Disclaimer: You should try to solve all problems by yourself. The solutions provided in these slides should serve only as hints and not as a study guide.

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Disclaimer You should try to solve all problems
by yourself. The solutions provided in these
slides should serve only as hints and not as a
study guide.
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Problem 1 f(x) is O ((x2) log(x2 1) ) . For
sufficiently large x, (x2) lt 4x (4 is chosen
arbitrarily, i.e., it can be 5 or 6.) and x21 lt
x4. That is, for sufficiently large x, f(x) lt
K ( 4x log x4 ) 16K (x log x), for some
constant K. That is, f(x) O(x log x).
Problem 2 Compute f( Ai ) for i from 1 to m.
You need to come up with an algorithm that can
determine if the list f (A1), f(A2), f(Am)
contains duplicated elements. There is a O(m2)
algorithm (the one we worked out in class).
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Problem 4 3100
Problem 5 Use induction. (K-1)2K 1 (K1) 2K
2K 2K 1 K 2K1 1
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Problem 6 There are 2100 subsets. There is 1
subset with zero element, 100 subsets with one
element and (10099)/2 5099 4950 subsets
with two elements 2100 101 4950 2100 -
5051
Problem 7 (a) 210 1024. (b) C(10, 2)45.
(c) C(10, 0) C(10, 1) C(10, 2) C(10, 3)
1 10 45 120. (d) There are exactly 5 heads
and 5 tails. The number of C(10, 5).
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Problem 8 count the numbers of committees with
4, 5, 6 women (or 2, 1, 0 men) (2 men) C(10,
2)xC(15, 4) (1 man) C(10, 1)xC(15, 5) (0 man)
C(15, 6). Add these numbers together and thats
the answer.
Problem 9 There are 8 pairs of 01 and two free
1s. - - 01 - - 01 - - 01 - - 01 - - 01 - - 01
- - 01 - - 01 - - There are 9 ways to put the
two free 1s next to each other ( 11 ) and there
are C(9, 2) 36 ways to put these two free 1s so
that they are NOT next to each other. Total is
369 45.
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Problem10 Each computer can connect to either 1,
2, 3, 4, or 5 computers. Use pigeonhole
principle, there are five boxes and six objects
a computer is put into box K ( K1, 2, 3, 4, 5)
if it is connected to K other computers. The
result follows.
Problem11 The hint (see the textbook page 354)
is to use the result that for a group of 10
people, there are either three mutual friends or
four mutual enemies, and there are either three
mutual enemies or four mutual friends. Now we are
given 20 people. Take one person A from the
group. A either (1) has at least 10 friends or
(2) at least 10 enemies. Case (1) we can find 4
mutual enemies among these 10 friends of A (we
are done) or we can find 3 mutual friends among
these 10 friends of A and this gives us 4 mutual
friends (adding A). Case (2) Similarly, among
the 10 enemies of A, we can find four mutual
friends (we are done) or we can find 3 mutual
enemies and this gives us 4 mutual enemies
(adding A).
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Problem12 Clearly, you can apply the pigeonhole
principle to solve the problem. We will define
50 boxes B500, B501, ., B549 For each street
address x, we can put x into box Ba, where a
floor(x/2). There are 50 boxes and 51 objects.
One box must have more than 1 street address
Problem15 There are n variables. Therefore, in
every truth table, we have 2n True/False entries
to fill. In this way, the number is the same as
the number of outcomes of flipping a coin 2n
times, which is 2(2n).
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Problem 13 The solution is quite simple. Lets
do a simple example (the general case is
straightforward). Take n 6.
u, v
Given any irrational number x, and j some
positive integer not exceeding n, the difference
between jx and the nearest integer to jx must
fall into the above six intervals (why?) If this
difference falls into interval C or D, we are
done (because its absolute value will be less
than 1/n1/6). So suppose the difference never
lands on interval C or D. For each positive
integer h not exceeding n, we place h into the
interval A, B, E, F depending on the difference
between hx and the nearest integer to hx. There
are 4 intervals and we have 6 numbers. By the
pigeonhole principle, one interval will have at
least two numbers, say a and b with a gt b (with
A, B integers and u, v the differences belonging
to the same interval u- v lt 1/6.) ax A
u, bx B v (a-b) x (A-B) u-v
a-b is an integer not exceeding 6 and
the absolute value of (a-b)x and its
nearest integer is lt 1/6 !
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Problem 14
From the problem statement, we are almost certain
that the problem can be solved using generalized
pigeonhole principle. We have 101 objects
(heights) and lets have 10 boxes, B1, B2 B10.
And we need to define how to put the 101 objects
into these 10 boxes. If we can do this, by the
pigeonhole principle, one box will have at least
11 objects and this will solve the problem. Lets
suppose that we cant find 11 people standing in
the line with decreasing height (this will allow
us to use 10 boxes to put all objects in).
First, the value of each box is initialized to
minus infinity for each object (the height of
the person in line) x, we go from the first box
to the last one. If the value of Box N is
smaller than x, we will put x into Box N and
increase the value of the Box N to x. Take a few
seconds to process the definition above. Notice
that each box contains an increasing sequence of
heights and if one box contains at least 11
objects (heights), then we are done. What is
left to show is that all objects an be put into
these 10 boxes (we need to use the assumption
that there is no 11 decreasing heights). Now, if
an x got put into Box N ( 1 lt N lt10), there
is a sequence of N decreasing heights (make sure
you see this). Because of the assumption, we
know that N can never be more than 10, hence all
objects can be put into one of the 10 boxes.
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Problem 16 The tree could be quite tedious to
draw. However, Example 20 on page 343 (5.1) is
very similar to this problem and make sure that
you know this example well.
Problem17 The total number of arrangements is
4!24. The number of arrangements with a followed
immediately by b is 3!. So the answer is 24-6
18.
Problem19 If Sum(n) is the algorithm that find
the sum of first n positive integers. Then the
recursive algorithm is Procedure Sum ( n
) if n 1 Sum(1) 1 else Sum(n)
n Sum(n-1). Remember that you need to define
the base case (n1) and the recursion (else
statement).
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Problem 20 Let L be a list of integers and L(n)
is the nth integer in the list L. Procedure Max
( L ) if length(L) 1 Max (L) L(1) else
if L(n) gt Max( L( 1n-1) ) Max(L) L(n)
else Max(L) Max (L(1n-1))
In the above, L(1n-1) is the sublist of L that
contains the first n-1 elements in L.
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Problem 21 The string has equal number of 0s and
1s. The string has 2n bits for some positive
integer n the first n bits are 0s followed by n
bits of 1s.
Problem 22 . All integers except 3 and 1.
Strong induction will go like this. Basis case
4 22 and 55. Suppose any integer 3 lt n
lt K can be written as a sum of 2s and 5s.
For K1, we know that K1 (K-1) 2 and K-1 can
be written as a sum of 2s and 5s. It follows
that K1 can be written as a sum of 2s and 5s as
well.
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Problem 23 The strong induction will go like
this The base case 1 20 and 2 21.
Now suppose the result is true for all integers
less than or equal to K. We need to show that
the result is also true for K1. If K1 is
odd then K is even and by induction hypothesis
it can be written as a sum of powers of 2
(without 20 term). Then we can writhe K1 as a
sum of powers of 2 by adding 20 to the sum for K.
If K1 is even then (K1)/2 is an
integer so according to the induction hypothesis,
it is a sum of powers of 2. From this, we can
that K1 is also a sum of powers of 2 by
increasing the powers in the sum for (K1)/2 by 1.
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Problem 24 You should be able to figure out
whats wrong with this proof. Note that the
statement of this theorem is outrageously
incorrect.
Problem 25 This is a simple induction. Check
the basis step and notice that
n(n1)(n2)(n3)/4 (n1)(n2)(n3)
(n1)(n2)(n3)(n4)/4 This equality will
complete the inductive step.
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Problem 26 Check the basis step when n1, 1
h lt (1h)n1 The inductive step assumes
that the above inequality is valid for all
integer 1 lt n lt K (note that because h gt -1,
1 h gt 0 ). 1 K h lt (1
h)K From this, we have (because 1 h gt 0 )
(1 h) (1 K h) lt (1 h)k1 The left
hand side is 1 (K1) h K h2 which is
greater than 1 (K1)h because K h2 is
positive. Putting these together, we have 1
(K1) h lt (1 h)k1
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Problem 27 You should be able to solve this
problem without difficulty. Problem 28 You can
ignore this problem. Problem 29
Stratightforward.
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Problem 30 The main idea is to show that ( n! )2
gt n n. If we can show this, then the result
follows easily by taking the log both sides 2
(log n! ) gt n log n, i.e. n log n is O( log
n!). Notice that ( n ! ) 2 n x (n-1) x (n-2)
xx 2 x 1 x
1 x 2 x 3 x x n-1 x n and each
vertical pair has product gt n ( 2 (n-1) gtn and
3(n-2) gt n, etc.) From this, we have that (
n ! ) 2 gt n n
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Problem 32 The previous problem shows that log
n! is O(n log n). To show that log n! is ?(n
log n), you need to show that log n! is also O(n
log n). This is quite straightforward and I will
leave it to you.
Problems 33 and 34. Make sure that you know how
to do these problems. Weve gone over several of
them in class before.