Table of Contents
Recursion in Python: A Guide to Understanding
Recursive Function for Summing a List of Integers
Tail Call Recursion
Passing Around Index
Inner Function Version
Default Parameters Version
Recursive Power Problem
Home Backend Development Python Tutorial How to Implement Tail Call Recursion for Efficient Summation in Python?

How to Implement Tail Call Recursion for Efficient Summation in Python?

Oct 21, 2024 pm 12:11 PM

How to Implement Tail Call Recursion for Efficient Summation in Python?

Recursion in Python: A Guide to Understanding

Recursive Function for Summing a List of Integers

Let's suppose we need to create a recursive function called "listSum" that calculates the sum of all integers in a list. The goal is to do this without using any built-in functions. First, we should think about how we can express the result of the function in terms of itself.

In this case, we can obtain the result by adding the first number with the result of calling the same function with the rest of the elements in the list. Recursively, this can be expressed as:

listSum([1, 3, 4, 5, 6]) = 1 + listSum([3, 4, 5, 6])
                         = 1 + (3 + listSum([4, 5, 6]))
                         = 1 + (3 + (4 + listSum([5, 6])))
                         = 1 + (3 + (4 + (5 + listSum([6]))))
                         = 1 + (3 + (4 + (5 + (6 + listSum([])))))
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The base case for the recursion is when the list becomes empty, an event necessitating a result of 0. Implementing this approach in Python code:

<code class="python">def listSum(ls):
    if not ls:
        return 0
    return ls[0] + listSum(ls[1:])</code>
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Tail Call Recursion

The previous implementation depends on the value of the previous function call to calculate the actual result. This can be improved using Tail Call Recursion:

<code class="python">def listSum(ls, result):
    if not ls:
        return result
    return listSum(ls[1:], result + ls[0])</code>
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By introducing an additional parameter result, we accumulate the sum in it and return it when the base condition is met.

Passing Around Index

To avoid creating multiple intermediate lists, we can pass around the index of the item to be processed:

<code class="python">def listSum(ls, index, result):
    if index == len(ls):
        return result
    return listSum(ls, index + 1, result + ls[index])</code>
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The base condition checks if the index has reached the length of the list.

Inner Function Version

To simplify parameter passing, we can use an inner function:

<code class="python">def listSum(ls):
    def recursion(index, result):
        if index == len(ls):
            return result
        return recursion(index + 1, result + ls[index])
    return recursion(0, 0)</code>
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The inner function recursion accepts the index and result parameters, and listSum returns the result of calling the inner function with initial values.

Default Parameters Version

Using default parameters is a further simplification:

<code class="python">def listSum(ls, index=0, result=0):
    if index == len(ls):
        return result
    return listSum(ls, index + 1, result + ls[index])</code>
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Default values are assigned to index and result if not explicitly specified by the caller.

Recursive Power Problem

Consider the problem of calculating power(base, exponent), which returns the value of base raised to the power of exponent. Recursively, we can define the solution:

power(2, 5) = 2 * power(2, 4)
            = 2 * (2 * power(2, 3))
            = 2 * (2 * (2 * power(2, 2)))
            = 2 * (2 * (2 * (2 * power(2, 1))))
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The base condition is when exponent becomes 1 or less, in which case the result is the base itself:

            = 2 * (2 * (2 * (2 * 2)))
            = 2 * (2 * (2 * 4))
            = 2 * (2 * 8)
            = 2 * 16
            = 32
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Implementation in Python:

<code class="python">def power(base, exponent):
    if exponent <= 1:
        return base
    return base * power(base, exponent - 1)</code>
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Using default parameters for the Tail Call Optimized version:

<code class="python">def power(base, exponent, result=1):
    if exponent <= 0:
        return result
    return power(base, exponent - 1, result * base)</code>
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