Transcript Lec. 8.1 - Homepages | The University of Aberdeen

CS1022
Computer Programming &
Principles
Lecture 1
Combinatorics
Plan of lecture
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Motivation
Addition principle
Multiplication principle
Analysing selection problems
k-Samples
k-Permutations
k-Combinations
k-Selections
Examples
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Why combinatorics?
• We sometimes need to consider the question
– How many elements are there in set A?
• If too many elements in a set, and we need to process
the set, then this might take too long
– What’s “too long”?
– Example: how many outfits can one wear?
• Blue trousers and yellow shirt
• Black trousers and yellow shirt
• Etc.
• When we are proposing a computational solution we
need to think of feasible solutions
– In realistic situations, will the program finish in “good time”?
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A scenario: timetabling
• Every term we organise your timetable
– Day/time for lectures, practicals, etc.
– Compulsory modules must be allowed
– Not all choices possible, though
• How many ways are there to
– Arrange lecture, practical and lab times
– For all modules, for the whole university,
– Using up all time slots, 9AM-5PM?
• Answer: many!
• A program to compute a timetable
needs to consider every option
– It is going to take a long time...
– International timetabling competition!
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Addition principle
• Disjoint choices/events don’t influence one another
– Choice of cakes in a shop – first choice influences second
one as there will be fewer cakes
– Throwing a die twice – 2nd time not influenced by 1st
time
• Suppose A and B disjoint choices/events
– The number of options/outcomes for A is n1
– The number of options/outcomes for B is n2
• Total number of options/outcomes for A or B is
n1  n2
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Multiplication principle
• Suppose a sequence of k choices/events
– With n1 options/outcomes for 1st choice/event
– With n2 options/outcomes for 2nd choice/event
– ...
– With nk options/outcomes for kth choice/event
• Total number of possible options/outcomes for the
sequence is
n1  n2  ...  nk
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Addition principle with sets
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Suppose A is the set of n1 options/outcomes
Suppose B is the set of n2 options/outcomes
Sets A and B are disjoint, that is, A  B  
So |A  B| |A|  |B|
That is, A  B contains n1  n2 elements
– There are n1  n2 possible outcomes for A or B
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Multiplication principle with sets
• We can “interpret” multiplication principle in terms
of sets too
– Let A1 be set of n1 options/outcomes for 1st choice/event
– Let A2 be setof n2 options/outcomes for 2nd choice/event
– ...
– Let Ak be set of nk options/outcomes for kth choice/event
• Sequence of k events is an element of Cartesian
product A1  A2  ...  Ak
– Set has cardinality |A1|  |A2|  ...  |Ak|
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Analysing information needs
• Suppose we need to create a way to identify cars
– What kind of licence plate would you propose?
• Some ways are better than others
– How many cars could we licence with a 6-digit licence?
– Answer: 1,000,000, that is from 000000 to 999999
– UK: 40,000,000 vehicles so we need more digits...
• Alternative – licence with 4 letters and 3 digits
– Examples: ABCD 123, GHHH 234, etc.
– How many cars can we licence?
– Answer: 26  26  26  26  10  10  10 = 456,976,000
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Analysing selection problems (1)
• Suppose we are offered 3 kinds of sweets
– Aniseed drops (A)
– Butter mints (B) and
– Cherry drops (C)
• How many ways can we choose 2 sweets?
– Can we choose the same sweet twice? (AA, BB or CC?)
– Does the order matter? (Is BA the same as AB?)
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Analysing selection problems (2)
• Cases to consider
– Repeats allowed and order matters
• 9 possibilities: AA, AB, AC, BA, BB, BC, CA, CB, CC
– Repeats allowed and order does not matter
• 6 possibilities: AA, AB, AC, BB, BC, CC
– Repeats not allowed and order matters
• 6 possibilities: AB, AC, BA, BC, CA, CB
– Repeats not allowed and order does not matter
• 3 possibilities: AB, AC, BC
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k-Samples
• Consider
– Selection of k objects from a set of n objects where
– Order matters and repetition is allowed
• Any such selection is called a k-sample
• Since repetition is allowed,
– There are n ways of choosing the 1st object, and
– There are n ways of choosing the 2nd object, and so on
– Until all k objects have been selected
• By the multiplication principle this gives
n  n  ...  n = nk possible k-samples
k
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k-Permutations (1)
• Consider
– Selection of k objects from a set of n objects where
– Order matters and repetition is not allowed
• Any such selection is called a k-permutation
– Total number of k-permutations is denoted by P(n, k)
• Since repetition is not allowed,
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There are n ways of choosing the 1st object
There are (n – 1) ways of choosing the 2nd object
There are (n – 2) ways of choosing the 3rd object, and so on
Up to (n – k  1) ways of choosing the kth object
• By the multiplication principle this gives
P(n, k)  n(n – 1)(n – k  1)  n!/ (n – k)!
possible k-permutations
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k-Permutations (2)
• How many 4-letter “words” can we write with
distinct letters from the set a, g, m, o, p, r?
• Solution:
– A “word” is any ordered selection of 4 different letters
P(n, k) 
n!
(n – k)!
We have in our scenario n  6 and k  4, so we have
P(6, 4)  6!/ (6 – 4)!  6!/2!
 (6  5  4  3  2  1)/(2  1)
 (6  5  4  3  2  1)/(2  1)  360
possible k-permutations
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k-Combinations (1)
• Consider
– Selection of k objects from a set of n objects where
– Order does not matter and repetition is not allowed
• Any such selection is called a k-combination
– Total number of k-combinations is denoted by C(n, k)
• By the multiplication principle:
– The number of permutations of k distinct objects selected
from n objects is
– The number of unordered ways to select the objects
multiplied by the number of ways to order them
• Hence, P(n, k)  C(n, k)  k! and so there are
C(n, k)  P(n, k) 
n!
k!
(n – k)! k!
possible k-combinations
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k-Selection (1)
• Consider
– Selection of k objects from a set of n objects where
– Order does not matter and repetition is allowed
• Any such selection is called a k-selection
• Since selection is unordered,
– We arrange the k objects so that like objects grouped
together and separate the groups with markers
– There are n ways of choosing the 2nd object, and so on
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k-Selection (2)
• Example: unordered selection of 5 letters from
collection a, b and c, with repetitions
– Selection of two a’s, one b and two c’s: aa|b|cc
– Selection of one a’s, and four c’s: a| |cccc
– Seven slots: five letters and two markers
– Different choices: ways to insert 2 markers into 7 slots
– That is, C(7, 2)
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k-Selection (3)
• Unordered selection of k objects from a set of n
objects with repetition allowed requires
(n – 1) markers and k objects
• Therefore, there are (n – 1)  k slots to fill
– The number of k-selections is the number of ways of
putting (n – 1) markers into the (n – 1)  k slots
• Therefore number of k-selections from n objects is
C((n  k – 1), (n – 1)) 
(n  k – 1)!
((n  k – 1) – (n – 1))! (n – 1)!
 (n  k – 1)!
k! (n – 1)!
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k-Selection (4)
• Five dice are thrown. How many different outcomes
are possible?
• Solution:
– Each of the dice shows one of six outcomes
– If 5 dice are thrown, the number of outcomes is the
unordered selection of 5 objects with repetition allowed
– C((n  k – 1), (n – 1)), with n  6 and k  5
– This gives
C((6  5 – 1), (6 – 1))  C(10, 5)  10!/(5! 5!)  252
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Summarising...
• What we have so far can be summarised as
Order matters
Order does not matter
(n  k – 1)!
Elements
k-sample: nk
k-selection:
repeated
k!(n  1)!
n!
n!
Elements not
k-combination:
k-permutation:
(n  k )!
(n  k )!k!
repeated
• We need to analyse the problem to choose formula
– They all need as input n and k
– If we choose the wrong formula, we still get a result...
– But the result will not be the right answer!
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Example: National Lottery (1)
• Twice-weekly draw selects 6 different numbers
– Random selection from {1, 2, ..., 49}
– People pick their numbers before draw (and pay for this)
• Let’s calculate the odds of hitting the jackpot
– Winning numbers: one of the unordered selection of six
numbers from 49 possibilities
– That is, C(49, 6) = 13,983,816 different ways
– The odds are 1 to 13,983,816
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Example: National Lottery (2)
• Smaller prizes for getting 5, 4 or 3 numbers right
– If you choose exactly 3 numbers right you get £10
• What are the odds of getting £10?
• Pre-selection of numbers means choosing three
correct and three incorrect numbers
– There are C(6,3) ways of selecting 3 correct numbers
– There are C(43,3) ways of selecting 3 incorrect numbers
– Total number of winning combinations is
6!
43!
C (6,3)  C (43,3) 

 246 ,820
3!3! 40!3!
– Odds of 13,983,816/246,820  57 to 1
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Example: Choosing people (1)
• 12 candidates for a committee of 5 people
• How many committees?
– Order does not matter
– Elements not repeated
Order matters
Elements
repeated
Elements not
repeated
nk
n!
(n  k )!
Order does not
matter
(n  k – 1)!
k!(n  1)!
n!
(n  k )!k!
• There are
12!
12!
C (12,5)

 792
(12  5)!5! 7!5!
possible committees
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Example: Choosing people (2)
• We want to know about two people, Mary & Peter
• How many committees contain both Mary & Peter?
• Reasoning
– If Mary & Peter are in the committee, then we need only
consider the remaining 3 people to complete the group
– There are 10 people to fill in the 3 places
– Order does not matter
– Elements not repeated
n!
n!
Order matters
Elements
repeated
• There are
Elements not
repeated
nk
(n  k )!
Order does not
matter
(n  k – 1)!
k!(n  1)!
(n  k )!k!
10!
10!
C (10,3)
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 120
(10  3)!3! 7!3!
possible committees with Mary and Peter in them
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Example: Choosing people (3)
• How many do not contain neither Mary or Peter?
• Reasoning
– If Mary & Peter are excluded, then we have to select 5
people from remaining 10 people
– Order does not matter
– Elements not repeated
Order matters
Elements
repeated
Elements not
repeated
nk
n!
(n  k )!
Order does not
matter
(n  k – 1)!
k!(n  1)!
n!
(n  k )!k!
• There are
10!
10!
C (10,5)

 253
(10  5)!5! 5!5!
possible committees without Mary and Peter
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Further reading
• R. Haggarty. “Discrete Mathematics for
Computing”. Pearson Education Ltd.
2002. (Chapter 6)
• Wikipedia’s entry
• Wikibooks entry
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