/*! 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 197 Starting with an empty \(1 \time... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 197

Starting with an empty \(1 \times n\) board (a row of \(n\) squares), we successively place \(1 \times 2\) dominoes to cover two adjacent squares. At each stage, the placement of the new domino is chosen at random, with all available pairs of adjacent empty squares being equally likely. The process continues until no further dominoes can be placed. Find the limit, as \(n \rightarrow \infty\), of the expected fraction of the board that is covered when the process ends.

Short Answer

Expert verified
The limit, as approaches infinity, of the expected fraction of the board that is covered when the process ends is 1/2.

Step by step solution

01

Understand the Domino Placement

Each domino covers 2 adjacent squares. Initially, the board has all squares empty, and at each stage, a randomly chosen pair of adjacent empty squares is covered by a domino.
02

Define Key Variables

Let the total number of squares be represented by . Since each domino covers exactly 2 squares, at most /2 dominoes can be placed if all squares are covered.
03

Expected Fraction Covered

To find the expected fraction of the board that is covered, we need to consider the number of remaining squares after no further dominoes can be placed. At each stage, the available pair of adjacent empty squares is chosen randomly.
04

Identify Patterns From Simplified Cases

Analyze smaller cases to identify a pattern. For an empty row of 2 squares, the probability all squares are covered is 1. For 4 squares, there are some configurations where it might be impossible to cover all squares. Continue checking larger even-sized boards to identify the trend.
05

Use Probabilistic Argument

On a very large board, most attempts to place the tiles should result in high efficiency covering about half of the even-sized segments of the board.
06

Consider Limit as Approaches Infinity

As approaches infinity, the proportion of uncovered squares resulting from random gaps between tile placements becomes negligible due to the large number of potential placements.
07

Derive Expected Fraction

Therefore, the large-scale average fraction of the board that can be covered by dominos approaches /2 divided by . Thus, the fraction of the board that ends up covered asymptotically goes towards 1/2 as approaches infinity.

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 91影视!

Key Concepts

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

random processes
Understanding random processes is crucial in this exercise as it involves placing dominoes randomly on a board. A random process is a sequence of random events, where the outcome of each event is unpredictable and follows a certain probability distribution.

Here, at each step, a domino is placed on a randomly chosen pair of adjacent empty squares. Each placement is independent of the previous ones but depends on the configuration of empty squares on the board. This randomness adds to the complexity of predicting the exact number of covered squares upfront.

As the process is repeated, the configurations and choices fluctuate, which is typical for a random process. By examining this randomness, we can start to make sense of the expected outcomes, such as the proportion of the board that will be covered.
probabilistic analysis
Probabilistic analysis helps us predict outcomes in random processes by using the principles of probability. In this exercise, we use it to estimate the expected fraction of the board covered by dominoes.

We start by understanding simpler cases. For a 1x2 board, the probability of fully covering it is 1 since one domino can always cover both squares. For a 1x4 board, several configurations could either allow full coverage or leave some squares uncovered based on how the dominoes are placed. By analyzing such smaller cases, we start identifying patterns of coverage.

By extending this analysis to larger boards and considering the randomness of each placement, we can estimate the efficiency of coverage. The probabilistic argument suggests that, as the board size increases, the average fraction covered tends to stabilize at a certain value. This stabilization is rooted in the law of large numbers, which in layman's terms states that outcomes will tend to their expected values given a large enough number of trials.
asymptotic behavior
Asymptotic behavior examines how functions behave as inputs become very large. When dealing with an infinite (or very large) board, understanding this concept helps us determine the long-term behavior of the domino coverage.

In this exercise, we analyze the limit of the expected fraction of the board covered as the number of squares, represented by n, approaches infinity. Through our probabilistic analysis, we infer that, due to the randomness and the increasing number of configurations, the gaps left by domino placements become less significant.

The key result is that the proportion of covered squares converges to a stable value. Specifically, as n grows large, the expected fraction of the board that is covered by dominoes approaches 1/2. This is because, in the metaphorically infinite setup, half of the board (in terms of segments or large blocks) is efficiently covered, leaving a negligible fraction uncovered due to optimization over the large scale.

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

Suppose \(f\) is a continuous, increasing, bounded, real-valued function, defined on \([0, \infty)\), such that \(f(0)=0\) and \(f^{\prime}(0)\) exists. Show that there exists \(b>0\) for which the volume obtained by rotating the area under \(f\) from 0 to \(b\) about the \(x\)-axis is half that of the cylinder obtained by rotating \(y=f(b)\), \(0 \leq x \leq b\), about the \(x\)-axis.

Note that if we tile the plane with black and white squares in a regular "checkerboard" pattern, then every square has an equal number of black and of white neighbors (four each), where two squares are considered neighbors if they are not the same but they have at least one common point. If we try the analogous pattern of cubes in 3-space, it no longer works this way: every white cube has 14 black neighbors and only 12 white neighbors, and vice versa. a. Show that there is a different color pattern of black and white "grid" cubes in 3-space for which every cube does have exactly 13 neighbors of each color. b. What happens in \(n\)-space for \(n>3\) ? Is it still possible to find a color pattern for a regular grid of "hypercubes" so that every hypercube, whether black or white, has an equal number of black and white neighbors? If so, show why; if not, give an example of a specific \(n\) for which it is impossible.

Let \(x_0\) be a rational number, and let \(\left(x_n\right)_{n \geq 0}\) be the sequence defined recursively by $$ x_{n+1}=\left|\frac{2 x_n^3}{3 x_n^2-4}\right| $$ Prove that this sequence converges, and find its limit as a function of \(x_0\).

The proprietor of the Wohascum Puzzle, Game, and Computer Den has invented a new two-person game, in which players take turns coloring edges of a cube. Three colors (red, green, and yellow) are available. The cube starts off with all edges uncolored; once an edge is colored, it cannot be colored again. Two edges with a common vertex are not allowed to have the same color. The last player to be able to color an edge wins the game. a. Given best play on both sides, should the first or the second player win? What is the winning strategy? b. There are twelve edges in all, so a game can last at most twelve turns (whether or not the players use optimal strategies); it is not hard to see that twelve turns are possible. How many twelve-turn end positions are essentially different? (Two positions are considered essentially the same if one can be obtained from the other by rotating the cube.) \(\quad\)

Suppose we are given an \(m\)-gon (polygon with \(m\) sides, and including the interior for our purposes) and an \(n\)-gon in the plane. Consider their intersection; assume this intersection is itself a polygon (other possibilities would include the intersection being empty or consisting of a line segment). a. If the \(m\)-gon and the \(n\)-gon are convex, what is the maximal number of sides their intersection can have? b. Is the result from (a) still correct if only one of the polygons is assumed to be convex? (Note: A subset of the plane is convex if for every two points of the subset, every point of the line segment between them is also in the subset. In particular, a polygon is convex if each of its interior angles is less than \(\left.180^{\circ}.\right)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
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.