Discrrete mathematics for computer science counting

Counting

First Principles

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Counting in Gambling

What fraction of poker hands
are “a pair of Jacks?”
(probability of a pair
of Jacks)

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Counting in Algorithms

# ops to update a data
structure (# comparisons

needed to sort n

items)
# steps in a computation (#
n

multiplies to compute d )

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Counting in Cryptography

# possible passwords
# possible keys

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lec 10W.4


Sum Rule

A

B

If sets A and B are disjoint, then
|A ∪ B| = |A| + |B|

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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

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lec 10W.6


The Product Rule

If there are 4 men and 3
women, there are

4⋅3 = 12
different man/woman couples

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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) }
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Product Rule: Counting Strings

# length-4 binary strings

= |B × B × B × B|
4

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

=2·2·2·2=2

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4

9

Product Rule: Counting Strings

# length n strings from an
alphabet of size m is

m
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n
10


Example: Counting Passwords
Password conditions:

•
•
•
•

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characters are digits & letters
between 6 & 8 characters long
starts with a letter
case sensitive

11

Counting Passwords

L ::= {a,b,…,z,A,B,…,Z}
D ::= {0,1,……,9}
Pn ::= length n words
starting w/letter

= L × (L ∪ D)
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n-1
12


Counting Passwords

|L × (L ∪ D)

n-1

= |L|·|(L ∪ D)|

|

n-1

= |L|·(|L| + |D|)
= 52·(52+10)

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n-1

n-1

13


Counting Passwords

set of passwords:

P ::= P6∪ P7∪P8
|P| = |P6|+|P7|+|P8|
5
6
7
= 52·(62 +62 +62 )
14
≈ 19·10

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# 4-digit nums w/ ≥ one 7
cases by 1st occurrence of 7:

x: any digit o: any digit ≠ 7
7xxx or o7xx or oo7x or ooo7

•

103
2
+ 9·10

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= 3439

2
3
+ 9 ·10 + 9

15


at least one 7: another way

|4-digit nums w/ ≥ one 7|

= |4-digit nums|
− |those w/ no 7|
=

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104 − 94 = 3439

16


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

•

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

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

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More than a day:

•
•
•
•

n-1
9
52∙62
> number of nanoseconds in a day = 24∙60∙60∙10
n-1
12
62
> 1.66∙10
n-1 > log (1.66∙10

12
)/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.

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FINIS

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