Review for Quiz 2

Review for Quiz 2

Great Theoretical Ideas In Computer Science

Quiz Review CS 15-251 Spring 2005

March 17, 2005

Carnegie Mellon University

StarringStarring Bonzo Jr…(251 student)

and Lucky(the turtle who introduced the magic square)

Disclaimer

This set of lecture slides is intended as an OVERVIEW of the things you should know; it is insufficient to simply glance over these slides 10 minutes before the quiz.

Moreover, this quiz review is NOT a substitute for the lectures; if you missed any lecture, watching the video is highly recommended.

Administrative Stuff

First, a quick review of the administrative stuff.

– 50 minutes long– Divided into 4 conceptual

sections– Practice quizzes have been/will

be uploaded.

Section 1: Repeat After me

Yay! Free points just for going to lecture and skimming over

the notes

This section contains about 6-8 questions based on quick facts covered in lecture. If

you went to each lecture and paid attention, you should be fine. It is recommended that you skim over the lecture notes before the quiz to refresh your memory on the subject

matter.

Section 2: Reading Solutions

You are expected to read the given solutions to all homework and recitation

problems - this section is comprised of one or two questions that are literally identical to questions that we've given you before. Note

that the problems we ask can come from any previous assignment. (its even possible

that we'll ask for the same problem's solution on all three quizzes).

Section 3: Basic Techniques

These problems are based on techniques taught during class, and practiced in

HWs/recitations. If you practice the ideas taught in lecture, you should have very little trouble with this section (problems vary only

slightly from results derived in lecture).

HINT: Look at the stuff from Monday’s recitation and the practice quizzes.

Section 4: A Moment’s Thought

These problems are similar to problems given on homework assignments, albeit

easier. If you do well on homework assignments, you shouldn't have too much

trouble with this section.

Who named this section…?

Time managementDon’t forget about time management!

The first two sections are designed to be able to be

completed relatively quickly - don’t spend all your time on a

Repeat After Me question that is worth only 5 points!

Beginning of Material Review

Enough; I know all this! What’s going to be on the quiz?

Stuff on quizThe new material for Quiz 2 is from Lecture 9: Counting III

to Lecture 17: On Time and

Input Size.

We are assuming that you already have an understanding of the

material from lectures 1-8. While we won’t explicitly test on that, you may need the knowledge

you’ve learned from those lectures in answering the

questions.

Key Areas

The following areas are rated as very likely to appear on your quiz.

– Finite State Machines (DFAs).– Counting and Binomials.– Matchings (including the TMA).– GCDs and Continued Fractions– Fibonacci and Fibonacci-like

series.– Asymptotic notation and runtime

analysis.



analysis.

Table of contents

Finite set of states

A start state

A set of accepting states

A finite alphabet a b #

x 1

State transition instructions

1 2{ , , , , }o kQ q q q q

Finite Automaton – quick review

oq

1 2, , ,

ri i iF q q q

:

( , )i j

Q Q

q a q

iq jq

a

Pumping Lemma

In simple terms, the pumping lemma says that for any string that’s long enough, there’s a

partof it we can repeat as many times as we want (or even

delete, if we let i = 0), and still get

a string that’s accepted.

Formal definition of pumping lemma

The pumping lemma states that if there exists a DFA accepting a certain language, thenthere is a positive integer k (called the pumping length) so that any string s whose lengthis greater than or equal to k can be split up into three strings, u, v, and w, with certain properties:

1. v is not empty; 2. |uv| k, and 3. The string uviw is also accepted by the DFA.

What strings are accepted by the following regexes?

– Ø– ε– 1001– 10*1– (0+10+110+111)*– 1(0+1)*0(0+1)*1

Regular Expressions

Important ConceptsCan you do the following? - Construct a DFA that accepts all strings with a length that is a multiple of 2 or 3.- Prove that the language of palindromes is not regular (use of pumping lemma is sufficient but not necessary).- Convert a simple regex to a DFA and vice versa?



analysis.

Table of contents

1 X 1 X1 X 1 X

1 X 1 X

1 X

Choice tree for terms of (1+X)3

1 X X X2 X X2 X2 X3

Combine like terms to get 1 + 3X + 3X2 + X3

The Binomial Formula

nkn xn

nx

k

nx

nx

nnx

......

110)1( 2

binomial expression

Binomial Coefficients

0

(1 )n

n k

k

nx x

k

The Binomial Formula

Be sure you can handle variants on this (remember the

problem from Monday’s recitation)

Pascal’s Triangle

11

1 11 1

1 2 11 2 1

1 3 3 11 3 3 1

1 4 6 4 11 4 6 4 1

1 5 10 10 5 11 5 10 10 5 1

1 6 15 20 15 6 11 6 15 20 15 6 1

So very many

properties…

Manhattan

k’th Avenue01

23

4

j’th Street 01

23

4



analysis.

Table of contents

Rogue Couples

Suppose we pair off all the boys and girls. Now suppose that some boy and some girl prefer each other to the people to whom they are paired. They will be called a rogue couple.

Stable Pairings

A pairing of boys and girls is called stable if it contains no rogue couples.

3,5,2,1,4

15,2,1,4,3

4,3,5,1,2

31,2,3,4,5

42,3,4,1,5

5

1

3,2,5,1,4

2

1,2,5,3,4

3

4,3,2,1,5

4

1,3,4,2,5

5

1,2,4,5,3

2

Traditional Marriage Algorithm

For each day that some boy gets a “No” do:• MorningMorning

– Each girl stands on her balcony– Each boy proposes under the balcony of the best

girl whom he has not yet crossed off• Afternoon (for those girls with at least one Afternoon (for those girls with at least one

suitor)suitor)– To today’s best suitor: “Maybe, come back “Maybe, come back

tomorrow”tomorrow”– To any others: “No, I will never marry you” “No, I will never marry you”

• EveningEvening– Any rejected boy crosses the girl off his list

Each girl marries the boy to whom she just said “maybe”

Opinion Poll

Who is better off

in traditio

nal

dating, the boys

or the girls

?



analysis.

Table of contents

Euclid’s GCD Formula

Note: GCD(67, 29) = 1

Euclid(67,29) 67 mod 29 = 9Euclid(29,9) 29 mod 9 = 2

Euclid(9,2) 9 mod 2 = 1 Euclid(2,1) 2 mod 1

= 0 Euclid(1,0) outputs 1

Euclid(A,B) // requires AB0 If B=0 then return A

else return Euclid(B, A mod B)

Extended GCD Algorithm

The lecture also covered Euclid’s Extended GCD

Algorithm – that might be something you should take a

look at.

The multiplicative inverse of x ≡ Zn* is

the unique y є Zn* such that

x *n y ≡ n 1. TO QUICKLY COMPUTE Y FROM X:

Run Extended_Euclid(x,n). It returns a,b, and d such that ax+bn = dBut d = GCD(x,n) = 1, so ax + bn = 1Hence MODULO n: ax = 1 (mod n)Thus, a is the multiplicative inverse of x.

A (Simple) Continued Fraction Is Any Expression Of The Form:

111111111

....

ab

cd

ef

gh

ij

where a, b, c, … are whole numbers.

A Finite Continued Fraction

12

13

142

Denoted by [2,3,4,2,0,0,0,…]

An Infinite Continued Fraction

11

12

12

12

12

12

12

12

122 ....

Denoted by [1,2,2,2,…]

An infinite continued fraction

12 1

12

12

12

12

12

12

12

122 ....

Phi!

Φ



analysis.

Table of contents

Check HW/Recitation

The best way to review for this particular section is to look over the Fibonacci homework.

Also, the recitation problems from Monday on Fibonacci are EXCELLENT practice.

Beginning of Material Review

How did we get the closed form for Fibonacci-like series?

A technique to derive the formula for the Fibonacci

numbersFn is defined by two conditions:

Base condition: F0=0, F1=1

Inductive condition: Fn=Fn-1+Fn-2

Forget the base condition and concentrate on satisfying the inductive condition

Inductive condition: Fn=Fn-1+Fn-2

Consider solutions of the form: Fn= cn for some complex constant c

C must satisfy: cn - cn-1 - cn-2 = 0

cn - cn-1 - cn-2 = 0

iff cn-2(c2 - c1 - 1) = 0

iff c=0 or c2 - c1 - 1 = 0

Iff c = 0, c = , or c = -(1/)

c = 0, c = , or c = -(1/)

So for all these values of c the inductive condition is satisfied:

cn - cn-1 - cn-2 = 0

Do any of them happen to satisfy the base condition as well? c0=0 and c1=1?

ROTTEN LUCK

Insight: if 2 functions g(n) and h(n) satisfy the inductive condition then so does

a g(n) + b h(n) for all complex a and b

(a g(n) + b h(n)) + (a g(n-1) + b h(n-1)) + (a g(n-2) + b h(n-2)) = 0

g(n)-g(n-1)-g(n-2)=0ag(n)-ag(n-1)-ag(n-2)=0

h(n)-h(n-1)-h(n-2)=0bh(n)-bh(n-1)-bh(n-2)=0

a,b a n + b (-1/ )n satisfies the inductive condition

Set a and b to fit the base conditions.

n=0 : a + b = 0n=1 : a 1 + b (-1/ )1 = 1

Two equalities in two unknowns (a and b).Now solve for a and b:

this gives a = 1/5 b = -1/5



analysis.

Table of contents

Useful notation to discuss growth rates

For any two monotonic functions f and g from the positive integers to the positive integers, we say

“f = O(g)” or “f is O(g)”if

Some constant times g eventually dominates f

[Formally: there exists a constant c such that for all sufficiently large n: f(n) ≤ c g(n) ]

O(n) graph

# of bits in numbers

time

f = O(g) means that there is some constant c such that c g(n) stays above f(n) from some point

on.

fg

1.5g

More useful notation: Ω


“f = Ω(g)” or “f is Ω(g)”if:

f eventually dominates some constant times g

[Formally: there exists a constant c such that for all sufficiently large n: f(n) ≥ c g(n) ]

Yet more useful notation: Θ


“f = Θ(g)” or “f is Θ(g)”if:

f = O(g) and f = Ω(g)

Quickies

• n = O(n2) ?– YES

• n = O(√n) ?– NO

• 3n2 + 4n + = O(n2) ?– YES

• 3n2 + 4n + = Ω(n2) ?– YES

• n2 = Ω(n log n) ?– YES

• n2 log n = Θ(n2)– NO

3n2 + 4n + = Θ(n2)

Final Thoughts

Do the practice quizzes (or at least look over them).

Don’t forget that the processor lab is due on Tuesday at the beginning of class; make sure to schedule everything appropriately.

Good luck!

Fin

Documents

Review for Quiz 2