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).

$p(1234)=\sum\limits_{i=0}^{\lfloor\log_{10}{1234}\rfloor}\Big[\big(\lfloor\frac{1234}{10^i}\rfloor-\lfloor\frac{1234}{10^{i+1}}\rfloor\cdot 10\big)\cdot 10^{\lfloor\log_{10}{1234}\rfloor-i}\Big]$

$p(1234)=\sum\limits_{i=0}^{3}\Big[\big(\lfloor\frac{1234}{10^i}\rfloor-\lfloor\frac{1234}{10^{i+1}}\rfloor\cdot 10\big)\cdot 10^{3-i}\Big]$

$p(1234)=d_0(1234)\cdot 10^3+d_1(1234)\cdot 10^2+d_2(1234)\cdot 10^1+d_3(1234)\cdot 10^0$

Jimon

2016-06-25, post № 130

games, programming, Pygame, Python, #color, #color memory, #memeory, #memory game, #remember, #sequence, #Simon, #Simon Says

This game is a recreation of the famous game Simon. In the game there are four colors which form a sequence that is expanding every cycle. The aim of the game is to memorize said sequence as far as possible.For more information on the Simon game visit this Wikipedia entry.

Controls

Click on the colored buttons to press them.

Source code: jimon.py

Numerals

2016-06-18, post № 129

programming, Python, #number words, #numbers, #numbers to words, #numeral, #words

This program takes a number and calculate it’s linguistic numeral.The number 𝟥 returns the numeral ‘three’, 𝟧𝟪 returns ‘fifty-eight’ and 𝟥𝟣𝟦𝟣𝟧.𝟫𝟤𝟨𝟧𝟥𝟧 returns ‘thirty-one thousand four hundred fifteen point nine two six five three five’.

Source code: numerals.py