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Palindrome Function

2016-07-02, post № 131

mathematics, programming, Python, #p(n), #sum

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])
4321

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=\lfloor 3.09\rfloor+1=4

Secondly, I need a way to isolate a specific digit. Using the floor function, this function returns the 𝑖-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