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Chapter 3: Problem 16

What is the largest \(n\) for which one can solve within a day using an algorithm that requires \(f(n)\) bit operations, where each bit operation is carried out in \(10^{-11}\) seconds, with these functions \(f(n) ?\) $$ \begin{array}{llll}{\text { a) } \log n} & {\text { b) } 1000 n} & {\text { c) } n^{2}} \\ {\text { d) } 1000 n^{2}} & {\text { e) } n^{3}} & {\text { f) } 2^{n}} \\ {\text { g) } 2^{2 n}} & {\text { h) } 2^{2^{n}}}\end{array} $$

Short Answer

Expert verified
Largest values for each case: a) Extemely large, b) 8.64 \times 10^{12}, c) 9.3 \times 10^{7}, d) 9.3 \times 10^{6}, e) 4.4 \times 10^{5}, f) 53.1, g) 26.5, h) 5.7.

Step by step solution

01

- Understand the Total Time Allowed

A day has 24 hours. Converting this into seconds, we get \(24 \times 60 \times 60 = 86,400\) seconds.
02

- Calculate the Total Number of Bit Operations

Each second allows for \(10^{11}\) bit operations, hence in 86,400 seconds, the total number of bit operations that can be performed is \(86,400 \times 10^{11} = 8.64 \times 10^{15}\) bit operations.
03

- Evaluate Each Function

Next, carry out the following evaluations for each function to find the largest \(n\) such that \(f(n) \leq 8.64 \times 10^{15}\):
04

- Case of \(\log n\)

For \(f(n) = \log n\), solve \(\log n = 8.64 \times 10^{15}\). Thus, \(n = 2^{8.64 \times 10^{15}}\). This value is extremely large.
05

- Case of \(1000 n\)

For \(f(n) = 1000 n\), solve \(1000 n = 8.64 \times 10^{15}\). Hence, \(n = \frac{8.64 \times 10^{15}}{1000} = 8.64 \times 10^{12}\).
06

- Case of \(n^{2}\)

For \(f(n) = n^{2}\), solve \(n^{2} = 8.64 \times 10^{15}\). Thus, \(n = \sqrt{8.64 \times 10^{15}} \approx 9.3 \times 10^{7}\).
07

- Case of \(1000 n^{2}\)

For \(f(n) = 1000 n^{2}\), solve \(1000 n^{2} = 8.64 \times 10^{15}\). Thus, \(n = \sqrt{\frac{8.64 \times 10^{15}}{1000}} \approx 9.3 \times 10^{6}\).
08

- Case of \(n^{3}\)

For \(f(n) = n^{3}\), solve \(n^{3} = 8.64 \times 10^{15}\). Hence, \(n = \sqrt[3]{8.64 \times 10^{15}} = 4.4 \times 10^{5}\).
09

- Case of \(2^{n}\)

For \(f(n) = 2^{n}\), solve \(2^{n} = 8.64 \times 10^{15}\). Taking logarithm, \(n \approx \log_2(8.64 \times 10^{15})\), \(n \approx 53.1\).
10

- Case of \(2^{2n}\)

For \(f(n) = 2^{2n}\), solve \(2^{2n} = 8.64 \times 10^{15}\). \(2n = \log_2(8.64 \times 10^{15})\), \(2n = 53.1 \), \( n = \frac{53.1}{2} = 26.5\).
11

- Case of \(2^{2^{n}}\)

For \(f(n) = 2^{2^{n}}\), solve \(2^{2^{n}} = 8.64 \times 10^{15}\). \(2^{n} = \log_2(8.64 \times 10^{15})\), \(2^{n} = 53.1\), \(n = \log_2(53.1) \approx 5.7\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Time Complexity
Understanding time complexity is crucial when analyzing the efficiency of algorithms. It tells us how the runtime of an algorithm grows with the size of the input. For example, O(n) means that if the input size doubles, the time it takes to complete the algorithm also doubles. Different types of time complexities include O(1) (constant time), O(log n) (logarithmic time), O(n) (linear time), O(n log n) (linearithmic time), O(n虏) (quadratic time), and O(2^n) (exponential time). This can help you determine the practicality and feasibility of using a certain algorithm.
Bit Operations
Bit operations are fundamental to computer algorithms. They involve the manipulation of individual bits within a binary representation of data. Common bit operations include AND, OR, XOR, NOT, bit shifts, and bit masking. Each of these operations can be used to efficiently solve problems related to arithmetic operations, data compression, cryptography, and error detection. For instance, the solution to the exercise involves counting the number of bit operations that can be carried out in a given time frame, demonstrating practical use in performance analysis.
Exponential Algorithms
Exponential algorithms have the time complexity of the form O(2^n). These algorithms have runtimes that double with each additional input element. As a result, they become impractical very quickly as the input size grows. For example, in the exercise, we have exponentially increasing functions such as 2^n and 2^(2n). These functions grow so rapidly that even small input sizes can be infeasible to calculate within reasonable time limits. Understanding when an algorithm is exponential helps in finding more efficient alternatives if possible.
Logarithmic Functions
Algorithms that have a time complexity of O(log n) are generally very efficient, especially for large input sizes. Logarithmic functions grow very slowly, meaning they can handle much larger inputs compared to linear or polynomial algorithms. In the exercise, the function log n allows solving extremely large values of n within a single day, showcasing the efficiency of logarithmic time complexity. In real-world applications, logarithmic algorithms are used in binary search, tree operations, and some sorting algorithms like quicksort and mergesort.
Polynomial Time
Polynomial time algorithms have a time complexity of O(n^k), where k is a constant. These algorithms are considered efficient and feasible for most practical purposes. Polynomial time can range from linear (O(n)) to quadratic (O(n虏)) or cubic (O(n鲁)), as seen in the exercise. For instance, the function n虏 is a polynomial time function, and we can calculate the largest n it can handle in a day. Polynomial algorithms are common in everyday applications like sorting, searching, and graph algorithms. Understanding polynomial time helps in discerning between feasible and infeasible algorithmic solutions based on input size.

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Most popular questions from this chapter

The following problems deal with another type of asymptotic notation, called little-o notation. Because little-o notation is based on the concept of limits, a knowledge of calculus is needed for these problems. We say that \(f(x)\) is \(o(g(x))\) [read \(f(x)\) is "little-oh" of \(g(x) ]\) , when $$\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=0$$ (Requires calculus) Show that a) \(x^{2}\) is \(o\left(x^{3}\right)\) b) \(x \log x\) is \(o\left(x^{2}\right)\) c) \(x^{2}\) is \(o\left(2^{x}\right)\) d) \(x^{2}+x+1\) is not \(o\left(x^{2}\right)\)

Two strings are anagrams if each can be formed from the other string by rearranging its characters. Devise an algorithm to determine whether two strings are anagrams a) by first finding the frequency of each character that appears in the strings. b) by first sorting the characters in both strings.

How much time does an algorithm take to solve a problem of size \(n\) if this algorithm uses \(2 n^{2}+2^{n}\) operations, each requiring \(10^{-9}\) seconds, with these values of \(n ?\) $$ \begin{array}{lllll}{\text { a) } 10} & {\text { b) } 20} & {\text { c) } 50} & {\text { d) } 100}\end{array} $$

Describe an algorithm based on the linear search for determining the correct position in which to insert a new element in an already sorted list.

(Requires calculus) Let \(H_{n}\) be the \(n\) th harmonic number $$H_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$ Show that \(H_{n}\) is \(O(\log n)\) [Hint: First establish the inequality $$\sum_{j=2}^{n} \frac{1}{j}<\int_{1}^{n} \frac{1}{x} d x$$ by showing that the sum of the areas of the rectangles of height 1\(/ j\) with base from \(j-1\) to \(j,\) for \(j=2,3, \ldots, n,\) is less than the area under the curve \(y=1 / x\) from 2 to \(n .1\)
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