Jetris CE

2016-08-20, post № 138

curses, games, programming, Python, #bag, #bag method, #Tetris

This game’s challenge is
to arrange the falling blocks.
Stay alive forever.

Jetris Console Edition is my second Tetris clone (see Jetris for the first one). It is programmed in Python and uses the curses module to display all its graphics on the shell.
Despite it running purely on the shell, I managed to eliminate any noticeable graphics update bugs. Furthermore the code to clear lines differs immensely from Jetris and actually works properly.
To enhance gameplay I implemented the so-called bag method. Instead of choosing pieces at random, a bag gets filled with all seven possible pieces. Each time you get a new piece, that piece gets randomly chosen out of the bag. If the bag is empty, it gets refilled. That way it is ensured that there will be no more than twelve pieces between two identical pieces (worst-case scenario). Also there will no more than four ‘S’ or ‘Z’ pieces in a row which makes the game fairer.
The pieces initially fall at one pps (pixel per second) and the game runs at twelve tps (ticks per seconds). Every ten cleared lines the speed increases by one tick or one twelfth of a second until it stays at the minimum speed, one twelfth of a second.^[1]
The game’s clock is handled by a thread and graphics updates are made when they are needed (there are no fps).
To achieve the pieces blocky look I used two spaces and a color pair whose background color is the piece’s color. On the shell two fully filled characters look like a square.
You also have the option to change your key bindings — which get saved on disk —, pause the game and there is a high score list — which also gets saved on disk. The files are located where the Python code file resides.

Source code: jetris-ce.py

Web Sudoku Solver

2016-08-13, post № 137

Auto Hotkey, games, programming, Python, #AHK, #automatic, #fast times, #good times, #web, #Web Sudoku

In my last post I used Web Sudoku to get a Sudoku as an example for my solver.

After that I wanted to automate the process of looking up a Sudoku, solving it and typing it in. But while trying to get the Sudoku’s numbers, I noticed that the whole, solved Sudoku was stored in plaintext! (Look at this page’s source code.)
So I just needed to get that information, open the Web Sudoku page in a browser and type in the already solved Sudoku.

To accomplish said goal I used the python module urllib to get the Web Sudoku page’s source code and the module webbrowser to open the page in a browser. To type in the Sudoku I used AutoHotkey.

The finished program takes a level (easy, medium, hard or evil) and an id (the Sudoku’s identification number) to get a Sudoku, create an AHK file, execute it and open a web browser.
All you have to do is to click into the first box, press a key (‘F1’ in this case) and the Sudoku gets solved! You then just need to wait a minute, which is the minimum time Web Sudoku wants you to take to solve a Sudoku, and the AHK script hits enter.

You can get really good times with this:

Source code: web-sudoku-solver.py

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.

Source code: sudoku-solver.py

Extra assets: sudoku-solver-2.png, sudoku-solver_bugfix-2017-12-02.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 $k\cdot(\cos{\varphi}+\sin{\varphi}\cdot i)=a+b\cdot i$ .

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.

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.

Source code: rgb-jallenge.py

Extra assets: rgb-jallenge-0.png, rgb-jallenge-1.png, rgb-jallenge-2.png, rgb-jallenge-3.png, rgb-jallenge-4.png, rgb-jallenge-5.png, rgb-jallenge-6.png, rgb-jallenge_colors-fixed.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).

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

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

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

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