Wait, so what? You only need the dark matter so that you can get to the planet faster than the other aliens. You can use your regular fuel to fly off after you collect the stuff.
That's what I was thinking, you'll just be a sitting duck waiting for the aliens from other species to arrive, take the artifact from you, and leave you stranded.
@Manuel yeah, this one's insultingly simple being told you've got three full tanks worth of fuel means three trips, which means two dropoffs and one real journey, then it's just about deciding how many parsecs away you'd need what quantity of fuel, and as mentioned there's multiple options there. honestly i'm just angrily typing because i forgot these weren't worth clicking on.
Ted Ed: Can you solve it?? Me: Yeah I'm trying hard..!! (Meanwhile other species from galaxy reached there and took all the precursor technology..) Many days past....... Me:----- Still trying to figure out the solution....!!!!!
...Sometime later Precursors wake up from their hidden hibernation chambers Track down all the civilizations that took their stuff, wipe them out Humans slip beneath the radar Win!!
Interesting. I did it in 3 separate trips. The start was the same: at the 3 parsec mark, drop off 9 units, then return. However, I deposited no more in that cache. I went to that cache, picked up 3 units, went forwards 5 units, deposited 5 units, then returned to the cache, where I got the 3 units to get back to the station. Then I went to each caches and refueled, leaving me with 15 units at the 8 parsec mark, like you did.
That was my solution as well, after deciding I liked it more than what I came up with first: leaving 2 at 6 parsecs and 3 at 8 parsecs instead of all 5 at 8 parsecs.
You need fuel to go faster than light though. Assuming you can go >90% of the speed of light without fuel it would take you a little over 25 years to coast the last 8 parsecs.
@@lordofthecats6397 that's not the way it works, you won't slow down in outer space. Period. Well unless you hit a planet or a an asteroid or whatever. 1 unit of fuel sends you flying at a speed higher than that of light, but due to the fact you don't slow down you can just cut the engine.
Wait...so while we were doing this, a bunch of other species were headed for this planet. Even if we managed to get there first, we’d be stuck there without fuel, waiting for the other species to arrive. And then when the other species DID arrive, we’d be at their mercy with no means of escape
My solution was a bit different. Still works though. My approach was to carry everything to the next parsec one step at a time. You carry everything to parsec 1, losing 5 units on the way. Repeat from 1 to 2, 2 to 3. From now on you only need 3 units for transferring all left units. Repeat until you're at parsec 8 and you have exactly 15 units left. Takes waaaay more trips than the solution here tho haha
Concur. I thought the cost-per-distance was non-linear, paying -d to advance (15-2d) fuel by d positions, so we minimize cost by minimizing d. But it turns out that the granularity is actually full-tanks, not individual fuel. Graphically, we can quickly see that the two solutions are equivalent, and furthermore generalize to other configurations of waystations. Consider a single full tank of 15 fuel. Think of it as 9 cargo + 6 self-fuel to move itself. Then you can pay 3+3 round-trip to advance the 9 cargo to position #3 in one jump-pair, as in the video, or (1+1)*3 to advance to position 3 in 3 jump-pairs. It follows that we could do it in two jump-pairs, of distance 1 and 2 or vice versa. It also follows that this generalizes to the isomorphic problems of carrying groceries up a staircase in 1 trip or several trips, or mowing a lawn in horizontal or vertical stripes. In every case, you're covering the same total area, and all partitions of an area sum to the same area.
Yeah but you are saving units too. I solved by your method and saw that when you reach 8 parsec you actually have 18 units left which is 3 more than required to reach 23 parsec. There is one problem though which is you take more time and by the story that they told you need to reach their as soon as possible.
@@feartheghus That's not true. If you don't have any unused fuel from the 45 units anywhere by the end of the trip, then you've still travelled 45 parsecs no matter what way you do it.
I actually got this one, I came at it slightly different than they did. I figured the trip would take 2 fuel dumps and 1 full trip. So for the first trip I wanted to dump enough fuel to refuel at the first dump site 3 times. A dump site at 3 lets you use 15 fuel to dump 9 units of fuel at point 3 and return. Next I wanted to dump enough fuel at the second point to refuel at the second site once. You can take 15 units of fuels, drive to the first dump site, take 3 units from the first dump site, drive out to 8(now the second dump site), leave 5 units, return to the first dump site, pick up 3 units, and then return home. Next, you have 3 left over at the first dump site, 5 at the second dump site, and 15 in your base. Refuel at each dump site and you can make it to the end with no problem.
By gods mercy it took me 15 min to solve the problem the points are same at 3 and 5 but the process is different The idea if.anybody conciders.is Place 9 fuels at the 3 point marks and now go back. And place 5 at 8 marks by taking 3 from the 3rd mark.of the 9 placed and use the 3 while returning also so at point 3 we have 3 and at 8 we have 5 now we at point 0 fill and move on
It also works with cashes at each of the first 8 parsec marks. Drop off 3 units at stops 1-3 on the first run and drop off 1 unit at stops 4-8 on the second run. Then you'll have 1 unit of fuel at each of the first 8 parsecs for your final journey from the start to the alien base
Try moving as much dark matter forward one unit at a time: distance from original base: 0->1->2->3->4->5->6->7->8->23 fuel stored after depleting previous location: 45->40->35->30->27->24->21->18->15->0
You can also do it with the same points in a different way. Put 9 cells at 3 parsecs and go back so base. Refuel fully in base, then go to 3 parsecs, refuell the with 3 cells, go to 8 parsecs and deposit 5 cells there. Use the last 5 cells you have left to return to 3 parsecs and use three more of the cells stashed there to return home. Now you can refuel fully to 15 there, pickup the last 3 cells at 3 parsecs and the 5 cells at 8 parsecs to be fully fueled up at 8 parsecs
You can cache at every single parsec and end up with the same amount of fuel being used. Carrying 31-45 fuel one parsec forward takes 5 fuel (2 round trips and 1 one way). Carrying 16-30 fuel one parsec takes 3 fuel (1 round trip and 1 one way). You can go 3 parsecs in and have 30 fuel cached (5*3=15) and go a further 5 parsecs and have 15 fuel (3*5=15). Now that you are 8 parsecs in with 15 fuel, you can complete the total trip (8+15=23). Where you cache doesn't matter as long as you have enough fuel to make the trip, since it's a linear problem.
I love these not only because I’m a puzzle nerd and I *love* them, but also because my teaches love to play them, and I seem interesting when I tell them I watch them in my spare time.
Problem is, unless you found dark matter or an alternative propulsion system, how do you get back? Yeah sure you can wait and get rescued reasonable guy now shush /s
Can you save enough fuel to get back? Well, definitely not... but here's another riddle: How much fuel is initially needed to make a round trip? And, what is your strategy? Hint: you need more than 46 units. I could say even more than 60, but what's the actual amount needed?
In interstellar space, there would be almost nothing to slow you down. Once you leave a particular solar system, you should be able to plot a course that should leave you very free of anything to slow down for, and ergo, you need no fuel to keep going. You will keep going at the same speed until something slows you down. Actually, the fuel in question would be most necessary when slowing you back down, especially in combination with a gravitational assist.
Huh. I got something different. 1st trip: we go to parsec 5, drop off 5 off, and return with 0 (45 remaining at the spot) 2nd trip: repeat step 1 (30 left at location) 3rd trip: repeat again (15 left at location) There is now 15 fuel at parsec 5 Now we return to the location and refuel fully. We travel to parsec 5 (10 fuel left) and take 5 out of our cashe (fully fueled now). We then travel to parsec 8 using 3 fuel (12 left) and drop 9 off at parsec 8 With the remaining 3 fuel, we travel back to parsec 5 and take all of the remaining fuel at parsec 5. With 10 fuel, we use 3 to go to parsec 8 (now with 7 remaining) and get out all of our fuel (9 btw) and manage to get to the destination with 1 extra fuel.
Your solution uses 59 fuel (45 at base + 15 in tank assuming tank is full at start -1 fuel remaining getting to goal), while the solution on the video only uses 45 fuel. Your assumption that the tank is full at the start is incorrect.
Or instead with the same exact cache points you could... Alternatively go 3 parsecs, leave 9 fuel there, and go back like they said. But on your second trip go 3 parsecs, and instead refuel with 3 of the 9 fuel stashed there, leaving 6 left. Then go 5 parsecs to the 8 parsec cache as they stated, and leave 5 fuel, and return back to the 3 parsec cache as they said, but having no fuel left, you use 3 of the left over 6 fuel stashed at the 3 parsec cache, leaving 3 left. Then with your final trip, go 3 parsecs, and refuel with the 3 left stashed at the 3 parsec cache. Repeat this with the 5 fuel left at the 8 parsec cache. Then make it with exactly 15 left.
In specifying the problem, it left out the very important fact that you apparently didn't need to find a solution which left you with enough fuel to return home... You only needed to make a one way trip. And everything about "getting there first" is pure fluff, the problem only involves getting there at all.
I solved it, with the 3-8 points, but a bit differently. Like this: *First, come to the point 3, release 9 units fuel, go back. [3 forward, 3 back, 9 units in 3] *Go to 3(15-3=12), take 3 units(6 units left there), go point 8, 5 units drained, 10 units left. Cash 5 units, use other 5 to go back point 3. Take 3 units there and 3 units fuel are left at point 3. Go to start with 3 units. *Now that you have 3 units in 3 and 5 units in 8. Go to 3, you have 12 units, take 3 units there, you have full tank, go point 8, you consumed 5 units and 5 units are there. Fill your tank and go to alien space station. [8+15=23]
Load in 15, fly one step, drop 13, go back. Repeat. Load in Rest fly to point 1. Now you sit here with 40 units. Do the same again for point 2, you end up in point 2 with 35 units. Do the same again for point 3, you end up in point 3 with 30 units. On your way to point 4, you only have to load up your ship twice (2*15). So you only use up 3 units to carry everything to point 4. You end up in point 4 with 27 untis. Do the same again for point 5, you end up in point 5 with 24 units. Do the same again for point 6, you end up in point 6 with 21 units. Do the same again for point 7, you end up in point 7 with 18 units. Do the same again for point 8, you end up in point 8 with 15 units. Use up the remaining 15 units to fly to point 23, the alien base. Voila!
I came up with a similar solution. It used the same cache points, but left less fuel in the vacuum of space, so I’d consider it a more elegant solution. Jump to 3, cache 9, return 3. Pickup 15, jump 3, pickup 3 (leaving point 3 with 6 fuel left on it), jump 5, cache 5 (at point 8), return 5 (to point 3, which has 6 fuel cached), pickup 3, return 3. Pickup 15, jump 3, pickup the 3 fuel left at 3 to return to full, jump 5, pickup the 5 cached at 8 to return to full. Jump 15 to goal.
Similar but you could also have done this: 1.Take 15 units, go to parsec 3, drop off 9, go back to station 2.Take 15 more units, at parsec 3 pick up 3 of the 9 units, travel to parsec 8, drop off 5 units, go back to parsec 3 and pick up 3 of the 6 remaining units there, head back to station 3.Take the final 15 units from the station, pick up the last 3 units at parsec 3, travel to parsec 8, pick up the 5 units there, and head to 23.
I had an interesting solution which used a TON of cache points. I figured it doesn't matter _where_ the fuel gets dumped, just so long as we have the storage capacity to pick it up later (for example, if you just dumped 8 fuel cells at Parsec 1, you would only be able to pick up 1 cell on your next voyage). I realized that any point you pass on your first trip will be passed a total of 5 times (3 going forwards, 2 going back). So when we reach Parsec 1, we will be back here exactly 4 more times. So we leave 4 fuel cells. Same for Parsecs 2 & 3. At this point we're out of fuel, but we've left enough cashes behind that we can use them to fly back, picking up 1 fuel cell at each stop. For our next journey, we'll pick up one fuel cell at each parsec, leaving 2 behind. Once we get to Parsec 4, start leaving 2 behind at each point. Eventually we'll find ourselves at Parsec 8 with no fuel, but with 2 fuel cells at each Parsec 1-8. From here, it's simple to just grab one fuel cell per Parsec to get home, fuel up, fly to Parsec 8 (picking up one fuel cell per Parsec as we go) and make it all the way to the finish line. TED-Ed's solution has fewer stops, though.
my solution is similar but different first trip i go for 3, deposit 9, come back for 3 second trip i go for 3, refill 3 (leaving 6), go for another 5, deposit 5 (point 8 in space now), come back 5, refill 3 (leaving 3), come back for 3 last trip i go for 3, refill 3 (leaving 0), go for 5 and refill 5 (leaving 0), go for 15!
i found quicker way to do it. go to 3 drop 9 and go back. instead of doing that again go to 3 pick up 3. go to 8 drop 5 and go back to 3 to pick up three remaining that will bring you to ship. Then just finish the trip while picking up the 3 at 3 and 5 at 8 along the way
Let the distance travel in the three trips be a, a+b,a+b+c respectively. This makes a+b+c=23. Let the fuel drop at the 1st point, and 2nd point x and y. 2a+x=15...(1) 2a+2b-x+y=15...(2) a+b+c-y=15 => 23-y=15 => y=8...(3) Combining the three eqt, we get 2b-2x=-8 => x-b=4 Now we have many combinations, such as 5,1/6,2/7,3...13,9. (xx since the ship can't carry fuel greater than 15, only one combination is possible; x=5, b=1, such that a=5 when put into (1). Ans: a=5, b=1, c=17, x=5, y=8. (People might think a+b, the furthest distance travelled in the 2nd trip, should be 8, since 8+15=23; the truth is that it does not have to. The answer is 6, leaving c=17 which is more than the most fuel carried in a ship.)
This guy who narrates riddles should do audiobooks, I swear he has the most calming voice in the world
His voice is good but you cannot ignore the soothing voice kurzgesagt narrator.
Kurzgesagt narrator is the best
.
Go see jerryrigeverything
Uhm no he doesn't
Addison Anderson is my religion
And now you're stuck on that planet without dark matter... RIP
Exactly!! That's why I couldn't solve it because it was not a good option
Wait, so what? You only need the dark matter so that you can get to the planet faster than the other aliens. You can use your regular fuel to fly off after you collect the stuff.
@@blankname1 rewatched it i guess they don't but i feel it's implied.
Or... Or you could hitch a ride with the other ships in exchange for probably information
But now you have yarn
Ted Ed: Can you sol-
Me: No, I can’t.
We believe in you!
Loool
@@TEDEd i don't
@@TEDEd : l
TED-Ed you have too much faith in all of us
There’s a solution to this, and it’s really easy:
Get a better ship
But you're on a government funding, and broke.
Ozo
But where's the fun in that?
Don’t care
Fun doesn’t equal efficiency
ulu
This may come in handy when raiding area 51
So where we droppin,bois?
Just pretend you're a lost farmer.
Tedd M
Invite skinny medic to the raid
Dress as an Alien that guard will capture u.
What is wrong with you
There’s a problem with this:
How will you leave the space station without any fuel?
Well, once you get there first, you can wait for someone to come and give you a lift.
There might be fuel on the space station.
That's what I was thinking, you'll just be a sitting duck waiting for the aliens from other species to arrive, take the artifact from you, and leave you stranded.
@@MrEel-dc4kh Exactly. It's a space station, what else would it do?
The story says: as you go in you find some upgrades for the maximum dark matter fuel and 70 dark matter fuel
Whenever I click on these videos, I know I'm not going to be able to solve the riddle. I just want to know the answer
Same
I as well
Same
@Manuel yeah, this one's insultingly simple being told you've got three full tanks worth of fuel means three trips, which means two dropoffs and one real journey, then it's just about deciding how many parsecs away you'd need what quantity of fuel, and as mentioned there's multiple options there.
honestly i'm just angrily typing because i forgot these weren't worth clicking on.
I’m just here cuz they make me feel smart
Ted Ed: Can you solve it??
Me: Yeah I'm trying hard..!!
(Meanwhile other species from galaxy reached there and took all the precursor technology..)
Many days past.......
Me:----- Still trying to figure out the solution....!!!!!
This comment is just LEGEND 😂😂 lol
Clarence Viñalon
But really I m yet thinking for solution....
@@sanketpatil6711 lol
...Sometime later
Precursors wake up from their hidden hibernation chambers
Track down all the civilizations that took their stuff, wipe them out
Humans slip beneath the radar
Win!!
How about just build a ship with a goddamn bigger tank
Bigger tank means more mass and that will decrease the fuel-distance efficiency.
Are you the real elon musk
@@atulkumarashish584 It's not rocket science!
_wait_
Or have a fuel storage on board so you can refuel without leaving the ship
perfect that your profile name is Elon Musk hahah
Interesting. I did it in 3 separate trips.
The start was the same: at the 3 parsec mark, drop off 9 units, then return.
However, I deposited no more in that cache.
I went to that cache, picked up 3 units, went forwards 5 units, deposited 5 units, then returned to the cache, where I got the 3 units to get back to the station.
Then I went to each caches and refueled, leaving me with 15 units at the 8 parsec mark, like you did.
I did that as well
That was my solution as well, after deciding I liked it more than what I came up with first: leaving 2 at 6 parsecs and 3 at 8 parsecs instead of all 5 at 8 parsecs.
me too. both is similar
Mhm, got that too
Nerds
Probably not
More like a 100% not
Ay ur here
Impossible
One of you least liked comments
Where are the likes?
I Understood Nothing, Except..
“Universe’s Biggest Ball Of Yarn”
I was waiting for long for a riddle .
Thank you
ditto
Me: Use one thing of fuel. There's no friction in space.
Genius😂
Yeah, and they specifically mentioned it's completely empty space lol
You need fuel to go faster than light though. Assuming you can go >90% of the speed of light without fuel it would take you a little over 25 years to coast the last 8 parsecs.
@@lordofthecats6397 that's not the way it works, you won't slow down in outer space. Period. Well unless you hit a planet or a an asteroid or whatever. 1 unit of fuel sends you flying at a speed higher than that of light, but due to the fact you don't slow down you can just cut the engine.
No because these are JUMPS. So basically with 1 unit, you jump to 1 parsec, but you have 0 momentum so you're actually stuck
Wait...so while we were doing this, a bunch of other species were headed for this planet. Even if we managed to get there first, we’d be stuck there without fuel, waiting for the other species to arrive. And then when the other species DID arrive, we’d be at their mercy with no means of escape
Human: "Finders keep..."
(Aliens laser gun to death.)
Precursor technology includes teleporters
There, you find a better ship that can hold all 23 units of fuel... and then you remember you forgot to take all the other precursor technology
TedEd: “Can you solve the riddle?”
Me at 3am, with 3 Braincells left “‘(crying) nO i cANt”
@Protoxicalbedwarsguy he lost some along the way, apparently.
You actually have 3 braincells left? Wow, lucky
I literally see the word riddle in a ted-ed video and smile :)
Who doesn't?
I think that these riddles are the best thing about the channel(and how it's executed plays a major role too, so thanks😊).
Other educational youtube channels: makes viewers feel smarter.
Ted ed: makes viewers feel like morons
The premium quality we subscribed for!
When it comes to riddles, yes.😂
But TedEd is a very good channel.👍👍
Come on dude this one was easy
Agree to disagree. 😑
Yes..
Step 1: Confirm dark matter has green eyes
Step 2: Ask all the aliens to leave
Step 3 : Make bigger fuel tanks
Step 4: Find a way to get off the planet now that you have a larger fuel tank but no fuel
Step 7:Sell the Minecraft and make money and buy a antimatter girlfriend and make babies with her.
AMEER ROCKZZ Step 8: ozo
This joke never gets old
My solution was a bit different. Still works though. My approach was to carry everything to the next parsec one step at a time. You carry everything to parsec 1, losing 5 units on the way. Repeat from 1 to 2, 2 to 3. From now on you only need 3 units for transferring all left units. Repeat until you're at parsec 8 and you have exactly 15 units left.
Takes waaaay more trips than the solution here tho haha
I did the same thing
Concur. I thought the cost-per-distance was non-linear, paying -d to advance (15-2d) fuel by d positions, so we minimize cost by minimizing d. But it turns out that the granularity is actually full-tanks, not individual fuel.
Graphically, we can quickly see that the two solutions are equivalent, and furthermore generalize to other configurations of waystations. Consider a single full tank of 15 fuel. Think of it as 9 cargo + 6 self-fuel to move itself. Then you can pay 3+3 round-trip to advance the 9 cargo to position #3 in one jump-pair, as in the video, or (1+1)*3 to advance to position 3 in 3 jump-pairs.
It follows that we could do it in two jump-pairs, of distance 1 and 2 or vice versa. It also follows that this generalizes to the isomorphic problems of carrying groceries up a staircase in 1 trip or several trips, or mowing a lawn in horizontal or vertical stripes. In every case, you're covering the same total area, and all partitions of an area sum to the same area.
the problem is that the total distance traveled in order to get 23 parsecs away would be so much more under that method.
Yeah but you are saving units too. I solved by your method and saw that when you reach 8 parsec you actually have 18 units left which is 3 more than required to reach 23 parsec. There is one problem though which is you take more time and by the story that they told you need to reach their as soon as possible.
@@feartheghus That's not true. If you don't have any unused fuel from the 45 units anywhere by the end of the trip, then you've still travelled 45 parsecs no matter what way you do it.
4:05 but now you can't get back home because you're out of fuel :o
Yeah. Maybe it’s time to start thinking of how to make a space ship that can carry more fuel.
@@rottenroads1982you only need the dark matter fuel to get to the space station
Quote at the beginning of this video: by Dan Brown
My ad before this video: Dan Brown's Masterclass
Me: I see you universe.
I actually got this one, I came at it slightly different than they did. I figured the trip would take 2 fuel dumps and 1 full trip. So for the first trip I wanted to dump enough fuel to refuel at the first dump site 3 times. A dump site at 3 lets you use 15 fuel to dump 9 units of fuel at point 3 and return. Next I wanted to dump enough fuel at the second point to refuel at the second site once. You can take 15 units of fuels, drive to the first dump site, take 3 units from the first dump site, drive out to 8(now the second dump site), leave 5 units, return to the first dump site, pick up 3 units, and then return home. Next, you have 3 left over at the first dump site, 5 at the second dump site, and 15 in your base. Refuel at each dump site and you can make it to the end with no problem.
Exactly my solution
By gods mercy it took me 15 min to solve the problem the points are same at 3 and 5 but the process is different
The idea if.anybody conciders.is
Place 9 fuels at the 3 point marks and now go back. And place 5 at 8 marks by taking 3 from the 3rd mark.of the 9 placed and use the 3 while returning also so at point 3 we have 3 and at 8 we have 5 now we at point 0 fill and move on
Well thought!
Same solution as in the video just a different order, but yeah... also took me way too long to find x / the 3 point mark ^^
how I worked it out, took me a little longer before it occurred to me that on my second trip I could refuel twice from cache a.
I had the same idea! It is actually better because it is a shorter process I think...
I did the same
Spaceship girl : I wasted 60 units
Of jump fuel only for creating biggest ball of yarn in universe
Me: wait what??!!!
Not 60, but forty-five.
It also works with cashes at each of the first 8 parsec marks. Drop off 3 units at stops 1-3 on the first run and drop off 1 unit at stops 4-8 on the second run. Then you'll have 1 unit of fuel at each of the first 8 parsecs for your final journey from the start to the alien base
Universes biggest ball of yarn lmao I’m dying🤣 4:10
Try moving as much dark matter forward one unit at a time:
distance from original base: 0->1->2->3->4->5->6->7->8->23
fuel stored after depleting previous location: 45->40->35->30->27->24->21->18->15->0
*makes it to space station with no fuel and only finds yarn inside*
WELP
*uses yarn to make a noose*
You can also do it with the same points in a different way. Put 9 cells at 3 parsecs and go back so base. Refuel fully in base, then go to 3 parsecs, refuell the with 3 cells, go to 8 parsecs and deposit 5 cells there. Use the last 5 cells you have left to return to 3 parsecs and use three more of the cells stashed there to return home. Now you can refuel fully to 15 there, pickup the last 3 cells at 3 parsecs and the 5 cells at 8 parsecs to be fully fueled up at 8 parsecs
1:15
solution
Use a unit of fuel
then wait
the ship will continue moving because of inertia, says newton
I admit that Ted ed riddles are difficult... But this one is an exception...
This one was *IMPOSSIBLE* !!!
Step 1: load 15 units
Step 2: grab all that’s left and dump it on the floor of ur ship
Step 3: replace when u run out
I really love this animation style!
You can cache at every single parsec and end up with the same amount of fuel being used. Carrying 31-45 fuel one parsec forward takes 5 fuel (2 round trips and 1 one way). Carrying 16-30 fuel one parsec takes 3 fuel (1 round trip and 1 one way). You can go 3 parsecs in and have 30 fuel cached (5*3=15) and go a further 5 parsecs and have 15 fuel (3*5=15). Now that you are 8 parsecs in with 15 fuel, you can complete the total trip (8+15=23). Where you cache doesn't matter as long as you have enough fuel to make the trip, since it's a linear problem.
Am I gonna solve this? No.
Will I continue watching these riddles? Heck yeah!
I love these not only because I’m a puzzle nerd and I *love* them, but also because my teaches love to play them, and I seem interesting when I tell them I watch them in my spare time.
first riddle I actually solved on my own :P
Congrats The Seperator!
Well done!
Lies
@@stunnerkumah what
Cats of Wisdom he’s lying
I feel like the examinations paper questions should be like this and we're all will excited to study and everything lol😂
Wow the first hit sounds like a very popular meme
hmmmm
I wonder which meme is that
Melody Thieme it’s a bit like Area 51, everyone it rushing to get alien things
I read that in Dr Phil’s voice
Problem is, unless you found dark matter or an alternative propulsion system, how do you get back?
Yeah sure you can wait and get rescued reasonable guy now shush /s
You wanna know how I did it? This is how I did it Anton.
I never saved anything for the jump back.
Alien help, duh.
Use the precursor technology
Can you save enough fuel to get back? Well, definitely not... but here's another riddle: How much fuel is initially needed to make a round trip? And, what is your strategy? Hint: you need more than 46 units. I could say even more than 60, but what's the actual amount needed?
Hi ted-ed
Another interesting episode..
Thanks for educating us...🙏👍😊
In interstellar space, there would be almost nothing to slow you down. Once you leave a particular solar system, you should be able to plot a course that should leave you very free of anything to slow down for, and ergo, you need no fuel to keep going. You will keep going at the same speed until something slows you down. Actually, the fuel in question would be most necessary when slowing you back down, especially in combination with a gravitational assist.
Huh. I got something different.
1st trip: we go to parsec 5, drop off 5 off, and return with 0 (45 remaining at the spot)
2nd trip: repeat step 1 (30 left at location)
3rd trip: repeat again (15 left at location)
There is now 15 fuel at parsec 5
Now we return to the location and refuel fully. We travel to parsec 5 (10 fuel left) and take 5 out of our cashe (fully fueled now).
We then travel to parsec 8 using 3 fuel (12 left) and drop 9 off at parsec 8
With the remaining 3 fuel, we travel back to parsec 5 and take all of the remaining fuel at parsec 5. With 10 fuel, we use 3 to go to parsec 8 (now with 7 remaining) and get out all of our fuel (9 btw) and manage to get to the destination with 1 extra fuel.
Your solution uses 59 fuel (45 at base + 15 in tank assuming tank is full at start -1 fuel remaining getting to goal), while the solution on the video only uses 45 fuel. Your assumption that the tank is full at the start is incorrect.
Ahhhh. I see what I did wrong. Thanks for the tip
I got it in about 15 minutes or so by putting 18 on the 3, but instead of dumping 5 on the 8 spot I put 2 on 5 spot, 2 on 7 spot, and 1 on 8 spot.
I subscribed your channel because your riddles are good for brain exercise.
Even, if I can't solve it I atleast try.
And that gives me pleasure 😇
Good job!
Now solve this bonus riddle!
How the Fuddly do you get back to your base?
Why would I need extra fuel if there is nothing to slow me down in space...
Now, we need to go back
The answer to all of these titles?
No, I can't
But I'll still watch
Or instead with the same exact cache points you could...
Alternatively go 3 parsecs, leave 9 fuel there, and go back like they said. But on your second trip go 3 parsecs, and instead refuel with 3 of the 9 fuel stashed there, leaving 6 left. Then go 5 parsecs to the 8 parsec cache as they stated, and leave 5 fuel, and return back to the 3 parsec cache as they said, but having no fuel left, you use 3 of the left over 6 fuel stashed at the 3 parsec cache, leaving 3 left.
Then with your final trip, go 3 parsecs, and refuel with the 3 left stashed at the 3 parsec cache. Repeat this with the 5 fuel left at the 8 parsec cache. Then make it with exactly 15 left.
Alchemi713 this is exactly the solution I came up with !!
Just add an additional tank.
I solved this but the question was difficult to understand. It was not obvious that the fuel you need and the dark matter were 1 in the same.
In specifying the problem, it left out the very important fact that you apparently didn't need to find a solution which left you with enough fuel to return home... You only needed to make a one way trip. And everything about "getting there first" is pure fluff, the problem only involves getting there at all.
I solved it, with the 3-8 points, but a bit differently. Like this:
*First, come to the point 3, release 9 units fuel, go back. [3 forward, 3 back, 9 units in 3]
*Go to 3(15-3=12), take 3 units(6 units left there), go point 8, 5 units drained, 10 units left. Cash 5 units, use other 5 to go back point 3. Take 3 units there and 3 units fuel are left at point 3. Go to start with 3 units.
*Now that you have 3 units in 3 and 5 units in 8. Go to 3, you have 12 units, take 3 units there, you have full tank, go point 8, you consumed 5 units and 5 units are there. Fill your tank and go to alien space station. [8+15=23]
Wow...distances in parsecs are being travelled....
I hope one day, such space travels would come to reality...
This animation makes the video so easy to understand
*Ted-Ed:* Can you solve this riddle?
*Me:* _o_O_
0:22
Of course I’ve got a problem!
Load in 15, fly one step, drop 13, go back. Repeat. Load in Rest fly to point 1. Now you sit here with 40 units.
Do the same again for point 2, you end up in point 2 with 35 units.
Do the same again for point 3, you end up in point 3 with 30 units.
On your way to point 4, you only have to load up your ship twice (2*15). So you only use up 3 units to carry everything to point 4. You end up in point 4 with 27 untis.
Do the same again for point 5, you end up in point 5 with 24 units.
Do the same again for point 6, you end up in point 6 with 21 units.
Do the same again for point 7, you end up in point 7 with 18 units.
Do the same again for point 8, you end up in point 8 with 15 units.
Use up the remaining 15 units to fly to point 23, the alien base.
Voila!
Yeah since there's no cost to dump fuel, this one can be brute forced and just constantly move the stash up one at a time until reaching 8.
I love this!it helps me so much with science and problem solving in school!I LOVE THIS CHANNEL!!!!
I Solved it
Damn
The very first riddle from this channel I was able to solve
Feels So GOOD 😌
I came up with a similar solution. It used the same cache points, but left less fuel in the vacuum of space, so I’d consider it a more elegant solution. Jump to 3, cache 9, return 3. Pickup 15, jump 3, pickup 3 (leaving point 3 with 6 fuel left on it), jump 5, cache 5 (at point 8), return 5 (to point 3, which has 6 fuel cached), pickup 3, return 3. Pickup 15, jump 3, pickup the 3 fuel left at 3 to return to full, jump 5, pickup the 5 cached at 8 to return to full. Jump 15 to goal.
Is it just me or was that orange costume similar to Gordon Freeman's from *Half-Life,*
ONLY on St3roids.
Just you man!
Nope. Just you
If all the fuel is in those capsules just bring extra in the cabin
Ofcourse, what's next, The Einstein field and Topology of Hyperspace riddle?
I praise the people that can actually figure out these riddles
0:11 Looks like the Protheans weren't that clever...
Very nice, but a pity that you have no way to go back now. kkkkkkkkkk love this riddle!
Extension: What's the minimum fuel needed for 25 parsecs? How about 30?
My guess is 59 units for 25 parsecs and around 115.1 for 30 parsecs
@@enderyu
Correct on 25. 30 parsecs requires 118.
just store fuel in the hold. If it's stable in Deep space, it's stable in a barrel on the floor of the ship.
Bruh I was watching Hamilton animatics how tf did I end up here...
**confused speech**
I first read this comment and thought it said Hamilton animaniacs and was wondering when that show rebooted to have a reference to the Hamilton play
Similar but you could also have done this:
1.Take 15 units, go to parsec 3, drop off 9, go back to station
2.Take 15 more units, at parsec 3 pick up 3 of the 9 units, travel to parsec 8, drop off 5 units, go back to parsec 3 and pick up 3 of the 6 remaining units there, head back to station
3.Take the final 15 units from the station, pick up the last 3 units at parsec 3, travel to parsec 8, pick up the 5 units there, and head to 23.
Ted ed: can you solve it
Me : no but my friends can't solve this too😂💪
Finally!! This is the first ted ed riddles I can solve 😭😭
Ted Ed: can you solve the -
Me: no
Don’t you need just 1 unit of fuel since in space their is no resistance of any kind so your speed would be constant
I wear my Dan Brown Origins shirt to school!
Btw TedEd: Antimatter is not Dark Matter ~ Yours Truly, Convicted Science Geek
@Adam Filinovich Haha, though they used the quote on Anti instead of the actual Dark
I had an interesting solution which used a TON of cache points. I figured it doesn't matter _where_ the fuel gets dumped, just so long as we have the storage capacity to pick it up later (for example, if you just dumped 8 fuel cells at Parsec 1, you would only be able to pick up 1 cell on your next voyage).
I realized that any point you pass on your first trip will be passed a total of 5 times (3 going forwards, 2 going back). So when we reach Parsec 1, we will be back here exactly 4 more times. So we leave 4 fuel cells. Same for Parsecs 2 & 3. At this point we're out of fuel, but we've left enough cashes behind that we can use them to fly back, picking up 1 fuel cell at each stop.
For our next journey, we'll pick up one fuel cell at each parsec, leaving 2 behind. Once we get to Parsec 4, start leaving 2 behind at each point. Eventually we'll find ourselves at Parsec 8 with no fuel, but with 2 fuel cells at each Parsec 1-8.
From here, it's simple to just grab one fuel cell per Parsec to get home, fuel up, fly to Parsec 8 (picking up one fuel cell per Parsec as we go) and make it all the way to the finish line.
TED-Ed's solution has fewer stops, though.
Yay problem solved
now she is stuck in an ancient abandoned alien space station with no fuel to go back
my solution is similar but different
first trip i go for 3, deposit 9, come back for 3
second trip i go for 3, refill 3 (leaving 6), go for another 5, deposit 5 (point 8 in space now), come back 5, refill 3 (leaving 3), come back for 3
last trip i go for 3, refill 3 (leaving 0), go for 5 and refill 5 (leaving 0), go for 15!
I SOLVED IT!!!! THIS WAS MY FIRST TIME BEING ABLE TOO!!! YYYYYAAAAAYYYYY!!!!! IM SO HAPPY;!!!!!
i found quicker way to do it. go to 3 drop 9 and go back. instead of doing that again go to 3 pick up 3. go to 8 drop 5 and go back to 3 to pick up three remaining that will bring you to ship. Then just finish the trip while picking up the 3 at 3 and 5 at 8 along the way
Or you can just blast off, then shut off the engine because space is frictionless. 😏
You are traveling parsec's, many, many, many light years, you would hit the planet, and be in a universe, trillion's of year in the future,
Karthikeya 5638 *BECAUSE SPACE IS FRICTIONLESS.*
3rd riddle (I think) that I’ve gotten right from here. I feel so accomplished finishing these, thanks for making them!
“Alien facility filled with Precursor technology” Is this subnatica?
Finally someone said it. +1 point Xd
How about just placing the other units of fuel on the floor of the ship?
I'm just happy that we're finally admitting that the speed of light is not a limit.
Um...
Star Wars already did that 😁😁😁
Always excited for a riddle
We: How many unsolvable riddles do need for u to stop?
TEDed: YES!
What about getting back? Fuel up at the station?
3:45
XO?
This has become un-advertisement friendly...
Wdym?
DO MORE RIDDLES!!!
Congratulations, you're now stuck on an alien planet with no fuel to get home!
What if that was the plan? What if they didn't want to go home? They planned on a one way trip for colonization purposes. Terraforming they say.
If only they had the millennium falcon, otherwise they could have made the journey in 12 parsecs
*Dark matter matters!*
See what I did there💪😏
Let the distance travel in the three trips be a, a+b,a+b+c respectively. This makes a+b+c=23.
Let the fuel drop at the 1st point, and 2nd point x and y.
2a+x=15...(1)
2a+2b-x+y=15...(2)
a+b+c-y=15 => 23-y=15 => y=8...(3)
Combining the three eqt, we get
2b-2x=-8 => x-b=4
Now we have many combinations, such as 5,1/6,2/7,3...13,9. (xx since the ship can't carry fuel greater than 15, only one combination is possible; x=5, b=1, such that a=5 when put into (1).
Ans: a=5, b=1, c=17, x=5, y=8.
(People might think a+b, the furthest distance travelled in the 2nd trip, should be 8, since 8+15=23; the truth is that it does not have to. The answer is 6, leaving c=17 which is more than the most fuel carried in a ship.)
Ted Ed I just love your videos
Your finaly pumping out riddles again
Good luck coming back