Download 98 16. (a) Proof. Assume first that f = O(g) and f = Ω(g). Since f = Ω(g

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16. (a) Proof. Assume first that f = O(g) and f = Ω(g). Since f = Ω(g), there exist a positive
constant C1 and a positive integer k1 such that f (n) ≥ C1 g(n) for every integer n ≥ k1 .
Since f = O(g), there exist a positive constant C2 and a positive integer k2 such that
f (n) ≤ C2 g(n) for every integer n ≥ k2 . Let k = max(k1 , k2 ). Then for every integer
n ≥ k, C1 g(n) ≤ f (n) ≤ C2 g(n). Therefore, f = Θ(g).
We now verify the converse. Assume that f = Θ(g). Then there are positive constants
C1 and C2 and a positive integer k such that C1 g(n) ≤ f (n) ≤ C2 g(n) for every integer
n ≥ k. Since C1 g(n) ≤ f (n) for every integer n ≥ k, it follows that f = Ω(g); while since
f (n) ≤ C2 g(n) for every integer n ≥ k, it follows that f = O(g).
(b) Proof. Assume first that f = Ω(g). Then there exist a positive constant C and a positive
integer k such that f (n) ≥ Cg(n) for every integer n ≥ k. This, however, implies that
g(n) ≤ C1 f (n) for every integer n ≥ k; so g = O(f ).
For the converse, assume that g = O(f ). Then there exist a positive constant C ′ and a
positive integer k ′ such that g(n) ≤ C ′ f (n) for every integer n ≥ k ′ . Hence f (n) ≥ C1′ g(n)
for every integer n ≥ k ′ . Therefore, f = Ω(g).
Exercises for Section 6.3. Analysis of Algorithms
1. Algorithm: Find a Number in a List s : a1 , a2 , . . . , an of n ≥ 3 Distinct Numbers That is
Neither the Maximum nor the Minimum Number in the List.
Input: An integer n ≥ 3 and a sequence s : a1 , a2 , . . . , an of n distinct numbers.
Output: A number y in s that is neither the maximum nor the minimum number in s.
1. x := a1
2. z := a2
3. if x > z then t := x, x := z and z := t
[If x > z, then the values of x and z are interchanged so that in every case, x < z.]
4. if x < a3 < z then y := a3
5. if a3 < x then y := x
6. if a3 > z then y := z
7. output y
Since there is one comparison in each of Steps 3, 5 and 6 and two comparisons in Step 4, the
time complexity of this algorithm (regardless of the value of n) is 5 = O(1).
2. Algorithm: Compute a1 a2 · · · an + a1 a2 · · · an−1 + · · · + a1 a2 + a1 for the Sequence s :
a1 , a2 , . . . , an of n Numbers
Input: A positive integer n and a sequence s : a1 , a2 , . . . , an of n numbers.
Output: s = a1 a2 · · · an + a1 a2 · · · an−1 + · · · + a1 a2 + a1 .
1. s := an
2. for i := 1 to n − 1 do
3.
s := an−i + an−i s
4. output s
Since there is one addition and one multiplication in Step 3 and this step is performed n − 1
times, the time complexity of this algorithm is 2(n − 1) = 2n − 2 = Θ(n).
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3. Algorithm: Compute (a1 + a2 + · · · + an )(a1 + a2 + · · · + an−1 ) · · · (a1 + a2 )a1 for the Sequence
s : a1 , a2 , . . . , an of n Numbers
Input: A positive integer n and a sequence s : a1 , a2 , . . . , an of n numbers.
Output: x = (a1 + a2 + · · · + an )(a1 + a2 + · · · + an−1 ) · · · (a1 + a2 )a1
1. i := 1
2. x := a1
3. s := a1
4. while i < n do
begin
i := i + 1
s := s + ai
x := sx
end
5. output x
Since there are n comparisons in Step 3, the time complexity of this algorithm is n = Θ(n).
4. Algorithm: Compute a1 a2 +a2 a3 +a3 a4 +· · ·+an−1 an +an a1 for the Sequence s : a1 , a2 , . . . , an
of n Numbers
Input: A positive integer n and a sequence s : a1 , a2 , . . . , an of n numbers.
Output: s = a1 a2 + a2 a3 + a3 a4 + · · · + an−1 an + an a1
1. s := 0
2. an+1 := a1
3. for i := 1 to n do
begin
4.
x := ai ai+1
s := x + s
end
4. output s
There is one multiplication in Step 4 and one addition in Step 5. Since each step is performed
n times, the time complexity of this algorithm is 2n = Θ(n).
5. Algorithm: Compute |a1 − a2 | + |a1 − a3 | + · · · + |a1 − an | + |a2 − a3 | + · · · + |an−1 − an | for
the Sequence s : a1 , a2 , . . . , an of n ≥ 2 Numbers
Input: An integer n ≥ 2 and a sequence s : a1 , a2 , . . . , an of n numbers.
Output: s = |a1 − a2 | + |a1 − a3 | + · · · + |a1 − an | + |a2 − a3 | + · · · + |an−1 − an |.
1. s := 0
2. for i := 1 to n − 1 do
begin
3.
for j := i + 1 to n do
begin
4.
x := |ai − aj |
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5.
s := s + x
end
end
6. output s
There is one subtraction in Step 4 and one addition in Step 5. These two steps are performed
n − i times for each value of i (1 ≤ i ≤ n − 1). Therefore, the time complexity of this algorithm
is
2(n − 1) + 2(n − 2) + · · · + 2 · 2 + 2 · 1
= 2[1 + 2 + · · · + (n − 1)]
n(n − 1)
= n2 − n = Θ(n2 ).
= 2
2
6. Algorithm: Compute a1 a22 a33 · · · ann for a Sequence s : a1 , a2 , . . . , an of n ≥ 2 Numbers
Input: An integer n ≥ 2 and a sequence s : a1 , a2 , . . . , an of n numbers.
Output: y = a1 a22 a33 · · · ann .
1. x := an
2. y := an
3. for i := 1 to n − 1 do
begin
4.
x := an−i x
y := xy
end
5. output y
There is one multiplication in Step 4 and one multiplication in Step 5. These steps are
performed n−1 times and so the time complexity of this algorithm is 2(n−1) = 2n−2 = Θ(n).
7. (a) Algorithm: Compute an For a Real Number a and a Nonnegative Integer n.
Input: A real number a and a nonnegative integer n.
Output: p = an
1. p := 1
2. for i := 1 to n do
3. p := ap
4. output y
(b) There is one multiplication in Step 3. Since Step 3 is performed n times, the time
complexity of this algorithm is n = Θ(n).
8. (a) Algorithm: Compute f (a) = cn an + cn−1 an−1 + · · · + c1 a + c0 for Real Numbers
cn , cn−1 , . . . , c1 , c0 , a, Where an , an−1 , . . ., a1 , a0 Are Computed by the Algorithm in
Exercise 7.
Input: A positive integer n, a real number a and sequences s : c0 , c1 , . . . , cn−1 , cn and
s′ : a0 , a1 , . . . , an , where the numbers in s′ are computed using the algorithm in Exercise 7.
Output: s = f (a)
1. s := 0
2. for i := 0 to n do
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3. s := ci ai + s
4. output s
(b) The algorithm in Exercise 7 has the time complexity i to compute ai for 1 ≤ i ≤ n. The
time complexity to compute all of a0 , a1 , . . . , an is therefore 1 + 2 + · · · + n = n(n + 1)/2.
Step 3 in the algorithm in (a) has one addition and one multiplication. Since Step 3 is
performed n+1 times, the time complexity of the algorithm in (a) is n(n+1)/2+2(n+1) =
(n + 1)(n + 4)/2 = Θ(n2 ).
9. (a) First, we write Horner’s algorithm.
Input: A positive integer n and real numbers cn , cn−1 , . . . , c1 , c0 , a.
Output: s = f (a)
1. s := cn
2. for i := 1 to n do
3. s := sa + cn−i
4. output s
There is one multiplication, one addition and one subtraction in Step 3. Since this step
is performed n times, the time complexity of this algorithm is 3n = Θ(n).
(b) Horner’s algorithm is more efficient than the algorithm in Exercise 8.
Exercises for Section 6.4. Searching and Sorting
1. Solution. In this case, k = 17, n = 9 and a1 = 3, a2 = 6, a3 = 7, a4 = 8, a5 = 15, a6 = 17,
a7 = 19, a8 = 23, a9 = 24. By Steps 1 and 2, a = 1 and b = 9. Since 1 = a ≤ b = 9, Steps 4–8
are executed. By Step 4, p = 5. Among Steps 5–7, only a5 < k is true and so only the
conclusion of Step 7 is performed, where a is assigned the value 5 + 1 = 6. The subsequence
a6 , a7 , a8 , a9 is now considered. Since 6 = a ̸= b + 1 = 10, the conclusion of Step 8 is not
performed. Since 6 = a ≤ b = 9, we go through the while loop at Step 3 again. In Step 4, we
have p = 7. Among Steps 5–7, only a7 > k is true in Step 6 and the conclusion of Step 6 is
performed. So b is assigned the value p − 1 = 6. Since 6 = a ̸= b + 1 = 7, we go through the
while loop at Step 3 again. In Step 4, we have p = 6. Since a6 = k, the following is output: 17
is in position 6. Furthermore, a is assigned the number b + 2 = 8. Since none of the conditions
in Steps 6–8 are satisfied, we return to the while loop at Step 3. However, 8 = a ≤ b = 6 is
false and the algorithm ends.
!
a=1
a1
a2
a3
a4
a5
a6
a7
a8
a9
k = 17
b=9
3
6
7
8
15
17
19
23
24
n=9
... ..
......
.
a=6
b=9
a6
17
a7
a8
19
23
a9
24
... ...
......
.
a=6
a6
b=6
17
“17 is in position 9” is output
Figure 30: Applying the Binary Search Algorithm to Exercise 1