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RGB Jallenge

2016-07-09, post № 132

games, programming, Pygame, Python, #blue, #color, #colors, #colour, #colours, #green, #guess, #guessing, #red

This is a clone of The Great RGB Guessing Challenge [1]. The challenge works like this: You are presented three numbers ranging from 𝟢 to 𝟤𝟧𝟧 representing a rgb color and three color bubbles. To get a point you must choose the color bubble corresponding to the rgb values. The more points you get, the higher your score.

Controls

  • Click on the bubble to choose it.
rgb-jallenge-7.png
rgb-jallenge-9.png
rgb-jallenge-8.png
Source code: rgb-jallenge.py

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

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.
jimon-0.png
jimon-1.png
jimon-2.png
Source code: jimon.py
Jonathan Frech's blog; built 2021/04/16 21:21:49 CEST