Week-4 (Heap/Heap Sort)

CE100 Algorithms and Programming II¶

Week-4 (Heap/Heap Sort)¶

Spring Semester, 2021-2022¶

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Heap/Heap Sort¶

Outline (1)¶

Heaps
Max / Min Heap
Heap Data Structure
Heapify
- Iterative
- Recursive

Outline (2)¶

Extract-Max
Build Heap

Outline (3)¶

Heap Sort
Priority Queues
Linked Lists
Radix Sort
Counting Sort

Heapsort¶

Worst-case runtime: $O(nlgn)$
Sorts in-place
Uses a special data structure (heap) to manage information during execution of the algorithm
Another design paradigm

Heap Data Structure (1)¶

Nearly complete binary tree
Completely filled on all levels except possibly the lowest level

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Heap Data Structure (2)¶

Height of node i: Length of the longest simple downward path from i to a leaf
Height of the tree: height of the root

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Heap Data Structures (3)¶

Depth of node i: Length of the simple downward path from the root to node i

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Heap Property: Min-Heap¶

The smallest element in any subtree is the root element in a min-heap
Min heap: For every node i other than root, $A[parent(i)] \leq A[i]$
Parent node is always smaller than the child nodes

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Heap Property: Max-Heap¶

The largest element in any subtree is the root element in a max-heap
We will focus on max-heaps
Max heap: For every node i other than root, $A[parent(i)] ≥ A[i]$
Parent node is always larger than the child nodes

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Heap Data Structures (4)¶

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Heap Data Structures (5)¶

Computing left child, right child, and parent indices very fast
left(i) = 2i $\Longrightarrow$ binary left shift
right(i) = 2i+1 $\Longrightarrow$ binary left shift, then set the lowest bit to 1
parent(i) = floor(i/2) $\Longrightarrow$ right shift in binary
$A[1]$ is always the root element
Array $A$ has two attributes:
length(A): The number of elements in $A$
n = heap-size(A): The number elements in $heap$
- $n \leq length(A)$

Heap Operations : EXTRACT-MAX (1)¶

EXTRACT-MAX(A, n)
  max = A[1]
  A[1] = A[n]
  n = n - 1
  HEAPIFY(A, 1,n)
  return max

Heap Operations : EXTRACT-MAX (2)¶

Return the max element,and reorganize the heap to maintain heap property

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Heap Operations: HEAPIFY (1)¶

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Heap Operations: HEAPIFY (2)¶

Maintaining heap property:
Subtrees rooted at $left[i]$ and $right[i]$ are already heaps.
But, $A[i]$ may violate the heap property (i.e., may be smaller than its children)
Idea: Float down the value at $A[i]$ in the heap so that subtree rooted at $i$ becomes a heap.

Heap Operations: HEAPIFY (2)¶

HEAPIFY(A, i, n)
  largest = i 

  if 2i <= n and A[2i] > A[i] then 
    largest = 2i;
  endif

  if 2i+1 <= n and A[2i+1] > A[largest] then 
    largest = 2i+1;
  endif

  if largest != i  then
    exchange A[i] with A[largest];
    HEAPIFY(A, largest, n);
  endif

Heap Operations: HEAPIFY (3)¶

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Heap Operations: HEAPIFY (4)¶

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Heap Operations: HEAPIFY (5)¶

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Heap Operations: HEAPIFY (6)¶

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Heap Operations: HEAPIFY (7)¶

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Heap Operations: HEAPIFY (8)¶

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Intuitive Analysis of HEAPIFY¶

Consider $HEAPIFY(A, i, n)$
let $h(i)$ be the height of node $i$
at most $h(i)$ recursion levels
- Constant work at each level: $\Theta(1)$
Therefore $T(i)=O(h(i))$
Heap is almost-complete binary tree
$h(n)=O(lgn)$
Thus $T(n)=O(lgn)$

Formal Analysis of HEAPIFY¶

What is the recurrence?
Depends on the size of the subtree on which recursive call is made
- In the next, we try to compute an upper bound for this subtree.

Reminder: Binary trees¶

For a complete binary tree:
$\#$ of nodes at depth $d$: $2^d$
$\#$ of nodes with depths less than $d$: $2^d-1$

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Formal Analysis of HEAPIFY (1)¶

Worst case occurs when last row of the subtree $S_i$ rooted at node $i$ is half full
$T(n) \leq T(|S_{L(i)}|) + \Theta(1)$
$S_{L(i)}$ and $S_{R(i)}$ are complete binary trees of heights $h(i)-1$ and $h(i)-2$, respectively

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Formal Analysis of HEAPIFY (2)¶

Let $m$ be the number of leaf nodes in $S_{L(i)}$
$|S_{L(i)}|=\overbrace{m}^{ext.}+\overbrace{(m–1)}^{int.}=2m–1$
$|S_{R(i)}|=\overbrace{\frac{m}{2}}^{ext.}+\overbrace{(\frac{m}{2}–1)}^{int.}=m–1$
$|S_{L(i)}|+|S_{R(i)}|+1=n$

Formal Analysis of HEAPIFY (2)¶

\[ \begin{align*} (2m–1)+(m–1)+1 &=n \\ m &= (n+1)/3 \\ |S_{L(i)}| &= 2m – 1 \\ &=2(n+1)/3 – 1 \\ &=(2n/3+2/3) –1 \\ &=\frac{2n}{3}-\frac{1}{3} \leq \frac{2n}{3} \\ T(n) & \leq T(2n/3) + \Theta(1) \\ T(n) &= O(lgn) \end{align*} \]

By CASE-2 of Master Theorem $\Longrightarrow$ $T(n)=\Theta(n^{log_b^a}lgn)$

Formal Analysis of HEAPIFY (2)¶

Recurrence: $T(n) = aT(n/b) + f(n)$
Case 2: $\frac{f(n)}{n^{log_b^a}}=\Theta(1)$
i.e., $f(n)$ and $n^{log_b^a}$ grow at similar rates
Solution: $T(n)=\Theta(n^{log_b^a}lgn)$
$T(n) \leq T(2n/3) + \Theta(1)$ (drop constants.)
$T(n) \leq \Theta(n^{log_3^1}lgn)$
$T(n) \leq \Theta(n^0lgn)$
$T(n) = O(lgn)$

HEAPIFY: Efficiency Issues¶

Recursion vs Iteration:
In the absence of tail recursion, iterative version is in general more efficient because of the pop/push operations to/from stack at each level of recursion.

Heap Operations: HEAPIFY (1)¶

Recursive

  HEAPIFY(A, i, n)
  largest = i 

  if 2i <= n and A[2i] > A[i] then 
    largest = 2i

  if 2i+1 <= n and A[2i+1] > A[largest] then 
    largest = 2i+1

  if largest != i  then
    exchange A[i] with A[largest]
    HEAPIFY(A, largest, n)

Heap Operations: HEAPIFY (2)¶

Iterative

HEAPIFY(A, i, n)
  j = i
  while(true) do
    largest = j 

  if 2j <= n and A[2j] > A[j] then 
    largest = 2j

  if 2j+1 <= n and A[2j+1] > A[largest] then 
    largest = 2j+1

  if largest != j  then
    exchange A[j] with A[largest]
    j = largest
  else return

Heap Operations: HEAPIFY (3)¶

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Heap Operations: Building Heap¶

Given an arbitrary array, how to build a heap from scratch?
Basic idea: Call $HEAPIFY$ on each node bottom up
Start from the leaves (which trivially satisfy the heap property)
Process nodes in bottom up order.
When $HEAPIFY$ is called on node $i$, the subtrees connected to the $left$ and $right$ subtrees already satisfy the heap property.

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Storage of the leaves (Lemma)¶

Lemma: The last $\lceil \frac{n}{2} \rceil$ nodes of a heap are all leaves.

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Storage of the leaves (Proof of Lemma) (1)¶

Lemma: last $\lceil n/2 \rceil$ nodes of a heap are all leaves
Proof :
$m=2^{d-1}$: $\#$ nodes at level $d-1$
$f$: $\#$ nodes at level $d$ (last level)
$\#$ of nodes with depth $d-1$ : $m$
$\#$ of nodes with depth $<d-1$ : $m-1$
$\#$ of nodes with depth $d$ : $f$
Total $\#$ of nodes :$n=f+2m-1$

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Storage of the leaves (Proof of Lemma) (2)¶

Total $\#$ of nodes : $f=n-2m+1$

\[ \begin{align*} \text{\# of leaves: }&=f+m-\lceil f/2 \rceil \\ &= m+\lfloor f/2 \rfloor \\ &= m+\lfloor (n-2m+1)/2 \rfloor \\ &= \lfloor (n+1)/2 \rfloor \\ &= \lceil n/2 \rceil \end{align*} \]

Proof is Completed

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Heap Operations: Building Heap¶

BUILD-HEAP (A, n)
  for i = ceil(n/2) downto 1 do
    HEAPIFY(A, i, n)

Reminder: The last $\lceil n/2 \rceil$ nodes of a heap are all leaves, which trivially satisfy the heap property

Build-Heap Example (Step-1)¶

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Build-Heap Example (Step-2)¶

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Build-Heap Example (Step-3)¶

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Build-Heap Example (Step-4)¶

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Build-Heap Example (Step-5)¶

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Build-Heap Example (Step-6)¶

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Build-Heap Example (Step-7)¶

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Build-Heap Example (Step-8)¶

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Build-Heap Example (Step-9)¶

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Build-Heap: Runtime Analysis¶

Simple analysis:
$O(n)$ calls to $HEAPIFY$, each of which takes $O(lgn)$ time
$O(nlgn)$ $\Longrightarrow$ loose bound
In general, a good approach:
Start by proving an easy bound
Then, try to tighten it
Is there a tighter bound?

Build-Heap: Tighter Running Time Analysis¶

If the heap is complete binary tree then $h_{\ell} = d – \ell$
Otherwise, nodes at a given level do not all have the same height, But we have $d – \ell – 1 \leq h_{\ell} \leq d – \ell$

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Build-Heap: Tighter Running Time Analysis¶

Assume that all nodes at level $\ell= d – 1$ are processed

\[ \begin{align*} T(n) &=\sum \limits_{\ell=0}^{d-1}n_{\ell}O(h_{\ell})=O(\sum \limits_{\ell=0}^{d-1}n_{\ell}h_{\ell}) \begin{cases} n_{\ell}=2^{\ell} = \# \text{ of nodes at level }\ell \\ h_{\ell}=\text{height of nodes at level } \ell \end{cases} \\ \therefore T(n) &= O \bigg( \sum \limits_{\ell=0}^{d-1}2^{\ell}(d-\ell) \bigg) \\ \text{Let } & h=d-\ell \Longrightarrow \ell = d-h \text{ change of variables} \\ T(n) &= O\bigg(\sum \limits_{h=1}^{d}h2^{d-h} \bigg)=O\bigg(\sum \limits_{h=1}^{d}h \frac{2^d}{2^h} \bigg) = O\bigg(2^d\sum \limits_{h=1}^{d}h (1/2)^h\bigg) \\ \text{ but } & 2^d = \Theta(n) \Longrightarrow O\bigg(n\sum \limits_{h=1}^{d}h (1/2)^h \bigg) \end{align*} \]

Build-Heap: Tighter Running Time Analysis¶

\[ \sum \limits_{h=1}^{d}h(1/2)^h \leq \sum \limits_{h=0}^{d}h(1/2)^h \leq \sum \limits_{h=0}^{\infty}h(1/2)^h \]

recall infinite decreasing geometric series

\[ \sum \limits_{k=0}^{\infty} x^k = \frac{1}{1-x} \text{ where } |x|<1 \]

differentiate both sides

\[ \sum \limits_{k=0}^{\infty}kx^{k-1} = \frac{1}{(1-x)^2} \]

Build-Heap: Tighter Running Time Analysis¶

\[ \sum \limits_{k=0}^{\infty}kx^{k-1} = \frac{1}{(1-x)^2} \]

then, multiply both sides by $x$

\[ \sum \limits_{k=0}^{\infty}kx^k = \frac{x}{(1-x)^2} \]

in our case: $x = 1/2$ and $k = h$

\[ \therefore \sum \limits_{h=0}^{\infty}h(1/2)^h = \frac{1/2}{(1-(1/2))^2}=2=O(1) \\ \therefore T(n)=O(n\sum \limits_{h=1}^{d}h(1/2)^h)=O(n) \]

Heapsort Algorithm Steps¶

(1) Build a heap on array $A[1 \dots n]$ by calling $BUILD-HEAP(A, n)$
(2) The largest element is stored at the root $A[1]$
Put it into its correct final position $A[n]$ by $A[1] \longleftrightarrow A[n]$
(3) Discard node $n$ from the heap
(4) Subtrees $(S2 \& S3)$ rooted at children of root remain as heaps, but the new root element may violate the heap property.
Make $A[1 \dots n-1]$ a heap by calling $HEAPIFY(A,1,n-1)$
(5) $n \leftarrow n-1$
(6) Repeat steps (2-4) until $n=2$

Heapsort Algorithm Example (Step-1)¶

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Heapsort Algorithm Example (Step-2)¶

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Heapsort Algorithm Example (Step-3)¶

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Heapsort Algorithm Example (Step-4)¶

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Heapsort Algorithm Example (Step-5)¶

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Heapsort Algorithm Example (Step-6)¶

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Heapsort Algorithm Example (Step-7)¶

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Heapsort Algorithm Example (Step-8)¶

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Heapsort Algorithm Example (Step-9)¶

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Heapsort Algorithm Example (Step-10)¶

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Heapsort Algorithm Example (Step-11)¶

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Heapsort Algorithm Example (Step-12)¶

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Heapsort Algorithm Example (Step-13)¶

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Heapsort Algorithm Example (Step-14)¶

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Heapsort Algorithm Example (Step-15)¶

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Heapsort Algorithm Example (Step-16)¶

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Heapsort Algorithm Example (Step-17)¶

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Heapsort Algorithm Example (Step-18)¶

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Heapsort Algorithm Example (Step-19)¶

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Heapsort Algorithm: Runtime Analysis¶

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\[ \begin{align*} T(n) &= \Theta(n)+\sum \limits_{i=2}^{n}O(lgi) \\ &= \Theta(n)+O\bigg( \sum \limits_{i=2}^{n}O(lgn) \bigg) \\ &= O(nlgn) \end{align*} \]

Heapsort - Notes¶

Heapsort is a very good algorithm but, a good implementation of quicksort always beats heapsort in practice
However, heap data structure has many popular applications, and it can be efficiently used for implementing priority queues

Data structures for Dynamic Sets¶

Consider sets of records having key and satellite data

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Operations on Dynamic Sets¶

Queries: Simply return info;
$MAX(S) / MIN(S):$ (Query) return $x \in S$ with the largest/smallest $key$
$SEARCH(S, k):$ (Query) return $x \in S$ with $key[x]= k$
$SUCCESSOR(S, x) / PREDECESSOR(S, x):$ (Query) return $y \in S$ which is the next larger/smaller element after $x$
Modifying operations: Change the set
$INSERT(S, x):$ (Modifying) $S \leftarrow S \cup \{x\}$
$DELETE(S, x):$ (Modifying) $S \leftarrow S - \{x\}$
$\text{EXTRACT-MAX}(S) / \text{EXTRACT-MIN}(S):$ (Modifying) return and delete $x \in S$ with the largest/smallest $key$
Different data structures support/optimize different operations

Priority Queues (PQ)¶

Supports
$INSERT$
$MAX / MIN$
$\text{EXTRACT-MAX} / \text{EXTRACT-MIN}$

Priority Queues (PQ)¶

One application: Schedule jobs on a shared resource
PQ keeps track of jobs and their relative priorities
When a job is finished or interrupted, highest priority job is selected from those pending using $\text{EXTRACT-MAX}$
A new job can be added at any time using $INSERT$

Priority Queues (PQ)¶

Another application: Event-driven simulation
Events to be simulated are the items in the PQ
Each event is associated with a time of occurrence which serves as a $key$
Simulation of an event can cause other events to be simulated in the future
Use $\text{EXTRACT-MIN}$ at each step to choose the next event to simulate
As new events are produced insert them into the PQ using $INSERT$

Implementation of Priority Queue¶

Sorted linked list: Simplest implementation
$INSERT$
- $O(n)$ time
- Scan the list to find place and splice in the new item
$\text{EXTRACT-MAX}$
- $O(1)$ time
- Take the first element
Fast extraction but slow insertion.

Implementation of Priority Queue¶

Unsorted linked list: Simplest implementation
$INSERT$
- $O(1)$ time
- Put the new item at front
$\text{EXTRACT-MAX}$
- $O(n)$ time
- Scan the whole list
Fast insertion but slow extraction.
Sorted linked list is better on the average
Sorted list: on the average, scans $n/2$ element per insertion
Unsorted list: always scans $n$ element at each extraction

Heap Implementation of PQ¶

$INSERT$ and $\text{EXTRACT-MAX}$ are both $O(lgn)$
good compromise between fast insertion but slow extraction and vice versa
$\text{EXTRACT-MAX}$: already discussed $\text{HEAP-EXTRACT-MAX}$
$INSERT$: Insertion is like that of Insertion-Sort.

HEAP-INSERT(A, key, n)
  n = n+1
  i=n  
  while i>1 and A[floor(i/2)] < key do
    A[i]=A[floor(i/2)] 
    i= floor(i/2)
  A[i]=key

Heap Implementation of PQ¶

Traverses $O(lgn)$ nodes, as $HEAPIFY$ does but makes fewer comparisons and assignments
$HEAPIFY$: compares parent with both children
$HEAP-INSERT$: with only one

HEAP-INSERT Example (Step-1)¶

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HEAP-INSERT Example (Step-2)¶

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HEAP-INSERT Example (Step-3)¶

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HEAP-INSERT Example (Step-4)¶

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HEAP-INSERT Example (Step-5)¶

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Heap Increase Key¶

Key value of $i^{th}$ element of heap is increased from $A[i]$ to $key$

HEAP-INCREASE-KEY(A, i, key)

  if key < A[i] then
    return error

  while i > 1 and A[floor(i/2)] < key do
    A[i] = A[floor(i/2)] 
    i = floor(i/2)

  A[i] = key

HEAP-INCREASE-KEY Example (Step-1)¶

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HEAP-INCREASE-KEY Example (Step-2)¶

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HEAP-INCREASE-KEY Example (Step-3)¶

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HEAP-INCREASE-KEY Example (Step-4)¶

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HEAP-INCREASE-KEY Example (Step-5)¶

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Heap Implementation of Priority Queue (PQ)¶

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Summary: Max Heap¶

Heapify(A, i)
Works when both child subtrees of node i are heaps
"Floats down" node i to satisfy the heap property
Runtime: $O(lgn)$
Max(A, n)
Returns the max element of the heap (no modification)
Runtime: $O(1)$
Extract-Max(A, n)
Returns and removes the max element of the heap
Fills the gap in $A[1]$ with $A[n]$, then calls Heapify(A,1)
Runtime: $O(lgn)$

Summary: Max Heap¶

Build-Heap(A, n)
Given an arbitrary array, builds a heap from scratch
Runtime: $O(n)$
Min(A, n)
How to return the min element in a max-heap?
Worst case runtime: $O(n)$
- because ~half of the heap elements are leaf nodes
Instead, use a min-heap for efficient min operations
Search(A, x)
For an arbitrary $x$ value, the worst-case runtime: $O(n)$
Use a sorted array instead for efficient search operations

Summary: Max Heap¶

Increase-Key(A, i, x)
Increase the key of node $i$ (from $A[i]$ to $x$)
“Float up” $x$ until heap property is satisfied
Runtime: $O(lgn)$
Decrease-Key(A, i, x)
Decrease the key of node $i$ (from $A[i]$ to $x$)
Call Heapify(A, i)
Runtime: $O(lgn)$

Phone Operator Problem¶

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A phone operator answering $n$ phones
Each phone $i$ has $x_i$ people waiting in line for their calls to be answered.
Phone operator needs to answer the phone with the largest number of people waiting in line.
New calls come continuously, and some people hang up after waiting.

Phone Operator Solution¶

Step 1: Define the following array:
$A[i]$: the ith element in heap
$A[i].id$: the index of the corresponding phone
$A[i].key$: $\#$ of people waiting in line for phone with index $A[i].id$

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Phone Operator Solution¶

Step 2: $\text{Build-Max-Heap}(A, n)$
Execution:
- When the operator wants to answer a phone:
- $id = A[1].id$
  - $\text{Decrease-Key}(A, 1, A[1].key-1)$
  - answer phone with index $id$
- When a new call comes in to phone i:
  - $\text{Increase-Key}(A, i, A[i].key+1)$
- When a call drops from phone i:
  - $\text{Decrease-Key}(A, i, A[i].key-1)$

Linked Lists¶

Like arrays, Linked List is a linear data structure.
Unlike arrays, linked list elements are not stored at a contiguous location; the elements are linked using pointers.

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Linked Lists - C Definition¶

C

// A linked list node
struct Node {
  int data;
  struct Node* next;
};

Linked Lists - Cpp Definition¶

Cpp

class Node {
public:
  int data;
  Node* next;
};

Linked Lists - Java Definition¶

Java

class LinkedList {
  Node head; // head of the list

  /* Linked list Node*/
  class Node {
      int data;
      Node next;

      // Constructor to create a new node
      // Next is by default initialized
      // as null
      Node(int d) { data = d; }
  }
}

Linked Lists - Csharp Definition¶

Csharp

class LinkedList {
  // The first node(head) of the linked list
  // Will be an object of type Node (null by default)
  Node head;

  class Node {
      int data;
      Node next;

      // Constructor to create a new node
      Node(int d) { data = d; }
  }
}

Priority Queue using Linked List Methods¶

Implement Priority Queue using Linked Lists.
push(): This function is used to insert a new data into the queue.
pop(): This function removes the element with the highest priority from the queue.
peek()/top(): This function is used to get the highest priority element in the queue without removing it from the queue.

Priority Queue using Linked List Algorithm¶

PUSH(HEAD, DATA, PRIORITY)
  Create NEW.Data = DATA & NEW.Priority = PRIORITY
  If HEAD.priority < NEW.Priority 
    NEW -> NEXT = HEAD
    HEAD = NEW 
  Else
    Set TEMP to head of the list 
  Endif

  WHILE TEMP -> NEXT != NULL and TEMP -> NEXT ->PRIORITY > PRIORITY THEN
    TEMP = TEMP -> NEXT 
  ENDWHILE

  NEW -> NEXT = TEMP -> NEXT 
  TEMP -> NEXT = NEW

Priority Queue using Linked List Algorithm¶

POP(HEAD)
//Set the head of the list to the next node in the list.
HEAD = HEAD -> NEXT.
Free the node at the head of the list

PEEK(HEAD): 
Return HEAD -> DATA

Priority Queue using Linked List Notes¶

LinkedList is already sorted.
Time Complexities and Comparison with Binary Heap

	peek()	push()	pop()
Linked List	$O(1)$	$O(n)$	$O(1)$
Binary Heap	$O(1)$	$O(lgn)$	$O(lgn)$

Sorting in Linear Time¶

How Fast Can We Sort?¶

The algorithms we have seen so far:
Based on comparison of elements
We only care about the relative ordering between the elements (not the actual values)
The smallest worst-case runtime we have seen so far: $O(nlgn)$
Is $O(nlgn)$ the best we can do?
Comparison sorts: Only use comparisons to determine the relative order of elements.

Decision Trees for Comparison Sorts¶

Represent a sorting algorithm abstractly in terms of a decision tree
A binary tree that represents the comparisons between elements in the sorting algorithm
Control, data movement, and other aspects are ignored
One decision tree corresponds to one sorting algorithm and one value of $n$ (input size)

Reminder: Insertion Sort Step-By-Step Description (1)¶

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Reminder: Insertion Sort Step-By-Step Description (2)¶

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Reminder: Insertion Sort Step-By-Step Description (3)¶

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Different Outcomes for Insertion Sort and n=3¶

Input : $<a_1,a_2,a_3>$

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Decision Tree for Insertion Sort and n=3¶

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Decision Tree Model for Comparison Sorts¶

Internal node $(i:j)$: Comparison between elements $a_i$ and $a_j$
Leaf node: An output of the sorting algorithm
Path from root to a leaf: The execution of the sorting algorithm for a given input
All possible executions are captured by the decision tree
All possible outcomes (permutations) are in the leaf nodes

Decision Tree for Insertion Sort and n=3¶

Input: $<9, 4, 6>$

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Decision Tree Model¶

A decision tree can model the execution of any comparison sort:
One tree for each input size $n$
View the algorithm as splitting whenever it compares two elements
The tree contains the comparisons along all possible instruction traces
The running time of the algorithm $=$ the length of the path taken
Worst case running time $=$ height of the tree

Counting Sort¶

Lower Bound for Comparison Sorts¶

Let $n$ be the number of elements in the input array.
What is the $min$ number of leaves in the decision tree?
$n!$ (because there are n! permutations of the input array, and all possible outputs must be captured in the leaves)
What is the max number of leaves in a binary tree of height $h$? $\Longrightarrow$ $2^h$
So, we must have:

$$ 2^h \geq n! $$

Lower Bound for Decision Tree Sorting¶

Theorem: Any comparison sort algorithm requires $\Omega(nlgn)$ comparisons in the worst case.
Proof: We’ll prove that any decision tree corresponding to a comparison sort algorithm must have height $\Omega(nlgn)$

\[ \begin{align*} 2^h & \geq n! \\ h & \geq lg(n!) \\ & \geq lg((n/e)^n) (Stirling Approximation) \\ & = nlgn - nlge \\ & = \Omega(nlgn) \end{align*} \]

Lower Bound for Decision Tree Sorting¶

Corollary: Heapsort and merge sort are asymptotically optimal comparison sorts.

Proof: The $O(nlgn)$ upper bounds on the runtimes for heapsort and merge sort match the $\Omega(nlgn)$ worst-case lower bound from the previous theorem.

Sorting in Linear Time¶

Counting sort: No comparisons between elements
Input: $A[1 \dots n]$, where $A[j] \in \{1, 2,\dots, k\}$
Output: $B[1 \dots n]$, sorted
Auxiliary storage: $C[1 \dots k]$

Counting Sort-1¶

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Counting Sort-2¶

Step 1: Initialize all counts to 0

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Counting Sort-3¶

Step 2: Count the number of occurrences of each value in the input array

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Counting Sort-4¶

Step 3: Compute the number of elements less than or equal to each value

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Counting Sort-5¶

Step 4: Populate the output array
There are $C[3] = 3$ elements that are $\leq 3$

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Counting Sort-6¶

Step 4: Populate the output array
There are $C[4]=5$ elements that are $\leq 4$

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Counting Sort-7¶

Step 4: Populate the output array
There are $C[3]=2$ elements that are $\leq 3$

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Counting Sort-8¶

Step 4: Populate the output array
There are $C[1]=1$ elements that are $\leq 1$

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Counting Sort-9¶

Step 4: Populate the output array
There are $C[4]=4$ elements that are $\leq 4$

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Counting Sort: Runtime Analysis¶

Total Runtime: $\Theta(n+k)$
$n$ : size of the input array
$k$ : the range of input values

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Counting Sort: Runtime¶

Runtime is $\Theta(n+k)$
If $k=O(n)$, then counting sort takes $\Theta(n)$
Question: We proved a lower bound of $\Theta(nlgn)$ before! Where is the fallacy?
Answer:
$\Theta(nlgn)$ lower bound is for comparison-based sorting
Counting sort is not a comparison sort
In fact, not a single comparison between elements occurs!

Stable Sorting¶

Counting sort is a stable sort: It preserves the input order among equal elements.
i.e. The numbers with the same value appear in the output array in the same order as they do in the input array.
Note: Which other sorting algorithms have this property?

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Radix Sort¶

Origin: Herman Hollerith’s card-sorting machine for the 1890 US Census.
Basic idea: Digit-by-digit sorting
Two variations:
Sort from MSD to LSD (bad idea)
Sort from LSD to MSD (good idea)

(LSD/MSD: Least/most significant digit)

Herman Hollerith (1860-1929)¶

The 1880 U.S. Census took almost 10 years to process.
While a lecturer at MIT, Hollerith prototyped punched-card technology.
His machines, including a card sorter, allowed the 1890 census total to be reported in 6 weeks.
He founded the Tabulating Machine Company in 1911, which merged with other companies in 1924 to form International Business Machines(IBM).

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Hollerith Punched Card¶

Punched card: A piece of stiff paper that contains digital information represented by the presence or absence of holes.
12 rows and 24 columns
coded for age, state of residency, gender, etc.

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Modern IBM card¶

One character per column
So, that’s why text windows have 80 columns!

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for more samples visit https://en.wikipedia.org/wiki/Punched_card

Hollerith Tabulating Machine and Sorter¶

Mechanically sorts the cards based on the hole locations.
Sorting performed for one column at a time
Human operator needed to load/retrieve/move cards at each stage

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Hollerith’s MSD-First Radix Sort¶

Sort starting from the most significant digit (MSD)
Then, sort each of the resulting bins recursively
At the end, combine the decks in order

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Hollerith’s MSD-First Radix Sort¶

To sort a subset of cards recursively:
All the other cards need to be removed from the machine, because the machine can handle only one sorting problem at a time.
The human operator needs to keep track of the intermediate card piles

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Hollerith’s MSD-First Radix Sort¶

MSD-first sorting may require:
very large number of sorting passes
very large number of intermediate card piles to maintain
S(d):
$\#$ of passes needed to sort d-digit numbers (worst-case)
Recurrence:
$S(d)=10S(d-1)+1$ with $S(1)=1$
- Reminder: Recursive call made to each subset with the same most significant digit(MSD)

Hollerith’s MSD-First Radix Sort¶

Recurrence: $S(d)=10S(d-1)+1$

\[ \begin{align*} S(d) &= 10 S(d-1) + 1 \\ & = 10 \bigg(10 S(d-2) + 1 \bigg) + 1 \\ & = 10 \Big(10 \bigg(10 S(d-3) + 1\bigg) + 1 \Big) + 1 \\ & = 10i S(d-i) + 10i-1 + 10i-2 + \dots + 101 + 100 \\ &=\sum \limits_{i=0}^{d-1}10^i \end{align*} \]

Iteration terminates when $i = d-1$ with $S(d-(d-1)) = S(1) = 1$

Hollerith’s MSD-First Radix Sort¶

Recurrence: $S(d)=10S(d-1)+1$

\[ \begin{align*} S(d) &=\sum \limits_{i=0}^{d-1}10^i \\ & = \frac{10^d-1}{10-1} \\ & = \frac{1}{9}(10^d-1)\\ & \Downarrow \\ S(d)&=\frac{1}{9}(10^d-1) \end{align*} \]

Hollerith’s MSD-First Radix Sort¶

$P(d)$: $\#$ of intermediate card piles maintained (worst-case)
Reminder: Each routing pass generates 9 intermediate piles except the sorting passes on least significant digits (LSDs)
There are $10^{d-1}$ sorting calls to LSDs

\[ \begin{align*} P(d) &= 9(S(d)–10^{d-1}) \\ &= 9\frac{(10^{d–1})}{9– 10^{d-1}} \\ &= (10^{d–1}–9 * 10^{d-1}) \\ &= 10^{d-1} - 1 \end{align*} \]

Hollerith’s MSD-First Radix Sort¶

\[ \begin{align*} P(d) &= 10^{d-1} - 1 \end{align*} \]

Alternative solution: Solve the recurrence

\[ \begin{align*} P(d) &= 10P(d-1)+9 \\ P(1) &= 0 \\ \end{align*} \]

Hollerith’s MSD-First Radix Sort¶

Example: To sort $3$ digit numbers, in the worst case:
$S(d) = (1/9) (103-1) = 111$ sorting passes needed
$P(d) = 10d-1-1 = 99$ intermediate card piles generated
MSD-first approach has more recursive calls and intermediate storage requirement
Expensive for a tabulating machine to sort punched cards
Overhead of recursive calls in a modern computer

LSD-First Radix Sort¶

Least significant digit (LSD)-first radix sort seems to be a folk invention originated by machine operators.
It is the counter-intuitive, but the better algorithm.
Basic Algorithm:

Sort numbers on their LSD first   (Stable Sorting Needed)
Combine the cards into a single deck in order 
Continue this sorting process for the other digits
  from the LSD to MSD

Requires only $d$ sorting passes
No intermediate card pile generated

LSD-first Radix Sort Example¶

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Correctness of Radix Sort (LSD-first)¶

Proof by induction:
Base case: $d=1$ is correct (trivial)
Inductive hyp: Assume the first $d-1$ digits are sorted correctly
Prove that all $d$ digits are sorted correctly after sorting digit $d$
Two numbers that differ in digit $d$ are correctly sorted (e.g. 355 and 657)
Two numbers equal in digit d are put in the same order as the input
(correct order)

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Radix Sort Runtime¶

Use counting-sort to sort each digit
Reminder: Counting sort complexity: $\Theta(n+k)$
$n$: size of input array
$k$: the range of the values
Radix sort runtime: $\Theta(d(n+k))$
$d$: $\#$ of digits

How to choose the $d$ and $k$?

Radix Sort: Runtime – Example 1¶

We have flexibility in choosing $d$ and $k$
Assume we are trying to sort 32-bit words
We can define each digit to be 4 bits
Then, the range for each digit $k=2^4=16$
- So, counting sort will take $\Theta(n+16)$
The number of digits $d =32/4=8$
Radix sort runtime: $\Theta(8(n+16)) = \Theta(n)$
$\overbrace{[4bits|4bits|4bits|4bits|4bits|4bits|4bits|4bits]}^{\text{32-bits}}$

Radix Sort: Runtime – Example 2¶

We have flexibility in choosing $d$ and $k$
Assume we are trying to sort 32-bit words
Or, we can define each digit to be 8 bits
Then, the range for each digit $k = 2^8 = 256$
- So, counting sort will take $\Theta(n+256)$
The number of digits $d = 32/8 = 4$
Radix sort runtime: $\Theta(4(n+256)) = \Theta(n)$
$\overbrace{[8bits|8bits|8bits|8bits]}^{\text{32-bits}}$

Radix Sort: Runtime¶

Assume we are trying to sort $b$-bit words
Define each digit to be $r$ bits
Then, the range for each digit $k = 2^r$
- So, counting sort will take $\Theta(n+2^r)$
The number of digits $d = b/r$
- Radix sort runtime:

\[ \begin{align*} T(n,b)&=\Theta \bigg( \frac{b}{r}(n+2^r) \bigg) \end{align*} \]

$\overbrace{[rbits|rbits|rbits|rbits]}^{b/r \text{ bits}}$

Radix Sort: Runtime Analysis¶

\[ \begin{align*} T(n,b)&=\Theta \bigg( \frac{b}{r}(n+2^r) \bigg) \end{align*} \]

Minimize $T(n,b)$ by differentiating and setting to $0$
Or, intuitively:
We want to balance the terms $(b/r)$ and $(n + 2^r)$
Choose $r \approx lgn$
- If we choose $r << lgn \Longrightarrow (n + 2^r)$ term doesn’t improve
- If we choose $r >> lgn \Longrightarrow (n + 2^r)$ increases exponentially

Radix Sort: Runtime Analysis¶

\[ \begin{align*} T(n,b)&=\Theta \bigg( \frac{b}{r}(n+2^r) \bigg) \end{align*} \]

\[ \begin{align*} \text{Choose } r=lgn \Longrightarrow T(n,b)=\Theta(bn/lgn) \end{align*} \]

For numbers in the range from $0$ to $n^d – 1$, we have:
The number of bits $b = lg(nd ) = d lgn$
- Radix sort runs in $\Theta(dn)$

Radix Sort: Conclusions¶

\[ \begin{align*} \text{Choose } r=lgn \Longrightarrow T(n,b)=\Theta(bn/lgn) \end{align*} \]

Example: Compare radix sort with merge sort/heapsort
$1$ million ($2^{20}$), $32$-bit numbers $(n = 2^{20}, b = 32)$
- Radix sort: $\lfloor 32/20 \rfloor = 2$ passes
- Merge sort/heap sort: $lgn = 20$ passes
Downsides:
Radix sort has little locality of reference (more cache misses)
The version that uses counting sort is not in-place
On modern processors, a well-tuned quicksort implementation typically runs faster.

References¶

$-End-Of-Week-4-Course-Module-$

Son Güncelleme: 13 Mayıs 2022

	peek()	push()	pop()
Linked List	\(O(1)\)	\(O(n)\)	\(O(1)\)
Binary Heap	\(O(1)\)	\(O(lgn)\)	\(O(lgn)\)