Get the solution to the bonus riddle here: brilliant.org/TedEdTimeTravel/! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, riddlers!
An intern no more! He's playing Jurassic park and your writing your thank you speech for the Nobel prize you are about to receive for time travel technology.
Or you take your time thinking, and, when you figure it out, you make another potal to travel back to when the portal connecting to the professor was open, and then save him
You just grab all of them and make a bunch of triangles no matter what at least one of them is going to have a pure triangle then you take the rest back in the past so that way it's going to be very simple to get there and back seriously you bring all of them there's a good chance that if you just use three of them for each portal you're going to get one eventually
Yea he’s my favorite Ted Ed narrator. He has a very clear pronunciation and his voice has the perfect tone for educating. Not too cold and not too friendly.
@@AbhishekKumar-uu4uj it is from one of the older riddles. The riddle consisted of people with green eyes and to escape. They needed to have green eyes but they didn't know they had green eyes
I went from the opposite and tried to find a way to AVOID same-color triangles. The fact that a hexagon is symmetric made it easier. So I tried different combinations of the two colors on the outline (there are only three). All the times I stumbled across two rays of the same color from the same vertex, I added the third line of the different color (worst case scenario). And all the times it eventually led to a triangle. With a triangle, a rectangle and a pentagon the same strategy led to "winning" (there is a way not to form a triangle), so I increased the quantity of vertexes until I found out that a hexagon doesn't work like that, so it should be the correct answer. The bonus riddle works the same way. I see that this way is less logical and more intuitive, but for me it was easier to actually see how it can play out.
Keep in mind that the professor could be dead already when you step through, although a slim chance, making your rescue mission pointless, even if you can get through.
Yeah my thought process was just "if there's a 50% chance of getting one colour and you need three of that one colour, 6 would work" I know that's almost certainly completely wrong, but I got the right answer :D
Actually, because of the bonus riddle and also needing to get the professor back, you'd need 12. 6 for yourself and 6 for the professor since 1 portal teleports 1 person before it disappears.
0:09 if he’s called Ramsey, I assume he rants at his workers Also, here’s the solution: 1. Bring all the nodules, 2. Make tons of 3 node triangles till you get what you want
@@zForce4 assuming each line has a 50% chance to be red or blue, then there is a 1/4 chance you get a working triangle. If there are, say, 100 nodes in a box, then you would get about 25 working portals. At least that’s the estimate.
The bonus riddle: there will always be at least 2 triangles with same color sides. Take the possible colors extending from 1 nodule as in video. (1) If there are 5 or 4 blue sides (or 0 or 1 as that is just 5 or 4 red sides with the same result just with opposing colors), then there are at least 2 sets of 3 sides that will result in a portal like in the first riddle. (2) For the case of 2 or 3 blue/red sides, there will be 1 set of 3 sides that will create a portal as in the first riddle, and there will be sides between the nodules that could be either color and not create a portal. Pick a side as either blue or red (which doesn’t matter), and then fill in the remainder to try to prevent a portal. Another portal will always form no matter which initial side you pick.
Or: Start with the triangle we know forms, let’s say it’s red. We know any ‘out of triangle’ node can’t be connected to the 3 ‘triangle nodes’ with 2 red lines, because that would make a red triangle, so at least 2 of their connections to the triangle are blue. If every out of triangle node has 2 blue connections to the triangle, 2 out of triangle nodes must share at least one vertex of the triangle that they both have a blue connection to. If this is true, out of triangle nodes can’t connect to each other with blue lines, or they’d make a blue triangle with that shared node. But if all out of triangle nodes are connected with red lines, they’d make a red triangle between the three of them. So there is no solution without at least 2 triangles.
cant we apply the argument in the video to all the points and say that by symmetry, there will always be 2 portals form, and going futher and say that at least 6 portals always form?
@@jatarokemuri4549 We can't say 6, cause the triangles formed by some of the points may be ones we already counted as formed from the first point. But we can make sure of at least 2 by using one of the nodules not used in the first 4 and doing a symmetry argument.
Or we could just throw the 6 chrononodules through the portal (without actually stepping through it) and have the professor connect them himself. Skipped the bonus riddle, only need 1 portal to work.
Me: * grabs six chrono nodules * Me: * opens portal * Me: * gets splinched because there *was* a chance of the portal collapsing, no matter how few you bring * Me: Well, fml.
@Brianna Sloves Me: realizes it would just be easier to just accio the guy through the portal Also me: realizes i don't have wand... cries for the next week
@Brianna Sloves yes but we can't apparate through it because it's a portal. You have to go through it apparation would be like walking around instantly or only parts of you going through portal.
00:15 =professor has fallen into a time hole and i have to go in through the same time hole to save him 03:52= A Given portal can transport ONE PERSON before it closes
lil chris tuten But the character doesn’t need to explain it. Are you telling me that you can’t figure something out faster than you can explain it to someone? Because I definitely can.
no use rushing through the portal... just perfect the time machine first before rescuing the professor. don't want to get trapped in prehistoric era... no wifi, no pizza delivery.
To maximise the chance of survival, you might want to bring 5 though, depending on the chance of the portal collapsing vs the chance of the nodules giving a dead arrangement
@@Thomas14717There's only a 1% chance of failure to make a triangle with only five nodules, and since each additional nodule increases the likelihood of the portal collapsing while you're going through it, you're less likely to have the portal collapse on you if you only bring five nodules instead of six.
I know right, this one was actually kind of obvious though i guess? That or we're both just really geared towards understanding this sort of problem >.
I hate to be *that* guy, but still: Maybe you both lucked out, as the range of sensible options is pretty small here. If you didn't solve it the way the video explains, then make that a probably. So how did you come up with 6 as your solution? I got it wrong. I intuitively thought the answer was 5 (as 3 out of 5 lines have to be the same color). I had the right idea, but didn't think it through (you require a sixth nodule to take advantage of this circumstance).
My instinct was that 6 would essentially give you a 6 pointed star, given that there were only 2 possible colour combinations 6 seemed like the mathematical choice given 2x3 and the layout giving you multiple points of contact
@@XCM666 by noticing that you'll need at least 3 edge of the same color to know it's possible (this one...is rather instict), then by php you'll need a minimum 5 edges connected to each vertex. Since it's a complete graph, hence n = 6 is the answer
Hey! I personally love this video cause I am a undergraduate math student and currently am basing off my research on this exact field of math (Ramsey Numbers of Graphs)!
I'm super happy that my super quick solution turned out to be the correct one. I reasoned that if you had one stone as origin point and had random lines coming out, you needed to add enough stones for there to be 3 of a colour, since even if they didn't get the original color, they would still make a triangle. 6 it was
Step 1: Confirm you have green eyes. Step 2: Throw a replica of the egg from 100th floor. Step 3: Ask, " If i asked you whether 2+2=4, would you say ozo?" Step 4: Divide the triangle into acute angles triangles with 7 cuts. Step 5: Give the box of pearls to sea monsters. Step 6: Cheat death by playing snakes and ladders. Step 7: Count the no of birthday candles. Step 8: Say, "You will release my brother" to save your bro and the creatures of paradox. Step 9: Say the name of the 5th house and the founder. Step 10: Press the button to save you and your friends from the 11 dimensions.
Other idea, throw 6 chrono nodules through instead of you going through yourself. Presumably, the lab would have some sort of protocol for this sort of thing on the offchance of an accidental exposure, and assuming the portal always exits at the same place, all the doctor would have to do is wait for his return trip to arrive.
What happens if the professor walks away? We already know he's got a tendency to wander without looking where he's going, and he was reading a paper when the accident happened. Who's to say that he didn't realize what happened and just kept walking? Besides, the nodules are very small, and could be crushed by large dino feet very easily. It's not a bad idea, but there's too many things that could go wrong. Better to be there to mitigate as many problems as you can.
As a theoretical math problem it works, but in practice you can just bring the whole box and activate one at a time until you generate the portal. It may happen by chance with 3, 4, or 5 nodules, and if it does activate with a lower number then it would be safer than activating 6.
This is not a bad way to get an upper bound on the minimum for any kind or combination of polygons ("portals") formed by nodes ("chrono nodules") either. It's just a coincidence that for two triangles, it's also the lower bound and an achievable number. However for greater numbers of polygons or greater sizes of polygon, it quickly becomes extremely suboptimal, and hence Ramsey theory (amongst other areas of graph theory) that try to give better bounds or exact results. Or at least show that you can do it at all, is often as far as we get sometimes, too
How many nodules will u take? Me: 6 cuz we can make two triangles with it (lol) The only possible answer is 6 Me jumping around screaming: MOM get in here!!!! Ur daughter is smart😎!!!
Fun fact: Ramsey theory is also the source of Graham’s number! The original question was similar to this, but instead of triangles, the goal was a square/rectangle with an X through it, and the question wasn’t the number of nodes, but the number of dimensions a hypercube would need to have enough nodes. At the time, the answer was “somewhere between 6 and Graham’s number.” Which is basically the least precise statement that’s ever been seriously made. Nowadays, we’ve got a better idea: it’s somewhere between 13 and (2^2^2^2…) with (2^2^2^2… with 5138 2s) 2s. Which seems like a lot but compared to Graham’s number it’s basically zero.
The approach shown is hit and trial. Better way to approach this question could be is looking at the number of lines you can draw between any two points starting from a square using the formula (n*(n-1))/2. If you see for square 4*3/2 = 6(even number) . For 5, it is 10(even). And for 6 it is 15(odd). Now you have to bring out every combination of red and blue lines occuring between any two points, if the number is even, there is a one possibility where the number of red lines might be equal to the blue lines. Eg in a pentagon, 5 red and 5 blue lines. In this situation we will have one such arrangment which will avoid us from producing a triangle arrangement. Which is when the outer shape of pentagon is all one colour and the inner lines of the Pentagon are another colour, this prevents from the formation of a triangle of single colour. To avoid this situation we need to have a geometric figure whose (n*(n-1))/2 will yield an odd number, to avoid a combination where equal number of red and blue line might have possibility of arising. This way we can choose a hexagon, i.e 6. So that is the least number, certain to give us what we need.
I've read this question but in a different description. It's in a Chinese encyclopedia and it says: "You are the restaurant manager and you need to let a group of customers to seat. All of them has got strong emotions either good or bad against each other. You won't want 3 of them with the same emotions in between to sit at the same table or they might make trouble. What is the maximum number of customers to seat at each table?" And the answer is 5. With 6 there is a sure triangle.
I thought about this a little differently, not sure if my thought process is correct, but it gave me the same answer. Odds of getting all red or all blue triangle with 3 nodules is 1/8 (0.5x0.5x0.5) and because extra nodules create colours mutually exclusively, all you have to do is create 8 triangles with the least number of nodules (bc 1/8 x 8 = 100%). So I was just thinking of shapes and thought octahedron has 8 triangles and just 6 nodules. Could’ve been luck, I didn’t put much thought into the shape, just first one that came to mind.
YOu can't win the lotteries if you don't buy the lotteries And the collapsing only happens when you throw them. Or go through them. so you argument is invalid.
I was able to solve this because I had successfully solved the famous "what's the minimum number of socks u need to take out a drawer at random to make a pair if the drawer contains an equal amount of blue and red socks"
I got 6 too but in a different way To get any color, the probability is 0.5 Therefore the chance to get 3 desires ones in 0.5^3 = 1/8 The minimum amount of nodules that it would take to form 8 triangles is 6, in a rectangular arrangement of 2 nodules as the width and 3 nodules as the height. This effectively creates 2 squares that share one line in common, and since each square can make 4 triangles, 2 squares will make 8. However, my method does not assure that a proper triangle will be formed 😅.
Never heard of Ramsey Theory before this, but I am somewhat familiar with basic Graph Theory. Looking at it one way, considering all the red edges yields a graph, while the blue edges form its complement. Unite them and you have a fully-connected graph, which forms the problem statement presented here. Coming to a similar conclusion to the video, for any given graph with 6 nodes, either the graph or it's complement will always contain a 3-node cycle, aka a triangle. Came to that result after a bit of brute-forcing some permutations, but hey! I got _something_ out of my Math lectures.
It should be 6, if I'm not mistaken... Label the nodules 1, 2, ...so, case 5 [(a, b) means: connection from a to b] (1, 2), (2,3), (3, 4), (4, 5), (5,1): all blue (the color doesn't matter, this is just an example to show it works, switch blue to red or vice-versa)... (1, 3), (1, 4), (2, 4), (2, 5), (3, 5): all red...so 5 works (one can explain this further, each nodule is connected to all 4 of the other ones, etc)... As for case 6, we have to first establish that: each nodule can have at most 2 red and 2 blue connections emanating from it...if it has any more than that, there will be a portal... Then since in the case of 6, there have to be 5 connections coming from each nodule, that means there have to more than 2 of one of those colors, etc, each module (any module) will have at least 3 connections of the same color coming from it...that will automatic mean that there will be a triangle of the same color in the configuration...
I solved it with diagrams, lol, I didn't do it with a bunch of coordinates, if I could post diagrams on TH-cam I wouldn't have to resort to explaining it with coordinates...the pictures shows things more intuitively...well, yeah, lol, coordinates are more systematic, but still...that's not how most people, perhaps, would solve a problem like that, most people would see it pictorially...
i would just grab as many as possible and then use them three at a time until a portal opened. this does create the chance of not getting a portal but using more than the minimum increases the chances that it immediately kills you when you try to use it.
"What's the minimum number of nodules you'll need to bring to guarantee you create a fully red or blue triangle?" Me: Well, Prof., it was nice knowin' ya.
I solved it!!! I’m jumping up and down in excitement. How I solved it: Step 1: take a whiteboard and 3 different color markers Step 2: draw a number of nodules and try to make it *not* work Step 3: if it doesn’t work, increase the number of nodules by 1 This was actually a fairly simple riddle, but I’m still happy with myself. Lol.
That's a terribly inefficient method, and I doubt you actually did this. By the time you reached the answer, you'll have drawn 2^(3) + 2^(6) + 2^(10) + 2^(15) = 8 + 64 + 1024 + 32768 = 33864 different diagrams. Probably a bit less, because you'll more quickly find an arrangement that doesn't have a monocolor triangle in the 3, 4 and even 5 nodule setup. But for the 6 nodule setup, you'll need to go through at least all the 32768 diagrams to rule all of them out. That's 15 lines per diagram and 6 nodules per diagram as well, so that's 15 * 32768 = 491520 lines and 6 * 32768 = 196608 nodules drawn. Even if you draw like 5 nodules/lines per second (that's inhumanly fast), that still takes you 688128 / 5 = 137625.6 seconds. That's 38 hours, 13 minutes, 45 seconds and 600 milliseconds. And this even assumes you are inhumanly fast with drawing, took no time to think between drawings, took no time to double-check, made absolutely zero mistakes, found a non-monocolor setup basically instantly in the 3, 4 and 5 nodule setup, took no time to rest at all for anything, took no time to erase anything on the whiteboard, and have an almost infinitely large whiteboard. And you're saying you did all that?
HOLY SHOOT I ACTUALLY GOT ONE OF THESE RIDDLES CORRECT I was like "okay need three to form red or blue, two colors times three modules , 6!" I was high key being sarcastic BUT IT WORKED This day will go down in history for me
Solution to Bonus riddle. - We already know that there is at least one same-color triangle. Suppose the triangle is compose of nodes A, B and C, and ABC is a red triangle, and the other three nodes are D, E and F. - Suppose there are no other same-color triangle, to show contradiction. - As there are 5 edges connected to D, with the same logic, we can get a same-color triangle. As there can be no other same-color triangles then ABC, this same-color triangle we get from D should be ABC. It means that DA, DB, DC should be blue. - For the same reason, EA, EB and EC, and FA, FB and FC are all blue. - If any of DE, DF or EF is blue, we get a blue triangle, so DE, DF, EF should be all red. But we get a red triangle DEF in that case. So we showed a contradiction.
Looking through the comments for one that explained the answer to the bonus riddle in a way I could easily picture. This was it. I know other people had already explained, but not in a way that that I could follow, so thank you for this!
4:05 TED-ED: This bonus riddle- Me:IS B.S. HOW COULD YOU HAVE GOTTEN TO THE PAST AFTER THE PROFESSOR STEPPED THROUGH IF IT CLOSES AFTER ONE OF YOU WALK THROUGH?!?!???!! This comment means no offense please don’t kill me.
the real way to solve this is to not save your professor, your already smarter than he was, he didnt put a barrier around the dang time portal, and walked into it accidentally when he was facing it
Get the solution to the bonus riddle here: brilliant.org/TedEdTimeTravel/! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, riddlers!
TED-Ed pls make more riddles
The answer is yes. It will always be 2.
I would take 7 and place it 1 at centre of the 6 nodules.!! Of cource 6 will be the minimum number.
TED-Ed is way too hard with their riddles!
Check my page for ideas to your next travel destination instagram.com/travellgoal/
*Bold of you to assume that I'm helping this guy out*
An intern no more! He's playing Jurassic park and your writing your thank you speech for the Nobel prize you are about to receive for time travel technology.
Anyone who “accidentally” falls through a time portal can’t be trusted
imagine how rich you'd be if you sell his laboratory
I want to like this comment but it has 420 likes. Please keep it at 420 likes.
*Well yes, my text IS bold*
"You only have one minute to go in the portal"
*alright then...lemme spend 3 minutes thinking about it*
Or you take your time thinking, and, when you figure it out, you make another potal to travel back to when the portal connecting to the professor was open, and then save him
@@ErmisSouldatos Epic🤣
Portal: Ok, I can wait for 30 seconds more
You just grab all of them and make a bunch of triangles no matter what at least one of them is going to have a pure triangle then you take the rest back in the past so that way it's going to be very simple to get there and back seriously you bring all of them there's a good chance that if you just use three of them for each portal you're going to get one eventually
Definitely caught in a time warp
This guy has an amazing voice
Ikr its so soothing
Yea he’s my favorite Ted Ed narrator. He has a very clear pronunciation and his voice has the perfect tone for educating. Not too cold and not too friendly.
Check him out, his name is Addison Anderson.
Dragon Slayer thanks, I will
IKR
0:14 love how I just casually watch as my boss "accidentally" steps through a time portal
This is the part where you just throw the 6 chrononodules through the portal for him, and don’t step through yourself.
Of course, the hexagon is still the bestagon.
hexagons are the bestagons baby!!
(watched that vid btw)
Naturally
I watched that vid
Bees!
righty you are
Now I understand why there were exactly 6 infinity stones
6 is the new 42
Perfectly balanced
@@JrKid-jb3vm but it cost Everything 🙃
Conclusion: Thanos will win against Captain Marvel.
lol xd
XD
How does one ACCIDENTALLY fall through a time vortex?
And how does one ACCIDENTALLY add deadly lasers to ants?
weird al once accidentally stepped into a alternate dimension
Hot tub time machine
@@justyouraveragecorgi
How does one not know what their eye color is?
Sain 7417 I ordered a pizza and got a time vortex instead, what's your point
3:38
Just grab him through, you have long enough arms
Hehe
Lmao
LMAOOOOO
you cant see where he is though
Rofl
Step one: confirm you have green eyes
Step two: ask the nodules to connect correctly so you can leave.
Reference please ?
@@AbhishekKumar-uu4uj it is from one of the older riddles. The riddle consisted of people with green eyes and to escape. They needed to have green eyes but they didn't know they had green eyes
@@agenti4734 thank you
This meme is getting really annoying
Step three cover yourself in oil
I went from the opposite and tried to find a way to AVOID same-color triangles. The fact that a hexagon is symmetric made it easier. So I tried different combinations of the two colors on the outline (there are only three). All the times I stumbled across two rays of the same color from the same vertex, I added the third line of the different color (worst case scenario). And all the times it eventually led to a triangle. With a triangle, a rectangle and a pentagon the same strategy led to "winning" (there is a way not to form a triangle), so I increased the quantity of vertexes until I found out that a hexagon doesn't work like that, so it should be the correct answer. The bonus riddle works the same way.
I see that this way is less logical and more intuitive, but for me it was easier to actually see how it can play out.
*tells riddle*
me: uh 6?
*says answer is 6*
me: IM SO FREAKING SMART
I just said 6 cuz 3x2
Txc rags same but I did a little more thinking
@@starboyjadenn me too
Omg, I am glad I have not wrote this comment yet because it was my first time solving this question successfully and getting correct.
Lina Li Same
Can't I just throw the box of nodules in the portal so that the professor figures out himself ?
Oooorrrrrr maybe the portal is kinda two way? If the portal isnt close yet maybe the professor can just jump back?
The assumption contradicts the solution to the problem, you guys are good.
Spotted the loophole.
@@kudoamv thanks
Keep in mind that the professor could be dead already when you step through, although a slim chance, making your rescue mission pointless, even if you can get through.
Can't just bring three modules and throw those together until colors match?
Me: five sounds nice-No 6, 3 for 2 triangles.
Ted-ed: the answer is 6
Me: yes big brain time
Annnnnd you most likely solved the bonus riddle
Basically same as me. I was like "50% chance of red or blue for one", 3*2 = 6. Might be six.
Big brain gang.
400 like
Yeah my thought process was just
"if there's a 50% chance of getting one colour and you need three of that one colour, 6 would work"
I know that's almost certainly completely wrong, but I got the right answer :D
Actually, because of the bonus riddle and also needing to get the professor back, you'd need 12. 6 for yourself and 6 for the professor since 1 portal teleports 1 person before it disappears.
0:09 if he’s called Ramsey, I assume he rants at his workers
Also, here’s the solution:
1. Bring all the nodules,
2. Make tons of 3 node triangles till you get what you want
1:04
@@LocalPlagueDoc never said they one portal affects others
3. Say you have green eyes.
4. Ask the professor to leave.
5. If you have been questioned about his whereabouts, just reply "OZO"
Or you could just never get one and stuck in history forever
Just like how I never get good teammates when solo Q for anygame
@@zForce4 assuming each line has a 50% chance to be red or blue, then there is a 1/4 chance you get a working triangle. If there are, say, 100 nodes in a box, then you would get about 25 working portals.
At least that’s the estimate.
Me: Wait so how many nachos does he need?
get a notification. watch the video again. nice comment.
The bonus riddle: there will always be at least 2 triangles with same color sides. Take the possible colors extending from 1 nodule as in video.
(1) If there are 5 or 4 blue sides (or 0 or 1 as that is just 5 or 4 red sides with the same result just with opposing colors), then there are at least 2 sets of 3 sides that will result in a portal like in the first riddle.
(2) For the case of 2 or 3 blue/red sides, there will be 1 set of 3 sides that will create a portal as in the first riddle, and there will be sides between the nodules that could be either color and not create a portal. Pick a side as either blue or red (which doesn’t matter), and then fill in the remainder to try to prevent a portal. Another portal will always form no matter which initial side you pick.
Im not going to read that im just gonna assume you're correct
Or:
Start with the triangle we know forms, let’s say it’s red.
We know any ‘out of triangle’ node can’t be connected to the 3 ‘triangle nodes’ with 2 red lines, because that would make a red triangle, so at least 2 of their connections to the triangle are blue.
If every out of triangle node has 2 blue connections to the triangle, 2 out of triangle nodes must share at least one vertex of the triangle that they both have a blue connection to.
If this is true, out of triangle nodes can’t connect to each other with blue lines, or they’d make a blue triangle with that shared node.
But if all out of triangle nodes are connected with red lines, they’d make a red triangle between the three of them.
So there is no solution without at least 2 triangles.
cant we apply the argument in the video to all the points and say that by symmetry, there will always be 2 portals form, and going futher and say that at least 6 portals always form?
@@jatarokemuri4549 We can't say 6, cause the triangles formed by some of the points may be ones we already counted as formed from the first point. But we can make sure of at least 2 by using one of the nodules not used in the first 4 and doing a symmetry argument.
Or we could just throw the 6 chrononodules through the portal (without actually stepping through it) and have the professor connect them himself. Skipped the bonus riddle, only need 1 portal to work.
Me: * grabs six chrono nodules *
Me: * opens portal *
Me: * gets splinched because there *was* a chance of the portal collapsing, no matter how few you bring *
Me: Well, fml.
Hehe nice... splinched
@Brianna Sloves haha... Potterhead :D
@Brianna Sloves Me: realizes it would just be easier to just accio the guy through the portal
Also me: realizes i don't have wand... cries for the next week
@Brianna Sloves yes but we can't apparate through it because it's a portal. You have to go through it apparation would be like walking around instantly or only parts of you going through portal.
Mujahid Syed same
00:15 =professor has fallen into a time hole and i have to go in through the same time hole to save him
03:52= A Given portal can transport ONE PERSON before it closes
bruhhhh
Srjan Shetty In the BONUS QUESTION, not the MAIN QUESTION. Pay attention.
00.17 you have just a minute to go through the portal.
Explanation takes more than a minute.
Yeah really, you should have just thrown the 6 modules through the hole and stayed in the future yourself.
lil chris tuten But the character doesn’t need to explain it. Are you telling me that you can’t figure something out faster than you can explain it to someone? Because I definitely can.
1:41
now he knows where the lamb sauce went
Finally
"Whats the minimum number of nodules"
Me: *Take 3 and a risk also* .
2:15 I thought we were traveling through time not summoning Satan
Lol
XD
😂
We're gonna resurect our mom, Al.
lol
Just leave him and take credit for the time machine , its a win win
Sounds good to me
Found the psychopath lol
Lmao 😂😂
orange man bad
no use rushing through the portal... just perfect the time machine first before rescuing the professor.
don't want to get trapped in prehistoric era... no wifi, no pizza delivery.
I randomly guessed that 6 is the answer so that means im a legit scientist
Me too
I am the sixth liker
42 :V Same XD
Lol same i was like um 3 x 2 is 6 so 6 is the answer....?
My favourite kookies are suga kookies. THAT IS WHAT I Did XD and also HI ARMY
I like how Ted ed says time travel riddle and just makes us learn geometry
I like how the return portal just conveniently takes us back to the right time xD
Can you solve th-
Me: no but I'll watch
I solved it :D
I'm kinda smart but I always just like to see how to solve it.
Hahahahaha, yeah😉
When people describe you as smart, but you can’t solve the TedEd riddles...
Me and you,we have a lot in common.
0:20 YO this scene is GORGEOUS!
Oh wow it is
Reminds me of a game.
*When you finally got a riddle right even though you guessed*
lmao same
Yea
You feel so big brained
*me tooo*
lol 😂
Ted-Ed: " Can you-"
Me : " NO!"
Ted-Ed : " But- "
Me : " Well I'm gonna see the video anyway "
Thanks TedEd! Now I know how much modules I need when this problem arises.
The nodules are fkin *raw* !
Okay wrong Ramsay
WHEREZ THE LAMB SAAAAAAUUUUUCEEEEEEE !???
Thanks for all the likes😁
WHERES THE LAMB SAUCE
COME ON SUSAN
Nino would've made sure it wasn't raw, and have pictures to prove it.
Answer: bring 12 nodules and use 6 per portal so both can escape time
Easy peasy
Lemon sandwich
Pokefan Alolagaming but u only brought with u 6 or u could just go through at the same time
Lemon sandwich??? I've never heard of that!
Me too
LegendAdam09 if they brung 2 portals then they both go in each with 6 nodules giving them the same chance
@@spoonythegamer21 brung
I wonder what brung means
To maximise the chance of survival, you might want to bring 5 though, depending on the chance of the portal collapsing vs the chance of the nodules giving a dead arrangement
Yep, the chance of 5 failing is only 12/1024
Explain please?
@@Thomas14717There's only a 1% chance of failure to make a triangle with only five nodules, and since each additional nodule increases the likelihood of the portal collapsing while you're going through it, you're less likely to have the portal collapse on you if you only bring five nodules instead of six.
with the bonus riddle just bring 12 nodules and leave the other 6 for yourself or the professor, depending on who goes first
The argument can always be that you were too late and your rushed decision came to this
OMG I can’t believe I finally got one of these riddles right. This was a good one!
I know right, this one was actually kind of obvious though i guess?
That or we're both just really geared towards understanding this sort of problem >.
I hate to be *that* guy, but still: Maybe you both lucked out, as the range of sensible options is pretty small here. If you didn't solve it the way the video explains, then make that a probably.
So how did you come up with 6 as your solution?
I got it wrong. I intuitively thought the answer was 5 (as 3 out of 5 lines have to be the same color). I had the right idea, but didn't think it through (you require a sixth nodule to take advantage of this circumstance).
My instinct was that 6 would essentially give you a 6 pointed star, given that there were only 2 possible colour combinations 6 seemed like the mathematical choice given 2x3 and the layout giving you multiple points of contact
@@XCM666 by noticing that you'll need at least 3 edge of the same color to know it's possible (this one...is rather instict), then by php you'll need a minimum 5 edges connected to each vertex. Since it's a complete graph, hence n = 6 is the answer
GamerX705 wait so you didn’t solve the robo ants riddle it had the most obvious answer you dont even need to guess it
Hey! I personally love this video cause I am a undergraduate math student and currently am basing off my research on this exact field of math (Ramsey Numbers of Graphs)!
I think it's generally known as coloring problems
Xzcouter Sukaiti that’s cool
I'm super happy that my super quick solution turned out to be the correct one. I reasoned that if you had one stone as origin point and had random lines coming out, you needed to add enough stones for there to be 3 of a colour, since even if they didn't get the original color, they would still make a triangle. 6 it was
Step 1: Confirm you have green eyes.
Step 2: Throw a replica of the egg from 100th floor.
Step 3: Ask, " If i asked you whether 2+2=4, would you say ozo?"
Step 4: Divide the triangle into acute angles triangles with 7 cuts.
Step 5: Give the box of pearls to sea monsters.
Step 6: Cheat death by playing snakes and ladders.
Step 7: Count the no of birthday candles.
Step 8: Say, "You will release my brother" to save your bro and the creatures of paradox.
Step 9: Say the name of the 5th house and the founder.
Step 10: Press the button to save you and your friends from the 11 dimensions.
Ted ed:your boss is trapped in time
Me:🎵arms of an angel🎵
No, but still i'll watch and not understand the solution too.
Yea I can't think of an answer
Never though Gordon Ramsay was a time traveler
His dishes even satisfied them for the stone age.
Been looking for this comment lol
His first name and last name, both contain six letters..
@@SreyaSanghai Hey Ted Ed I found a much quicker solution over here
*It took me some **_time_** to solve this riddle*
True, it took me an hour but I did solve it!
Wow, i went through each case to solve, elegant solution!
_Bob McCoy *lisa Guerrero would confront the time*
I spelled that wrong didn’t I
no *get out* 😂
Bud Dum CRASH
Other idea, throw 6 chrono nodules through instead of you going through yourself. Presumably, the lab would have some sort of protocol for this sort of thing on the offchance of an accidental exposure, and assuming the portal always exits at the same place, all the doctor would have to do is wait for his return trip to arrive.
What happens if the professor walks away? We already know he's got a tendency to wander without looking where he's going, and he was reading a paper when the accident happened. Who's to say that he didn't realize what happened and just kept walking? Besides, the nodules are very small, and could be crushed by large dino feet very easily. It's not a bad idea, but there's too many things that could go wrong. Better to be there to mitigate as many problems as you can.
For a second I thought this was a brilliant sponsor 0:39
Can you solve the time travel riddle? Off course I can't, but let's give it a try!
He went through the portal to figure out where the lamb sauce is.
y
Woosh
Ohhh That Explains It
I love TedEd riddles
@Unbreakable Patches what
Take a bunch and calculate when you get there, bring the extra back.
My first riddle i ever solved on this channel.
Why are you scrolling through comments? We both know you should be studying.
😭
Im at school rn lol
Stooopp, I have two essays I have to do that I'm trying not to think about
i _am_ studying
At school (lunch) rn
As a theoretical math problem it works, but in practice you can just bring the whole box and activate one at a time until you generate the portal. It may happen by chance with 3, 4, or 5 nodules, and if it does activate with a lower number then it would be safer than activating 6.
Ahaha I just chose 6 because it's two triangles together
surprised i was correct 😂
we all did
Same
exactly like a cat
This is not a bad way to get an upper bound on the minimum for any kind or combination of polygons ("portals") formed by nodes ("chrono nodules") either.
It's just a coincidence that for two triangles, it's also the lower bound and an achievable number.
However for greater numbers of polygons or greater sizes of polygon, it quickly becomes extremely suboptimal, and hence Ramsey theory (amongst other areas of graph theory) that try to give better bounds or exact results.
Or at least show that you can do it at all, is often as far as we get sometimes, too
“You have one minute to go through the portal” Me calculating wether or not I can call ted Ed in that amount of time
How many nodules will u take?
Me: 6 cuz we can make two triangles with it (lol)
The only possible answer is 6
Me jumping around screaming: MOM get in here!!!! Ur daughter is smart😎!!!
I guess 6 randomly does that still count
Me too do we get prizes
bob berry no
Thanks a lot for making riddle videos more often ❤❤
Make this a series called Ted-dle
Or call it T series
I can only do the tissue thing haha lame
jk
Me going through probability this year:
finally knows the answer to a ted ed riddle
Fun fact: Ramsey theory is also the source of Graham’s number!
The original question was similar to this, but instead of triangles, the goal was a square/rectangle with an X through it, and the question wasn’t the number of nodes, but the number of dimensions a hypercube would need to have enough nodes.
At the time, the answer was “somewhere between 6 and Graham’s number.” Which is basically the least precise statement that’s ever been seriously made.
Nowadays, we’ve got a better idea: it’s somewhere between 13 and (2^2^2^2…) with (2^2^2^2… with 5138 2s) 2s. Which seems like a lot but compared to Graham’s number it’s basically zero.
0:12
Until, that is, he called you a donkey and said the meat is undercooked!
😂
where’s the lamb sau- THE ANSWER
This is awesome! I love new riddles.
By the time i finished watching this video the window would have already closed leaving my professor stuck in history :D
The approach shown is hit and trial. Better way to approach this question could be is looking at the number of lines you can draw between any two points starting from a square using the formula (n*(n-1))/2. If you see for square 4*3/2 = 6(even number) . For 5, it is 10(even). And for 6 it is 15(odd). Now you have to bring out every combination of red and blue lines occuring between any two points, if the number is even, there is a one possibility where the number of red lines might be equal to the blue lines. Eg in a pentagon, 5 red and 5 blue lines. In this situation we will have one such arrangment which will avoid us from producing a triangle arrangement. Which is when the outer shape of pentagon is all one colour and the inner lines of the Pentagon are another colour, this prevents from the formation of a triangle of single colour. To avoid this situation we need to have a geometric figure whose (n*(n-1))/2 will yield an odd number, to avoid a combination where equal number of red and blue line might have possibility of arising. This way we can choose a hexagon, i.e 6. So that is the least number, certain to give us what we need.
I've read this question but in a different description. It's in a Chinese encyclopedia and it says:
"You are the restaurant manager and you need to let a group of customers to seat. All of them has got strong emotions either good or bad against each other. You won't want 3 of them with the same emotions in between to sit at the same table or they might make trouble. What is the maximum number of customers to seat at each table?" And the answer is 5. With 6 there is a sure triangle.
I thought about this a little differently, not sure if my thought process is correct, but it gave me the same answer. Odds of getting all red or all blue triangle with 3 nodules is 1/8 (0.5x0.5x0.5) and because extra nodules create colours mutually exclusively, all you have to do is create 8 triangles with the least number of nodules (bc 1/8 x 8 = 100%). So I was just thinking of shapes and thought octahedron has 8 triangles and just 6 nodules. Could’ve been luck, I didn’t put much thought into the shape, just first one that came to mind.
0:00 the way I type is phenomenal.
Or just bring all of them and keep throwing them until they work
The more you bring the higher chance they have of collapsing while you are going through then.
YOu can't win the lotteries if you don't buy the lotteries
And the collapsing only happens when you throw them.
Or go through them. so you argument is invalid.
...
Alternative: bring as many as you can and then calculate how many you need to use once youre through
First Ted ed riddle I solved without pausing the video, so proud of myself
I think the lesson here is that if you have an active time machine turned on in your lab, you should be more careful while wandering around.
Just throw the whole damn box in and make the professor do the damn work of going back
Me: Ugghhhh 6?
Ted Ed: The answer is 6
Me: Lol this is the first riddle I got right.
I was able to solve this because I had successfully solved the famous "what's the minimum number of socks u need to take out a drawer at random to make a pair if the drawer contains an equal amount of blue and red socks"
I got 6 too but in a different way
To get any color, the probability is 0.5
Therefore the chance to get 3 desires ones in 0.5^3 = 1/8
The minimum amount of nodules that it would take to form 8 triangles is 6, in a rectangular arrangement of 2 nodules as the width and 3 nodules as the height.
This effectively creates 2 squares that share one line in common, and since each square can make 4 triangles, 2 squares will make 8.
However, my method does not assure that a proper triangle will be formed 😅.
Never heard of Ramsey Theory before this, but I am somewhat familiar with basic Graph Theory. Looking at it one way, considering all the red edges yields a graph, while the blue edges form its complement. Unite them and you have a fully-connected graph, which forms the problem statement presented here. Coming to a similar conclusion to the video, for any given graph with 6 nodes, either the graph or it's complement will always contain a 3-node cycle, aka a triangle. Came to that result after a bit of brute-forcing some permutations, but hey! I got _something_ out of my Math lectures.
neat
Bring as many as you can and keep forming triangular portals.
EEAASSSYYY
I was hoping for some kind of paradox riddle :(
Is there such thing??
You just predicted the future!!! th-cam.com/video/mS5eEhLN57s/w-d-xo.html
Are you a time traveler?
Here th-cam.com/video/mS5eEhLN57s/w-d-xo.html
It should be 6, if I'm not mistaken...
Label the nodules 1, 2, ...so, case 5 [(a, b) means: connection from a to b]
(1, 2), (2,3), (3, 4), (4, 5), (5,1): all blue (the color doesn't matter, this is just an example to show it works, switch blue to red or vice-versa)...
(1, 3), (1, 4), (2, 4), (2, 5), (3, 5): all red...so 5 works (one can explain this further, each nodule is connected to all 4 of the other ones, etc)...
As for case 6, we have to first establish that: each nodule can have at most 2 red and 2 blue connections emanating from it...if it has any more than that, there will be a portal...
Then since in the case of 6, there have to be 5 connections coming from each nodule, that means there have to more than 2 of one of those colors, etc, each module (any module) will have at least 3 connections of the same color coming from it...that will automatic mean that there will be a triangle of the same color in the configuration...
I solved it with diagrams, lol, I didn't do it with a bunch of coordinates, if I could post diagrams on TH-cam I wouldn't have to resort to explaining it with coordinates...the pictures shows things more intuitively...well, yeah, lol, coordinates are more systematic, but still...that's not how most people, perhaps, would solve a problem like that, most people would see it pictorially...
i would just grab as many as possible and then use them three at a time until a portal opened. this does create the chance of not getting a portal but using more than the minimum increases the chances that it immediately kills you when you try to use it.
"What's the minimum number of nodules you'll need to bring to guarantee you create a fully red or blue triangle?"
Me: Well, Prof., it was nice knowin' ya.
I solved it!!!
I’m jumping up and down in excitement.
How I solved it:
Step 1: take a whiteboard and 3 different color markers
Step 2: draw a number of nodules and try to make it *not* work
Step 3: if it doesn’t work, increase the number of nodules by 1
This was actually a fairly simple riddle, but I’m still happy with myself. Lol.
That's a terribly inefficient method, and I doubt you actually did this. By the time you reached the answer, you'll have drawn 2^(3) + 2^(6) + 2^(10) + 2^(15) = 8 + 64 + 1024 + 32768 = 33864 different diagrams.
Probably a bit less, because you'll more quickly find an arrangement that doesn't have a monocolor triangle in the 3, 4 and even 5 nodule setup. But for the 6 nodule setup, you'll need to go through at least all the 32768 diagrams to rule all of them out.
That's 15 lines per diagram and 6 nodules per diagram as well, so that's 15 * 32768 = 491520 lines and 6 * 32768 = 196608 nodules drawn.
Even if you draw like 5 nodules/lines per second (that's inhumanly fast), that still takes you 688128 / 5 = 137625.6 seconds. That's 38 hours, 13 minutes, 45 seconds and 600 milliseconds.
And this even assumes you are inhumanly fast with drawing, took no time to think between drawings, took no time to double-check, made absolutely zero mistakes, found a non-monocolor setup basically instantly in the 3, 4 and 5 nodule setup, took no time to rest at all for anything, took no time to erase anything on the whiteboard, and have an almost infinitely large whiteboard.
And you're saying you did all that?
Quickly the portal is about to close
Gives four days to figure it out
I love these videos SOOOOOOOOOOOOOOOOOOOOO MUCH!!!!!!! I love how in each video there is a different art style!
Tachyons are hypothetical particles that is known to move faster than the light. Meaning that its v > c, where c is the speed of light.
HOLY SHOOT I ACTUALLY GOT ONE OF THESE RIDDLES CORRECT
I was like "okay need three to form red or blue, two colors times three modules , 6!" I was high key being sarcastic BUT IT WORKED
This day will go down in history for me
Who else thinks the scream at 1:28 kinda sounds like toad from super mario
You bring a lot of them but only use them one at a time until a portal opens. You could get lucky and only have to activate as few as three.
Music
explanation: calm peaceful
Timer: sounds like the suffer of a wildfire
I solved almost every riddle on your channel easily and all it took was time and I figured it out
I’ll get it, *It’s only a matter of tiiiime*
I just guessed six... i figured 3 for red, 3 for blue... Couldn't think of a way to prove me right, though.
Who else is never able to solve any of these?
This one was the only Ted ed riddle I got right cuz by brain was like "it's a 50 chance to get either a red or blue line so 3 times 2 is 6" solved it
@@carastark9163 that was me too
Me but I watch cause it’s interesting
Me
ME not kno da wae to solve
Me: *grabs entire box*
Also Me: *gets back*
Solution to Bonus riddle.
- We already know that there is at least one same-color triangle. Suppose the triangle is compose of nodes A, B and C, and ABC is a red triangle, and the other three nodes are D, E and F.
- Suppose there are no other same-color triangle, to show contradiction.
- As there are 5 edges connected to D, with the same logic, we can get a same-color triangle. As there can be no other same-color triangles then ABC, this same-color triangle we get from D should be ABC. It means that DA, DB, DC should be blue.
- For the same reason, EA, EB and EC, and FA, FB and FC are all blue.
- If any of DE, DF or EF is blue, we get a blue triangle, so DE, DF, EF should be all red. But we get a red triangle DEF in that case. So we showed a contradiction.
Looking through the comments for one that explained the answer to the bonus riddle in a way I could easily picture. This was it. I know other people had already explained, but not in a way that that I could follow, so thank you for this!
4:05
TED-ED: This bonus riddle-
Me:IS B.S. HOW COULD YOU HAVE GOTTEN TO THE PAST AFTER THE PROFESSOR STEPPED THROUGH IF IT CLOSES AFTER ONE OF YOU WALK THROUGH?!?!???!!
This comment means no offense please don’t kill me.
Oh it’s just because it’s a new one
ANSWER:
bring all of the chrono nodules
Shandel Dela Rosa innnnnniiiitttt
wellllllll
Questions on Instagram and FB "how many triangle's are there" turns useful.
First TED riddle I got... Yay life💜💜💜💜
the real way to solve this is to not save your professor, your already smarter than he was, he didnt put a barrier around the dang time portal, and walked into it accidentally when he was facing it