Exercises from Chapter 2
                  
2.2
2.6
2.10
2.17
2.22
2.23
2.27

                                   Solutions

2.2 (a) For any a, a+a = 2a is even, so the relation is reflexive.
a+b = b+a, so the relation is symmetric. If a+b and b+c are even, then
(a+b) + (b+c) = a+c+2b is even, which implies a+c is even, so the
relation is transitive.

(b) Not an equivalence relation because it is not reflexive: a+a is not odd.

(c) Equivalence relation.  You can check the details.

(d)  Not an equivalence relation because it is not symmetric.

(e) For any a, a-a=0, which is an integer, so the relation is reflexive.
If a-b is an integer, then since b-a = -(a-b), it is also an integer, so
the relation is symmetric.
If a-b and b-c are integers, then since a-c = a-b + b-c, it
is also an integer, and the relation is transitive.
Therefor the relation is an equivalence relation.

(f) The relation is not an equivalence relation because it is not
transitive. Here is a counterexample showing it is not transitive:
a=3, b=1, c=0. Then |a-b|<=2 and |b-c|<=2, but |a-c|>2.

2.6 void clear();
    void insert(int);
    void remove(int);
    void sizeof();
    bool isEmpty();
    // return count of integers with a given value
    int countInBag(int);
    // add the integers in another bag to this bag
    // If an integer occurs a times in this bag and b times in the other
    // bag, then it will occur a+b times after performing the operation
    void union(bag);
    // intersect this bag with another bag.
    // If an integer occurs a times in this bag and b times in the other
    // bag, then it will occur min(a,b) times after performing the operation
    void intersect(bag);
    // take the integers in another bag from this bag.
    // If an integer occurs a times in this bag and b times in the other
    // bag, then it will occur max(a-b,0) times after performing the operation
    void difference(bag);

2.10 (a) Probably fibr, unless you don't understand recursion.

(b) It can be shown that the running time of fibr is 2#(Omega(n)),
while the running time of fibi is Omega(n), which is much faster.
(2#k means 2 to the power k). It's not hard to show the running time
of fibi is Omega(n).

To estimate the running time of fibr, let S(n) be its running time.
Then S(1) = S(2) = 1, and
     S(n) = S(n-1) + S(n-2) + 1 for n > 2.
(Strictly speaking, I should use constants instead of the 1's,
but since we are trying to get an estimate, that's OK.) Then, using
induction on n, we can show that
        S(n) >= 2#(n/2 - 1).
(use strong induction with two base cases: n = 1 and n = 2.)
Therefore S(n) = 2#(Omega(n)).

2.17 The first sum is just 1+2+...+n.
The second sum is (n-1+1)+(n-2+1)+...+(n-n+1).
The third sum is (n-0)+(n-1)+...+(n-n+1).
Obviously they are all the same.

2.22 (a) Assume 1 + 3 + ... + (2n-1) = n#2.
The sum of the first n even numbers is
2 + 4 + ... + 2n = (1+1) + (3+1) + ... + (2n-1+1)
                 = 1 + 3 + ... + (2n-1) + 1 + 1 + ... + 1 (n times)
                 = n#2 + n

(b) Base case: n = 1. The sum of the first even number is 2.
1#2 + 1 = 2, so the claim is true for n = 1.

Induction step: assume the sum of the first n even numbers is n#2 + n.
The sum of the first n+1 even numbers is
2 + 4 + ... + 2n + (2n+2) = n#2 + n + (2n+2) by the induction assumption
                          = n#2 + 2n + 1 + n+1
                          = (n+1)#2 + (n+1)

2.23 Use induction on n.
Base cases: n=1: Fib(1) = 1, and (5/3)#1 = 5/3 > 1.
n=2: Fib(2) =1 < (5/3)#2.

Induction step: Assume Fib(m) < (5/3)#m for all m<n, where n>2.  We will prove
Fib(n) < (5/3)#n.
By definition, Fib(n) = Fib(n-1) + Fibn(n-2)
                      < (5/3)#(n-1) + (5/3)#(n-2) by induction assumption
                      = (5/3)#(n-2)(5/3 + 1)
                      = (5/3)#(n-2)8/3
                      < (5/3)#n
since 8/3 < 25/9 = (5/3)#2.

2.27 Base case: n=1: T(1) = 1 by defintion, and 1(1+1)/2 = 1.

Induction step: assume T(n-1) = (n-1)n/2. To prove T(n) = n(n+1)/2,
T(n) = T(n-1) + n by definition
     = (n-1)n/2 + n by induction assumption
     = n(n+1)/2.