For people who don't know, it is an Euler Path. Basically Euler studied the math behind pathways connecting points. From what I remember (the math class in question was about 18 years ago) there are Euler Paths and Euler Circuits. A circuit you can start from any point, draw the entire thing, and come back to your starting point. A path you can draw over all of the lines but won't come back to the start, and the start can only be in certain locations. To identify if a diagram is an Euler Circuit, Path, or neither, you count the number of odd intersections (a intersection with an odd number of lines touching it). If there are zero odd intersections, it is a circuit. If there are 2 odd intersections, it is a path. If there are more than 2 it can't be done. This diagram has 2 odd points. To draw an Euler Path, you must start from one odd intersection, and end at the other. I can't remember if there are any tricks for the path in between though.
The trick is fleury algorithm Also in general what Fleury algorithm says is just "don't turn into a corner you can't come out of" if you do that u will definitely find the Euler circuit If you want to find path just make one point joining both the start and finish. Then find circuit and remove the point
Specifically, there must be either zero or two nodes with odd number of edges! If there’s zero, you can start anywhere and win; it’s usually very easy once you know this trick and you can solve it almost automatically
Всегда начинаешь с точки, где количество путей нечетное. Если в фигуре таких точек больше двух, ее нельзя обвести одной линией без повторения пути или прерывания его.
@@tapalagealexandru6805 ну все правильно, я лишь озвучил алгоритм решения. Контретно в этой задаче решение есть и это понятно уже потому, что точек с нечетным количеством путей 2. Не обязательно ее даже решать при этом.
I absolutely support your opinion.The principle of this kind of problems has been mentioned in the graph theory.Before I learnt this theory,I even can't imagined that these abstract problems can be solved in a scientific approach.
If there's two spots where an odd number of lines meet, you start at one and end at the other. If there's more than two, it's impossible. For shapes with only even-lined intersections, you have a way in and a way out (or two ways in and two out, etc.) So it doesn't matter where you start or where you end. When you have the three-line intersections, every line coming in allows for a line going back out so that deals with two of them but then for the third line, you get stuck unless you start at one of the odd nodes and that, in effect turns it into an even-numbered node. (thus leaving two more that are in and out) and end at the other. (if you already did two, that leaves a single line for the finish. This also expands out to intersections with four or six or eight and also the odd-lines extends to five or seven or nine, etc. But because odd-nodes must always have pairs removed till you're left with one, then the last line would end at that odd-node and never escape back out. Thus "start at one odd-number node" turns it in to an even node and do all other lines before ending at the other, And, curiously enough, there'll always bw an even number of odd nodes and if there;s 4, 6, etc of there, you can never reach/finish more than two of them
@@josealexanderdeleonfernand7613As long as there's no odd-numbered nodes at all, you can start at any node and you'll end at the same node. The actual path you do it in isn't that important. So if you took the simplest shape, a triangle, you can start at any corner so there's three solutions there. And you can go either clockwise or counter-clockwise so that doubles the solutions to 6. And it's obvious that you have to end at the node you started at. If you had a square and ran a single diagonal, then you'd have to start at one end of the diagonal and end at the other and I'm too lazy right now to figure out exactly how many solutions. I'm sure there's a formula for it somewhere. But if you added a second diagonal, then there's no solution because you have more than two nodes with odd # of lines, And, btw, the number of odd nodes can be limitless but it's always an even # of them. So the above shapes have 0, 2 and 4 odd nodes respectively. The number of even nodes can be any where from 3 and up (or if you allow curved lines, you could do two nodes with a shape like an American football.
Надо начинать с точки, в которой нечётное количество линий и заканчивать в этой точке или в такой же. В которой нечётное количество линий. И количество этих точек должно быть чётным.
U start from a point with 3 lines connected to it and end at the other point with 3 lines connected( Vertex with odd degree and end with the second vertex with odd degree)
So the secret is to check each connection/corner point. Count how many lines are in each connection point and label them even or odd. There can only be 2 solvable scenarios: 1) have all points even: you can start anywhere but must end at the same point. 2) only 2 odd points but everything is even: you must start at one of the odd points and end in the other. The shape in this video is Case 2
А на третьем участникн звучит смех моего маленького сынишки... Я его записывала на кассетный магнитофон для свадебного конкурса. Какой то идиот взял у меня эту кассету , и потом,через много лет, когда появился иртернет - выложил его в сеть... Никогда ни с каким его не спутаю... Помню лействия этой записи до мелочей... Сейчас сыну 45. Хакеры, вы талантлиаые люди.... Как можно убрать это смех из сети? ....
You start on the lower right line. Diagonally right, diagonally left, vertical up, diagonally left, diagonally right, vertical up, diagionally left, diagonally right. This took me five seconds.
The trick to puzzles like these are to start at an intersection with an odd number of paths. If the drawing is possible, there will be up to 2 odd intersections.
Ada yang bilang, diatas langit ada langit, di atas orang pintar ada yang lebih pintar . Tapi bagi saya, siapa pun yang tidak bisa menjawab permainan ini dengan benar . Maka dia adalah manusia terbodoh yang pernah ada didunia ini 😅🤭
Always start where there’s an odd number of edges - unless there isn’t any such part, in which case you can start anywhere. If there’s not 0 or 2 odd-edge spots, it’s impossible and you’re being trolled
Bonjour, commencer au sommet de la base inférieur du losange, puis suivre le périmètre du triangle bas à gauche, continuer le périmètre du losange vers la droite et remonter vers la gauche puis finir par les deux côtés du triangle du haut. Mer
For those who want to learn - You're looking at a type of problem that involves graphs. The graph is made up of vertices and paths (edges) that connect the vertices. The degree of the individual vertices is the count of the paths (edges) touching that vertex. A Euler Path is a path that goes through every edge of a graph exactly once. A Euler Circuit is an Euler Path that begins and ends at the same vertex. The solution for these problems depends on either trial and error and exhausting possible solutions OR knowing how Euler's Theorem works. Euler's Theorem: 1. If a graph has more than 2 vertices of odd degree then it has no Euler paths. 2. If a graph is connected and has 0 or exactly 2 vertices of odd degree (odd number of edges), then it has at least one Euler path. 3. If a graph is connected and has 0 vertices of odd degree, then it has at least one Euler Circuit. So break the image up into nodes (vertices) and edges (paths between the nodes) At each vertex place a number or if its even or odd at the vertex. Count the number of odd vertices. If you find 2 odd vertices then you must start at one of the two odd vertices. If you find 1 odd vertex then you will either start or end there. Enjoy.
Solved into in just one try
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❤❤❤😂😂😂🎉🎉Yfhgugh
Me too
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This is literally just like one of those "99% of people cannot do this" ads
Many as this should be get a report
And it wofked.. You commented
What ? Demo luke ?
@@jennyfernandabernalgaravit7552 ?
@@DemoLuke what 💀💀💀💀💀
For people who don't know, it is an Euler Path. Basically Euler studied the math behind pathways connecting points. From what I remember (the math class in question was about 18 years ago) there are Euler Paths and Euler Circuits. A circuit you can start from any point, draw the entire thing, and come back to your starting point. A path you can draw over all of the lines but won't come back to the start, and the start can only be in certain locations.
To identify if a diagram is an Euler Circuit, Path, or neither, you count the number of odd intersections (a intersection with an odd number of lines touching it). If there are zero odd intersections, it is a circuit. If there are 2 odd intersections, it is a path. If there are more than 2 it can't be done.
This diagram has 2 odd points.
To draw an Euler Path, you must start from one odd intersection, and end at the other. I can't remember if there are any tricks for the path in between though.
❤😂❤😂❤😂❤😂❤
Gracias por tan relevante información 👏👏👏🙏🙏🙏
@@Hill-chanilvhj
bro ez way just do try some weird combinations then look at then think and u will figure out
The trick is fleury algorithm
Also in general what Fleury algorithm says is just "don't turn into a corner you can't come out of" if you do that u will definitely find the Euler circuit
If you want to find path just make one point joining both the start and finish. Then find circuit and remove the point
It became very simple after seeing the last person do it 😅😉
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*AYO WHAT THE FUCKKKK*
Мужчина в бардовой кофте был близок к победе😊
I figured it out after the first person did it.
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Very simple and easy start from topup
Yes there is an answer to this one
The formula to solving this kind of puzzle. You always start at a location that has an odd number of spokes.
This is going to sound weird but I found it right on my first thought.
Seven Bridges of Königsberg!
Simple Euler advice: Always start in a node with odd number of edges.
also, if there are more than 2 nodes with odd number of edges, it can't be solved.
可以一筆畫，先畫兩個外边小三角的各2條邊线，后畫4邊菱形
Эта самая простая задача.
Graph theory
Specifically, there must be either zero or two nodes with odd number of edges! If there’s zero, you can start anywhere and win; it’s usually very easy once you know this trick and you can solve it almost automatically
that bag was empty 🤣
Всегда начинаешь с точки, где количество путей нечетное. Если в фигуре таких точек больше двух, ее нельзя обвести одной линией без повторения пути или прерывания его.
А я решил задачу
@@tapalagealexandru6805 ну все правильно, я лишь озвучил алгоритм решения. Контретно в этой задаче решение есть и это понятно уже потому, что точек с нечетным количеством путей 2. Не обязательно ее даже решать при этом.
I absolutely support your opinion.The principle of this kind of problems has been mentioned in the graph theory.Before I learnt this theory,I even can't imagined that these abstract problems can be solved in a scientific approach.
If there's two spots where an odd number of lines meet, you start at one and end at the other. If there's more than two, it's impossible.
For shapes with only even-lined intersections, you have a way in and a way out (or two ways in and two out, etc.) So it doesn't matter where you start or where you end.
When you have the three-line intersections, every line coming in allows for a line going back out so that deals with two of them but then for the third line, you get stuck unless you start at one of the odd nodes and that, in effect turns it into an even-numbered node. (thus leaving two more that are in and out) and end at the other. (if you already did two, that leaves a single line for the finish. This also expands out to intersections with four or six or eight and also the odd-lines extends to five or seven or nine, etc.
But because odd-nodes must always have pairs removed till you're left with one, then the last line would end at that odd-node and never escape back out. Thus "start at one odd-number node" turns it in to an even node and do all other lines before ending at the other,
And, curiously enough, there'll always bw an even number of odd nodes and if there;s 4, 6, etc of there, you can never reach/finish more than two of them
Además de la solución mostrada encontré otra forma de hacerlo.
Ghki🎉😊
@@josealexanderdeleonfernand7613As long as there's no odd-numbered nodes at all, you can start at any node and you'll end at the same node. The actual path you do it in isn't that important. So if you took the simplest shape, a triangle, you can start at any corner so there's three solutions there. And you can go either clockwise or counter-clockwise so that doubles the solutions to 6. And it's obvious that you have to end at the node you started at.
If you had a square and ran a single diagonal, then you'd have to start at one end of the diagonal and end at the other and I'm too lazy right now to figure out exactly how many solutions. I'm sure there's a formula for it somewhere.
But if you added a second diagonal, then there's no solution because you have more than two nodes with odd # of lines,
And, btw, the number of odd nodes can be limitless but it's always an even # of them. So the above shapes have 0, 2 and 4 odd nodes respectively. The number of even nodes can be any where from 3 and up (or if you allow curved lines, you could do two nodes with a shape like an American football.
🎉🎉😮😅😢😢😅😊
ĐJjeuuhrl
Надо начинать с точки, в которой нечётное количество линий и заканчивать в этой точке или в такой же. В которой нечётное количество линий. И количество этих точек должно быть чётным.
Точно. Это фигура даже проще чем пресловутый "конверт".👍
More people need to understand the basic principle of a Semi-Eulerian Graph.
Drawing that rectangle from center would be a easy step
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I didn’t need to wait for the answer I did it right away. Most of these are fairly easy.
😢
ЛГУН 😷
Yeah, I did it straight away but not the way he did it.... There is another way too
@@easygarden777 most of these puzzles have more than one way to solve it
Браво😊😊😊😊😊😊😊😊😊🎉🎉🎉🎉🎉🎉🎉🎉😂
❤❤❤❤❤❤❤❤❤❤❤❤🎉🎉🎉🎉🎉🎉🎉
U start from a point with 3 lines connected to it and end at the other point with 3 lines connected( Vertex with odd degree and end with the second vertex with odd degree)
Song name? 🙏
X2
小城夏天 xiao cheng xia tian
It is so much 🤣🤣😂
@@berylslim tysm 🙏🙏
Like If you got it first try
I don't want to be that person but TYSM for the likes ❤️
Bisa
Should start at the first inner point from below
@@rioaben363 ?
@@imsmiley23 Or first inner point above.
@@frankt... i stand corrected 🤣🤣🤣
Начинаем с одного крыла , переходим на второе , все это время не задевая сам ромб , после крыльев уже идет ромб
Можно и с ромба, тоже получится
@@user-jd3og6ji9z
Да, тут много вариантов, в отличии от пресловутого "конверта".
@@user-jd3og6ji9z Ну, или так! (с)
😂😂😂😂😂😅😅😅😅😅😅
So the secret is to check each connection/corner point. Count how many lines are in each connection point and label them even or odd. There can only be 2 solvable scenarios:
1) have all points even: you can start anywhere but must end at the same point.
2) only 2 odd points but everything is even: you must start at one of the odd points and end in the other.
The shape in this video is Case 2
*The most beautiful and natural model I ever seen. Her vibes, her eyes smile*❤🔥
홀숫점에서 시작하면 됩니다
홀숫점이란 점에 연결된 선이 홀수개인 점을 말하죠
세로선 두개보이죠 저기서, 안쪽점 에서 시작해야 합니다.
Solved easy.
The trick is to start at a point where three lines connect.
А на третьем участникн звучит смех моего маленького сынишки... Я его записывала на кассетный магнитофон для свадебного конкурса. Какой то идиот взял у меня эту кассету , и потом,через много лет, когда появился иртернет - выложил его в сеть...
Никогда ни с каким его не спутаю... Помню лействия этой записи до мелочей... Сейчас сыну 45.
Хакеры, вы талантлиаые люди.... Как можно убрать это смех из сети? ....
Wow Awesome 👍 Amazing 👍
cringe
Isko to ham Bhai 1 minut mein solve kar sakte hain
Ah the good old "99% of people are paid to act stupid" ads
この種の形では、線が奇数集まる点から始めた時に成功する。
Gampang banget 😊😊😊😅😅
done 💯 ❤
Song is XiaoCheng XiaTian by LBI
It’s a graph with 2 odd nodes so an Euler path starts from an odd node and ending on the other odd node exists
Нижний треугольник ромба , затем выход на оба крыла и завершение ромба внутри
O ALLAH, BLESS mouhammd and the family for mouhammd ❤
Muy bueno,además de divertirnos aprendemos,felicitaciones al que que lo pensó👍👍👍😊😊los chinos,son unos genios!!🎉🎉🎉
Should start with the node with odd number of lines joined to it
한번에 그리기는 무조건 홀수 꼭지점에서 시작해야 합니다
Simple
Qoyil ijodilaga omat ❤❤
You start on the lower right line. Diagonally right, diagonally left, vertical up, diagonally left, diagonally right, vertical up, diagionally left, diagonally right.
This took me five seconds.
1st center one draw chesi ,next one side draw chesi second side kuda draw cheyali.🙋🙋👍👍
😂😂😂
Excellent
This is obviously an eulerian path with 2 odd degrees, so you start at one of them and end at the other.
Start at the top or bottom of the diamond. Follow the perimeter, starting towards the right,
Good, great interesting activity🎉🎉🎉🎉
It's easy I know another method
Me too
😂😮😂😂😂😂
Already figured it out as during the video. It's so easy.
The trick to puzzles like these are to start at an intersection with an odd number of paths. If the drawing is possible, there will be up to 2 odd intersections.
If you already know the right drawing before the end gather here 👇👇👇
I puss it to check the right drawing and I got it 💖💯💯
😂😂😂
sa ibaba ng palikpik mo umpisahan tas derecho tas balik pataas then baba
Here
Simple
Ada yang bilang, diatas langit ada langit, di atas orang pintar ada yang lebih pintar .
Tapi bagi saya, siapa pun yang tidak bisa menjawab permainan ini dengan benar .
Maka dia adalah manusia terbodoh yang pernah ada didunia ini 😅🤭
Solved in less than one second
진짜틀리는게 더힘들어
@@user-yv7xi3mm8j
Exactly
@@user-yv7xi3mm8j
Exactly… kid stuff
The winner had enough time to rehearse
Dragon Age Inquisition has prepared me well.
Start @ top of left half of diamond; left, right, down, left, right, down, right, left; done.
Always start where there’s an odd number of edges - unless there isn’t any such part, in which case you can start anywhere. If there’s not 0 or 2 odd-edge spots, it’s impossible and you’re being trolled
😂😂😂😂😅😅😅 ❤❤❤
That dance at the end was just hilarious.🤣🤣🤣🤣
It's transversable if you start at an odd vertex. Network theory. Very simple.
These puzzles are trivial. You have to start at a vertex connected to an odd number of lines.
Amazing, next more ❤😊
😅
ruclips.net/user/shortssR06T2hIKXw?si=DWDr9GhSP2RS670r
@SherazSunpal-ok3kwo ko
Okk was not in😊
La clave es empiezas ir las.lineas de adentro y luego te pasas a la de afuera o de viceversa
Start at a junction with 3 lines and end at then other. Super simple
this is so easy. this is literally those "99.99999 percent of people cant do this" ads.
I solved easily
❤❤
😂😂😂😂😂😂😂🎉❤
dog
Uncle bob will replace him 😂😂😂😂
Kaun kaun ise banane ka try kiya 😂😂😂
I solved
Nice of you
same same
Amazing😊
gringe u bot
Lo descubrí en otra forma
Начинать надо всегда с нечётного количества пересечений. Если таких два, то закончится на втором таком пересечении. Если больше двух - решения нет.
I have 3 ways to solve this puzzle
Способа решения минимум 4, пробуйте!
You can literally do it from the bottom or the top
Bro took the hard way
Bonjour, commencer au sommet de la base inférieur du losange, puis suivre le périmètre du triangle bas à gauche, continuer le périmètre du losange vers la droite et remonter vers la gauche puis finir par les deux côtés du triangle du haut. Mer
So much for the results 😸😸
Wow amazing ❤️
Solved it too in one second and my answer is different than the video 😉
Сначало центральный ромб обводишь и потом с верхней точки ромба закрываешь остальное😊
L❤❤❤❤❤❤❤❤
Я еще одно решение нашел
very easy to solve
Есть ещё один вариант, начать также, но очертить ромб полностью, а затем по внешней стороне, в общем зеркально .
Start at a junction with an odd number of lines and do anything
U can also start with the square atan angle near to top move toward left then😉
From Timor-Leste 🇹🇱🇹🇱👍👍
I love your video's ❤
Start from upper middle down to the left then right upper etc
Too easy under 3 sec
👍👇easy
So easy, there is only tow points to start, and if you do it will end up correctly for any place you go.
Nice one 🎉
Wow nice great ❤❤❤❤
Team easy
👇🏼
There are a few ways to solve it. Pretty easily
Or, just look at where the chalk line begins-
For those who want to learn -
You're looking at a type of problem that involves graphs. The graph is made up of vertices and paths (edges) that connect the vertices. The degree of the individual vertices is the count of the paths (edges) touching that vertex.
A Euler Path is a path that goes through every edge of a graph exactly once.
A Euler Circuit is an Euler Path that begins and ends at the same vertex.
The solution for these problems depends on either trial and error and exhausting possible solutions OR knowing how Euler's Theorem works.
Euler's Theorem:
1. If a graph has more than 2 vertices of odd degree then it has no Euler paths.
2. If a graph is connected and has 0 or exactly 2 vertices of odd degree (odd number of edges), then it has at least one Euler path.
3. If a graph is connected and has 0 vertices of odd degree, then it has at least one Euler Circuit.
So break the image up into nodes (vertices) and edges (paths between the nodes)
At each vertex place a number or if its even or odd at the vertex.
Count the number of odd vertices.
If you find 2 odd vertices then you must start at one of the two odd vertices.
If you find 1 odd vertex then you will either start or end there.
Enjoy.
Exactly
Leonhard Euler， one of the most revered mathematicians in the history
There are several paths to get it done correctly
Solved into in just one try .
Solve into in one try 🙌🙌🙌🙌