Palindrome Function

To get a number’s palindrome in a programming language like python is easy. There are ways to swap between integer and string and strings can be manipulated.

>>> n = 1234
>>> int(str(n)[::-1])

But I wanted to create a mathematical function p(n), which returns an integer’s palindrome. Thus p(1234) = 4321.

Firstly I needed a way of determining the number’s size. In base 10 the length is calculated using the logarithm to said base.
l(n) = \lfloor \log_{10}{n} \rfloor + 1
l(1234) = \lfloor \log_{10}{1234} \rfloor = \lfloor3.09 \rfloor + 1 = 4

Secondly I need a way to isolate a specific digit. Using the floor function, this function returns the i\text{-th} digit (starting on the right with i=0).
d_i(n) = \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10
d_2(1234) = \lfloor \frac{1234}{10^2} \rfloor - \lfloor \frac{1234}{10^{2+1}} \rfloor \cdot 10 = \lfloor 12.34 \rfloor - \lfloor 1.23 \rfloor \cdot 10 = 12 - 1 \cdot 10 = 2

Thirdly both of these functions can be used to split up the number into a sum.
n = \sum\limits_{i=0}^{l(n)-1} \Big[ d_i(n) \cdot 10^{i} \Big] = \sum\limits_{i=0}^{\lfloor \log_{10}{n} \rfloor} \Big[ \big( \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{i} \Big]

Fourthly I only need to swap the power of ten at the end to get my palindrome function.
p(n) = \sum\limits_{i=0}^{l(n)-1} \Big[ d_i(n) \cdot 10^{l(n) - 1 - i} \Big] = \sum\limits_{i=0}^{\lfloor \log_{10}{n} \rfloor} \Big[ \big( \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{\lfloor \log_{10}{n} \rfloor - i} \Big]

Thus the final function p(n) is defined.
p(n) = \sum\limits_{i=0}^{\lfloor \log_{10}{n} \rfloor} \Big[ \big( \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{\lfloor \log_{10}{n} \rfloor - i} \Big]

To check if the formula is correct, I use 1234 (as seen above).
p(1234) = \sum\limits_{i=0}^{\lfloor \log_{10}{1234} \rfloor} \Big[ \big( \lfloor \frac{1234}{10^i} \rfloor - \lfloor \frac{1234}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{\lfloor \log_{10}{1234} \rfloor - i} \Big]
p(1234) = \sum\limits_{i=0}^{3} \Big[ \big( \lfloor \frac{1234}{10^i} \rfloor - \lfloor \frac{1234}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{3 - i} \Big]
p(1234) = d_0(1234) \cdot 10^3 + d_1(1234) \cdot 10^2 + d_2(1234) \cdot 10^1 + d_3(1234) \cdot 10^0
p(1234) = 4000 + 300 + 20 + 1 = 4321

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