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Consider the following recursive function: void recrun (int x) \(f\)\\[\begin{array}{l}\text { If }(x>0) \\\\\text { cout }

Short Answer

Expert verified
258, 25, 2; 7; 36, 3; (no output)

Step by step solution

01

Understand the Recursive Function

The function 'recrun' takes an integer 'x' as input. If 'x' is greater than 0, it outputs the value of 'x' if it is not 10 and then recursively calls itself with 'x' divided by 10. If 'x' is not 0 (as an 'else if' condition), it outputs 'x'. If both conditions fail, which means 'x' is 0, the function simply ends.
02

Analyze recrun(258)

Start by evaluating if '258' is greater than 0, which it is. So, the function outputs '258,' and calls itself with 'x/10', which is '25'. For '25', it is also greater than 0, so it outputs '25,' again calls itself with '2'. Finally, the function outputs '2,' and calls recrun(0), which results in stopping the recursion.
03

Analyze recrun(7)

Since '7' is greater than 0, the function outputs '7,' and then calls itself with '7/10', which is 0.7, rounded down to '0', stopping the recursion due to the base case.
04

Analyze recrun(36)

Start with '36', which is greater than 0, so it outputs '36,' before calling itself with '3'. '3' is also greater than 0, thus it outputs '3,' then calls recrun(0), which halts further recursion.
05

Analyze recrun(-85)

The input '-85' is neither greater than 0 nor not 0 in terms of positive integer division by 10's conditions, so the function ends with no output as the conditions of the recursive function are not met.

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Key Concepts

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

Integer Division
In mathematics and programming, integer division is a type of division in which the result is an integer, discarding any remainder. When this operation is performed, you divide two integers to get the quotient without the fractional part. Let's break it down with the recursion function 'recrun': when it calls itself with x divided by 10, it effectively reduces the number by removing its last digit. This is because each division operation by 10 shifts the decimal point left (in a decimal system) which eliminates the least significant digit.

For example, with recrun(258), dividing by 10 gives:
  • 258/10 = 25 (integer division discards the 0.8)
  • 25/10 = 2
  • 2/10 = 0
As the number reduces by each recursion, reaching eventually zero, integer division is crucial in controlling the recursion depth.
Base Case
The base case is an essential part of recursion that defines the condition under which the recursive function stops calling itself. It's like setting a ground rule to prevent endless repetition. Within 'recrun', the base case is reached when 'x' becomes zero.

When analyzing the recursion, the sequence where 'x' is repeatedly divided by 10 eventually leads to 'x = 0'.
This acts as a stop signal for further recursive calls and assures that the recursion concludes. A base case ensures that every recursive function has an end, preventing it from going infinitely and causing a stack overflow.
Recursion Termination
Recursion termination happens when a function doesn’t call itself anymore. This is usually achieved when the base case condition is met. In the function 'recrun', recursion termination is evident when 'x' is zero after successively dividing by 10.

Here’s how it works:
  • The function keeps calling itself until 'x' equals zero.
  • When 'x' becomes zero, the recursion stops as no further calls are needed.
This careful structuring ensures that the code safely exits the recursion loop, returning control back to the original call. Without proper recursion termination, a function could crash due to excessive memory use or an endless loop.
Conditional Statements
Conditional statements are the backbone of decision-making in programming. They allow a program to choose between different paths based on certain conditions. In 'recrun', several conditional statements are employed to control logic flow.

Key parts to consider:
  • The first `if` condition checks if 'x' is greater than 0. If true, it prints 'x' and recurses with `x/10`.
  • The `else if` handles the case when 'x' is not zero and serves additional logic, but 'x' being zero naturally terminates recursion.
These conditions ensure that 'recrun' behaves correctly under varying circumstances, dictating how and when the recursion continues or ends effectively. Using well-placed conditional statements helps in achieving desired program outcomes with precision.

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Most popular questions from this chapter

Suppose that intArray is an array of integers, and length specifies the number of elements in intarray. Also, suppose that 1 ow and high are two integers such that \(0<=1\) ow \(<\) length, \(0<=\) high \(<\) length, and 1 ow \(<\) high. That is, low and high are two indices in intarray. Write a recursive definition that reverses the elements in intArray between low and high.

What is indirect recursion?

Consider the following function: int func (int x)\\{\\[\text { if } \quad(x=0)\\]return 2 else if \((x=1)\) return 3 else return \((\text { func }(x-1)+\text { func }(x-2))\)} What is the output of the following statements? a. \(\quad\) cout \(\langle\langle\text { func }(0)<<\text { end } 1\) b. \(\quad\) cout \(\langle\langle\text { func }(1)<<\text { end } 1\) c. cout \(<<\) func \((2)<<\) endl d. \(\quad\) cout \(<\langle\text { func }(5)<<\text { end } 1\)

Consider the following recursive function: int mystery (int number) / / Line 1 \(f\) If (number = 0) / / Line 2 return number; else return (mystery (number +1 ) \(-\) number) \(;\) //Line 5\\} a. Identify the base case. b. Identify the general case. c. What valid values can be passed as parameters to the function mystery? d. If mystery (0) is a valid call, what is its value? If not, explain why. e. If mystery (10) is a valid call, what is its value? If not, explain why. f. If mystery (-3) is a valid call, what is its value? If not, explain why.

Consider the following problem: How many ways can a committee of four people be selected from a group of 10 people? There are many other similar problems in which you are asked to find the number of ways to select a set of items from a given set of items. The general problem can be stated as follows: Find the number of ways \(r\) different things can be chosen from a set of \(n\) items, in which \(r\) and \(n\) are nonnegative integers and \(r \leq n .\) Suppose \(C(n, r)\) denotes the number of ways \(r\) different things can be chosen from a set of \(n\) items. Then, \(C(n, r)\) is given by the following formula: \(C(n, r)=\frac{n !}{r !(n-r) !}\) in which the exclamation point denotes the factorial function. Moreover, \(C(n, 0)=C(n, n)=1 .\) It is also known that \(C(n, r)=C(n-1, r-1)+C(n-1, r)\) a. Write a recursive algorithm to determine \(C(n, r)\). Identify the base case \((\mathrm{s})\) and the general case \((\mathrm{s})\) b. Using your recursive algorithm, determine \(C(5,3)\) and \(C(9,4)\)

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