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Chapter 5: Problem 37

Give a recursive definition of the reversal of a string. [Hint: First define the reversal of the empty string. Then write a string \(w\) of length \(n+1\) as \(x y,\) where \(x\) is a string of length \(n,\) and express the reversal of \(w\) in terms of \(x^{R}\) and \(y . \)]

Short Answer

Expert verified
Reverse the empty string as itself: \({R}(\varepsilon) = \varepsilon\). For a non-empty string \(w = xy\), reverse as \(w^{R} = y + x^{R}\).

Step by step solution

01

- Define the Reversal of the Empty String

The base case for the reversal of a string is the empty string. The reversal of an empty string is defined as the empty string itself: type: 'string'\textend{document}\({R}(\varepsilon) = \varepsilon\).
02

- Define the Reversal of a Non-Empty String

Consider a non-empty string \(w\) of length \(n+1\). This string can be split into two parts: a string \(x\) of length \(n\), and a single character \(y\). Thus, we can write \(w = xy\). The reversal of \(w\), denoted \(w^{R}\), can be expressed in terms of the reversal of \(x\) (denoted \(x^{R}\)) and \(y\): \textend{document}\(w^{R} = (xy)^{R} = y + x^{R}\).

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Key Concepts

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

base case
The base case is critical in recursion as it defines the simplest instance of the problem, which can be solved directly without further recursion.
In the context of reversing a string, the base case is the empty string. The reversal of an empty string is the empty string itself:
\textend{document}\[R(\varepsilon) = \varepsilon\]
We establish this base case because the empty string has no characters, so there's nothing to reverse. This step ensures that our recursive function has a stopping condition, preventing an infinite loop.
Remember:
  • Base case: The simplest, smallest problem instance.
  • Provides a direct answer: Here, the empty string reversed is still an empty string.
  • Stops further recursion: Ensures the function terminates.
string manipulation
String manipulation involves altering, combining, or analyzing strings in various ways.
In the problem of reversing a string, we need to split and reconstruct strings.

When handling the non-empty string of length>0, we consider breaking it into a smaller string and a character. For a given string \(w\) of length v+1, we define it as \(w = xy\), where \(x\) is a substring and \(y\) is the last character. The goal is to rearrange these components:
  • Split string \(w\): If \(w\) = 'hello鈥�, then \(x\) = 'hell' and \(y\) = 'o'
  • Reverse and concatenate: To reverse 'hello', reverse 'hell' and add 'o' at the beginning.
    You get \(o\) + \(R('hell')\)
This approach efficiently breaks down the problem, allowing the smaller problem (reversing the smaller string) to be solved first.
recursion
Recursion is a method where a problem is solved by solving smaller instances of the same problem.
It involves a function calling itself until reaching the base case.

For reversing a string recursively, let鈥檚 break it down:
  • Base case: When the string is empty, return the empty string.
  • Recursive case: For a string \(w\) of length \(n+1\), split it into \(x\) (length \(n\)) and \(y\), then reverse \(x\) and attach \(y\) at the front: \(R(w) = y + R(x)\),
    For example,\[R('hello') = 'o' + R('hell')\]
This approach works by solving each smaller sub-problem (reversing progressively shorter strings) until it hits the base case.
Key concepts of recursion:
  • Solve simpler instances first.
  • Combine results to solve the whole problem.
  • Use base case to stop recursion.
Recursion makes complex problems simpler and is essential for many algorithms.

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Most popular questions from this chapter

Suppose that \(P\) is a simple polygon with vertices \(v_{1}, v_{2}, \ldots, v_{n}\) listed so that consecutive vertices are con- nected by an edge, and \(v_{1}\) and \(v_{n}\) are connected by an edge. A vertex \(v_{i}\) is called an ear if the line segment connecting the two vertices adjacent to \(v_{i}\) is an interior diagonal of the simple polygon. Two ears \(v_{i}\) and \(v_{j}\) are called nonoverlapping if the interiors of the triangles with vertices \(v_{i}\) and its two adjacent vertices and \(v_{j}\) and its two adjacent vertices do not intersect. Prove that every simple polygon with at least four vertices has at least two nonoverlapping ears.

Let \(S\) be the set of positive integers defined by Basis step: 1\(\in S\) . Recursive step: If \(n \in S,\) then \(3 n+2 \in S\) and \(n^{2} \in S\) . a) Show that if \(n \in S,\) then \(n \equiv 1(\bmod 4)\) . b) Show that there exists an integer \(m \equiv 1(\bmod 4)\) that does not belong to \(S .\)

A jigsaw puzzle is put together by successively joining pieces that fit together into blocks. A move is made each time a piece is added to a block, or when two blocks are joined. Use strong induction to prove that no matter how the moves are carried out, exactly \(n-1\) moves are required to assemble a puzzle with \(n\) pieces.

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?

Suppose that \(P(n)\) is a propositional function. Determine for which positive integers \(n\) the statement \(P(n)\) must be true, and justify your answer, if a) \(P(1)\) is true; for all positive integers \(n,\) if \(P(n)\) is true, then \(P(n+2)\) is true. b) \(P(1)\) and \(P(2)\) are true; for all positive integers \(n,\) if \(P(n)\) and \(P(n+1)\) are true, then \(P(n+2)\) is true. c) \(P(1)\) is true; for all positive integers \(n,\) if \(P(n)\) is true, then \(P(2 n)\) is true. d) \(P(1)\) is true; for all positive integers \(n,\) if \(P(n)\) is true, then \(P(n+1)\) is true.
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