/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 In a popular computer game the c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a popular computer game the computer picks an integer from 1 to \(n\) at random. The player is given \(k\) chances to guess the number. After each guess the computer responds "correct," "too small," or "too big." (a) Show that if \(n \leq 2^{k}-1,\) then there is a strategy that guarantees you will correctly guess the number in \(k\) tries. (b) Show that if \(n \geq 2^{k}-1,\) there is a strategy that assures you of identifying one of \(2^{k}-1\) numbers and hence gives a probability of \(\left(2^{k}-1\right) / n\) of winning. Why is this an optimal strategy? Illustrate your result in terms of the case \(n=9\) and \(k=3\).

Short Answer

Expert verified
(a) Use binary search for \(n \leq 2^k - 1\) numbers. (b) With \(n = 9, k = 3\), distinguish \(2^k - 1 = 7\) numbers, so win probability is \(\frac{7}{9}\). Strategy is optimal by minimizing guess attempts.

Step by step solution

01

Define the Problem

To guarantee the identification of a randomly chosen integer by the computer from 1 to \(n\) in \(k\) tries, we need to assess if \(n\) numbers can be uniquely identified by \(k\) guesses. This is achievable if \(n\) is less than or equal to \(2^k - 1\) via binary search logic.
02

Use Binary Search Strategy

The binary search strategy is efficient in finding numbers within a range. In binary search, each guess narrows down the possible numbers by half. For instance, with \(k = 3\), you start by guessing the middle number. Each response determines whether to search above or below the middle, guaranteeing a find within \(2^k - 1\) numbers.
03

Analyze the Case (a) with \(n \leq 2^k - 1\)

Since each guess can logically split the options in half, for \(k\) guesses, you can identify up to \(2^k - 1\) numbers. For example, if \(k = 3\), then \(2^3 - 1 = 7\). Thus, if \(n \leq 7\), we can guarantee identifying the number in 3 or fewer guesses.
04

Illustrate the Case (b) with \(n = 9\) and \(k = 3\)

If \(n = 9\) and \(k = 3\), \(2^3 - 1 = 7\) implies a probability of identifying one of 7 numbers out of 9. With binary search, you can identify the number among 7 possibilities fully, but 2 numbers from 1 to 9 will remain indistinguishable after 3 guesses. Thus, probability of winning is \(\frac{7}{9}\), as explained by partitioning the possible numbers into 3 groups: less than midpoint, at midpoint, greater than midpoint.
05

Conclusion on Optimality

The search strategy is optimal as it minimizes the number of guesses by maximum division of the search space, thereby efficiently determining the number within the constraints of attempts. Given that each guess narrows the field, no fewer than \(k\) guesses will cover the most possibilities efficiently, providing the highest winning probability by maximizing coverage of possibilities.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability
Probability is a measure that quantifies the likelihood of a certain event happening. In the context of the game described in the exercise, probability plays a vital role in determining the chances of successfully guessing the correct number chosen by the computer. The success probability is calculated based on how many outcomes can be accurately guessed given the changes you have.

In our exercise, the player has a set number of guesses, or "chances," to find the randomly chosen integer out of the possible numbers from 1 to \(n\). If \(n\) is greater than \(2^k - 1\), the total possible numbers exceed the range that can be covered efficiently, reducing the likelihood of a successful guess. The probability of winning, when \(n = 9\) and \(k = 3\), for instance, is \(\frac{7}{9}\), as only 7 out of the 9 numbers can be discerned with certainty with 3 guesses. This fraction is found by recognizing that, although 9 numbers exist, optimal guessing only distinguishes 7 groups vividly before running out of chances.

Understanding probability ensures players are aware of their chances and can plan their strategy accordingly, maximizing their ability to predict the correct number.
Game Theory
Game theory is the study of mathematical models of strategic interaction among rational decision-makers. It is fundamentally about understanding choices and the strategies behind those choices based on a rational understanding of other players' choices.

In this guessing game, the player is engaged in a kind of strategic battle against the computer's random choice of number. The player aims to optimize their guesses based on the feedback they receive with each attempt. Game theory elements appear here, where the player must use the responses "too small" or "too big" to adjust their strategy and account for each possible outcome.

When you apply game theory, you want to minimize your potential losses while maximizing your potential gains, even under uncertainty. For example, with \(n = 9\) and \(k = 3\), the strategy involves choosing guesses that effectively halve the range based on logical partitioning of the numbers. This strategic thinking exemplifies game theory principles, where players weigh each decision’s risks and rewards. An efficient strategy acknowledges every feedback, and this systematic approach reduces remaining possibilities in the smartest way possible.
Optimal Search
Optimal search is about finding the most efficient way to locate an answer or result, usually minimizing resources like time or attempts. In binary search, the optimal strategy involves repeatedly dividing the search interval in half, allowing the player to narrow down possible numbers quickly and efficiently.

The structure of the game ensures that the player's strategy is optimal if they always aim to cut the number of remaining possibilities in half. In the scenario where \(n \leq 2^k - 1\), such as \(n = 7\) with \(k = 3\), the player can be assured of finding the correct number by methodically reducing the range of options each time. Each guess gives a directive of either narrowing down the range above or below the previous median choice, which was strategically picked to maximize elimination of unneeded choices.

For scenarios where \(n \geq 2^k - 1\), like \(n = 9\) when \(k = 3\), the player can still apply this optimal search strategy, but with the knowledge that not all numbers can be precisely distinguished. Here, the player's aim is to cover as much ground as possible in those 3 guesses, ensuring they skim through the largest feasible set of possibilities effectively, achieving a fair chance of winning by maximizing their guessing power.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A coin is tossed three times. Let \(X\) be the number of heads that turn up. Find \(V(X)\) and \(D(X)\)

Assume that the lifetime of a diesel engine part is a random variable \(X\) with density \(f_{X}\). When the part wears out, it is replaced by another with the same density. Let \(N(t)\) be the number of parts that are used in time \(t\). We want to study the random variable \(N(t) / t\). Since parts are replaced on the average every \(E(X)\) time units, we expect about \(t / E(X)\) parts to be used in time \(t\). That is, we expect that $$\lim _{t \rightarrow \infty} E\left(\frac{N(t)}{t}\right)=\frac{1}{E(X)}$$ This result is correct but quite difficult to prove. Write a program that will allow you to specify the density \(f_{X},\) and the time \(t,\) and simulate this experiment to find \(N(t) / t\). Have your program repeat the experiment 500 times and plot a bar graph for the random outcomes of \(N(t) / t\). From this data, estimate \(E(N(t) / t)\) and compare this with \(1 / E(X) .\) In particular, do this for \(t=100\) with the following two densities: (a) \(f_{X}=e^{-t}\) (b) \(f_{X}=t e^{-t}\)

Let \(X\) be a random variable that takes on nonnegative values and has distribution function \(F(x) .\) Show that $$E(X)=\int_{0}^{\infty}(1-F(x)) d x$$ Hint: Integrate by parts. Illustrate this result by calculating \(E(X)\) by this method if \(X\) has an exponential distribution \(F(x)=1-e^{-\lambda x}\) for \(x \geq 0,\) and \(F(x)=0\) otherwise.

A multiple choice exam is given. A problem has four possible answers, and exactly one answer is correct. The student is allowed to choose a subset of the four possible answers as his answer. If his chosen subset contains the correct answer, the student receives three points, but he loses one point for each wrong answer in his chosen subset. Show that if he just guesses a subset uniformly and randomly his expected score is zero.

(from Pittel \(^{17}\) ) Telephone books, \(n\) in number, are kept in a stack. The probability that the book numbered \(i\) (where \(1 \leq i \leq n\) ) is consulted for a given phone call is \(p_{i}>0,\) where the \(p_{i}\) 's sum to 1. After a book is used, it is placed at the top of the stack. Assume that the calls are independent and evenly spaced, and that the system has been employed indefinitely far into the past. Let \(d_{i}\) be the average depth of book \(i\) in the stack. Show that \(d_{i} \leq d_{j}\) whenever \(p_{i} \geq p_{j}\). Thus, on the average, the more popular books have a tendency to be closer to the top of the stack. Hint: Let \(p_{i j}\) denote the probability that book \(i\) is above book \(j\). Show that \(p_{i j}=p_{i j}\left(1-p_{j}\right)+p_{j i} p_{i}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.