Why the area of a square of side L is LxL?

Recently I watched a video talking about the difference of the volume of the sphere of radius 1 and the cube of side 1 in N-dimensions, and I asked myself why the cube volume is always 1 in any dimension, or, which is an equivalent question:

Why the area of a square of side L is LxL?

Because the unit to measure areas is the square of side 1.

Why the unit to measure areas is the square of side 1?

Because it's convenient, since we usually measure areas by dividing them in rectangles and then measuring the sides of those rectangles. But it's not mandatory!

What happens if we change the unit to measure areas?

Well, you can continue doing maths, but then, the area formulas will be all wrong. Yes, they only work if you first state that the area of a square of side 1 is 1. Anyway, the resulting formulas will only be different by a multiplying constant.

It's an interesting exercise to derivate the main area formulas by choosing, for example, the equilateral triangle of side 1 as the area unit if you like square roots and Pythagoras, or the circle of radius 1 as the area unit if you like π and how to measure circles with triangles and vice versa.

As with the common use of our arbitrary base 10, it's important to note that much of our mathematical knowledge, as the formula of the area of the rectangle, is just conventional. Think about it before talking to any aliens, or perhaps you will not be able to understand each other...

Add or subtract seconds to a subtitles .SRT file

Do you have a subtitles .SRT file not synchronized with the video?

Recently I recovered an old program that I wrote and I made it better. Now it supports milliseconds and negative numbers: add_seconds_to_srt.pl

> perl add_seconds_to_srt.pl
usage: add_seconds_to_srt.pl SECONDS[,MILLISEC] <INPUTFILE >OUTPUTFILE
> perl add_seconds_to_srt.pl -17,995
00:00:18,994 --> 00:00:24,594
00:00:00,999 --> 00:00:06,599
>

Enjoy it!

Generate in Perl all the Dyck words of length 2N

 The Dyck words are the elements that form the Dyck language and can be defined as any sequence of parenthesis correctly balanced, for example, these are the Dyck words having 3 pairs of parenthesis:

((()))  (()())  (())()  ()(())  ()()()

Also, the number of Dyck words of length 2N is equal to the Nth Catalan number, registered in OEIS as A000108, and the Dyck words can be converted to many geometric representations of the Catalan numbers.

The following program in Perl generates all the Dyck words of length 2N using the characters "0" and "1", receiving N as argument:

#!/usr/bin/perl
# dyck.pl - prints all dyck words of length 2N given N as argument
my $s=('0'x$ARGV[0]).('1'x$ARGV[0]); my $t;
do{
    print $s, "\n";
}while($s=~s/(.*)01(1.*)/$1.'10'.join('',($t=$2)=~m'0'g).join('',$t=~m'1'g)/e);

On each iteration, the last instruction generates the next word transforming the current word by finding the last "011", replacing "01" with "10" and then reordering the other "1" and its following characters (if any) placing first the 0's and then the 1's.

Code explanation: The variables $1 and $2 store the begining and the end of the word found by the regular expression (.*)01(1.*) and the variable $t is used to save the content of $2 because $1 and $2 are rewritten when the other regular expression operators =~ are executed for collect and separate the 0's and the 1's of the end.

For example:

> perl dyck.pl 4
00001111
00010111
00011011
00011101
00100111
00101011
00101101
00110011
00110101
01000111
01001011
01001101
01010011
01010101

Enjoy it!

Generate in Perl the numbers with digits in order

If you try to count using only numbers with all its digits in ascending order, you will get this:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122, 123, ...

This sequence is registered in OEIS as A009994 - Numbers with digits in nondecreasing order.

Note that all the jumps are located after the last digit becomes 9, because the digit 0 cannot appear after the first occurrence. Considering this, I have created a little Perl program to generate it:

#!/usr/bin/perl
for($_="";;) {
    s/([0-8]?)(9*)$/$2?($1?$1+1:1)x(length($2)+1):length($1)?$1+1:0/e;
    print "$_\n";
}

On each iteration, the regular expression is applied to the variable $_ generating the next number, following this algorithm:

• If there are one or more 9's at the end then:
• If there is a digit X<9 just before the 9's then that digit X and those 9's are all replaced with the digit X+1 repeated as many times as the number of 9's plus one, for example 22499=>22555
• If there is no digit just before the 9's then is considered that the digit is 0 and the same operation is done, for example 999=>1111
• If there are no 9's at the end then:
• If there is a digit X at the end then the digit X is replaced with the digit X+1, for example 133=>134
• If there is no digit in the number then the digit 0 is inserted

I tried to do the code as easy to understand as I could, so you can use it if you need without problem ;-). Thanks for reading!

Program to find the maximum multiplicative persistence

After watching an interesting video from Numberphile about multiplicative persistence, one of the basic operations in Persistence of numbers, I thought that I wanted to make it by myself in Perl, so here you have the program multdigits.pl:

$ perl multdigits.pl -h

Usage: perl multdigits.pl [OPTION]...
Multiply the digits of each integer recursively until get one digit,
printing all the products obtained and counting the number of them.
See https://oeis.org/A003001 for more information about the sequence.

-r, --radix=NUM   use the NUM radix or base, 10 by default
-s, --sorted      use only numbers with sorted digits as optimization
-m, --mindigit=N  when sorted, minimum digit to optimize, 1 by default
-a, --all         prints all instead of only those with more products
-o, --others      prints others with same product number, not 1st only
-d, --dots        print dots when reaches a number with one digit more
-i, --initial=NUM initial number to begin the search, 0 by default
-h, --help        print the help and exits

Faster options: perl multdigits.pl --sorted --mindigit=2 --indicator
Copyright 2021 Carlos Rica <jasampler AT gmail DOT com>

Using the faster options it finds pretty fast the smallest integer (277777788888899) having the maximum multiplicative persistence known (11) for base 10, and then the program cannot reach numbers much bigger than that, but it can do it in any base, and also it can show other numbers with the same maximum multiplicative persistence, which give interesting information about the repetition of the longest sequences of products. Enjoy!

How do I solve the puzzle 2048

The game 2048 is an entertaining one, despite its simple rules. You can play it from your internet browser in many websites by searching 2048 in any search engine, or by installing an app in your smartphone.

I'm sure that better strategies can be found, but this is what worked for me: Maintain always the biggest numbers in a line at one side of the board, in ascending order, for example:

Having this, all you have to do is increase the numbers of this line in a progressive way, in ascending order, avoiding separate the line from the border and maintaining the biggest number in the corner, so you can use the rest of the space to work.

In order to maintain this structure, is important to have this line complete with different numbers all the time, so you can move the rest of the numbers without affecting it. If the line is altered, as when is temporarily shortened at the minor side after merging two numbers, or if you are forced to separate the line from the border, the structure must be recovered as soon as possible, filing the gaps and restoring the order to make it easier.

I hope this advice helps you to enjoy it even more. Happy game!

Magic squares generator in C

2021-11-10: NOTE: I rewrote the program to make it faster in magic-square.

A magic square is a table with equal number of rows and columns that is filled with all the numbers from 1 to NxN (being N the number of rows) in a way that verifies that the sum of the numbers in any row, in any column and in any of the two diagonals gives the same result in all cases, which is called the magic constant. If you choose a different initial number than 1 for filling the square (even zero or negative) and a different increment than 1 to get the following numbers, you will also get the same number of magic squares although the magic constant will change.

Because by rotation or reflection of a magic square you can get other 8 magic squares, only one of those variations (or trivial solutions) is counted. One magic square of size 1x1 exists, zero magic squares of size 2x2 exist and one magic square of size 3x3 exist (with 8 rotations and/or reflections) whose lines sum 15. If you choose the magic square of size 3x3 with the minimum corners, you get this one (printed by my magic square generator):

 2 | 9 | 4 
---+---+---
 7 | 5 | 3 
---+---+---
 6 | 1 | 8 

Exist 880 magic squares of size 4x4 (7040 if you count the trivial solutions) with sum 34, and the magic squares of size 5x5 with sum 65 are many many more, so I created this magic square generator to count them and verify the number of magic squares that others found. Because that number is so big and the program cannot generate all magic squares in a short time, you can can restart the counting from any point by choosing the numbers of the corners with the -c option, although it only accepts ordered numbers for the initial corners. The minimum corners that I have found that generate solutions is this:

$ ./magic-square -c1,2,5,22 5
  1 | 18 | 20 | 24 |  2 
----+----+----+----+----
 23 |  8 |  6 | 12 | 16 
----+----+----+----+----
 19 |  3 | 25 |  7 | 11 
----+----+----+----+----
 17 | 21 |  4 |  9 | 14 
----+----+----+----+----
  5 | 15 | 10 | 13 | 22 
...

To stop the program you can hit Control+C or wait until it reaches the last corners 22,23,24,25. The available options are shown by executing the program without arguments. The -q option counts the solutions without printing them. Happy searching!