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Sudoku Solver

2016-08-06, post № 136

programming, Python, #666, #evil, #logic, #solve, #solving, #Web Sudoku

This program solves a given Sudoku.
It uses a simple strategy, looking at each box and determining those numbers that are not in its row, column and square. If that list has length 𝟣, the box’s number is determined.
After going through each box, the program generates child Sudokus in which the first empty box get filled with one of the possible numbers for that box. Sudokus with an error get eliminated.
Using this simple but effective strategy this program can solve hard Sudokus in under a second. As an example, I used Evil Puzzle 666 from Web Sudoku.
For more information on Sudokus, visit this Wikipedia entry.

sudoku-solver-1.png
Source code: sudoku-solver.py

JClock VIII

2016-07-30, post № 135

programming, Pygame, Python, #clock, #complex number, #hour, #minute, #plane, #time

Interpreting the hour hand on a clock as a two-dimensional object on a plane, the hand’s tip can be seen as a complex number.
This clock converts the hour hand’s position into a complex number, sets the number’s length to the current minutes and displays it in the form a+b\cdot i.
The angle 𝜑 is determined by the hours passed (\frac{2\cdot\pi\cdot\text{hour}}{12}=\frac{\pi\cdot\text{hour}}{6}) but has to be slightly modified because a complex number starts at the horizontal axis and turns anti-clockwise whilst an hour hand starts at the vertical axis and turns — as the name implies — clockwise. Thus, \varphi=(2\cdot\pi-\frac{\pi\cdot\text{hour}}{6})+\frac{\pi}{2}=(\frac{15-\text{hour}}{6})\cdot\pi.

The complex number’s length is simply determined by the minutes passed. Because the length must not be equal to 𝟢, I simply add 𝟣: |z|=k=\text{minute}+1.
Lastly, to convert a complex number of the form k\cdot e^{\varphi\cdot i} into the form a+b\cdot i, I use the formula k\cdot(\cos{\varphi}+\sin{\varphi}\cdot i)=a+b\cdot i.

jclock-viii_17h04m.png
Source code: jclock-viii.py

Weekday

2016-07-23, post № 134

programming, Python, #calculation, #date, #day, #month, #time, #year

Determining the weekday based on a date composed of day, month and year.
The program counts up all the days from the 1st of January 1 to the given date, divides it by 𝟩, looks at the remainder and returns the weekday.

weekday.png
Source code: weekday.py

Triangular Squares

2016-07-16, post № 133

programming, Python, Wolfram Language, #equation, #number, #number theory, #numbers, #OEIS, #square, #triangle, #triangles

In a recent video, Matt Parker showed a triangular number that also is a square number, 𝟨, and asked if there were more.

A triangular number has the form \frac{n^2+n}{2} — shown by Euler — and a square number has the form m^2.
Triangular squares are those numbers for which \frac{n^2+n}{2}=m^2 with n,m\in\mathbb{N}.
Examples are \{0,1,6,35,204,1189,6930,\dots\} (sequence A001109 in OEIS).

To check if triangular numbers are square numbers is easy (code listed below), but a mathematical function would be nicer.
The first thing I tried was to define the triangular number’s square root as a whole number, \sqrt{\frac{n^2+n}{2}}=\lfloor\sqrt{\frac{n^2+n}{2}}\rfloor. This function does not return the square numbers that are triangular but the triangular numbers that are square.
The resulting sequence is \{0,1,8,49,288,1681,9800,\dots\} (sequence A001108 in OEIS).

Source code: triangular-squares.py

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

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’.

numerals.png
Source code: numerals.py

Leaf

2016-06-11, post № 128

art, haiku, poetry, #anim gif, #animated, #animated gif, #cliff, #fall, #foliage, #gif

A leaf on a tree
gets blown away by the wind
and falls down a cliff.

leaf.gif

Cycloids

2016-06-04, post № 127

mathematics, programming, Pygame, Python, #circle, #circles, #curve, #cycloid, #epi, #epicycloid, #geometry, #hypo, #hypocycloid, #rolling, #trace

A cycloid is the curve generated by tracing a point on a smaller circle rolling around an bigger circle.
The word cycloid comes from the greek word “κύκλος” meaning “circle”. There are epicycloids and hypocycloids (smaller circle located above and beneath the bigger circle).
The cycloid is determined by the ratio between the two circle’s radii k=\frac{R}{r}.
More information can be found on this Wikipedia entry.

cycloids-8_k2.1_hypo.png
cycloids-9_k2.1_epi.png
cycloids-14_k7.2_hypo.png
Source code: cycloids.py

Sierpiński TIrangle

2016-05-28, post № 126

BASIC, mathematics, programming,

Using the same method used in my previous Sierpiński Triangle program, which is written in Python, I wrote a fractal generator for my graphing calculator TI-84 Plus in BASIC.

sierpinski-tirangle-2.png
sierpinski-tirangle-4.png
sierpinski-tirangle-9.png
Source code: sierpinski-tirangle.basic