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If you are here directly, here is the link with the context: https://www.facebook.com/reel/375727531350653/ since the video is in Chinese, here are the translations of the instructions given by GaoYuTian to get your age: 1) pick a number from 1 to 10 2) multiply it by 2, then add 5 to the result 3) take this result and multiply it by 50 4) if you had your birthday, add 1772 to the result, else, add 1771 to the result 5) subtract the year when you were born from the result 6) in this case, we will have an answer with 3 digits, the first digit is the number you have picked at the beginning, and the last two digits represent your age. this blog post will explain why this method works <中文> 如果你是直接来到这里的，原视频在这里 https://www.facebook.com/reel/375727531350653/ 这一个帖文会讲解“为什么这个方法行得通” Why does it work? / 为什么行得通? let's say the number we have picked at the beginning is "t", we have the following table 假设我们选择的数字是 "t"，我们就可以获得以下表格： so far we have 2 equations, let's focus on the one which considers that you had your birthday, which is : 目前为止我们有两个方程式，我们先正对如果你已将过了生日的情况: (2t + 5) * 50 + 1772 - y = 100t + a here we can try to rearrange the equation into something straightforward. 重新整理一下方程式，我们会获得: (2t + 5) * 50 + 1772 - y = 100t + a (100t + 250) + 1772 - y = 100t + a as you can see the number you pick multiplied by 100 will be canceled on both sides, meaning whatever number you picked, it will never affect "your age". what we have here now is : 由此可见你开始选择的数字乘于100 将会被抵消， 也就是说不管你选择什么数字，“你的年龄” 不会被影响。 250 + 1772 - y = a 2022 - y = a and yes, the numbers were picked carefully, and the equation above makes simple sense: take 2022, minus the year you were born gets your age, which makes total sense 当然，这些数字是被小心筛选的，以上的方程式也非常明显说明：“拿2022，减去你的 出生年份，就能获得你的年龄”。 Can we use other numbers ? / 我们可以用其他数字吗 ? let's try to come up with a formula to simulate the situation given we will turn the given instructions into a set of "universal" instructions, meaning not sticking with a specific input 我们可以尝试找出这一系列动作的公式： by rearranging the fomular, we have the following 重新整理方程式，我们可以获得: (kt + d)s + j - y = 100t + a kst + ds + j - y = 100t + a kst - 100t + ds + j - y = a (ks - 100)t + (ds + j) - y = a from the example given by GaoYuTian, we know that both 100t must be canceled out, or the coefficient of t must be 0, so we know that the first criterion is 从 高雨田大魔王 的例子，我们可以知道 100t 将会被抵消，也就是说 "t" 的系数 一定是0，所以我们可以知道第一个条件是： ks - 100 = 0 ks = 100 we also noticed that 2022 just pop up "from no where", and 2022 is the year when this post is posted, so we know that the second criterion is 当然我们也有注意到 2022 这个数字 “无缘无故” 的出现， 2022 是此贴发布的年份，所以我们可以推断出 ds + j - y = a ds + j = y + a ds + j = this year 今年年份 and that's all we need to know, let's quickly come up with a new set of instructions to guess someone's age: 有了这些材料，我们就可以开始制作出一套新的“年龄计算配套”： A quick Example / 例子 Step 1 \\ 步骤 1) assign "k" and "s" a specific value, note they have to be a factor of 100, which can be 给 "k" 和 "s" 一个值，必须注意的是 "k" 和 "s" 必须是 100 的因数 for the sake of demonstration, let's pick k = 5, s = 20 为了示范，我们可以选择让 k = 5, 让 s = 20 Step 2 \\ 步骤2) assign "d" a specific value, this can be any natural number, let's pick 6 for example 给 “d” 一个值，可以是任何自然数，在这里我们选择 6 Step 3 \\ 步骤3) find the value of "j", which we know ds + j = this year, so let's plug in the numbers and we will get 找出 "j" 的值，既然我们已将知道 ds + j = 今年年份，把 "d" 和 "s" 代入式中，我们可以获得 ds + j = this year 今年年份 (6)(20) + j = 2022 120 + j = 2022 j = 2022 - 120 j = 1902 Step 4 \\ 步骤4) and now we will put everything into a set of instructions: 整理出一系列的步骤： (1) pick a number from 1 to 10 从 1-10 选一个数字 (2) multiply it by 5 (=k), then add 6 (=d) to the result 乘于 5(=k) ,然后加上 6(=d) (3) take this result and multiply by 20 (=s) 用这个结果乘于 20(=s) (4) add 1902 (=j) to the result (post birthday), or add 1901 to the result (pre birthday) 如果生日过了，加上 1902(=j) 如果生日还没过，加上 1901(=j) (5) take the result, minus the year you were born (=y) 减去 出生年份(=y) (6) now we have a result of 3 digits, the 1st one is the number you picked and the rest is your age 我们会获得一个三位数，百位数是你选择的数字，后面的两位数是你的年龄。 and here we go, a new set of instructions just to "guess someone's age", have fun. 新的步骤出炉了，玩得开心点~

If you are here from Facebook, hi, if you are here directly, here is the link to the original video : https://www.facebook.com/watch/?ref=saved&v=986837095369965 如果你是从面子书来的，嗨， 如果你是在这里看到这个帖子的， 原视频在这里： https://www.facebook.com/watch/?ref=saved&v=986837095369965 如果你是從面子書來的，嗨， 如果你是在這裡看到這個帖子的， 原視頻在這裡： https://www.facebook.com/watch/?ref=saved&v=986837095369965 here is the original problem: 以下为原题： 以下為原題： If we have a rope of 56cm, what is the biggest rectangle it can make ? 一条周长为56公分的绳子，可以为成的最大矩形面积是多少？ 一條周長為56公分的繩子，可以為成的最大矩形面積是多少？ Of course from the previous part of this post, we can know that the biggest rectangle it can make is actually a square. This problem is proposed by a father to a Science/Math Teacher 佑来了, note that the father also knows the answer as well, but this problem is actually a math problem from his 3rd-grade son. This father is trying to find a way to explain "why using the same length to make a rectangle, the biggest rectangle you can have is a square". 当然从这个帖子的上一个帖子， 我们可以知道最大的矩形其实是一个正方形。 这个题目其实是由一位父亲向理科老师 佑来了 提出的， 当然这位父亲也知道答案， 问题就出在于这是他小学三年级的儿子的数学题。 这位父亲正在尝试解释 “为什么在周长固定的情况下，最大的矩形就是正方形？” 當然從這個帖子的上一個帖子， 我們可以知道最大的矩形其實是一個正方形。 這個題目其實是由一位父親向理科老師 佑來了 提出的， 當然這位父親也知道答案， 問題就出在於這是他小學三年級的兒子的數學題。 這位父親正在嘗試解釋 “為什麼在周長固定的情況下，最大的矩形就是正方形？” Here I have a solution, but before jumping right in, he has to know a bit of algebra. by using visualization, things can be easily done. 以下是我的方法，但是在直接证明之前，需要先让他了解一个代数概念。 使用视觉证明，可以使证明容易理解。 以下是我的方法，但是在直接證明之前，需要先讓他了解一個代數概念。 使用視覺證明，可以使證明容易理解。 Step 1 第一步: (S+d)(S-d) = ? of course, we all know that is thing equals s^2 - d^2, but using direct algebra to explain this thing to him would be hard for him, so here we are going to use a diagram to help solve the problem. 当然我们知道这堆东西会是 "s^2 - d^2", 但是使用直接的代数证明对他来说太困难了， 所以我们可以用几何来证明这个公式。 當然我們知道這堆東西會是 "s^2 - d^2", 但是使用直接的代數證明對他來說太困難了， 所以我們可以用幾何來證明這個公式。 first, let's have a square with its width and height S 首先，我们先画个正方形，边长为 S。 首先，我們先畫個正方形，邊長為 S。 then select a length "d" where d < S (d must be smaller than S) then divide the square into 4 regions 然后选择一个长度“d”，要注意的是 d < S （d 必须小于 S） 然后这个正方形可以被分割成4块区域。 然後選擇一個長度“d”，要注意的是 d < S （d 必須小於 S） 然後這個正方形可以被分割成4塊區域。 then adding a small rectangle on the top 然后在上方加多一个小长方形 然後在上方加多一個小長方形 when we ask (S+d)(S-d) = ? we can understand the question as "What is the area of the red rectangle if we know what S and d are ?" 当我们问“（S+d）（S-d）=?”时，我们可以把问题理解为 “红色的长方形面积是多少？如果我们知道 S 和 d 的话。” 當我們問“（S+d）（S-d）=?”時，我們可以把問題理解為 “紅色的長方形面積是多少？如果我們知道 S 和 d 的話。” we can actually (secretly) move that small rectangle on the top and fit it inside the square 其实我们可以（偷偷）的把上面的那一块塞到白色的正方形里面去。 其實我們可以（偷偷）的把上面的那一塊塞到白色的正方形裡面去。 we can see that the little square inside the big square has an area of d^2 我们也会发现到这是白色正方形里也出现了一个小的正方形，面积为d^2。 我們也會發現到這是白色正方形裡也出現了一個小的正方形，面積為d^2。 if the red region is what we want, then we know that the red region is the big square minus the small square, which is 如果红色区域是我们要的目标， 我们可以知道红色的区域其实是大正方形减小正方形，就是： 如果紅色區域是我們要的目標， 我們可以知道紅色的區域其實是大正方形減小正方形，就是： S^2 - d^2 and so we finally know that 那我们就可以证明 那我們就可以證明 (S+d)(S-d) = S^2 - d^2 Step 2 第二步 : Proving the real thing 开始正式证明 Say we have a rope forming a square with a side length of "f" colored white. 假设我们有一条题目中说到的绳子，但是不一定是56公分，围成一个正方形，边长为"f"，被标为白色。 假設我們有一條題目中說到的繩子，但是不一定是56公分，圍成一個正方形，邊長為"f"，被標為白色。 Let's squash the white square (not for long) such that its width decreases by a length say "g" and so the new rope would be colored yellow, so we know that the new width is 我们可以对正方形进行挤压，以至让宽度较少某个长度，“g” 新的长方形被标为黄色， 所以新的宽度是 我們可以對正方形進行擠壓，以至讓寬度較少某個長度，“g” 新的長方形被標為黃色， 所以新的寬度是 f-g but what is the new height? we know that the rope length doesn't change, so we can know that : 不过新的高度是什么？我们知道绳子的长度不变，所以： 不過新的高度是什麼？我們知道繩子的長度不變，所以： the perimeter of the previous square (white) = the perimeter of the new rectangle (yellow) half of the white perimeter = half of the yellow perimeter 正方形周长 = 长方形周长 正方形周长的一半 = 长方形周长的一半 正方形周長 = 長方形周長 正方形周長的一半 = 長方形周長的一半 f + f = (f-g) + ? //here, "?" represents the height of the new rectangle， //在这里，“？”代表新长方形的高。 //在這裡，“？”代表新長方形的高 f + f = f - g + ? f + g = ? so we know that the height of the new rectangle is "f+g", and we can know that the little rectangle on the top has a height of "g" 所以我们可以知道新长方形的高为“f+g”， 当然我们也可以知道那个小长方形的高其实是"g" 所以我們可以知道新長方形的高為“f+g”， 當然我們也可以知道那個小長方形的高其實是"g" now we can ask: what is the area of the new rectangle? from the diagram, we can see that the new area is 现在我们可以问：新的长方形的面积为何? 从图中我们可以知道，新的长方形面积为 現在我們可以問：新的長方形的面積為何? 從圖中我們可以知道，新的長方形面積為 (f+g)(f-g) looks familiar? we just saw this and proved it visually so that we can know that 熟悉吗？我们刚才才看过，所以我们可以知道 熟悉嗎？我們剛才才看過，所以我們可以知道 (f+g)(f-g) = f^2 - g^2 but what does f^2 - g^2 means ? we can interpret the equation as 不过 f^2 - g^2 是什么意思? 我们可以把这个等式理解为 不過 f^2 - g^2 是什麼意思? 我們可以把這個等式理解為 " take the original square we start with (which is f^2), remove a small square from it (which is g^2), the rest is the new rectangle (which is (f+g)(f-g)) " “ 拿出我们刚开始有的正方形(f^2), 切除一小块正方形(g^2), 那么剩下的就是我们的新长方形((f+g)(f-g)) ” “ 拿出我們剛開始有的正方形(f^2), 切除一小塊正方形(g^2), 那麼剩下的就是我們的新長方形((f+g)(f-g)) ” with a bit of logic, we can conclude that 再加上一点点逻辑思考，我们可以知道 再加上一點點邏輯思考，我們可以知道 "Since we take the square minus something and it becomes the rectangle, the rectangle MUST be smaller than the square" “既然我们是拿一个正方形切除掉一个小正方形然后成为新的长方形， 那么新的长方形 一定 比正方形小。” “既然我們是拿一個正方形切除掉一個小正方形然後成為新的長方形， 那麼新的長方形 一定 比正方形小。 ” and proof is done. 证明完毕。 證明完畢。

Since Astralica is based on Logic and Computer Science as well, Astralica has its own set of mathematical symbols, some of them are easy one by one substitution, some of them requires a bit of explanation because some symbols work in a slightly different way than how we normally would use them. And here are the list of them : You may notice that some of them would share the same symbol, because in some way, they are connected. Power, roots, logarithm and natural logarithm share the same symbol. Factorial, nPr, nCr share the same symbol because they are almost the same. You might think addition , subtraction, multiplication, division, modulation is simple substitution, but it also can be used in a story-like behavior. Soon we will direct each group and understand how are the symbols made, and how to use them. Addition, Subtraction, Multiplication, Division, & Modulation. These 5 operations are the 5 basic operators, they modify the quantity of an environment. So here are the concept of each of them. if you look at the symbols, they all share the "┐" part， which represents the box, when you add things into it, it goes in from the top, that's why the "+" symbol has the "⌠" part to show something is going inside from the top. when you minus things from it, it drops from the bottom, that's why the "-" symbol has the "⌡" part to show something is dropping outside from the bottom. multiplication is the slight superior version of addition, if you compare the add symbol and the multiply symbol, you can find that the multiply symbol has an extra stroke, that stroke is used to say "this is the superior version of it". A similar concept goes for division, by comparing the minus symbol and divide symbol, we can see that the divide symbol has one more stroke than the minus symbol. Modulation is a modification of division, that is why the symbol is derived from the divide symbol and an extra stroke is given to say "this is the modified version of division." EG01: as you can see, there are 2 ways to use the symbols, either we can use it as a string, or we can use "Story Script". From the 1st example, we can see that 3 is being added to the box with 2 sitting inside the box, and the result is 5. From the 2nd example, we can see that 4 is taken away from the box which has 7 inside of it before, and the result is 3. From the 3rd example, we can combine the number with the same operation and separate the numbers with a "math comma", which is something like a lighting bolt. From the 4th example, we can combine both addition and subtraction together as well. EG02: here we can see a similar pattern exists. still, Astralica follows the "MultiplyDivision first, then AdditionSubtraction" rule, that's why sometimes we need parenthesis. we also can use the "Story Script" for multiplication and division. we can interpret multiplication as : it will scale the box upwards by how many times. we can also interpret division as : it will slice the box downwards into several pieces. you can notice that when we are writing negative numbers, we don't use the minus symbol to represent the negativity, instead, we have its specific symbol for that, and since we have the symbol for negativity, we also have the symbol to represent the positivity of a number, but most of the time, we don't use it. Power, Root, Logarithm. here we have a slightly complicated situation, let's say we have a magical triangle, the left corner will be a slot for the base, the slot on the right would be the slot for a power, and the slot on the top would be the slot for a product. the main idea for this symbol is it acts like a function that takes 2 inputs, and the output is the missing corner. base on the different output, the triangle will act as different operations or functions here we have a table for the inputs and the outputs of the different operations: we can also choose to use the string script, let's say we use "Δ" to represent the triangle, and "," to represent the "math comma", and "_" to represent "empty space" the main syntax would be: Δ(base, power, product) it will be something like this : EG03: aq

Each language has its own number system, and so does Astralica. Most languages use the decimal system, or base ten, which has ten symbols to represent the quantity of different powers of ten, which are the normal 0,1,2,3,4,5,6,7,8,9 we know, but decimal is not the only number system we have. Take Binary for example, or base 2, which only has two symbols to represent the quantity of different powers of two, which is just 0 and 1. Since Astralica is based on logic and Computer Science, Astralica uses hexadecimal or base16. Base16 is commonly used in computers because 16 is a power of 2, which can be converted easily between them. As time evolved, Astralica has different scripts to write the same set of 16 symbols, which are : Quadrant Script, Binary Script, Prime Factor Script & Stick Script. Each of them has its own purpose. Quadrant Script The main concept here is that the numbers are arranged around the origin point with the same direction when an angle is increasing back when we were learning trigonometry. Quadrant script is suitable or addition and subtraction, which is not quite enough. one thing to clarify, to write "0", just simply write "L". Binary Script Since Astralica is Base16, hexadecimal can be converted into binary easily. here are the hexadecimal numbers getting converted into binary: PS: we use "0x" to say "hey, this is a hexadecimal number~" And this down here is the Binary Script in Astralica Prime Factor Script The prime factor script would be one of the complicated script, the main idea is: we have different symbols for different factors which are prime, which are 2, 3, 5, 7, 11, 13. although 0 and 1 are not prime, they still have their own symbols. here is the main logic behind the Prime Factor Script: > both 0 and 1 can be grouped together, they are designed to look like "off" and "on". > the base symbol for 2 is just 2 vertical strokes, the more horizontals, the more power it has. > the base symbol for 3 is derived from a triangle missing one of its sides for simplification. > the base symbol for 5 is originally a 5-pointed star, but drawing a 5-pointed star is cumbersome, so the first 2 strokes remain. > you can notice that some numbers are products of two factors, like 10, 6, 12, 14, 15, their symbols are made by combining two symbols together. > but for 7,11,13, they will be considered as "big prime factors, and cumbersome", so they will share the same base symbol, with additional strokes to represent the "level of cumbersomeness" the prime factor script will allow us to do multiplication and division easier, and here is why: the strokes in the middle show the shared power across the bases, if a certain base still requires it's own private power, private strokes are given, here's an example: With this, calculating fractions and areas and scales are much easier somehow. And since huge numbers can be written as one symbol, Astralica can be base64, which is not mandatory. one thing to clarify here is the cumbersome prime factors, the horizontal strokes on the right represent how many 16s it is bigger than. Stick Script The Stick Script is heavily inspired by the IK writing system, the IK counting system is base20, here we take the concept and apply it to Astralica, this makes addition, subtraction and sometimes multiplication and division easier relative easier to other scripts. you might be thinking since the stick script is so superior, why do we need the other scripts ? here, the stick script will be or can be used in mathematical calculations, but the grammatical part of Astralica will use the Prime Factor Script, the prime factor script will be combined with other linguistic characters to make other characters afterwards. Saying numbers just knowing how to write numbers is not quite helpful, so here is how we say numbers in Astralica: here, we are not going to use Astralica letters otherwise you may have a hard time reading this. each number is made by combining different numbers of 4s and 1s. Normally we would just literally say the digits from left to right, take 0xCA256 for example, it would be "to - dæ - næ - li - læ", but of course we can also add the bases of 16s into the string. so 0xCA256 can also be "to fras - dæ pras - næ bras - li mras - læ", one thing which would be a little bit complicated is Astralica also sometimes uses the "Chinese Logic", which is when a base is not specified, it is expected such that it is 1 base power lower than the previous said base. Here are some English, Chinese, Astralica examples so in Astralica, we do the same things, but in base16, here are some examples: you might be thinking about how we are going to say cardinal numbers and ordinal numbers, we will talk about this in Volume 2.

The main logic of the letters is the vowels and the consonants with the same family will share a same radical (a radical is a part of a charracter). here are the different families of vowels and consonants: (here we don't use IPA here) you might be thinking what does the following vowel sounds like, here we are going to use some examples: if you are not a native Chinese speaker, most likely you are going to confuse between i, ė, ë. Here are some ways to understand them more: > to pronounce "ė", try starting with "e", which is something like "uuuh...", while saying it, close your teeth until your lower teeth touch the upper teeth. that sound is "ė". > to pronounce "ë", start with ė, while saying it, pull your jaw backward, when doing this, both sides of your lower jaw should touch both sides of the upper jaw. > one example from KungFu Panda, Po will address his master as "Master Shifu", in terms of Chinese, it is spelled correctly, but pronounced in a very English way, instead of "Shifu", Jack Black would pronounce that as "Sheefu", which is not accurate in terms of Chinese pronunciation. > to distinguish between "i" and "æ:", if I were to say "ræ - bit" in Astralica, in English it would sound like "ræ - beet" Astralica was started back at 2020, along the time, many different scripts has been made and discarded, during the process, new letters are added but here (this post) are the final types of scripts. one amazing feature here is the letters can be arranged in such a way, a pattern exists. just for reference, here are all the letters we are going to have here: you may notice that some letters written here are slightly different from the ones we mentioned earlier, so here's a table to tell you what is what: 1) Ancient Script The first and the most original form of the Astralica Letters, nowadays used for Caligraphy and art. 2) Formal Script If the Ancient script is somehow equivalent to the Greek letters or Runes, the the Formal Script is somehow equivalent to TimesNewRoman, this is used for Formal Letters or reports or declarations. 3) Casual Script If you were to look at both Ancient script and Formal script, you can see there are way too many strokes, such that it will take a long time to write the whole sentence. That's why the casual script is the simplified version of the letters, it is meant to be simple, cute, but still elegant at the same time. you might notice that some letters can be written in both ways, the one on the left is the slight previous version and the one on the right should be the newest version of the Casual Script. there are 2 ways to arrange the letters in Astralica, one of them is the Block Script, which is used mostly when Ancient script is used, another way to arrange the letters is the String Script, which is extremely simple for beginners because it is the way we normally write, which is from left to right, from up to down. The block script is inspired by Chinese, notice that Chinese characters often can be separated into 2 parts, usually the part on the left is called the radical, which is used to give you the main idea of what the character has something to do with, the other part will give you a hint on how will it sound like. take "跑" for example, it can be broken down into 2 parts, the left one which is "足", which has the meaning of "leg", for the right part "包", which is pronounced as "bau", so from this, we know that this word has something to do with leg and it somehow sounds alike with "bau", infact, this character is pronounced as "pau" which means "run". Astralica Block Script has almost the same concept here, each block is one syllable, the left part which is the radical part of the syllable is the vowel, that is going to be the main characteristic of the block, to modify the vowel, we use consonants, we have the consonant come before the vowel, which is the start consonant, and the consonant come after the vowel, which is the end consonant. below is a diagram to show how they are arranged in a block, and we have the english world "lemon" as our example. so we have two directions to write in Astralica: the String Script and the Block Script:

Astralica is a language that I have been designing since 2020. The main idea for Astralica is it is a universal logical language, it is going to be as logical as possible. Astralica is inspired by English, Chinese, Japanese, Natural Science, and Computer Science. This series is going to record all the data about Astralica. Below are the contents of the series: please note that Astralica is still not complete yet, and this will be a long journey.

We know that it is impossible to divide a number by zero. And in the past, it is impossible to square root a negative number, but afterward, we defined that the square root of negative 1 is I. So we can almost do the same thing here. Let's define that 1/0 = ζ so the arithmetic calculations would look something like: here we have to treat 1/0 as a unit instead of a value and also, since division is the opposite of multiplication, we also need to define 1*0, so let's define that 1*0 = ξ and so we would have: here we need to notice that 1*0 is 0, but not any 0, is the 0 made by using 1*0 2*0 is 0, but not any 0, is the 0 made by using 2*0, it is not the same as 1*0 or 3*0 etc. n*0 is 0, but not any 0, it is the 0 made by using n*0, it is not the same as m*0 , where m ≠ n this will be mentioned later. but what will happen if we multiply ζ and ξ together? we would have the following : ζ * ξ = (1/0)ξ = (1/0)*0 = 1 ζξ = 1 so we know that ζ and ξ will cancel each other with the result above, we can easily derive the following n*(1/ξ) = nζ n*(1/ζ) = nξ then we can easily solve the following : ζ / ξ = ζ * (1/ξ) = ζζ = ζ^2 ξ / ζ = ξ * (1/ζ) = ξξ = ξ^2 note that 1ζ is not the same as 2ζ, instead, 2ζ is 2 times bigger than 1ζ, we can also say that Kζ is K times bigger than 1ζ with the same idea, 1ξ is not the same as 2ξ, instead, 2ξ is 2 times bigger than 1ξ, we can also say that Kξ is K times bigger than 1ξ another way that we can understand the concept is Aζ and Bζ are not the same, they maintain some ratio between themselves. The same goes for Aξ and Bξ, they are not exactly the same, they maintain some ratio between themselves as well. we can treat ζ and ξ as a special unit or vector. Since we have the real number line and the imaginary number line, it would be fun if we add a "Zerota Ksive line" (derived from the words : Zero + Zeta + Ksi + Vector + Unit) with this we can easily explain one of the famous proof that is used to say "you are no allowed to divide by zero", at some point you may have seen the following: 0 = 0 1*0 = 2*0 1*0/0 = 2*0/0 1 = 2 and they will say this does not make sense so you cannot divide but 0, but according to what we have defined earlier, the second line has already broken the rules, we have defined that "A*0 ≠ B*0", and so the second line "1*0 = 2*0" is wrong, it should have been "1*0 ≠ 2*0"

I tried to played with a modified version of Tower of Hanoi : instead of just having 3 Rods, what will happen when we have 4 rods instead ? my naive first move (Caution : it is wrong) If I know how to solve a n-disk puzzle, then I can solve a (n+1)-disk puzzle, move n-1 disks to a temporary rod, move the largest disk, then move the n-1 disks again so T{n} = 2T{n-1} + 1 as well the least number of disk to use all 4 rods to solve the puzzle should be 3-disks, we will need state 0 : [1,2,3][_,_,_][_,_,_][_,_,_] state 1 : [_,2,3][_,_,1][_,_,_][_,_,_] state 2 : [_,_,3][_,_,1][_,_,2][_,_,_] state 3 : [_,_,_][_,_,1][_,_,2][_,_,3] state 4 : [_,_,_][_,_,1][_,_,_][_,2,3] state 5 : [_,_,_][_,_,_][_,_,_][1,2,3] so I knew that T{3} = 5 and I have used the same approach and I ended up with: T{n} = 2^(n-2)*6 - 1 = 2^(n-1)*3 - 1 which is TECHNICALLY doable, but wrong this formula will only work ONLY if I will use all the 4 rods for the (n-1) disks and treat the whole stack as a 3-rod puzzle. I thought that T{7} = 2^(7-1)*3 - 1 = 2^6*3-1 = 64*3-1=192-1= 191 but turns out one of the ways to do it can be: state 00 : [1,2,3,4,5,6,7][_,_,_,_,_,_,_][_,_,_,_,_,_,_][_,_,_,_,_,_,_] 5 steps to move 1,2,3 from rodA to rodB by using 4 rods state 05 : [_,_,_,4,5,6,7][_,_,_,_,1,2,3][_,_,_,_,_,_,_][_,_,_,_,_,_,_] 15 steps to move 4,5,6,7 from rodA to rodD by only using rodA,CandD state 20 : [_,_,_,_,_,_,_][_,_,_,_,1,2,3][_,_,_,_,_,_,_][_,_,_,4,5,6,7] 5 steps again by using all 4 rods state 25 : [_,_,_,_,_,_,_][_,_,_,_,_,_,_][_,_,_,_,_,_,_][1,2,3,4,5,6,7] obvious enough 25 < 191 my second guess: my new strategy would be: divide the pile into 3 groups, say Alpha, Beta and Gamma note that Alpha.AnyDiameter < Beta.AnyDiameter < Gamma.AnyDiameter if n is odd step 1: move Alpha from rodA to rodB, since Alpha has the smallest diameters, it is free to use all 4 rods step 2: move Beta from rodA to rodC, but rodB cannot be used, so it is actually a 3rod puzzle step 3: move Gamma from rodA to rodD, not that Qty(Gamma) = 0 or 1 if n is even step 1: move Alpha from rodA to rodB by using all 4 rods. step 2: move Beta from rodA to rodD by not using rodB (3rods). but how are the groups are going to get their disks ? if n is odd Alpha : 1st ~ [(n-1)/2]th Beta : [(n-1)/2+1]the ~ [n-1]th Gamma : [nth] for T{n} : Alpha and Beta and Gamma for T{7} : 1,2,3 and 4,5,6 and 7 for T{9} : 1,2,3,4 and 5,6,7,8 and 9 for T{11} : 1,2,3,4,5 and 6,7,8,9,10 and 11 and so on for T{2z+1} : z and z and 1 if n is even Alpha: 1st ~ [n/2]th Beta: [n/2+1]the ~ [n]th Gamma : none for T{14} = 1,2,3,4,5,6,7 and 8,9,10,11,12,13,14 and none so for the example T{7}, the solving process would be like pile = {1,2,3,4,5,6,7} alpha = {1,2,3} beta = {4,5,6} gamma = {7} state 00 : [alpha,beta,gamma][][][] //use 5 steps to move alpha from rodA to rodB state 05 : [beta,gamma][alpha][][] //use 7 steps to move beta from rodA to rodC, 3 rods available state 12 : [gamma][alpha][beta][] //use 1 step to move the only disk (gamma) from rodA to rodD state 13 : [][alpha][beta][gamma] //use 7 steps to move beta from rodC to rodD, 3 rods available state 20 : [][alpha][][beta,gamma] //use 5 steps to move alpha from rodB to rodD state 25 : [][][][alpha,beta,gamma] so we have T{7} = 25 since we are now having 2 systems (3rod and 4rod), let's redefine the syntax as T{p,d} = least number of steps required to solve a d-disks p-rod Tower of Hanoi puzzle from the examples, we know that: if d is odd T{4,d} = 2T{4,(d-1)/2} + 2T{3,(d-1)/2} + T(1,1) if d is even T{4,d} = 2T{4,d/2} + 2T{3,d/2} + T(1,0) PS: T{3,d} = 2^d - 1 this is an exploration of craziness, I will pause here... for now

Now since we know the background of the puzzle let's start proving the formula. 2 vital keys Let's start with what we definitely know: To solve an n-disk puzzle step 1: use multiple steps to move n-1 disks from RodA to RodB step 2: move the largest disk from RodA to RodC step 3: use multiple steps to move n-1 disk from RodB to RodC let's say the number of steps required to solve an n-disk Hanoi puzzle is T{n}, then we definitely know that T{n} = T{n-1} + 1 + T{n-1} = 2T{n-1} + 1 /* in this case, we use "{}" to state the subscript total steps = step1(move n-1 disks) + step2(move the largest disk) + step3(move the n-1 disks again) */ another thing that we definitely know that is T{0} = 0 // you will need 0 steps to finish a 0-disk Hanoi puzzle the Proof : The main idea is to keep expanding T{n} until we reach the term T{0}. so here we go~ T{n} = 2 T{n-1} +1 = 2 (2 T{n-2} +1)+1 = 2 (2 (2 T{n-3} +1)+1)+1 = 2 (2 (2 (2 T{n-4} +1)+1)+1)+1 . . . = 2(2(2(2 ... 2(2(2(2T{n-n=0}+1 )+1 )+1 )+1 ...+1 )+1 )+1 )+1 = 2(2(2(2 ... 2(2(2(2T{n-n=0}+2^0)+2^0)+2^0)+2^0...+2^0)+2^0)+2^0)+2^0 = 2(2(2(2 ... 2(2(2(2^1 T{0} +2^0)+2^0)+2^0)+2^0...+2^0)+2^0)+2^0)+2^0 = 2(2(2(2 ... 2(2( 2^2 T{0} +2^1 +2^0)+2^0)+2^0...+2^0)+2^0)+2^0)+2^0 = 2(2(2(2 ... 2( 2^3 T{0} +2^2 +2^1 +2^0)+2^0...+2^0)+2^0)+2^0)+2^0 = 2(2(2(2 ... 2^4 T{0} +2^3 +2^2 +2^1 +2^0...+2^0)+2^0)+2^0)+2^0 . . . = 2^n T{0} + 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^2 + 2^1 + 2^0 since we know that T{0} = 0 what we have to deal with is the "2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 4 + 2 + 1" part first, we need to come up with a proof that we can "collapse" this string: ∀ x∈ℤ 2^x + 2^x = 2^x(1 + 1) = (2^x)(2) = 2^(x+1) with this we can try to focus on that string, say that string is S, so we would have S = 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^2 + 2^1 + 2^0 S + 2^0 = 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^2 + 2^1 + 2^0 + 2^0 = 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^2 + 2^1 + 2^1 = 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^2 + 2^2 = 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^3 . . . S + 2^0 = 2^n S = 2^n - 1 so if we plug this thing back in where we left off, we would have T{n} = 2^n T{0} + 2^(n-1) + 2^(n-2) + 2^(n-3) + ... + 2^2 + 2^1 + 2^0 = 2^0 (0) + S = 0 + 2^n - 1 and so finally we have proved that the least number of required steps to solve an n-disks Tower of Hanoi puzzle is : T{n} = 2^n - 1, ∀ x∈ℤ, x≥0 more stuff for Part 3

For those who really don't know what is the Tower of Hanoi, you are suggested to read from here. If you know what is the Tower of Hanoi, this series of posts is going to prove the formula of the least number of steps required to finish an n-disk Tower of Hanoi puzzle WITHOUT USING INDUCTION (I will explain why.). What is the Tower of Hanoi? It is a mathematical puzzle that perfectly demonstrates induction and recursion. It consists of 3 rods and several disks with different diameters which can slide into every rod. let's name the rods rodA, rodB, rodC. The set up of the puzzle is to stack all the disks in rodA, with the largest disk at the bottom, such that no disk with a larger diameter is above a disk with a shorter diameter, i.e. bottom to top, largest to shortest. The goal of the puzzle is to move all the disk from RodA to RodC (or RodB, both are "topologically" the same.) by obeying the following rules: (1) you are only allowed to move only 1 disk from a rod to a different one at once, and this will be considered as a step (2) no disk with a larger diameter can appear above a disk with a shorter diameter. Get 3 Coins, or anything legit ~ Let's start with something simple, let's start with a 2-disk Tower of Hanoi: [1,2][][] //each [] is a rod, the element in the [] is listed from top to bottom (on the rod), left to right (in written form) So the steps required to solve that would be something like: [1,2][_,_][_,_] [_,2][_,1][_,_] [_,_][_,1][_,2] [_,_][_,_][1,2] Now add a new disk and go give it a try. Did you notice something interesting here? Didn't get it ? You somehow did a puzzle of 2disk twice during the process. state 0 : [1,2,3][_,_,_][_,_,_] state 1 : [_,2,3][_,_,_][_,_,1] state 2 : [_,_,3][_,_,2][_,_,1] state 3 : [_,_,3][_,1,2][_,_,_] state 4 : [_,_,_][_,1,2][_,_,3] state 5 : [_,_,1][_,_,2][_,_,3] state 6 : [_,_,1][_,_,_][_,2,3] state 7 : [_,_,_][_,_,_][1,2,3] you moved the 2disc tower from rodA to rodB, and then from rodB to rodC. we can get another disk (or maybe another bigger coin if that is what you are using or anything which is legit) and see that the 4disc puzzle also contains two 3disc puzzles which each of them contains two 2disc puzzles. state 00 : [1,2,3,4][_,_,_,_][_,_,_,_] ┐ ┐ state 00 : [_,2,3,4][_,_,_,1][_,_,_,_] │ 3disk │ 2disk state 00 : [_,_,3,4][_,_,_,1][_,_,_,2] │ A to B │ A to C state 00 : [_,_,3,4][_,_,_,_][_,_,1,2] │ ┘ state 00 : [_,_,_,4][_,_,_,3][_,_,1,2] │ ┐ state 00 : [_,_,1,4][_,_,_,3][_,_,_,2] │ │ 2disk state 00 : [_,_,1,4][_,_,2,3][_,_,_,_] │ │ C to B state 00 : [_,_,_,4][_,1,2,3][_,_,_,_] ┘ ┘ state 00 : [_,_,_,_][_,1,2,3][_,_,_,4] ┐ ┐ state 00 : [_,_,_,_][_,_,2,3][_,_,1,4] │ 3disk │ 2disk state 00 : [_,_,_,2][_,_,_,3][_,_,1,4] │ B to C │ B to A state 00 : [_,_,1,2][_,_,_,3][_,_,_,4] │ ┘ state 00 : [_,_,1,2][_,_,_,_][_,_,3,4] │ ┐ state 00 : [_,_,_,2][_,_,_,1][_,_,3,4] │ │ 2disk state 00 : [_,_,_,_][_,_,_,1][_,2,3,4] │ │ A to C state 00 : [_,_,_,_][_,_,_,_][1,2,3,4] ┘ ┘ we will start proving the formula in Part 2.

set Z = Tz ė = Ei ë = Eih [ = [= R = Rl (AS)(Rt5,Sh1,St3)(Rt1,Sh2,St0)(Rt2,Sh8,St3)(AS) commence_the_attack_on_their_left_flank_at_dawn 00000001111111111222222222233333333334444444444 34567890123456789012345678901234567890123456789 ÷=9ix?,w!"ecuZ"0yk.×qEzq8TeaEŊy2×?alZėbvFi_:[n3 === commence_the_attack_on_their_left_flank_at_dawn ÷=9ix?,w!"ecuZ"0yk.×qEzq8TeaEŊy2×?alZėbvFi_:[n3 (AS)("","")(Rt5,Sh2,St1)(Rt1,Sh5,St5)(Rt2,Sh6,St0)("","")(AS) constructing_dimension_structure_-_emeverter_green_in_unity 00000000001111111111222222222233333333334444444444555555555 01234567890123456789012345678901234567890123456789012345678 9bėz5kr?[zf(Rao,ëeauoR?gbCüDrnbAhA×ai.kfhC!EAyAqR?Dg(,v!mmü === constructing_dimension_structure_-_emeverter_green_in_unity 9bėz5kr?[zf(Rao,ëeauoR?gbCüDrnbAhA×ai.kfhC!EAyAqR?Dg(,v!mmü 9bėz5kr?[zf( ao,ëeauoR gbCüDrnbA A ai.kfhC!E yAqR? g( v!mmü EN,i,ii,iii,EN 2,0,2 4,0,5 computer_science_is_interesting yp&eh'q,m(%'+]d*i6y2m$4}5<:zugm