Completing Kosaraju’s Algorithm Introduction to Greedy Algorithms Scheduling Monday, July 7th

Completing Kosaraju’s Algorithm

Introduction to Greedy Algorithms

Scheduling

Monday, July 7th

Announcements

Scoryst: Annotate which pages answer which problems.

OH room B24A: A is the desk number. The room is

Gates B24.

Night office hours:

Yiming: Thursday 7-9pm

Billy: Wednesdays 7-9pm

Slides are posted on the website on the schedule tab.

Last Wednesday

1. DAGs

2. Topological Sorting

3. Strongly Connected Components

Outline For Today

1. Complete Kosaraju’s Algorithm

2. Interesting Phenomenon About Shortest Paths And

Connected Components

3. Introduction to Greedy Algorithms

4. Scheduling Problem Variant 1 & Greedy Algorithm 1


Outline For Today







Recap: DAGs

Job Scheduling in Operating Systems

grep

tee

wc

awk

cat

job1

job2

echo

DAG Terminology

grep

tee

wc

awk

cat

job1

job2

echo

source: no incoming edges

sink: no outgoing edges

Topological Sorting of A DAGInput:

CS 106A

CS 106B CS 107 CS 110

CS 103 CS 109 CS 161

CS 143

106A 106B 107 110 103 109 161 143

Output:

OR

106A 103 106B 109 161 107 143 110

OR…Output is not unique

Topological Sorting Algorithm #1

procedure topologicalSort1(DAG G): let result empty list while G is not empty: 1. let v be a source node in G 2. add v to result 3. remove v from G return result/reverse

Note: You can also pick a sink

iteratively!

Pseudocode TS Algorithm #2 (Using DFS) procedure topologicalSort2(DAG G): run DFS(G) & put each finished vertex into an array in reverse orderRuntime = Runtime of DFS = O(n + m)

Correctness due to Key lemma about finishing times:

If u v, then f[u] > f[v]

Recap: Strongly Connected Components

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SCC1 SCC2

SCC3

SCC4

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SCC1 SCC2

SCC3

SCC4

Claim: GSCC has to be a DAG!

Two-tiered Structure

Kosaraju’s Algorithm: High-level Goal

Use BFS/DFS to peel off the components by label propagation

by picking start nodes from sink SCCs iteratively.

Goal: Use BFS/DFS to Peel-off SCCs

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How To Pick The Source Nodes?

Run DFS on Grev and pick source nodes

in

decreasing order of finishing times

Kosaraju’s Algorithm

procedure kosarajusSCC(DAG G): 1. run DFS on Grev and compute finishing times f 2. run DFS/BFS in G:

pick start vertices by decreasing f times from the first DFS propagate the ID of each new start vertex during traversalRun-time: O(n + m)!

DFS & Finishing Times on Grev

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Why Does DFS on Grev Work?

Key Lemma:

In GSCC if SCCi SCCj, then ***f[SCCi] > f[SCCj]***

Define: f[SCCi] = maxv∈SCCi f[v]

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SCC4

SCC3

SCC2

SCC1 f[SCC1] = 7 f[SCC2] = 9

f[SCC3] =10

f[SCC4] = 12

SCC4 > SCC3 > SCC1SCC2 >

Topological Ordering of SCCs

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Let’s go back to the original graph G

SCC4

SCC3

SCC2

SCC1 f[SCC1] = 7 f[SCC2] = 9

f[SCC3] =10

f[SCC4] = 12SCC4 > SCC3 > SCC1SCC2 >

Kosaraju’s Algorithm Simulation

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First Step: DFS on GrevF

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Kosaraju’s Algorithm Simulation

Second Step: DFS/BFS on G

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Kosaraju’s Algorithm SimulationSecond Step: DFS/BFS on

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Kosaraju’s Algorithm SimulationSecond Step: DFS/BFS on G

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Proof of Key Lemma:In GSCC if SCCi SCCj, then f[SCCi] > f[SCCj] SCCi SCCj

Let u be the first vertex visited in SCCi or SCCj

Case 1: u is in SCCi =>DFS starts at umoves to SCCjfinishes

SCCjfinishes uf[u] > f[v] ∀v ∈SCCj

=> f[SCCi] ≥ f[u] ≥ f[SCCj]Case 2: u is in SCCj => DFS starts at ufinishes SCCjstarts SCCifinishes SCCi

f[u] < s[y] ∀y ∈SCCi => f[u] < f[y] ∀y ∈SCCi => f[SCCi] > f[SCCj]

u u

Proof of Correctness of Kosaraju

Claim: Kosaraju finds SCCs in topological order TO* of (Grev)SCC output by the first DFS.

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TO* For Our Example

SCC4

SCC3

SCC2

SCC1 f[SCC1] = 7 f[SCC2] = 9

f[SCC3] =10

f[SCC4] = 12

SCC4 > SCC3 > SCC1SCC2 >

Topological Ordering of SCCs

Correctness of Kosaraju (1)

Claim: Kosaraju finds SCCs in topological order TO* of (Grev)SCC output by the first DFS.

Proof Sketch: Induct on the SCCs in the topological order.

Let the TO* be SCC1 SCC2 … SCCm

Base Case: by definition of TO*, the highest ft vertex is in

SCC1 AND SCC1 is a sink SCC in G

BFS/DFS can only discover SCC1

Correctness of Kosaraju (2)

Induc. Hypothesis: Assume Kosaraju finds first k-1 SCCs

in TO*

By definition of TO* and inductive hypothesis, the next

highest ft vertex that is unvisited, say u, will be from

SCCk.

By key lemma: BFS/DFS discovers only vertices in

SCCk && SCC1 … SCCk-1

But (by inductive hypothesis) all vertices in the k-1 SCCs

are already “peeled-off”, i.e. finished

only SCCk will be discovered during a BFS/DFS from u.

Outline For Today







Interesting Shortest Paths PhenomenonMilgram’s 1967 “six-degrees of separation” experiment

Humans are extremely well connected

On average 6 hops apart

Latest studies range from ~4.7 to ~5.8

Surprising because we form an extremely sparse graph

On FB > 1 Billion users: 109

~300 average edges per user

Interesting fact: Same Milgram who performed the

“obedience to authority” experiment

Interesting Conn. Comp. PhenomenonReal Graphs: Very Skewed Distribution in component

sizes

Checkout Jure Leskovec’s Graph Data:

http://snap.stanford.edu/data/

Bow-tie web graph from Broder et. al. 2000

Outline For Today







Greedy Algorithms

Algorithms that iteratively make

“short-sighted”, “locally optimum looking” decisions

hoping to output a good solution (hopefully

optimum)

Example: Dijkstra’s SSSP Algorithm

Greedy vs Divide-And-Conquer Algorithms

Greedy Divide and Conquer

easy to design difficult to design




easy to analyze run-time difficult to analyze run-time




easy to analyze run-time difficult to analyze run-time

difficult to prove correctness

easy to prove correctness

Two Correctness Proof Techniques

Call greedy’s solution Sg, and let S be any other

solution

1. “Greedy stays ahead”

Argue Sg is optimal/better than S at each step

Proof by induction

Example: Dijkstra’s correctness proof

2. Exchange Arguments

Argue any S can be transformed into Sg step by step

and without getting worse along the way

Warning: They’re common but not applicable to every

greedy algorithm!

Outline For Today







Scheduling Problem 1 Input: A set of n jobs J. Each jobi has length li

l1Job 1

l2Job 2

lnJob n

…

Output: A schedule of the jobs on a processor

s.t:

is minimum over all possible n! schedules.

completion time of job

i

Completion Time of Job iDefinition: time when jobi finishes

i.e., sum of scheduled job lengths up to and including jobi

S1 1

3J15J2

1J4

1J3

J3 J2 J1 J45 3 1

time 1 6 9 10

Total Cost of S1:1 + 6 + 9 + 10 = 26

0

Another Example Schedule

S1 1

J3 J2 J1 J45 3 1

time 1 6 9 10

Total Cost of S1: 1 + 6 + 9 + 10 = 26

S2 1

J3 J25

J13

J41

time 1 4 5 10

Total Cost of S2:1 + 4 + 5 + 10 = 20

Goal is to find the min cost schedule!

Let’s Start Simple

3J15J2

What are all possible schedules?

S1

J25

J13

time 3 8

Total Cost:3 + 8 = 11

S2

J25

J13

time 5 8

Total Cost:5 + 8 = 13

Why Put One Job In Front of Another?

Observation:

Shorter jobs have less impact on the

completion times of future jobs

Greedy Algorithm 1

Schedule jobs by increasing lengths

3J15J2

1J4

1J3

Ex:

Sg 1

J3 J25

J13

J41

time 51 2 1

0

Total Cost of Sg:1 + 2 + 5 + 10 = 18

Run-time O(nlog(n))!

Comparing Sg to Previous Schedules

Sg 1

J3 J25

J13

J41

time 51 2 1

0

S1 1

J3 J2 J1 J45 3 1

time 1 6 9 10

S2 1

J3 J25

J13

J41

time 51 4 10

26

20

18

Proof of Correctness (1)

“Greedy stays ahead” proof:

Induct on the cost of the first k jobs executed

Argue Sg beats everyone else at each step

Let S[i]: the ith job that a schedule S executes

E.g., Sg[1] is the first job Sg executes

Let Cost(S, i): be the sum of the costs of the first i jobs

that schedule S executes.

E.g., Cost(Sg, 3) is the sum of completion times

Sg[1], Sg[2], Sg[3]: Sg[1] + (Sg[1]+Sg[2]) +

(Sg[1]+Sg[2]+Sg[3])

Goal: Argue ∀S, Cost(Sg, n) ≤ Cost(S, n) by inducting

on i


Base Case: ∀S, Cost(Sg, 1) = Sg[1] ≤ Cost(S, 1) since

Sg[1] is the shortest length job

Inductive Hypothesis: Cost(Sg, k-1) ≤ Cost(S, k-1)

By inductive hypothesis

By greedy criterion of Sg

QED

Outline For Today







Scheduling Problem 2 Input: Now each jobi has length li AND weight wi

l1, w1Job 1

l2, w2Job 2

ln,wnJob n

…

Output: A schedule of the jobs on a processor

s.t:

is minimum over all possible n! schedules.

weighted completion time of

job i

Example Schedule And Cost

S1

l=3 w=1J1l=5 w=2J2

l=1 w=1J4

l=1 w=3J3

J3 J2 J1 J4

time 1 6 9 10

Total Cost of S1:3*1 + 2*6 + 1*9 + 1*10 = 34

l=1 w=3 l=5 w=2 l=3 w=1

l=1 w=1

Q1: What To Do When Weights Are Same?

Same As Un-weighted Case Shorter lengths

first

Previous Greedy Algorithm is Optimal

l=3 w=1J1l=5 w=1J2

l=1 w=1J4

l=1 w=1J3

Q2: What To Do When Lengths Are Same?

Higher weights first

l=3 w=2J1l=3 w=3J2

l=3 w=4J4

l=3 w=1J3

minimize:

Q3: What To Do With Mixed L & W?

Say J1 is shorter and also has less weight than J2?

l=3 w=1J1l=5 w=2J2

minimize:

Unclear:

Intuition for Q1 says J1 should come first

Intuition for Q2 says J2 should come first

Ideal Scenario: Combine l and w into a

single score that combines both intuitions

and we could order by that score

Possible Combined Scores?

The combined score f(li, wi) should satisfy:

1. If weights same shorter lengths get smaller

scores

2. If lengths same larger weights get smaller

scores

Guess 1: f1(li, wi) = li – wi

Guess 2: f2(li, wi) = li / wi

Is either one correct?

Let’s First Try To Eliminate One Guess

l=3 w=1J1l=5 w=2J2

f1 = li - wi f2 = li/wi

J1 (l=3, w=1)

J2 (l=5, w=2)

SCHEDULE

TOTAL COST


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2

J2 (l=5, w=2)

SCHEDULE

TOTAL COST


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2

J2 (l=5, w=2)

3

SCHEDULE

TOTAL COST


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2

J2 (l=5, w=2)

3

SCHEDULE J1::J2TOTAL COST


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2

J2 (l=5, w=2)

3


1*3 + 2*8 = 19


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2 3

J2 (l=5, w=2)

3


1*3 + 2*8 = 19


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2 3

J2 (l=5, w=2)

3 2.5


1*3 + 2*8 = 19


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2 3

J2 (l=5, w=2)

3 2.5

SCHEDULE J1::J2 J2::J1TOTAL COST

1*3 + 2*8 = 19


l=3 w=1J1l=5 w=2J2


J1 (l=3, w=1)

2 3

J2 (l=5, w=2)

3 2.5


1*3 + 2*8 = 19

2*5 + 1*8 = 18

Guess 1 is certainly not optimal.

Is Guess 2 optimal?

Greedy Algorithm (Weighted Scheduling)Schedule jobs by increasing li/wi

scores Run-time

O(nlog(n))!

S1

l=3 w=1J1l=5 w=2J2

l=1 w=1J4

l=1 w=3J3

J3 J2J1J4

time 1 2 7 10

Total Cost of Sg:3*1 + 1*2 + 2*7 + 1*10 = 29

l=1 w=3 l=5 w=2 l=3 w=1

l=1 w=1

score:3

score:2.5

score:1/3

score:1


By Exchange Argument:

Argue any S can be transformed into Sg step by

step and without getting worse along the way

Let’s rename jobs so that Sg schedules jobs in order:

J1, J2, …, Jn

i.e., J1 happens to be the job with smallest l/w ratio

Therefore Sg = J1::J2::…::Jn


Consider any other schedule S ≠ Sg

Claim: In S=Js1::Js2::…::Jsn there is a job k, right

after a job i where k < i.

Sg

J1l1 w1

J2l2 w2

J3l3 w3

Jnln wn…

SJ1

l1 w1

J9l9 w9

……

J6l6 w6

Exchange J9 with J6


S`J1

l1 w1

J6l6 w6…

J9l9 w9

Exchange J11 with J8

J11

l11 w11 …

J8l8 w8

S``J1

l1 w1

J6l6 w6…

J9l9 w9

J8l8 w8

…J11

l11 w11

Exchange J8 with J9…

J1

l1 w1

J2

l2 w2

J3

l3 w3

Jn

ln wn…

Sg

…

Completing the Proof

Recall Claim: In S=Js1::Js2::…::Jsn there is a job k, right

after a job i where k < i.

Q: How does the cost of S change when we swap i

and j?

wi * lk

Jili wi

Jklk wk

Jili wi

Jklk wk

wk * li

Prefix Jobs

Prefix Jobs

Suffix Jobs

Suffix Jobs

By renaming of jobs and k < i => lk/wk ≤ li/wi => wilk ≤

wkliQED

On Wednesday We Will Look at Minimum Spanning Trees

BACKUP SLIDES


l=3 w=1J1l=5 w=2J2

f1(li, wi) = li – wi f1(J1)= 3 – 1 =2, f1(J2)= 5 – 2 =3

f1 schedules them as J1::J2

Score: 1*3 + 2*8 = 19

f2(li, wi) = li / wi f2(J1)= 3/1 =3, f1(J1)= 5/2 =2.5

f1 schedules them as J2::J1

Score: 2*5 + 1*8 = 18

f1=li - wi f2 = li/wi

J1 (l=3, w=1)

2 3

J2 (l=5, w=2)

3 2.5


1*3 + 2*8 = 19

2*5 + 1*8 = 18

Directed Acyclic Graphs (DAGs)

Graphs to model “dependency”/“prerequisite”

relationships

directed & acyclic

CS 106A

CS 106B CS 107 CS 110

CS 103 CS 109 CS 161

CS 143

Topological Sorting of a DAG

Input: A G(V, E) which is a DAG

Output: vertices in sorted order s.t

each vertex v appears after its dependencies

A More Formal Definition:

Given an input DAG G(V, E) order the nodes

such that

if (u, v) ∈ E, then u appears before v

ExampleInput:

CS 106A

CS 106B CS 107 CS 110

CS 103 CS 109 CS 161

CS 143

106A 106B 107 110 103 109 161 143

Output:

OR

106A 103 106B 109 161 107 143 110

OR…Output is not unique

Two Topological Sorting Algorithms (1)

procedure topologicalSort1(DAG G): let result empty list while G is not empty: 1. let v be a source node in G 2. add v to result 3. remove v from G return result/reverse

Two Topological Sorting Algorithms (2) procedure topologicalSort2(DAG G): run DFS(G) & put each finished vertex into an array in reverse orderRuntime = Runtime of DFS = O(n + m)

Correctness depended on Key Claim:

In a DAG if u v, then f[u] > f[v]

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Completing Kosaraju’s Algorithm Introduction to Greedy Algorithms Scheduling Monday, July 7th

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