CS 401 Midterm review Xiaorui Sun 1 Midterm Exam Midterm exam via - - PowerPoint PPT Presentation

CS 401

Midterm review

Xiaorui Sun

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Midterm Exam

Midterm exam via gradescope: October 16 (Friday)

• The exam will be available on Oct 16 0am
• You must submit no later than Oct 16 11:59pm
• Usually can be finished in 2 hours
• Open textbook, but no copy from internet and between each other
• No class on October 16, I will answer exam questions on blackboard

from 1pm to 4pm.

• The last problem is for graduate student only.
• Graduate students must finish it
• Undergraduate students DO NOT work on it (no extra credit).

Midterm Exam

True or false

• Only answer true or false, no justification
• Choose true/false via gradescope

Short answer

• Answer questions, no justification
• Either type your solution via gradescope or upload the pic of

your handwriting to gradescop

Algorithm design

• Either type your solution via gradescope or upload the pic of

your handwriting to gradescope

• Graph algorithm
• Greedy algorithm
• Davide and conquer
• Each problem have several questions, understand and answer

each question, no justification/correctness proof

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Topics

• Analysis of running time
• Graphs
• Greedy algorithms
• Divide and conquer

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Analysis of running time

Given two positive functions f and g f(N) is O(g(N)) iff there is a constant c>0 and N0 ³ 0 s.t., 0£ f(N) £ c⋅g(N) for all N ³ N0 f(N) is W(g(N)) iff there is a constant c>0 and N0 ³ 0 s.t., f(N) ³ c ⋅ g(N) ³ 0 for all N ³ N0 f(N) is Q(g(N)) iff there are c0>0, c1>0 and N0 ³ 0 s.t. c0 ⋅ g(N) £ f(N) £ c1 ⋅ g(N) for all N ³ N0

• f(N) is Q(g(N)) iff f(N) is both O(g(N)) and W(g(N)).

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Properties

• Reflexivity. f is O(f).
• Constants. If f is O(g) and c > 0, then c⋅f is O(g).
• Products. If f1 is O(g1) and f2 is O(g2), then f1⋅f2 is

O(g1⋅g2).

• Sums. If f1 is O(g1) and f2 is O(g2), then f1 + f2 is

O(max {g1, g2}).

• Transitivity. If f is O(g) and g is O(h), then f is O(h)

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Asymptotic Bounds for common fns

Polynomials: !" + !$% + ⋯ + !'%' is ( %' Logarithms: log, % = ((log/ %) for all constants !, 2 > 0 Logarithms: log grows slower than every polynomial For all 5 > 0, log % = ((%6) % log % = ( %$."$

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Exercise

Suppose ! " = "!, & " = 2( Is ! = ) & , ! = Ω & or ! = Θ(&)? Solution 1: Consider ! " & " = 1 2 2 2 … " 2 ≥ " 4 … " 2 ≥ " 4

(/3

Solution 2: Take log.

• log ! " = log 1 + log 2 + ⋯ + log " = Θ(" log ")
• log & " = " log 2 = Θ(")
• So, log ! " = Ω(log &(")), hence ! " = Ω(& " )

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" terms "/2 terms This method does not work when log ! " = Θ(log & " )

Undirected Graphs G=(V,E)

• Notation. G = (V, E)
• V = nodes (or vertices)
• E = edges between pairs of nodes
• Captures pairwise relationship between objects
• Graph size parameters: n = |V|, m = |E|

V = {1, 2, 3, 4, 5 ,6, 7, 8} E = {(1,2), (1,3), (2,3), (2,4), (2,5), (3,5), (3,7), aaaa(3,8), (4,5), (5,6), (7,8)} m=11, n=8 No self-loop, no multiedge

Terminology

Path: A sequence of vertices s.t. each vertex is connected to the next vertex with an edge Cycle: Path of length > 2 that has the same start and end Tree: A connected graph with no cycles

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3 4 5 6 2 10 1 2 5 1 3 4 6

Terminology

Degree of a vertex: # edges that touch that vertex deg(6)=3 Connected: Graph is connected if there is a path between every two vertices Connected component: Maximal set of connected vertices

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3 4 5 6 7 2 10 1

Graph representation

Adjacency matrix. n-by-n matrix with Auv = 1 if (u, v) is an edge. Space proportional to n2. Checking if (u, v) is an edge takes Q(1) time. Identifying all edges takes Q(n2) time.

Graph representation

Adjacency list. Node indexed array of lists. Space proportional to m+n. Checking if (u, v) is an edge takes O(deg(u)) time. Identifying all edges takes Q(m+n) time.

Graph Traversal

Walk (via edges) from a fixed starting vertex ! to all vertices reachable from !. Breadth First Search (BFS): Order nodes in successive layers based on distance from ! Depth First Search (DFS): More natural approach for exploring a maze;

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BFS

1 2 3 10 5 4 9 12 8 13 6 7 11

BFS Tree gives shortest paths from 1 to all vertices

1 2 3 4

All edges connect same

• r adjacent levels

BFS

Properties:

• Edges into then-undiscovered vertices define a tree –

the “Breadth First spanning tree” of !

• Level " in the tree are exactly all vertices # s.t., the

shortest path (in !) from the root $ to # is of length "

• All nontree edges join vertices on the same or adjacent

levels of the tree Applications:

• Find connected components
• Single source shortest part on unweighted undirected

graph

• Testing bipartiteness

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DFS

A,1 B,2 J,10 I,9 H,8 C,3 G,7 F,6 D,4 E,5 K,11 L,12 M,13

Edge code: Tree edge Back edge

DFS

Properties:

• Edges into then-undiscovered vertices define a “DFS

tree” of !

• All nontree edges {#, %}, one of # or % is an ancestor of

the other in the DFS tree.

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Directed Graphs

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1 2 10 9 8 3 4 5 6 7 11 12 13

No self-loop, no multiedge (8, 10) and (10, 8) are different edges

Directed Acyclic Graphs (DAG)

Def: A DAG is a directed acyclic graph, i.e.,

• ne that contains no directed cycles.

Def: A topological order of a directed graph G = (V, E) is an

• rdering of its nodes as !", !$, … , !& so that for every edge

(!(, !)) we have + < -.

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a DAG

2 3 6 5 4 7 1

a topological ordering of that DAG– all edges left-to-right

1 2 3 4 5 6 7

Single Source Shortest Path

Given an (un)directed connected graph ! = ($, &) with non- negative edge weights () ≥ 0 and a start vertex ,. Find length of shortest paths from , to each vertex in !

Dijkstra’s algorithm

Cost of path s-2-3-4-t = 9 + 23 + 6 + 6 = 44.

s 3 t 2 6 7 4 5 23 18 2 9 14 15 5 30 20 44 16 11 6 19 6 length of path = sum of edge weights in path

Spanning Tree

Given a connected undirected graph ! = #, % . We call & is a spanning tree of ! if All edges in & are from %. & includes all of the vertices of !.

Kruskal’s algorithm

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& !

Exercise

True or false: Given a graph G with n vertices and n-1 edges, G is a tree.

• No. G can be a disconnected graph. So G can be a cycle of

n-1 vertices + an isolated vertex

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Greedy Algorithms

‘Best’ current partial solution at each step

• Solution is built in small steps
• Decisions on how to build the solution are made to

maximize some criterion without looking to the future

• Want the ‘best’ current partial solution as if the

current step were the last step How to define each step? What is the strategy of each step?

Interval Scheduling

Interval Scheduling

• Job j starts at !(#) and finishes at %(#).
• Two jobs compatible if they don’t overlap.
• Goal: find maximum subset of mutually compatible jobs.

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Time 1 2 3 4 5 6 7 8 9 10 11

f g h e a b c d h e b

Interval Scheduling

Every step we consider a single job In each step, we decide if the job will be in the solution set

• Strategy: if the job is compatible with the current solution set, put it

into the solution set How do we order jobs?

• Sort in the ascending order of the finish times

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Problem 2 of Homework 2

Each step we select a job to execute Which job to execute?

• In the i-th step, if we select job j, then the contribution of

job j to C1+C2+…+Cn is (n-i+1)tj

• Because we want to minimize C1+C2+…+Cn, so we

should select the job with smallest running time in each step

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Divide and Conquer

Divide: We reduce a problem to several subproblems.

Typically, each sub-problem is at most a constant fraction of the size of the original problem

Conquer: Recursively solve each subproblem Combine: Merge the solutions Examples:

• Mergesort, Binary Search, Closest Pair of Points,

Strassen’s Algorithm, FFT Log n levels

n n/2 n/2 n/4

Master Theorem

Suppose ! " = $ !

% & + (") for all " > +. Then,

• If $ < +) then ! " = Θ ")
• If $ = +) then ! " = Θ ")log "
• If $ > +) then ! " = Θ "23456

Example: For mergesort algorithm we have ! " = 2! " 2 + 8 " . So, 9 = 1, $ = +) and ! " = Θ(" log ")

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Naïve idea: Just sort all the elements (O(nk log (nk)) time) Another idea: Merge 1st array and 2nd array, then merge with 3rd array, then merge with 4th array, and so on…

• Running time: O((k+k)+(2k+k)+(3k+k)+…+((n-1)k+k)) =

O(n2 k)

• Even worse

Problem 4 of Homework 2

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Divide and conquer: given n sorted arrays, sort them into a single array

• Divide: partition n arrays into two groups of size n/2 arrays
• Conquer: sort each group
• Combine: merge sort the solution of each group

Let T(n) denote the running time to sort n arrays T(n) = 2 T(n/2) + O(nk) ⟹ T(n) = O(nk log n)

Problem 4 of Homework 2

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Divide and conquer: give an array, find the maximum subarray sum

• Divide: partition the array into two halves
• Conquer: Find the solution of each halves
• Combine: What if the solution contains elements from

both halves? (for [6,-2, -3, 1,5], [6, -2] is from the first half, and [-3, 1, 5] is from the second half)

Problem 5 of Homework 2

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If the solution contains elements from both halves,

• The elements in the solution from the first half
• Form an interval contains the last element of the first half
• Have largest sum among all the intervals of the first half containing the

last element of the first half ([6, -2] is the interval of [-2, -5, -6, -2] that (1) contains the last element and (2) has the largest sum)

• The elements in the solution from the second half
• Form an interval contains the first element of the second half
• Have largest sum among all the intervals of the second half containing

the first element of the second half

Problem 5 of Homework 2

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Redefine problem: give an array, find (1)$ : the maximum subarray sum (2)& : the maximum of subarray sum among all subarrays containing the first element (3)( : the maximum of subarray sum among all subarrays containing the last element

Problem 5 of Homework 2

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Given !", $", %" for the first half, and !&, $&, %& for the second half, how to find the !, $, %?

• ! = max{!", !&, %" + $&}
• $ = max{$", !/0" + $&}
• % = max{%&, !/0& + %"}

T(n) = 2 T(n/2) + O(n) ⟹ T(n) = O(n log n)

Problem 5 of Homework 2

!/0": sum of all the elements in the first half !/0&: sum of all the elements in the second half