Post on 24-Mar-2018
Discrete Mathematics Lecture Notes
Prof. Yuh-Dauh Lyuu
Dept. Computer Science & Information Engineering
and
Department of Finance
National Taiwan University
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 1
Class Information
• Grimaldi. Discrete and Combinatorial Mathematics.
– An excellent book for undergraduate students.
– All odd-numbered exercises have an answer.
– We more or less follow the topics of the book.
– More “advanced” materials may be added.
• Subjects on probability theory, algorithms, boolean
circuits, and information theory will be skipped as they
are covered by other classes.
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Class Information (concluded)
• More information and lecture notes (in PDF format) can
be found at
www.csie.ntu.edu.tw/~lyuu/dm.html
– Homeworks, exams, solutions and teaching assistants
will be announced there.
• Please ask many questions in class.
– The best way for me to remember you in a large
class.a
a“[A] science concentrator [...] said that in his eighth semester of
[Harvard] college, there was not a single science professor who could
identify him by name.” (New York Times, September 3, 2003.)
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 3
Grading
• Homeworks.
– Do not copy others’ homeworks.
– Do not give your homeworks for others to copy.
• Two to three exams.
• You must show up for the exams in person.
• If you cannot make it to an exam, please email me or a
TA beforehand (there should be a legitimate reason).
• Missing the final exam will automatically earn a “fail”
grade.
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Fundamental Principles of Counting
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And though the holes were rather small,
they had to count them all.
— The Beatles,
A Day in the Life (1967)
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 6
Counting
• Capable of solving difficult problems.
• Sometimes useful in reproving difficult theorems in
mathematics in an elementary way.
• Very useful in establishing the existence of solutions.
• Occasionally helps design efficient algorithms.
• Essential for the analysis of algorithms.
• Exact counts may not be necessary in many applications.
• Maybe the only method available to solve some open
problems in complexity theory.
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Founder of Combinatoricsa
• Archimedes (287BC–212BC) is the founder.b
• Archimedes in his Stomachion tried to find the number
of ways we put the 14 pieces together to make a square
(see next page).
• The answer was 17,152.
aGina Kolata, “In Archimedes’ Puzzle, a New Eureka Moment,” New
York Times, December 14, 2003.bReviel Netz (2003).
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Permutation and Combination
• n! = n · (n− 1) · · · 1.
• 0! = 1.
•C(n, k) ≡
(n
k
)=
n!
k! (n− k)!
for 0 ≤ k ≤ n.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 10
Permutations with Repetition
• Suppose there n distinct objects and r is an integer,
0 ≤ r.
• The number of permutations (linear arrangements of
these objects) of size r is
nr
when repetitions are allowed.
– There are n choices for the 1st position, n for the
2nd, etc.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 11
The Proof (concluded)
?
nchoices
2
?
nchoices
2
?nchoices
2
?
nchoices
2
· · ·
?
nchoices
2
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Palindromes
• A palindrome is a sequence of symbols that reads the
same left to right as right to left.
– For example, abccba (even length) and abcdcba
(odd length).
– For numbers, the leading digit should be nonzero
(015 is not allowed).
• A key observation: The first half mirrors the second half.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 13
Palindromic Decimal Numbers
Lemma 1 The number of palindromic decimal numbers of
even length n is 9× 10(n/2)−1.
• A palindromic decimal number of length n = 2k looks
like1st half︷ ︸︸ ︷
a2ka2k−1 · · · ak+1
mirror half︷ ︸︸ ︷ak+1 · · · a2k−1a2k,
where 0 ≤ ak+1, ak+2, . . . , a2k−1 ≤ 9 and 1 ≤ a2k ≤ 9.
• The count is therefore 9× 10k−1 = 9× 10(n/2)−1.
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Palindromic Decimal Numbers (continued)
Corollary 2 A palindromic decimal number of even length
is a multiple of 11.
• A palindromic decimal number of length n = 2k looks
like1st half︷ ︸︸ ︷
a2ka2k−1 · · · ak+1
mirror half︷ ︸︸ ︷ak+1 · · · a2k−1a2k .
• It equals
a2k102k−1 + a2k−110
2k−2 + · · ·+ ak+110k
+ak+110k−1 + · · ·+ a2k−110 + a2k
= a2k(102k−1 + 100) + a2k−1(10
2k−2 + 10) + · · ·
+ak+1(10k + 10k−1).
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 15
Palindromic Decimal Numbers (continued)
Lemma 3 The number of palindromic decimal numbers of
odd length n is 9× 10(n−1)/2.
• A palindromic decimal number of length n = 2k + 1
looks like︷ ︸︸ ︷a2k+1a2k · · · ak+2 ak+1
︷ ︸︸ ︷ak+2 · · · a2ka2k+1,
where 0 ≤ ak+1, ak+2, . . . , a2k ≤ 9 and 1 ≤ a2k+1 ≤ 9.
• The count is therefore 9× 10k = 9× 10(n−1)/2.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 16
Palindromic Decimal Numbers (concluded)
Combine the above two lemmas to obtain
Theorem 4 The number of palindromic decimal numbers of
length n is 9× 10⌊(n−1)/2⌋.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 17
Permutations
• Suppose there are n distinct objects and r is an integer,
1 ≤ r ≤ n.
• The number of permutations (linear arrangements of
these objects) of size r is
P (n, r) = n(n− 1) · · · (n− r + 1) =n!
(n− r)!.
• In particular,
P (n, n) = n!.
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Permutations with Repeated Objects
• Suppose there n objects with n1 of a first type, n2 of a
second type, . . ., and nr of an rth type, where∑
i ni = n
and 1 ≤ r ≤ n.
• Objects of the same type are indistinguishable.
• The number of permutations of these objects is
n!
n1!n2! · · ·nr!. (1)
– 10!4!3!1!1! = 25, 200 permutations of MASSASAUGA.
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Combinatorial Proofs
Lemma 5 (2k)!2k
is an integer.
• Consider 2k symbols x1, x1, x2, x2, . . . , xk, xk.
• The number of ways they can be permuted is
(2k)!
2! 2! · · · 2!=
(2k)!
2k.
• It must be an integer.
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Arrangement around a Circle
• Consider n distinct objects.
• Consider two circular arrangements to be equivalent if
one can be obtained from the other by rotation.
• The number of circular arrangements is
n!
n= (n− 1)!.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 21
A
B
C
D
D
A
B
C
C
D
A
B
B
C
D
A
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 22
Two Proofs
• First proof:
– For each linear arrangement, n− 1 others can be
rotated to give the same linear arrangement.
• Second proof:
– Fix say the first object to a particular position.
– The problem becomes that of permuting n− 1
distinct objects.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 23
Circular Arrangement with Restrictions
• There are 2 types of objects.
• Consider n/2 distinct objects of type one and n/2
distinct objects of type two.
• The number of circular arrangements where object types
alternate is
(n/2)! [(n/2)− 1]!.
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 24
The Proof
• Fixing say the first object of type one to a particular
position.
• Clockwise:
– There are n/2 ways to fill the next position with a
type-two object.
– There are (n/2)− 1 ways to fill the next position
with a type-one object.
– And so on.
• So the desired number is
(n/2)[(n/2)− 1][(n/2)− 1] · · · 1 · 1 = (n/2)! [(n/2)− 1]!.
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Combinations
• Suppose there n distinct objects and r is an integer,
1 ≤ r ≤ n.
• The number of combinations (selections without
reference to order) of r of these objects is
C(n, r) =
(n
r
)=
n(n− 1) · · · (n− r + 1)
r(r − 1) · · · 1.
– C(n, r) = P (n, r)/r! because order is irrelevant.
• C(n, 0) = 1 for all n ≥ 0.
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Combinations (concluded)
• It is easy to check that
C(n, r) = C(n, n− r)
for all n ≥ 0.
• We shall adopt the convention that(n
i
)= 0
for i < 0 or i > n, where n is a positive integer.
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Basic Properties of Combinations
• A finite sequence (a1, a2, . . . , an) of real numbers is
unimodal if there exists a positive integer 1 < j < n
such that
a1 < a2 < · · · < aj−1 ≤ aj > aj+1 > · · · > an.
• (C(n, 0), C(n, 1), . . . , C(n, n)) is unimodal.
– C(n, r + 1)/C(n, r) = (n− r)/(r + 1).
– C(n, n/2) is the maximum element when n is even.
– C(n, (n− 1)/2) and C(n, (n+ 1)/2) are the
maximum elements when n is odd.
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An Example
How many ways are there to arrange TALLAHASSEE with
no adjacent As?
• Rearrange the characters as AAAEEHLLSST.
• AAAEEHLLSST has 11 characters among which there
are 3 As.
• There are 8!2! 2! 2! 1! 1! = 5, 040 ways to arrange the 8
non-A characters.
• For each such arrangement, there are 9 places to insert
the 3 As:
2 T 2 A 2 A 2 E 2 E 2 H 2 L 2 L 2 S 2 S 2 A 2.
• The desired number is hence 5, 040×(93
)= 423, 360.
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Pascal’sa Identity
Lemma 6(n+1r
)=
(nr
)+(
nr−1
).
• Algebraic proof.
• Combinatorial proof.
• Generating-function proof (p. 427).
aBlaise Pascal (1623–1662).
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Newton’sa Identity
Lemma 7(nr
)(rk
)=
(nk
)(n−kr−k
).
• Here is a combinatorial proof.
• A university has n professors.
• The faculty assembly requires r professors.
• Among the members of the assembly, k serve the
executive committee.
• Let us count the number of ways the executive
committee can be formed.
aIsaac Newton (1643–1727).
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The Proof (continued)
• We can first form the assembly in(nr
)ways.
• Then we pick the executive committee members from
the assembly in(rk
)ways.
• The total count is(nr
)(rk
).
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The Proof (concluded)
• Alternatively, we can pick the executive committee first,
in(nk
)ways.
• Then we pick the remaining r − k members of the
assembly in(n−kr−k
)ways.
• The total count is(nk
)(n−kr−k
).
• As we count the same thing twice, they must be equal.
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Combinatorial Proof: Counting It Twice
Lemma 8 For m,n ≥ 0,∑n
k=m
(km
)=
(n+1m+1
).
• We want to pick m+1 tickets from a set of n+1 tickets.
• There are(n+1m+1
)ways.
• Alternatively, label the n+ 1 tickets from 0 to n.
• There are(km
)ways to do the selection when the ticket
with the largest number is k (m ≤ k ≤ n).
• Alternative proof: Apply Lemma 6 (p. 30) iteratively.
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Combinatorial Proof: Counting It Twice (continued)
Corollary 9∑m
k=1 k(k + 1) = 2(m+23
).
Note that
m∑k=1
k(k + 1) = 2
m∑k=1
(k + 1
2
)= 2
(m+ 2
3
)by Lemma 8 (p. 34).
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Combinatorial Proof: Counting It Twice (continued)
Corollary 10 For m,n ≥ 0,∑m
k=0
(n+kk
)=
(n+m+1
m
).
Note that
m∑k=0
(n+ k
k
)=
m∑k=0
(n+ k
n
)
=n+m∑k=n
(k
n
)=
(n+m+ 1
n+ 1
)by Lemma 8 (p. 34)
=
(n+m+ 1
m
).
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Combinatorial Proof: Counting It Twice (continued)
Corollary 11 1 + 2 + · · ·+ n = n(n+ 1)/2.
• Set m = 1 in Lemma 8 (p. 34) to obtain
n∑k=1
(k
1
)=
(n+ 1
2
).
• But this is equivalent to
n∑k=1
k = n(n+ 1)/2,
as desired.
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Combinatorial Proof: Counting It Twice (continued)
Lemma 12(m+n
2
)−
(m2
)−(n2
)= mn.
• Consider m men and n women.
• The number of heterosexual marriages is mn.
• On the other hand, there are(m+n
2
)ways to choose 2
persons.
• Among them,(m2
)+(n2
)are same-sex.
Corollary 13(2n2
)= n2 + 2
(n2
).
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Combinatorial Proof: Counting It Twice (continued)
Corollary 14 12 + 22 + · · ·+ n2 = n(n+ 1)(2n+ 1)/6.
• From Corollary 13 (p. 38),
n∑k=1
k2 =n∑
k=1
(2k
2
)− 2
n∑k=2
(k
2
)
=n∑
k=1
k(2k − 1)− 2
(n+ 1
3
)by Lemma 8 (p. 34).
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Combinatorial Proof: Counting It Twice (concluded)
• Son∑
k=1
k2 = 2n∑
k=1
k2 −n∑
k=1
k − 2
(n+ 1
3
).
• We conclude that
n∑k=1
k2 =n∑
k=1
k + 2
(n+ 1
3
)=
n(n+ 1)
2+
n(n+ 1)(n− 1)
3
=n(n+ 1)(2n+ 1)
6.
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Binomial Random Walk
• A particle starting at the origin can move right (up) or
left (down) in each step.
+
−
• It is a standard model for stock price movements called
the binomial option pricing model.a
aCox, Ross, and Rubinstein (1979).
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 41
Dynamics of the Binomial Random Walk
Lemma 15 The number of ways the particle can move from
the origin to position k in n steps is(n
n+k2
). (2)
(Assume that n+ k is even.)
• To reach position k, the number of up moves must
exceed the number of down moves by exactly k.
– UUDUUDUUD reaches position 3 in 9 steps as there
are 6 Us and 3 Ds.
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The Proof (concluded)
• Now
(n+ k)/2 + (n− k)/2 = n,
(n+ k)/2− (n− k)/2 = k.
• So the desired number equals the number of ways
(n+k)/2︷ ︸︸ ︷UU · · ·U
(n−k)/2︷ ︸︸ ︷DD · · ·D
can be permuted.
• The desired number is
n!
[(n+ k)/2]! [(n− k)/2]!=
(n
n+k2
).
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Probability of Reaching a Position
• Suppose the binomial random walk has a probability of
p of going up and 1− p of going down.
• The number of ways it is at position k after n steps is(n
n+k2
)by Eq. (2) on p. 42.
• The probability for this to happen is(n
n+k2
)p
n+k2 (1− p)
n−k2 .
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Probability of Reaching a Position (concluded)
• Alternatively, suppose a position is the result of i up
moves and n− i down moves.
– Clearly, the position is i− (n− i) = 2i− n.
• The number of ways of reaching it after n steps is(n
i
).
• The probability for this to happen is(n
i
)pi(1− p)n−i. (3)
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 45
Vandermonde’s Convolutiona(n
i
)=
k∑l=0
(k
l
)(n− k
i− l
).
• Let stateb (i, j) be the result of j up moves and i− j
down moves.
• Suppose the walk starts at state (0, 0) and ends at (n, i).
• There are(ni
)such walks.
• State (k, l) is on (k
l
)(n− k
i− l
)walks that reach (n, i), where 0 ≤ k ≤ n and 0 ≤ l ≤ k.
aAlexandre-Theophile Vandermonde (1735–1796).bNot position. State (i, j) corresponds to position (i, 2j − i).
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 46
Vandermonde’s Convolution (concluded)
• Every walk that reaches (n, i) must go through a state
(k, l) for some 0 ≤ l ≤ k.
• Add up those walks that go through state (k, l) over
0 ≤ l ≤ k to obtaina(n
i
)=
k∑l=0
(k
l
)(n− k
i− l
).
• Applications in artificial neural networks.b
aTechnically, the summation should be over 0 ≤ l ≤ min(k, i). But
recall that(ni
)= 0 for i < 0 or i > n, where n is a positive integer.
bBaum and Lyuu (1991) and Lyuu and Rivin (1992).
c⃝2012 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 47
(0,0) ( k , l )=(7,3)
( n , i )=(16,7)
The labels are for states, however, not positions.
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