On the middle line, let's call the intersections A B C and D. Start from C, go to D, do the lower curve, go to C, do the upper curve, landing you into A. Do the full circle at once until you land back to A and from there, go to B. Congrats.
Me screaming at the end FINALLY!!!! because it was so EASY 😭 Ty for 200+ likes (I would say "not to be that person.." but everyone types that) I litteraly looked at the likes like 3 weeks ago when I replied to my replies and now we are 8 likes away from 1k 😭❤️ (I think I might be dramatic but maybe not, ty)
Dá sa to, začat v strede mala rovná čiara doprava,oblúčik doľava, čiarka doprava, spodný oblučik doprava,veľky kruh doľava a mala čiarka dokončiť doľava😉
Start at the either the top-left of the lower-middle arch or the bottom-right of the higher middle arch and draw arch to the outer circle. Then draw full outer circle (either direction), draw the straight line all the way across, then draw the remaining arch inward.
They're very easy. First, count all the intersections of the lines at the intersection. If it's possible than there can be no more than 2 intersections with odd numbers of lines. If it is all even numbers of lines then you can start anywhere but if there's one or two odd numbers of lines then you have to start at one of them and if there's two then you will end at the other. It's still possible to make mistakes when tracing out but doing the above will prevent you from starting somewhere that's impossible to complete. That algorithm was formulated by Leonard Euler when solving the 7 Bridges of Konigsberg problem.
You can't have diagram that has only 1 odd number of lines. Your logic is correct, so to complete it, it should be "the puzzle is solvable if there is only 0 or 2 odd-number intersection."
A little math: An undirected graph has an Eulerian trail if and only if exactly zero or two vertices have odd degree, and all of its vertices with nonzero degree belong to a single connected component.
No she couldn't have cause her starting point was an intersection of 4. With 2 odd intersections the only way to do this correctly is by starting at one and ending at the other. (the two intersections in the middle)
Setidaknya ada 4 alternatif: Setengah lingkaran dalam atas ke kiri --> lingkaran luar penuh searah atau berlawanan jarum jam --> garis tengah --> setengah lingkaran dalam bawah ke kiri. Atau sebaliknya Atau Setengah lingkaran dalam bawah ke kanan --> lingkaran luar penuh searah atau berlawanan jarum jam --> garis tengah --> setengah lingkaran dalam atas ke kanan. Atau sebaliknya
Start from inner top semicircle to the left, then thru the diameter to your right, then upwards around the circle anticlockwise, then draw the inner bottom semicircle
Draw in one stroke is always about identifying the two points with an odd number of lines coming in. You have to start and end in these. Every other point you will enter and leave an equal amount of times and have even number of lines connected. They constantly start at the outside points here with 4 lines. As long as you start at one of the middle points with 3 lines you basically can't get it wrong.
On the middle line, let's call the intersections A B C and D. Start from C, go to D, do the lower curve, go to C, do the upper curve, landing you into A. Do the full circle at once until you land back to A and from there, go to B. Congrats.
Yes, this is possible and there is also at least one other solution than the one shown last. Start at the left corner of the lower inner semicircle until you meet the outer circle, then go left until you meet the right corner of the upper inner semicircle, follow it until you meet the outer circle, then go up and once around the circle completely until you reach the point where the upper inner semicircle and the outer circle meet and fill up the missing line by going right. Done.
There are only two points from which an uneven number of lines emanate (here three). Those have to be chosen as the start and end points, otherwise you will fail. This is because at any point that is NOT your start or end point, in each pass of your attempt you will approach it and then leave it again, therefore adding two lines emanating from the point; in other words, any such point will have an even number of emanating lines.
Тут даже и теория не нужна. В каждую точку надо прийти и уйти , чтобы продолжить рисовать (то есть в ней должно сходиться чётное число линий). Кроме самой первой (в которую не надо приходить) и самой последней (из которой не надо уходить) .Значит, начинать надо с точки, в которой сходится нечётное число линий, и другой такой же точкой заканчивать. Если таких точек больше двух - задача неразрешима.А народ упорно начинал с точки, в которой сходятся 4 линии - и неизбежно проигрывал.
@@ikssafonвообще не поняла, о чем вы тут. Начинаем с верхней дуги из центра к краю, заканчиваем в точке где центр нижней дуги. Не так, как парень на видео, но тоже получилось
Trick is there will be two points/intersections with an odd number of lines and the rest will have an even number of lines. The two that are odd will be the start and stop points. Notice in this case 2 points have 4 lines and 2 have 3 lines. Its mathematically impossible to solve staring on an point with even lines.
Start with the top half moon at the right base line, center of picture. Go left drawing the top half moon to the outside circle, then draw up at that point clockwise all the way around the outside circle till you get back to the center line, then draw center line and the lower half moon. Done. Take a bow. 😁
The secret behind such kind of puzzles is to start from the nod that has an odd number of connecting lines. If the number of odd nodes is more than 2 ( actually it is either 2 or 0) then the diagram is impossible to solve.
Theres a flow to it, start from the inner end of the small half circle, flowout to the bigger circle, complete the big circle and draw a streight line through, then draw the last smaller half circle.
i feel like bro just cheated at the end, you can’t start in middle, you hit the left curve first take it around the top side of the circle take it straight and then loop the bottom and hit the other curve
Можно по-другому. Верхнюю дугу из центра к краю, из этой точки вниз вправо полукруг, от туда нижнюю дугу к центру, и часть центральной полосы до края, потом заканчиваем круг по верху, и из этой точки уже дорисовываем центральную полосу
simple math problem it's the same when you draw the house of nicolause the form is possible to draw if there are only even intersections or at max 2 uneven intersections. You have to start at an uneven intersection and have to stop at an uneven intersection. 4th class math problem
team possible
👇
Dễ vãi l
I Did it
Dhvkvjugfggh,,,,,,,, I love u I love you
Yess!! I did it!
On the middle line, let's call the intersections A B C and D.
Start from C, go to D, do the lower curve, go to C, do the upper curve, landing you into A. Do the full circle at once until you land back to A and from there, go to B. Congrats.
Me screaming at the end FINALLY!!!! because it was so EASY 😭
Ty for 200+ likes
(I would say "not to be that person.." but everyone types that)
I litteraly looked at the likes like 3 weeks ago when I replied to my replies and now we are 8 likes away from 1k 😭❤️
(I think I might be dramatic but maybe not, ty)
😂
É do começar pelo meio
Dá sa to, začat v strede mala rovná čiara doprava,oblúčik doľava, čiarka doprava, spodný oblučik doprava,veľky kruh doľava a mala čiarka dokončiť doľava😉
@@vladislavmajdan2807 Chcu napísať veľa vecí ale aj nie tak iba toto: *Prečo si to napísala pod môj komentár a nie ako vlastný komentár?*
Sim, realmente (usei traduzir)
一筆書きは常に「どこともつながってないところから始まり、どこともつながってないところで終わり」ます。
例外は終点と始点がつながってる☆とか。
从奇数个数连线的顶点开始到奇数个数连线顶点结束
この図のように「どこともつながってないところ」がない場合は、「つながってる線が奇数のところ」から始めるとよいです。
@@さすらいのクリスチャン ていうか奇数のとこから始めれば良いって覚えれば良い件
한붓그리기
奇数の点が2こか0こだと一筆書きできる
そこから始める
There’s two tri-nodal points. Start and finish at those and it’s pretty tough to fail.
Yes, thats right approach. Start and end nodes will have odd number of connections. There should be two or zero such nodes, otherwise it's impossible.
I was just about to post that
This is called "Euler way"
@@AltafKhan-qd1tk
😂 there's always one "I was thinking this" type person 😂
It was so easy
It just took me a sec.
Thanks for 1k likes
Me too
Mee too.
Yes it's very easy
Extremely easy
Same
Start at the either the top-left of the lower-middle arch or the bottom-right of the higher middle arch and draw arch to the outer circle. Then draw full outer circle (either direction), draw the straight line all the way across, then draw the remaining arch inward.
se comienza por un punto donde los recorridos sean impares y se termino por otro igual.
Bravo last one, there is another way of making it.
从中心往外转圈即可😂
More than one other solution. More than 2 starting points.
Yes
Yes..
Yes
國中科展研究過一筆畫，每個經過的交點都是有進有出，有進有出表示通過該點的線條數量是偶數，如果是單數表示有進無出，或者有出無進，那就一定是起點或終點，而且奇數線條的交點只能一個或兩個，超過就無解，這個圖案中間兩個點都是三線的交點，所以只能以中間兩點為起點跟終點
小学奥数
They're very easy. First, count all the intersections of the lines at the intersection. If it's possible than there can be no more than 2 intersections with odd numbers of lines.
If it is all even numbers of lines then you can start anywhere but if there's one or two odd numbers of lines then you have to start at one of them and if there's two then you will end at the other.
It's still possible to make mistakes when tracing out but doing the above will prevent you from starting somewhere that's impossible to complete.
That algorithm was formulated by Leonard Euler when solving the 7 Bridges of Konigsberg problem.
Oh OK pythagoras.
Cool, nice info!
You must be a math or computer science professor; I teach graph theory
I took away two things from my college math class, Euler circuits and fair division.
You can't have diagram that has only 1 odd number of lines. Your logic is correct, so to complete it, it should be "the puzzle is solvable if there is only 0 or 2 odd-number intersection."
A little math: An undirected graph has an Eulerian trail if and only if exactly zero or two vertices have odd degree, and all of its vertices with nonzero degree belong to a single connected component.
This is a semi Eulerian graph problem. You start from one odd valency vertex and end at the other valency vertex
Yeah. I’m not surprised that no one else mentioned Euler paths, but they’re a cool tidbit of knowledge.
If you are the best asian of the asians you have unlocked a new mastery level... 😎
first girl would've had it if she started from the arch
No she couldn't have cause her starting point was an intersection of 4. With 2 odd intersections the only way to do this correctly is by starting at one and ending at the other. (the two intersections in the middle)
@@kaiburrus3190 that's what the comment above wants to say.
Setidaknya ada 4 alternatif:
Setengah lingkaran dalam atas ke kiri --> lingkaran luar penuh searah atau berlawanan jarum jam --> garis tengah --> setengah lingkaran dalam bawah ke kiri. Atau sebaliknya
Atau
Setengah lingkaran dalam bawah ke kanan --> lingkaran luar penuh searah atau berlawanan jarum jam --> garis tengah --> setengah lingkaran dalam atas ke kanan. Atau sebaliknya
Start from inner top semicircle to the left, then thru the diameter to your right, then upwards around the circle anticlockwise, then draw the inner bottom semicircle
Yes I did the same way
I did the same
😮degrees turned
Did it almost same way. I just didnt do the first semi circle first, did them last except middle line did it 2/3 of way.
اباذردربربرب
Draw in one stroke is always about identifying the two points with an odd number of lines coming in. You have to start and end in these. Every other point you will enter and leave an equal amount of times and have even number of lines connected.
They constantly start at the outside points here with 4 lines. As long as you start at one of the middle points with 3 lines you basically can't get it wrong.
This called ‘graf’ in mathematics
On the middle line, let's call the intersections A B C and D.
Start from C, go to D, do the lower curve, go to C, do the upper curve, landing you into A. Do the full circle at once until you land back to A and from there, go to B. Congrats.
I love watching these kids crack up time and time again 😅😂
Calam
There's two viable start/end points. It only took two seconds to work that out.
What do you call an animal with 4 paws, tails that meows?
- herring
That's quite an example of kinda videos
Yes, this is possible and there is also at least one other solution than the one shown last. Start at the left corner of the lower inner semicircle until you meet the outer circle, then go left until you meet the right corner of the upper inner semicircle, follow it until you meet the outer circle, then go up and once around the circle completely until you reach the point where the upper inner semicircle and the outer circle meet and fill up the missing line by going right. Done.
This deserves way more views.
ง่ายมาก
1 ลากจากวงเล็กด้านในบนจากขวามือลากโค้งขึ้นบนมาซ้าย
2 ลากตรงกลางจากซ้ายไปขวาสุด
3 ลากโค้งบนวงนอกไปทางซ้าย
4 ลากไปโค้งล่างไปจนสุ่ด วกเข้าวงเล็ก
5 จบ
There are only two points from which an uneven number of lines emanate (here three). Those have to be chosen as the start and end points, otherwise you will fail. This is because at any point that is NOT your start or end point, in each pass of your attempt you will approach it and then leave it again, therefore adding two lines emanating from the point; in other words, any such point will have an even number of emanating lines.
You can see him puff up like a pufferfish when he gets angry
You have to start (and end) on a point with a odd number of connections
Теория графов.
Есть две точки куда приходят четыре линии и две точки куда приходят три линии. Последнее это начало и конец.
Тут даже и теория не нужна. В каждую точку надо прийти и уйти , чтобы продолжить рисовать (то есть в ней должно сходиться чётное число линий). Кроме самой первой (в которую не надо приходить) и самой последней (из которой не надо уходить) .Значит, начинать надо с точки, в которой сходится нечётное число линий, и другой такой же точкой заканчивать. Если таких точек больше двух - задача неразрешима.А народ упорно начинал с точки, в которой сходятся 4 линии - и неизбежно проигрывал.
@@ikssafonвообще не поняла, о чем вы тут. Начинаем с верхней дуги из центра к краю, заканчиваем в точке где центр нижней дуги. Не так, как парень на видео, но тоже получилось
Thank god the last one got it. It was so easy, yet they all couldn't see it.
Me skipping the shorts to end to get my pattern proved correct 😅
Jhhk
Kjj
Trick is there will be two points/intersections with an odd number of lines and the rest will have an even number of lines. The two that are odd will be the start and stop points. Notice in this case 2 points have 4 lines and 2 have 3 lines. Its mathematically impossible to solve staring on an point with even lines.
Another solution: start f4om inside the circle
That's what he did
Love the three people doing the exact same thing
Does anyone know the song????
?
Xiao chong xiao tian
@@ravindrabijarniya2646
Start with the top half moon at the right base line, center of picture. Go left drawing the top half moon to the outside circle, then draw up at that point clockwise all the way around the outside circle till you get back to the center line, then draw center line and the lower half moon. Done. Take a bow. 😁
❤❤❤❤❤❤❤j'aime beaucoup ce jeu 😅
مثل تماما. ياحبيبتي ❤❤❤❤❤❤
Ee❤❤❤❤❤❤
This is was the masterpiece of Omegle
一筆書は始点と終点から延びてる線の数が奇数の位置に定めると、一筆書が可能です。
ただし通過点から延びる線は全て偶数でなくてはならず、それを満たさない図形は一筆書不可能です。
It looks impossible until when a genus gets it right ✅️
Doesn't anybody learn about nodal networks anymore? You have to start on the odd node and end on the other odd node
and if they are all even, then you can start anyway
The secret behind such kind of puzzles is to start from the nod that has an odd number of connecting lines. If the number of odd nodes is more than 2 ( actually it is either 2 or 0) then the diagram is impossible to solve.
Song name pls
A summer in a small town LBI ratio
They are deliberately getting ridiculously easy quizzes wrong for the video.
What is this song
小城夏天
xiao cheng xia tian
Just sprating instant asbestos on eachother
저 쉬운 걸 모르다니.
교육 의 중요성.
홀수에서 시작하고
홀수에서 끝나야 한다
정답! 오일러의 법칙!
Theres a flow to it, start from the inner end of the small half circle, flowout to the bigger circle, complete the big circle and draw a streight line through, then draw the last smaller half circle.
Very easy
ですよね
I always knew Naruto was a genius!! He can only show his brilliance when he's deep undercover spying for Konoha!!
Done
👇
Very. Good 😮😂❤
I did it a very slightly different way than the last dude, but both ways work 😂👍
우리땐 중학교 교과 과정에서 배웠었는데 지금은 안 배우는것 같던데..
예전에 수학을 더 다양하게 배웠던것 같고 지금은 문제 유형이 다 털려서 중학교때 고등학교 일반수학때 배운것까지 미리 땡겨서 유형화해서 배우긴 하더라
뫼비우스띠 같은것도 예전엔 배웠던것 같은데
Absolutely beautiful!
Anyone tell me song name pls
xiao cheng xiao tian
That guy at last retraced 😂
i feel like bro just cheated at the end, you can’t start in middle, you hit the left curve first take it around the top side of the circle take it straight and then loop the bottom and hit the other curve
There is a glitch when that man tries to draw
Always start these at a point with an ODD number of lines from it
I think I lost some brain cells watching that😂
North Korean sense of humor.
Можно по-другому. Верхнюю дугу из центра к краю, из этой точки вниз вправо полукруг, от туда нижнюю дугу к центру, и часть центральной полосы до края, потом заканчиваем круг по верху, и из этой точки уже дорисовываем центральную полосу
I figured another method but this one is good as well
We have to complete a eulerian trail in which we start and end on vertices with odd degrees, using this information, this problem is very easy
Just imagine a square over the circle and draw it with horizontal or vertical lines
wow He get it , ... btw what background sound name ? I love it
That’s so easy. Just watch it for a few seconds before trying to solve it.
that box is empty
Got the clue too but with a different drawing pattern 😊
They all got it wrong. Even the guy at the end who went over a spot twice. It's so easy I can't believe they don't know how to do it.
Have to start and end at the points with three lines. A basic topology mathematics issue, If there are more than 2 "odd" points it is impossible.
Here is a TRICK,
Always start at a point with an odd number of lines👍
Two ways to do it. From inside out, go around and then back in for the rest of the uncovered line. 🤔👍
I like this type videos so much...❤
❤❤❤❤❤❤❤❤❤❤❤❤😂❤❤❤❤
Assign node 1 2 3 4
Step ...
Line 3-2-1
Inside upper curve 1-3
Line 3-4
Outside upper curve 4-1
Outside lower curve 1-4
Inside lower curve 4-2
Heck no my childhood?!
Bridges of Konigsburg inspired. You must start and end at the nodes with an odd number of edges.
Thanks ❤❤❤
Easy to solve once you understand the rules. Just follow the algorithm for success.
You have to start and end at one of the interior vertices to draw the Euler circuit.
Spray will make a blind someone
He overlapped at the start.
simple math problem it's the same when you draw the house of nicolause
the form is possible to draw if there are only even intersections or at max 2 uneven intersections. You have to start at an uneven intersection and have to stop at an uneven intersection.
4th class math problem
That must be the 2nd child
The pitch boy is super ❤
one more way to complete puzzle is to start from the lower arc
Always have to start and end at odd intersections if there are two in the pathway.
I like the background music
1 glance and figured all out
How is it possible for me to figure these out in less than 15 seconds? I'm smart but not that smart.
It was so easy!! They started with the wrong place. Support please 👇
I would never use this spray foam. 😮
Я сразу догадался: Надо начинать из середины.
You have to start from a point with 3 lines
Bro this is why TikTok is better
I thought you could only start at the corners
we start from a point with an odd number of intersections
It have many ways to solve
Starting from any point of the two points in the middle and then taking any path
That was fantastic 😍😍
Why is the colored lines edited in?
一筆書きは主に奇数の接点が２つある
奇数の接点からスタートする事と別の奇数の接点を行き止まりにしない事を注意して描くとクリアできる
First one just needed one more step at beginning