Discrrete mathematics for computer science counting

Counting in GamblingWhat fraction of poker hands are “a pair of Jacks?” of Jacks... Counting in Cryptography# possible passwords # possible keys lec 10W.4 3/22/19... If there are 4 men

First Principles

Counting in Gambling

What fraction of poker hands

are “a pair of Jacks?”

of Jacks)

# ops to update a data

structure (# comparisons needed to sort n

items)

# steps in a computation (#

multiplies to compute d n )

Counting in Algorithms

Counting in Cryptography

# possible passwords

# possible keys

lec 10W.4 3/22/19

Sum Rule

If sets A and B are disjoint , then

| A ∪ B | = | A | + | B |

lec 10W.6

The Sum Rule

• Class has 17 women, 25 men so

total enrollment = 17 + 25 = 42

• 26 lower case letters, 26 upper case letters, and 10 digits, so # characters = 26+26+10 = 62

3/22/19

If there are 4 men and 3

women, there are

4⋅3 = 12 different man/woman couples

The Product Rule

Product Rule

If |A| = m and | B | = n , then

|A × B | = m⋅ n

A = { a, b, c, d }, B = { 1, 2, 3 }

A × B = {( a , 1 ),( a , 2 ),( a , 3 ),

( b , 1 ),( b ,2 ),( b , 3 ), ( c , 1 ),( c , 2 ),( c , 3 ), ( d , 1 ),( d , 2 ),( d , 3 ) }

Product Rule: Counting Strings

# length- 4 binary strings

= |B × B × B × B|

= | B 4 | where B ::= { 0 , 1 }

= 2 · 2 · 2 · 2 = 2 4

Product Rule: Counting Strings

# length n strings from an

alphabet of size m is

m n

Example: Counting Passwords

• characters are digits & letters

• between 6 & 8 characters long

• starts with a letter

• case sensitive

Password conditions:

L ::= {a,b,…,z,A,B,…,Z}

D ::= {0,1,……,9}

P n ::= length n words

starting w/letter

= L × (L ∪ D) n-1

Counting Passwords

|L × (L ∪ D) n-1 |

= |L|·|(L ∪ D)| n-1

= |L|·(|L| + |D|) n-1

Counting Passwords

Counting Passwords

set of passwords:

P ::= P 6 ∪ P 7 ∪P 8

|P| = |P 6 |+|P 7 |+|P 8 | = 52·(625+626+627) ≈ 19·1014

cases by 1st occurrence of 7 :

x : any digit o : any digit ≠ 7

103

= 3439

# 4 -digit nums w/ ≥ one 7

+ 9·102 + 92·10 + 93

at least one 7 : another way

| 4 -digit nums w/ ≥ one 7 |

= | 4 -digit nums|

− |those w/ no 7 |

= 104 − 94 = 3439

Cracking Passwords by Exhaustive Search

• Suppose a password has to be a string of letters and digits beginning with a letter

• So there are 52∙62n-1 passwords of length n

• Suppose a computer can try a password every billionth of a second (one nanosecond per try)

• How big does n have to be for it to take more than a day to try them all?

More than a day:

• 52∙62n-1 > number of nanoseconds in a day = 24∙60∙60∙109

• 62n-1 > 1.66∙1012

• n-1 > log (1.66∙1012)/log 62 ≈ 6.8

• So if passwords are at least 8 characters long it would take more than a day to try them all by exhaustive search

FINIS