/*! 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 55 A knight on a chessboard can mov... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 55

A knight on a chessboard can move one space horizontally (in either direction) and two spaces vertically (in either direction) or two spaces horizontally (in either direction) and one space vertically (in either direction). Suppose that we have an infinite chessboard, made up of all squares \((m, n),\) where \(m\) and \(n\) are nonnegative integers that denote the row number and the column number of the square, respectively. Use mathematical induction to show that a knight starting at \((0,0)\) can visit every square using a finite sequence of moves. [Hint: Use induction on the variable \(s=m+n . ]\)

Short Answer

Expert verified
By induction, a knight starting at (0,0) can visit every square on an infinite chessboard.

Step by step solution

01

Base Case

Show that the statement is true for the base case where the sum of row (m) and column (n) indices, s, equals 0. Here, the knight is already at (0,0), so the statement is trivially true.
02

Inductive Hypothesis

Assume that for some integer k, the knight can reach all squares where the sum of row and column indices is equal to k. In other words, assume that for all squares (m, n) such that m + n = k, the knight can visit them using a finite number of moves.
03

Inductive Step

Prove that the knight can also reach squares where the sum of row and column indices is k + 1. Consider any square (m, n) such that m + n = k + 1. The goal is to show that the knight can move to (m, n) starting from some square where the sum of indices is k.
04

Possible Moves and Verification

Analyze the knight's possible moves:- Move from (m-2, n+1) if m 鈮� 2.- Move from (m+2, n-1) if n 鈮� 1.- Move from (m-1, n+2) if m 鈮� 1.- Move from (m+1, n-2) if n 鈮� 2.Each of these starting points (m-2, n+1), (m+2, n-1), (m-1, n+2), and (m+1, n-2) have indices that sum to k. By the inductive hypothesis, the knight can reach these starting points in a finite number of moves.
05

Conclusion of Inductive Step

Therefore, since the knight can reach any square (m, n) with a sum of indices k+1 from a square with a sum of indices k, by induction the knight can reach any square on the infinite chessboard.

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Key Concepts

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

Inductive Hypothesis
The heart of our proof lies in the concept of the Inductive Hypothesis. To understand it better, let's break it down. An Inductive Hypothesis is an assumption used in the method of mathematical induction. We assume that a statement is true for a certain case, often denoted by some integer k. For the problem at hand, we assume that a knight can reach all squares where the sum of row numbers m and column numbers n equals k. In symbols, for all squares \((m, n)\) such that m + n = k, the knight can visit them using a finite number of moves. This assumption, albeit unproven for now, forms the basis of our inductive proof. By presuming this to be true, we can make the leap to proving that the knight can also reach squares where m + n = k + 1. This crucial step helps us inch closer to our final goal through methodical and logical advancing steps.
Base Case
Every induction proof starts with a foundation called the Base Case. This serves as the simplest scenario for our assumption. For our exercise, the base case is where the sum of row and column indices, s, equals 0. This translates to the knight being at the origin \(0, 0\). Since the knight is already at this position, it requires no moves to fulfill our condition. The base case is therefore trivially true. Establishing this basic truth is crucial because it sets the stage for applying our inductive hypothesis to more complex cases. Without a solid base case, the entire structure of our proof could falter. So, in essence, proving the simplest case as true is key to the success of our induction method.
Chessboard Moves
A knight in chess has a unique way of moving, and understanding these Chessboard Moves is essential to grasping our proof. The knight moves in an 鈥楲鈥� shape: either two squares in one direction and one square perpendicular, or vice versa. Specifically, for each position \(m, n\), a knight can:
  • Move from \(m-2, n+1\) if \(m 鈮� 2\)
  • Move from \(m+2, n-1\) if \(n 鈮� 1\)
  • Move from \(m-1, n+2\) if \(m 鈮� 1\)
  • Move from \(m+1, n-2\) if \(n 鈮� 2\)
Each of these starting points has row and column indices that sum to k. By the inductive hypothesis, the knight can reach these points in a finite number of moves. From any of these points, the knight can then make exactly one move to reach the target square \(m, n\) such that \(m + n = k + 1\). This step ties together our base case and inductive hypothesis, forming a chain of logical steps that demonstrate the knight鈥檚 ability to reach any square on an infinite chessboard.

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Most popular questions from this chapter

Suppose you begin with a pile of \(n\) stones and split this pile into \(n\) piles of one stone each by successively split- ting a pile of stones into two smaller piles. Each time you split a pile you multiply the number of stones in each of the two smaller piles you form, so that if these piles have \(r\) and \(s\) stones in them, respectively, you compute \(r s .\) Show that no matter how you split the piles, the sum of the products computed at each step equals \(n(n-1) / 2\)

Exercises \(49-51\) present incorrect proofs using mathematical induction. You will need to identify an error in reasoning in each exercise. What is wrong with this "proof"? "Theorem" For every positive integer \(n,\) if \(x\) and \(y\) are positive integers with max \((x, y)=n\) , then \(x=y\) . Basis Step: Suppose that \(n=1 .\) If \(\max (x, y)=1\) and \(x\) and \(y\) are positive integers, we have \(x=1\) and \(y=1\) . Inductive Step: Let \(k\) be a positive integer. Assume that whenever max \((x, y)=k\) and \(x\) and \(y\) are positive integers, then \(x=y .\) Now let max \((x, y)=k+1,\) where \(x\) and \(y\) are positive integers. Then \(\max (x-1, y-1)=k,\) so by the inductive hypothesis, \(x-1=y-1 .\) It follows that \(x=y\) completing the inductive step.

Give iterative and recursive algorithms for finding the \(n\) th term of the sequence defined by \(a_{0}=1, a_{1}=3, a_{2}=5\) and \(a_{n}=a_{n-1} \cdot a_{n-2}^{2} \cdot a_{n-3}^{3} .\) Which is more efficient?

Exercises 22 and 23 present examples that show inductive loading can be used to prove results in computational geometry. Let \(P(n)\) be the statement that when non intersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two vertices of the polygon are not endpoints of any of these diagonals. a) Show that when we attempt to prove \(P(n)\) for all integers \(n\) with \(n \geq 3\) using strong induction, the in- ductive step does not go through. b) Show that we can prove that \(P(n)\) is true for all inte- gers \(n\) with \(n \geq 3\) by proving by strong induction the stronger assertion \(Q(n),\) for \(n \geq 4,\) where \(Q(n)\) states that whenever nonintersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two nonadjacent vertices are not endpoints of any of these diagonals.

Use strong induction to show that all dominoes fall in an infinite arrangement of dominoes if you know that the first three dominoes fall, and that when a domino falls, the domino three farther down in the arrangement also falls.
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