3.5.1: Recursive function: Writing the base case. Add an if branch to complete double_pennies()'s base case.Sample output with inputs: 1 10Number of pennies after 10 days: 1024# Returns number of pennies if pennies are doubled num_days timesdef double_pennies(num_pennies, num_days): total_pennies = 0 if "solution goes here" else: total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) return total_pennies# Program computes pennies if you have 1 penny today,# 2 pennies after one day, 4 after two days, and so onstarting_pennies = int(input())user_days = int(input())print('Number of pennies after', user_days, 'days: ', end="")print(double_pennies(starting_pennies, user_days)) ### CS 219T: Python Programming#### Section 3.5: Creating Recursion---### 3.5.1: Recursive Function: Writing the Base Case**Challenge Activity**Add an `if` branch to complete the `double_pennies()` base case.**Sample Output with Inputs: 1 10**```Number of pennies after 10 days: 1024```**Note:** If the submitted code has an infinite loop, the system will stop running the code after a few seconds and report "Program end never reached." The system doesn't print the test case that caused the reported message.---**Code:**```python# Returns number of pennies if pennies are doubled num_days timesdef double_pennies(num_pennies, num_days): total_pennies = 0 if ((num_pennies <= 0) or (num_days == 0)): total_pennies = num_pennies else: total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) return total_pennies# Program computes pennies if you have 1 penny today, # 2 pennies after one day, 4 after two days, and so onstarting_pennies = int(input())user_days = int(input())print('Number of pennies after', user_days, 'days: ', end='')print(double_pennies(starting_pennies, user_days))```**Explanation:**- This Python program calculates the number of pennies if the amount is doubled each day for a specified number of days.- The `double_pennies()` function uses recursion to double the pennies. It includes a base case that stops the recursion when the number of days is 0, returning the current number of pennies.- The function is tested with user inputs for the starting number of pennies and the number of days.**Graphical Representation:***(Note: Since the presented image doesn't contain any graphs or diagrams, this section is not applicable.)* ## Python Recursion: Doubling Pennies### Problem StatementThe goal is to create a Python function that returns the number of pennies if the pennies are doubled after a certain number of days. Specifically, starting with a given number of pennies, the pennies are doubled every day for a specified number of days.### Sample Output```Number of pennies after 10 days: 1024```**Note:**- If the submitted code has an infinite loop, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message.### Python Code```python# Returns number of pennies if pennies are doubled num_days timesdef double_pennies(num_pennies, num_days): total_pennies = 0 # Base case, if no days are left, return the current count of pennies if num_days == 0: total_pennies = num_pennies # Recursive case, double the pennies and reduce the number of days else: total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) return total_pennies# Program computes pennies if you have 1 penny today,# 2 pennies after one day, 4 after two days, and so on.starting_pennies = int(input())user_days = int(input())print('Number of pennies after', user_days, 'days: ', end='')print(double_pennies(starting_pennies, user_days))```### Explanation1. **Function Definition**: - The `double_pennies` function takes two arguments: `num_pennies` and `num_days`. - A base case is defined where if `num_days` is 0, the function returns the current count of pennies (`num_pennies`). - A recursive case is defined to call the function again, doubling the number of pennies and decrementing the number of days by 1.2. **Main Program**: - The user inputs the starting number of pennies and the number of days to double the pennies. - The `double_pennies` function is called with the user-provided inputs. - The result is printed, showing the number of pennies after the specified number of days.### Error

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3.5.1: Recursive function: Writing the base case.

Add an if branch to complete double_pennies()'s base case.

Sample output with inputs: 1 10

Number of pennies after 10 days: 1024

# Returns number of pennies if pennies are doubled num_days times
def double_pennies(num_pennies, num_days):
total_pennies = 0

if "solution goes here"

else:
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))

return total_pennies

# Program computes pennies if you have 1 penny today,
# 2 pennies after one day, 4 after two days, and so on
starting_pennies = int(input())
user_days = int(input())

print('Number of pennies after', user_days, 'days: ', end="")
print(double_pennies(starting_pennies, user_days))

### CS 219T: Python Programming

#### Section 3.5: Creating Recursion

---

### 3.5.1: Recursive Function: Writing the Base Case

**Challenge Activity**

Add an `if` branch to complete the `double_pennies()` base case.

**Sample Output with Inputs: 1 10**

```
Number of pennies after 10 days: 1024
```

**Note:**
If the submitted code has an infinite loop, the system will stop running the code after a few seconds and report "Program end never reached." The system doesn't print the test case that caused the reported message.

---

**Code:**

```python
# Returns number of pennies if pennies are doubled num_days times
def double_pennies(num_pennies, num_days):
total_pennies = 0

if ((num_pennies <= 0) or (num_days == 0)):
total_pennies = num_pennies
else:
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))

return total_pennies

# Program computes pennies if you have 1 penny today,
# 2 pennies after one day, 4 after two days, and so on
starting_pennies = int(input())
user_days = int(input())

print('Number of pennies after', user_days, 'days: ', end='')
print(double_pennies(starting_pennies, user_days))
```

**Explanation:**
- This Python program calculates the number of pennies if the amount is doubled each day for a specified number of days.
- The `double_pennies()` function uses recursion to double the pennies. It includes a base case that stops the recursion when the number of days is 0, returning the current number of pennies.
- The function is tested with user inputs for the starting number of pennies and the number of days.

**Graphical Representation:**

*(Note: Since the presented image doesn't contain any graphs or diagrams, this section is not applicable.)*

## Python Recursion: Doubling Pennies

### Problem Statement
The goal is to create a Python function that returns the number of pennies if the pennies are doubled after a certain number of days. Specifically, starting with a given number of pennies, the pennies are doubled every day for a specified number of days.

### Sample Output
```
Number of pennies after 10 days: 1024
```

**Note:**
- If the submitted code has an infinite loop, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message.

### Python Code
```python
# Returns number of pennies if pennies are doubled num_days times
def double_pennies(num_pennies, num_days):
total_pennies = 0

# Base case, if no days are left, return the current count of pennies
if num_days == 0:
total_pennies = num_pennies

# Recursive case, double the pennies and reduce the number of days
else:
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))

return total_pennies

# Program computes pennies if you have 1 penny today,
# 2 pennies after one day, 4 after two days, and so on.
starting_pennies = int(input())
user_days = int(input())

print('Number of pennies after', user_days, 'days: ', end='')
print(double_pennies(starting_pennies, user_days))
```

### Explanation
1. **Function Definition**:
- The `double_pennies` function takes two arguments: `num_pennies` and `num_days`.
- A base case is defined where if `num_days` is 0, the function returns the current count of pennies (`num_pennies`).
- A recursive case is defined to call the function again, doubling the number of pennies and decrementing the number of days by 1.

2. **Main Program**:
- The user inputs the starting number of pennies and the number of days to double the pennies.
- The `double_pennies` function is called with the user-provided inputs.
- The result is printed, showing the number of pennies after the specified number of days.

### Error

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After working this code, this is the error code that popped up resulted in half right. Traceback (most recent call last): File "main.py", line 292, in <module> print(double_pennies(starting_pennies, user_days)) File "main.py", line 264, in double_pennies total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) File "main.py", line 264, in double_pennies total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) File "main.py", line 264, in double_pennies total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) [Previous line repeated 995 more times] File "main.py", line 260, in double_pennies if num_days == 1: RecursionError: maximum recursion depth exceeded in comparison

### Creating Recursion for Doubling Pennies

#### Objective
To teach how to create a recursive function that computes the number of pennies when they are doubled for a given number of days.

#### Description

Below is a Python code snippet that demonstrates a recursive function called `double_pennies`. This function returns the number of pennies if they are doubled a specified number of times (`num_days`).

```python
# Returns number of pennies if pennies are doubled num_days times
def double_pennies(num_pennies, num_days):
total_pennies = 0
if num_days == 1:
return num_pennies * 2
else:
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))

return total_pennies

# Program computes pennies if you have 1 penny today,
# 2 pennies after one day, 4 after two days, and so on
starting_pennies = int(input())
user_days = int(input())

print('Number of pennies after', user_days, 'days: ', end="")
print(double_pennies(starting_pennies, user_days))
```

#### Explanation

1. **double_pennies function:**
- **Parameters:**
- `num_pennies`: Initial number of pennies.
- `num_days`: Number of days the pennies are doubled.
- **Logic:**
- If `num_days` is 1, it returns the number of pennies doubled (`num_pennies * 2`).
- Otherwise, it calls itself recursively with `num_pennies` doubled and `num_days` decremented by 1.
- **Return:**
- Returns the total number of pennies after doubling for the specified duration.

2. **Main Program:**
- **Inputs:**
- `starting_pennies`: Initial number of pennies entered by the user.
- `user_days`: Number of days entered by the user.
- **Output:**
- Displays the number of pennies after the given days using the `double_pennies` function.

#### Diagram

- **Flowchart:**

1. Start
2. Input `starting_pennies` and `user_days`
3. Call `double_pennies` with `starting_pennies` and `user_days

**Recursive Function Example with Pennies Doubling**

**Testing with inputs: 1 10**
- **Your output:** Number of pennies after 10 days: 1024

**Testing with inputs: 1 40**
- **Your output:** Number of pennies after 40 days: 1099511627776

**Testing with inputs: 1 1**
- **Your output:** Number of pennies after 1 day: 2

**Test aborted**
- **Error Details:**
```
File "main.py", line 292, in <module>
print(double_pennies(starting_pennies, user_days))
File "main.py", line 264, in double_pennies
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))
File "main.py", line 264, in double_pennies
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))
File "main.py", line 264, in double_pennies
total_pennies = double_pennies((num_pennies * 2), (num_days - 1))
...
[Previous line repeated 995 more times]
File "main.py", line 260, in double_pennies
```

The error log shows that the program encountered a "recursion depth exceeded" error. This suggests that the recursive function `double_pennies` called itself too many times without reaching a base case, causing the program to exceed the maximum recursion depth allowed in Python.

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