To remove an element
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- remove from the top, replace with the last element
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- to fix the first condition swap the top element with the highest of its children until fixed
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To Add an element
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- add to the next position
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- If its larger than its parent then swap them
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How deep is the tree?
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- each layer is twice as deep and the preceding one
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- layer k can hold $2^k$ elements
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- to store n elements we use k layers where $k = lg n$
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- so we need ϴ(lg n) layers
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- So any algorithm that 'walk along a branch' in while or in part will have Ο(n) complexity (assuming constant time work at each node)
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# 1 Overview
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[[notes/heap]]
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# 2 Operations
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## 2.1 Add element
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Assumptions
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- access first vacant position
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- set (or find) the value $H.q$ stored in any (occupied) position $q$
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- access parent of any given position
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- identify when we're at the root
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(all in constant time)
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Outcome: Change $H$ by adding x to it, while maintaining the heap conditions
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```
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p <- first vacancy, H.p <- x
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while p is not the root and H.parent(p) < H.p do
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swap H.parent(p) and H.p
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p <- parent(p)
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end while
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```
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## 2.2 Remove the maximum
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Outcome: Change H by removing its maximum (i.e., root) value wile maintaining the heap conditions
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```
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v <- H.root
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set H.root to be the value stored in the last occupied position
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p <- root
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while p has children
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if the largest value, H.c of a child of p is greater than H.p then
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swap H.c and H.p, p <-c
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else
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Break
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end if
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end while
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return v
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```
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## 2.3 Complexity
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In addition, we move along a branch from an added element up to the root, fixing violations as we go
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In removal, we move from the root down through some branch until all violations are fixed (can only occur if node has children)
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So both loops do most Ο(lg n)
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## 2.4 Storage
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- Array
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- root at position 0 and children at 1 and 2
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- children of 1 to in 3 and 4, children of 2 go in 5 and 6
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- first vacant pos --> heap[n]
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- value at pos q --> heap[q]
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- get parent of q --> parent(q) = (q-1)/2
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- get children of q --> (2 * q) ± 2
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- identify if q is root --> q == 0
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## 2.5 Implementation
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Use java.util.PriorityQueue
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