Download Counting

Topics
• Section 6.1 – 6.6
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Original author of the slides:
Vadim Bulitko
University of Alberta
http://www.cs.ualberta.ca/~bulitko/W04
Modified by T. Andrew Yang
([email protected])
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Counting
• A random process
The set of outcomes is known, but the specific outcome is
not predictable.
– Outcomes
– Sample space
The set of all possible outcomes of a random process (or
experiment).
– Events
An event is a subset of a sample space.
• Examples
– coins
– dice
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Counting and Probability
• The chance that a given event will occur
• The ratio of the number of outcomes in an exhaustive set of
equally likely outcomes that produce a given event to the total
number of possible outcomes
• Outcomes are assumed to be equally likely.
• E: an event.
• S: the sample space.
• N(E): the number of outcomes in E
• N(S): the total number of outcomes in S.
• P(E): the probability that E will occur.
• Then P(E) =
N (E)
N (S )
• Examples:
– Coins
– Dice
– Tournament
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Permutations, combinations, etc.
Attributes
Ordered
Unordered
Reps
No Reps
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Multiplication Rule
• p.208: Theorem 6.2.1
k steps (s1, s2, …, sk) of an operation
n1 ways in s1
n2 ways in s2
…
Then, the entire operation can be performed in
n1n2…nk ways.
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Multiplication Rule (cont.)
• Example:
– How many different PINs are possible? p.308
(repetition is allowed)
– How if repetition is not allowed? p.309 
permutations (p.313)
– Password-based authentication
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Passwords-based Authentication
•
A dictionary attack is the guessing of a
password by repeated trial and error.
•
The dictionary may be a set of strings in
random order, or a set of strings in
decreasing order of probability of
selection.
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Passwords-based Authentication
•
Countering dictionary attack
–
–
The goal: To maximize the time needed to
guess the password
TG
Anderson’s Formula: P 
N
P: The probability that an attacker guesses a password in
a specified period of time
G: The number of guesses that can be tested in one time
unit
T: The number of time units during which guessing occurs
N: The number of possible passwords
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Passwords-based Authentication
•
An example:
–
Let S be the length of the password.
–
Let A be the number of characters in the alphabet from which the
characters of the password are drawn.
Then N = AS.
–
Let E be the number of characters exchanged when logging in.
–
Let R be the number of bytes per minute that can be sent over a
communication link.
–
Let G be the number of guesses per minute. Then G = R / E.
–
If the attack extends over M months, T = 30 x 24 x 60 x M.
–
Let P be the probability that the attack would succeed.
Then
TG
P 
N
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Passwords-based Authentication
•
Analysis of the Anderson Formula:
TG
P 
N
–
The goal is to maximize the time (T) needed for the attacker to
guess the password.
–
That is, to decrease the chance that the attack may succeed (P).
•
Approaches:
–
To increase N, the set of possible passwords
–
To decrease the time allowed to guess the passwords, that is, to
reduce T
–
To decrease G
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Possibility Trees
• Used to count the number of outcomes
• Can be used to illustrate the multiplication
rule (e.g., toss a coin for three times)
• Useful when the multiplication rule is
difficult or impossible to apply
• Examples
– Possibilities for tournament play: p.306
– Election of officers: p.311
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Permutations
• A permutation of a set of objects is an ordering
of the objects in a row.
• Example:
S = {a, b}
Permutations: ab, ba
Order matters!
Distinct objects!
• Formula
Given n objects (n >= 1), the number of permutations is
n!
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r-permutations
• An ordered selection of r elements out of n elements
• Still, order matters and no repetition
P(n,r) = n(n-1)(n-2)…(n-r+1)
n!
=
( n  r )!
• Exercises: P(5,3), P(7,3), P(3,3)
• Example 6.2.11 (p.317)
• Q16 on p.319
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Set Operations & Counting
• The addition rule (p.321)
Example 6.3.1: number of passwords
Note: distinct, mutually disjoint sets
• To make sets disjoint:
intersection, symmetric difference
• Inclusion/exclusion, difference rules
– Example 6.3.6
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Combinations
• Order is not important (i.e., sets)
• c.f., Order is important in permutations
• So, the different combinations can be
considered as subsets of a given set
• Example 6.4.2 (p.335)
S = {0,1,2,3}
Q: How many unordered selections of two
elements can be made from S?
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r-combinations
• An r-combination of a set of n elements is a
subset of r of the n elements
• n choose r: the number of r-combinations that
can be chosen from a set of n elements
Note: Order is not important, No repetition of elements
• p.364: computing binomial coefficients
• Formula
n
n!
  
 r  r!( n  r )!
• Example 6.4.10: p.344
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r-permutations vs r-combinations
• Share: no repetitions, distinct elements
• Difference:
– Permutations: unordered
– Combinations: ordered
• Figure 6.4.1: p.336
•
n
P( n, r )    * r!
r 
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Be aware of double-counting!
• A false solution: p.346
• Another example: M={a,b}, F={c,d,e}. Form 2-person
teams, but one of them must be a woman.
• Questions to ask:
– Am I counting everything?
– Am I counting anything twice?
• Multiplication rule
– Am I looking at everything at the possibility tree?
– Does every outcome appear on a branch of tree?
• Addition rule:
– Does every outcome appear in some subset of the diagram?
– Are the subsets disjoint?
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r-permutations with repetition
• r-permutations without repetitions: order
matters
P(n,r)=n(n-1)(n-2)…(n-r+1)
• What if we allow to put elements back?
• How many ways can we choose r
elements from n types of elements?
– Order matters
– Repetitions are allowed
• Formula?
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Permutations of a set with repeated elements
• Theorem 6.4.2:
p.345
• Example 6.4.11
 n  n  n1  n  n1  n2   n  n1  n2  ...  nk 1 

...

 
 n1  n2
 n3
  nk

n!

n1! n2 ! n3 !...nk !
• Which to use? c.f.: nk
• Q1: How many different bit strings can 4 bits hold?
• Q2: What are the total number of transpositions for
the 4-bit bit string 0110b? That is, how many 4-bit bit
strings contain exactly 2 1’s? 0110, 0101, 1010, 1001
See example 6.4.10 (p.344)
• Exercise: Try the same with 5 bits
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Special case
 n  n  n1  n  n1  n2   n  n1  n2  ...  nk 1 

...

 
 n1  n2
 n3
  nk

n!

n1! n2 ! n3 !...nk !
•When k = 2, permutations with repeated elements is
reduced to r-combinations. True? False?
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r-combinations with repetition
• What if we allow repetitions?
• Choose r elements out of n but allow
repetitions (e.g., put the elements back
after drawing them)
• Order is not important
• The underlying construct is multiset
• Theorem 6.5.1: p.351
• Examples
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Summary
Attributes
Ordered
Unordered
Reps
r-permutations
r-combinations
No Reps
• Question: How about
 n  n  n1  n  n1  n2   n  n1  n2  ...  nk 1 

...

 
n
n
n
n
 1  2
 3
  k

n!

n1! n2 ! n3 !...nk !
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Questions?
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