Exactly 1 minute. Ants bumping into each other and turning around as described is essentially equivalent to the ants simply walking through each other. 1 comes from each direction, 1 leaves in each direction, and there is no transition time. Equivalent situations. Equivalent results. 1 minute.
if you think about it the answer is undetermined. the amount of ants going left and the amount going right are undetermined, the answer would change depending on that variable...
However, if the ants only have to be evenly spaced, then they can all be placed at the edge facing the edge at an even spacing of zero meters for a zero second solution. But if some of the ants have to be facing separate ways while overlapping that causes a paradox with the definition of collision. But the universe can't handle a paradox. Oh man you just destroyed the universe.
@antares5245 What the hell are you on dude? Watch the video. The ants are evenly spread over the stick. In order for 100 ants to be evenly spread in a 100cm they have to be in the middle of the cm marks. If you make the ants on the left look to the left and the ants on the right look to the right the longest way is 49,5cm. The ants walk 100cm/60seconds so it would take them 29,7 seconds. The longest way any ant would possibly walk is 99,5cm which would take 59,7 seconds.
Shortest time is incredibly hard to calculate...if the ants have to be positioned with the edge ants on the edge, then the shorted time is 49/101 minutes. If they have to be positioned equidistant from the edges as well as each other, then the shorted time is 51/103 minutes, slightly longer...
> if the ants have to be positioned with the edge ants on the edge, then the shorted time is 49/101 minutes
How did you get that? There are 99 gaps between the 100 ants. The most an ant has to cover is 49 gaps. That takes 49/99 of a minute, not 49/101. And 51/103? Where on earth did that come from?
assuming cane exactly 100cm and ant falls off at 100cm , Max is 60 seconds and depending how the ants are "spread out" the Min would about about 30 seconds (this depends where the middle ant is .. so between 29.5 and 30 seconds), I would say the spirit of the puzzle would be to realise the collisions mean nothing and the middle ant would take 30 seconds to reach the end... of course one could say that the longest is any number because ants just keep walking around cane and never would leave lol
It takes only as long as it takes Charlie Sheen to run his nose down the length of the cane... Charlie Sheen is still topical, right? Everybody still loves jokes that have his name in it.
well I'm not really thinking if I'm honest because it sounds like effort, but wouldn't it take longer if almost all ants collide with other ants than it would if there was only one collision at the furthest point, which at a glance seems to be how people are getting the longest time. I'm probably wrong but I don't know how to work it out quickly so I'll have to wait a few days I guess
You could assume that the last ant off will be the one in the middle, and the shortest that would take is half a minute if all ants started traveling toward the nearest end. Now lets look at travel distances for the middle ant. He first moves .5 cm to the right, then 1 cm left, then 1 right back and forth, until all the other ants are headed off. it takes .01 min for an ant near the end to turn around. 50 of them makes .5min. total max time= 1minute.
The last ants will be 0.5 cm away from the end and there will be a 1 cm gap between each ant the 50th ant will be 49.5 cm 100/49.5 = 2.0202....... and 1divided by that is 0.495 so the shortest is 0.495 sec
Well, if you think of it as when each ant hits each other and turns around, it is the same as the ants simply passing each other, so the longest time is 1 minute, the shortest being 30 seconds, if each ant was facing the closest end.
Either with every ant facing the opposite direction than those next to it, and with the ants at both ends facing inwards, or with all the ants facing the center
the imaginative min: 0 sec
All the ants jump off instantly
the imaginative max: infinite
This are mathematic ants, therfore smart ones, they can stay as long as they want if they decide to. Although I would expect a few missions to outer geometry
shortest time is 30s, if on the lest half every ant walk to the left and on the right side every ant walks to the right, no colisions, they just walk the 1/2meter in 1/2 minute
the longest time 12minuts and 15 seconds if you place the ants in pairs that face each other. the two middle ants would first hit each other, turn around and after a short walk hit their neighbours, turn around and go back and hit each other, turn around again go down a 1cm longer path, hit their neighbour and go back etc
you can think of the ants bouncing as a pass its easy to the maximum time is 1 minute if all of the ants are orientated so they all face away from the centre point then the minimum time will be distance from edge/1m m^-1
1 minute minimumif there not facing the nearest side because they are packed close enough that they dont move and keep spinning untill the stick is free to go thus meaning that they head towards thw nearest end meaning 1 minute. but if they all face the nearest end then it would be 30 seconds from the middle of the stick (half way) so half 1 minute? am i close?
just guessing off the top of my head...30s = shortest...
1m30s = longest (presuming the worst scenario is with all the ants facing toward the center) the pair of ants who start nearist the center will walk 1cm to each other, and then walk 1cm back to colide with next wave of ants, thus althogheter walking 2(50) + the 50 cm to the end, 150cm = 1m30s... PHEW! i hope i am right now, after all that..
Well, assuming that the left half is facing left, and the right half is facing right, then ½ an hour (still satisfying conditions). The oposite of that is if every other and is facing an oposite direction, with the two ends facing inward. In that case, each ant takes a ½ hour to go from the center to the end. Thus an ant bumping back and forth in the center must wait a ½ hour for EACH ant, then travel the entire length itself (it would bump back at that point). I can't calculate this myself.
You need a monocle if you're going to spin a cane like that!
There was a similar puzzle I remember from Puzzle Based Learning (you may remember me linking a video about it before) where an ant was travelling at a given speed on a piece of elastic. Each time the ant travelled a cm, the elastic will stretch a metre or something. The question was about if it could ever reach the end. I think I can get the exact question if interested.
let me see, fastest would be, each 50 looking in the right direction, never hitting anyone = 29.7 secs @ 1m/min. (because they only go 49.5cm)
Slowest should be disoptimal (great hint by Vellwander btw) being at least one outer ant taking 59.7 secs accross the complete 99,5cm (or meeting in the middle and walking back=99,5cm). Assuming they start at #1 at 0.5cm, #2 at 1.5cm ... #100 at 99.5cm.
So if they collide and go the other way that's the same as them passing through each other. So if one ant is pointing left but is on the right of the cane, it would take him a little less than* 1 minute. But if the left ants were pointing left and same with the right, that would mean that it would take slightly less than* 30 seconds to fall.
*It's slightly less than 30 seconds is because he said evenly spread, not the 2 middle ants are infinitely close together*I think this is the best one*WINK*
Shortest is 50 (50 ants each direction, facing the nearest end). Longest... I would assume is longer than 100 (one ant walks all the way to the end), but I cannot work out any reason why, so I'll just say 100.
so that's n ants walking left and 100-n ants walking right.
but i have math homework i need to do before class tomorrow and that is a bit more important than figuring this out. Although this problem will probably bug me later and if it does i'll end up trying to solve it anyway
I shall First assume there are 101 ants, as that then means they are all 1cm Spaced. the shortest time would be if the left side was facing left, and the right was facing right \\//. the middle ant would leave the stick in .5 of a Min. the left was facing right and the right was facing left (the middle is one of these directions doesn't matter) //\\ it would take Min for the middle ant to leave the stick . note in the case of /\\ the middle and the right leave the stick at the same time
Minimum is 30s, max is 1 min. when ants collide, since they are "mathematical" its the same as if they go straight through. therefore the longest it will take would be if an ant was positioned on the very end of the stick walking to the other end. this would take a minute
Lets see. Well two ants colliding and going back at the reverse is the same as them not colliding. Therefore the longest it will take is for one ant at one end to go off the other end, at 1m/min, about 1 minute. The shortest? well thats assuming all of them are facing the closest end, that would take roughly 30 seconds.
@iamdansin I figured the same with minor adjustments. My answer would be 29.7 seconds shortest and 59.7 longest as I am supposing initially no ant is not at the edge of the stick. So, the ant positions I am assuming are at 0.5, 1.5, 2.5.....99.5 cm from the edge.
@tiger55331 The process of bouncing is instantaneous. No matter how many collisions occur, the amount of time spent turning around is assumed to be too miniscule to take into account. The only issue with this problem is where the "point ants" start at each centimeter. Do they start in the middle of each centimeter? End? Or is 1 centimeter a point, so the ants would move 10 points every 6 seconds which would produce a more simple/exact answer of one minute (longest) . Just a guess BTW.
@iamdansin That's exactly what I was going to say. The longest is 1 minute, and the shortest, assuming the ants are equally spaced, and that they start with one on each end, is 49/99 minutes, or 29.6969... seconds.
@velocityeleven Right. Thats how I came up with 29.7 sec. To avoid spoiler for the rest of the people, I didn't put up this explanation. It would look like this:
| 0.5cm (* 1cm (* 1cm (*..............*) 1cm *) 1cm *) 0.5cm | ; where | is end of the canestick,(* an ant facing left, *) an ant facing right. (consider * to be an eye ;) )
When two ants collide they both switch their direction. This is equivalent to both ants just passing by each other, so we can ignore the collision factor. The longest it would take for the ants to walk off the ends of the cane is 1 minute - in case the furthermost ant(or two ants) on one side is facing the other direction.
The shortest amount of time it would take was already correctly answered.
Knew I should have come up with a solution for maximum possible time before reading any of the other comments.... no point in me trying to solve it now since your explanation makes perfect sense.
you are awesme btw!!!! I am very pleased to see somebody talking so passionately about maths, without any fear of getting any tag (nerd/geek etc etc etc). Reason why I called you awesome!
With this information we can compute that it will take 15 seconds for the ants at the very end to 'Walk Off'.
By now, basically every ant is now looking in its original direction, and initial position.
The time required for the final ants to fall off now will increase by 0.3 seconds as 1 cm of distance is increased for them to cover. This will form an AP with with 15.3 difference. Total time: 318.625 Minutes MAX
Well assuming the 50 on the left are facing left and the 50 on the right are facing right, (i.e. there are no collisions), it would take 30 seconds for all of them (specifically, for the innermost and final ants) to walk off their respective ends. That should be the minimum time unless I'm missing something obvious... I'll get back to you on the maximum.
Actually I take that back. The ants being evenly spaced is a little ambiguous. For the case of n=100 ants, the ants could be on the cm markers for 0, 1, 2, 3,... 99, with an empty cm on one end, or they could be on 1, 2, 3,... 99, 100, with an empty cm at the other. But the minimum time actually occurs when the ants are on 0.5, 1.5, 2.5, 3.5,... 99.5. The farthest any ant has to travel, then, is 49.5cm (from 49.5cm to 0 and from 50.5cm to 100). This actually gives min=29.7sec.
If there are fifty ants in each direction, all towards the right on the left, and all towards the left on the right, then:
Each and has a velocity of 0.6 cm/s. So for two ants coming towards each other, their relative velocity will be 1.2 cm/s.
From that we can calculate that in 0.3 s, it will cover a distance of 1m, hence causing a collision and resulting in the ant turning backwards. This ant will now cover some more distance before colliding with the ant that was behind it.
Walk off? So, if (or "when" because of a collision) an ant at an extreme end is walking outwards (i.e. in the direction of nearest end of the stick), it falls off right? So that suicidal mathematical ant will no longer be hindrance to others right?? PLEASE LET ME KNOW IF MY UNDERSTANDING IS RIGHT!
@robertmetaDOTcom Life is worth living when there is something to look forward to, there exist choices to experiment with. These ants dont even have 'choice' for direction, let alone the dimension!!! I would have been highly surprised had they not become suicidal. ;)
I agree with your answer for shortest, intuitively. Having said that, this is gonna sound stupid but... while your solution for "longest" seems (intuitively) that it'd make sense... I don't see any reason why it really would. I mean, can you prove that?
Also, can you give a specific amount of time for each of your answers?
For shortest time, no ant should collide with other ants, and since we will spread all ants every 1 cm, for last pair of ants it will take 30seconds.
For longest, I think I made mistake. it is 1min @iamdansin is right about longest time, it will take for the ant to go fro one end to other, that is all of them facing one end. If they face 50-50 to each other, it is same as shortest (given that changing direction takes zero seconds). All of them will face one side then will go out
For shortest time though, the last ants don't take 30 seconds to walk off the end. Consider the starting positions of each ant at 0.5cm, 1.5cm, 2.5cm,..., 49.5cm, 50.5cm,..., 98.5cm, 99.5cm. With this setup, all ants are equally spread across the cane (1cm apart), and the two ants that are furthest from their respective ends are at 49.5cm and 50.5cm. Both of those ants are 49.5cm away from the nearest end, meaning they have to travel slightly less than 1/2m... 49.5cm specifically.
This means it should take slightly less than 1/2 minutes to walk to their doom. The math happens to work out to be 29.7seconds, not 30.
As far as the longest goes... I believe 1 minute to be correct. Since 2 ants colliding with each other is mathematically equivalent to them just passing each other by, we can more or less ignore the whole collision aspect of the problem.
If we set the ants at 0, 1, 2,..., 99, all that matters is that the ant on the left faces right.
Or similarly we could set the ants @ 1, 2, 3,..., 100, and make sure that the ant @ 100cm is facing left. As long as either of those initial conditions is set, the longest it could take is 1 minute exactly (the time it'd take the final ant to travel all the way across the cane).
Are the ants moving in one single line? Or in multiple lines (meaning that are they going to collide if they are at the same distance from one end of the stick)?
And around how many ants would be moving in each direction at a time?
@PandeyNisheeth They are all in one line - so they are 1-dimensional ants as well! The ants are randomly facing left and right, but try it with 50 and 50 if you prefer.
Exactly 1 minute. Ants bumping into each other and turning around as described is essentially equivalent to the ants simply walking through each other. 1 comes from each direction, 1 leaves in each direction, and there is no transition time. Equivalent situations. Equivalent results. 1 minute.
gamegeekkl 2 days ago
0:36 just like the andriods in the Around the World music video.
deutschmaster21 3 days ago
oh god.....
DarthBane123 1 month ago
awesome puzzle! p.s you look like the twin weasley brothers in harry potter!
GeordieeBoyy 2 months ago
Shortest would be 30 seconds.
iGuruRevolution 4 months ago 3
if you think about it the answer is undetermined. the amount of ants going left and the amount going right are undetermined, the answer would change depending on that variable...
guppiemagic101 4 months ago
However, if the ants only have to be evenly spaced, then they can all be placed at the edge facing the edge at an even spacing of zero meters for a zero second solution. But if some of the ants have to be facing separate ways while overlapping that causes a paradox with the definition of collision. But the universe can't handle a paradox. Oh man you just destroyed the universe.
antares5245 5 months ago
@antares5245 What the hell are you on dude? Watch the video. The ants are evenly spread over the stick. In order for 100 ants to be evenly spread in a 100cm they have to be in the middle of the cm marks. If you make the ants on the left look to the left and the ants on the right look to the right the longest way is 49,5cm. The ants walk 100cm/60seconds so it would take them 29,7 seconds. The longest way any ant would possibly walk is 99,5cm which would take 59,7 seconds.
GammahooX 5 months ago
Shortest time is incredibly hard to calculate...if the ants have to be positioned with the edge ants on the edge, then the shorted time is 49/101 minutes. If they have to be positioned equidistant from the edges as well as each other, then the shorted time is 51/103 minutes, slightly longer...
antares5245 5 months ago
@antares5245
> if the ants have to be positioned with the edge ants on the edge, then the shorted time is 49/101 minutes
How did you get that? There are 99 gaps between the 100 ants. The most an ant has to cover is 49 gaps. That takes 49/99 of a minute, not 49/101. And 51/103? Where on earth did that come from?
WhiteHenny 5 months ago
Each ant after 1 cm
GSA14101996 5 months ago
Shortest 29.7 sec longest 59.7 sec Unless ( for longest u can have the ants one at the end and then
GSA14101996 5 months ago
Comment removed
CHTis87 5 months ago
assuming cane exactly 100cm and ant falls off at 100cm , Max is 60 seconds and depending how the ants are "spread out" the Min would about about 30 seconds (this depends where the middle ant is .. so between 29.5 and 30 seconds), I would say the spirit of the puzzle would be to realise the collisions mean nothing and the middle ant would take 30 seconds to reach the end... of course one could say that the longest is any number because ants just keep walking around cane and never would leave lol
TheLarryTangoShow 5 months ago
Min 30 Seconds , Max 60 Seconds
TheLarryTangoShow 5 months ago
minimum of 30 sec
TheSuperMagicTaco 5 months ago
I thought this was going to be the pattern-generating ants :p
FHomeBrew 5 months ago
"About" half a minute. They're all facing toward the closest end. ;^)
Kevbox2008 5 months ago
Comment removed
Kevbox2008 5 months ago
42 minutes and 30 seconds
jampk24 5 months ago
29.7 seconds scroll down for full explaination
GSA14101996 5 months ago
the shortest time is 30s, but the longest time sounds like a G.P. question to me... =\
JRocKinPENANG 5 months ago
29.7 seconds (tht im pretty sure of!) and a minute.....
my explanations for a min is too long to post here! myt jus b wrong!
geeta172 5 months ago
This reminds me a lot of the puzzle with the fly flying between two oncoming trains. I have a feeling the solution is also going to be similar.
GrimsditchImages 5 months ago
Chuck Norris would chop that stick in a hundered pcs in about 3 sec causing each ant to fall off .. so the answer is 3 seconds
pisspee73 5 months ago
i don't like the phrases "ABOUT 1 meter long" and "ABOUT 1 and per centimeter" in this question. :|
scarhead427 5 months ago
Ungh, not factorial. I meant 50 sec, + 49, + 48... etc. X 2
LoklarYsera 5 months ago
50 factorial x2 = longest ?
LoklarYsera 5 months ago
It takes only as long as it takes Charlie Sheen to run his nose down the length of the cane... Charlie Sheen is still topical, right? Everybody still loves jokes that have his name in it.
illustriouschin 5 months ago
Well if they are all facing toward the closest end: 30s
Longest. I'm stumped.
JustinRiedyk 5 months ago
does "pimping" mean the same thing in England?
knopf152 5 months ago
Is it 116 seconds?
TheWarcraftPro 5 months ago
well I'm not really thinking if I'm honest because it sounds like effort, but wouldn't it take longer if almost all ants collide with other ants than it would if there was only one collision at the furthest point, which at a glance seems to be how people are getting the longest time. I'm probably wrong but I don't know how to work it out quickly so I'll have to wait a few days I guess
IGaveMyNameToThatGuy 5 months ago
At least 30 seconds, at most 60 seconds. OH, you were talking about ants, not aunts! Not nearly as funny.
HaslamCorp 5 months ago
1 seconed if you flip it upside down :)
nicholishere 5 months ago 41
@nicholishere They're not walking off the ends then are they? Just falling off. ^.^
Qwubs 5 months ago
@nicholishere Have you ever tried flipping a stick with ants on upside down? They don't seem to notice, and just keep walking.
WhiteHenny 5 months ago
@nicholishere It says how long for them to walk off, not fall off. Also as has already been said, ants can cling on to a stick if it is upside down.
renardmigrant 2 months ago
30 seconds is the shortest time if the ant are facing the closest end and the two center ants are facing away from one another.
falconseye97 5 months ago
shortest: 29.7s
longest: 59.7s
LordSaradoc 5 months ago
Shortest time is 30 seconds. But longest... no idea how to go about doing that, aside from making thousands of different diagrams
elizze6 5 months ago
Shortest time = 30 seconds
Longest time = . . . not yet awake enough to figure out
sgtpoodlepants 5 months ago
You could assume that the last ant off will be the one in the middle, and the shortest that would take is half a minute if all ants started traveling toward the nearest end. Now lets look at travel distances for the middle ant. He first moves .5 cm to the right, then 1 cm left, then 1 right back and forth, until all the other ants are headed off. it takes .01 min for an ant near the end to turn around. 50 of them makes .5min. total max time= 1minute.
Thehuskarl 5 months ago
One APC as it's not known LOL at least the best you get in maths videos
GSA14101996 5 months ago
I wish that you where my math toucher. . .
Clapboardstudio 5 months ago
shortest 30 seconds
longest (guessing) 1minute 40 seconds??
the longest guess is kindof a longshot but i think i got the shortest one covered
CardMagicianJoeKing 5 months ago
0.495 min = 29.7 seconds is shortest time
GSA14101996 5 months ago
When I sai sec I meant minute
GSA14101996 5 months ago
The last ants will be 0.5 cm away from the end and there will be a 1 cm gap between each ant the 50th ant will be 49.5 cm 100/49.5 = 2.0202....... and 1divided by that is 0.495 so the shortest is 0.495 sec
GSA14101996 5 months ago
Well, if you think of it as when each ant hits each other and turns around, it is the same as the ants simply passing each other, so the longest time is 1 minute, the shortest being 30 seconds, if each ant was facing the closest end.
thekingcuber 5 months ago
the expected min: 30 sec
With every ant facing the nearest end
the expected max: 1 min
Either with every ant facing the opposite direction than those next to it, and with the ants at both ends facing inwards, or with all the ants facing the center
the imaginative min: 0 sec
All the ants jump off instantly
the imaginative max: infinite
This are mathematic ants, therfore smart ones, they can stay as long as they want if they decide to. Although I would expect a few missions to outer geometry
Theraot 5 months ago
This has been flagged as spam show
30 sec/12.5 min
Consider::
If the ants are at 0, and 100 and 1APC apart, we'd need 101 ants.
If both end ants are 1cm from their respective end, we'd need only 99 ants to maintain a 1cm distancing, but we have 100.
As the distancing is 1APC, there must be 0.5 cm from the left-most and right-most ants to the respective ends.
So, in the shortest situation, we arrive at 0.495 mins.
The longest scenario would be 12.495 mins
Spocknotch 5 months ago
30 sec/12.5 min
Consider::
If the ants are at 0, and 100 and 1APC apart, we'd need 101 ants.
If both end ants are 1cm from their respective end, we'd need only 99 ants to maintain a 1cm distancing, but we have 100.
As the distancing is 1APC, there must be 0.5 cm from the left-most and right-most ants to the respective ends.
So, in the shortest situation, we arrive at 0.495 mins.
The longest scenario would be 12.495 mins
Spocknotch 5 months ago
shortest time is 30s, if on the lest half every ant walk to the left and on the right side every ant walks to the right, no colisions, they just walk the 1/2meter in 1/2 minute
the longest time 12minuts and 15 seconds if you place the ants in pairs that face each other. the two middle ants would first hit each other, turn around and after a short walk hit their neighbours, turn around and go back and hit each other, turn around again go down a 1cm longer path, hit their neighbour and go back etc
WhiteDragonTile 5 months ago
@WhiteDragonTile or is it 24.5 minutes... I'm not sure anymore.... I'm gonna stick with 12.25 minutes
WhiteDragonTile 5 months ago
you can think of the ants bouncing as a pass its easy to the maximum time is 1 minute if all of the ants are orientated so they all face away from the centre point then the minimum time will be distance from edge/1m m^-1
cycloidpoi 5 months ago
I'm going to say it will take from 30 secs to 1 min.
oliethefolie 5 months ago
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oliethefolie 5 months ago
I may have made a mistake but I think the longest is 35minutes 41.66666seconds and the shortest is 30 seconds.
OliphanOracus 5 months ago
30 sec for shortest possible time
LewisAM37 5 months ago
30 sec. - 1 min
toon2912 5 months ago
Every single ant is facing the edge with a minimal distance and will go down thats as minimal as we want basicly 0 sec.
I havent got so much time but i have a feeling that 1 min will be enough every time
fodorfa 5 months ago
1 min? Im not a maths man lol
Lewis9547 5 months ago
Half an hour
Lewis9547 5 months ago
Miniute
Lewis9547 5 months ago
1 min? Im not a maths man lol
Lewis9547 5 months ago
1 minute minimumif there not facing the nearest side because they are packed close enough that they dont move and keep spinning untill the stick is free to go thus meaning that they head towards thw nearest end meaning 1 minute. but if they all face the nearest end then it would be 30 seconds from the middle of the stick (half way) so half 1 minute? am i close?
St34DyB0y1409 5 months ago
just guessing off the top of my head...30s = shortest...
1m30s = longest (presuming the worst scenario is with all the ants facing toward the center) the pair of ants who start nearist the center will walk 1cm to each other, and then walk 1cm back to colide with next wave of ants, thus althogheter walking 2(50) + the 50 cm to the end, 150cm = 1m30s... PHEW! i hope i am right now, after all that..
OisinCAR 5 months ago
Well, assuming that the left half is facing left, and the right half is facing right, then ½ an hour (still satisfying conditions). The oposite of that is if every other and is facing an oposite direction, with the two ends facing inward. In that case, each ant takes a ½ hour to go from the center to the end. Thus an ant bumping back and forth in the center must wait a ½ hour for EACH ant, then travel the entire length itself (it would bump back at that point). I can't calculate this myself.
EighteenCharacters 5 months ago
If exactly 50 are facing left, and 50 are facing right, should take half a minute for shortest time :)
Ryzie123 5 months ago
Longest 1 hour, shortest, half an hour. (I think I am right.)
Rockwarrior2004 5 months ago
You need a monocle if you're going to spin a cane like that!
There was a similar puzzle I remember from Puzzle Based Learning (you may remember me linking a video about it before) where an ant was travelling at a given speed on a piece of elastic. Each time the ant travelled a cm, the elastic will stretch a metre or something. The question was about if it could ever reach the end. I think I can get the exact question if interested.
WhiteRAZOR 5 months ago
let me see, fastest would be, each 50 looking in the right direction, never hitting anyone = 29.7 secs @ 1m/min. (because they only go 49.5cm)
Slowest should be disoptimal (great hint by Vellwander btw) being at least one outer ant taking 59.7 secs accross the complete 99,5cm (or meeting in the middle and walking back=99,5cm). Assuming they start at #1 at 0.5cm, #2 at 1.5cm ... #100 at 99.5cm.
Florence00pi 5 months ago
This has been flagged as spam show
wow. i filled the 500 characters remaining limit
NikWillOrStuff 5 months ago
So if they collide and go the other way that's the same as them passing through each other. So if one ant is pointing left but is on the right of the cane, it would take him a little less than* 1 minute. But if the left ants were pointing left and same with the right, that would mean that it would take slightly less than* 30 seconds to fall.
*It's slightly less than 30 seconds is because he said evenly spread, not the 2 middle ants are infinitely close together*I think this is the best one*WINK*
NikWillOrStuff 5 months ago
@NikWillOrStuff wow. i filled the 500 characters remaining limit
NikWillOrStuff 5 months ago
@NikWillOrStuff Oh yea i forgot to say, I was assuming the cane is exactly 1 meter
NikWillOrStuff 5 months ago
Shortest - 30 seconds
Longest - 1 minute
larxiz 5 months ago
Shortest is 50 (50 ants each direction, facing the nearest end). Longest... I would assume is longer than 100 (one ant walks all the way to the end), but I cannot work out any reason why, so I'll just say 100.
AdderSIG 5 months ago
@AdderSIG Scale error. Calculated it in some wave-number style unit. 30 seconds and 1 minute.
AdderSIG 5 months ago
so that's n ants walking left and 100-n ants walking right.
but i have math homework i need to do before class tomorrow and that is a bit more important than figuring this out. Although this problem will probably bug me later and if it does i'll end up trying to solve it anyway
LostJourneyer 5 months ago
I shall First assume there are 101 ants, as that then means they are all 1cm Spaced. the shortest time would be if the left side was facing left, and the right was facing right \\//. the middle ant would leave the stick in .5 of a Min. the left was facing right and the right was facing left (the middle is one of these directions doesn't matter) //\\ it would take Min for the middle ant to leave the stick . note in the case of /\\ the middle and the right leave the stick at the same time
nitrodavid 5 months ago
the sshortest is half a minute right?
123amsterdan456 5 months ago
100 minutes
lukazd99 5 months ago
i'm gonna say 2minutes as each ant on the end will only turn around max once then drop leaving the ant previous to turn once max and drop so 1 minute
burny1 5 months ago
Minimum is 30s, max is 1 min. when ants collide, since they are "mathematical" its the same as if they go straight through. therefore the longest it will take would be if an ant was positioned on the very end of the stick walking to the other end. this would take a minute
thekillerhjkhjk 5 months ago
shortest would be 1 minute.. the ant goes straight from end to end..
longest will be the ant will collide with each other on each cm..
so it will be a permutation & combination of collision..
which will take a hell lot of time for that ant to reach the end.. lol
demon7s 5 months ago
@demon7s shortest is 30 seconds. longest is 1 min
bscutajar 5 months ago
@bscutajar lol.. that means i didnt understand the question..
demon7s 5 months ago
Shortest is 30 seconds easy but the longest..... I havent a clue!
alewisgb 5 months ago
why would ants walk of this makes no sense i unsubbed this is fake
MrBabacus 5 months ago
@MrBabacus is a mathematical problem!!
alewisgb 5 months ago
Lets see. Well two ants colliding and going back at the reverse is the same as them not colliding. Therefore the longest it will take is for one ant at one end to go off the other end, at 1m/min, about 1 minute. The shortest? well thats assuming all of them are facing the closest end, that would take roughly 30 seconds.
iamdansin 5 months ago 40
@iamdansin I figured the same with minor adjustments. My answer would be 29.7 seconds shortest and 59.7 longest as I am supposing initially no ant is not at the edge of the stick. So, the ant positions I am assuming are at 0.5, 1.5, 2.5.....99.5 cm from the edge.
mihassan 5 months ago
@iamdansin What if many were bouncing there and back in the middle, waiting for others to fall off?
tiger55331 5 months ago
@tiger55331 The process of bouncing is instantaneous. No matter how many collisions occur, the amount of time spent turning around is assumed to be too miniscule to take into account. The only issue with this problem is where the "point ants" start at each centimeter. Do they start in the middle of each centimeter? End? Or is 1 centimeter a point, so the ants would move 10 points every 6 seconds which would produce a more simple/exact answer of one minute (longest) . Just a guess BTW.
TehMathWiztard 5 months ago
@iamdansin That's exactly what I was going to say. The longest is 1 minute, and the shortest, assuming the ants are equally spaced, and that they start with one on each end, is 49/99 minutes, or 29.6969... seconds.
WhiteHenny 5 months ago
If I remembered the maths I learned 5 years ago, I could do this... but I am an old man now..
Xerxes4242 5 months ago
@velocityeleven Right. Thats how I came up with 29.7 sec. To avoid spoiler for the rest of the people, I didn't put up this explanation. It would look like this:
| 0.5cm (* 1cm (* 1cm (*..............*) 1cm *) 1cm *) 0.5cm | ; where | is end of the canestick,(* an ant facing left, *) an ant facing right. (consider * to be an eye ;) )
vedhasp 5 months ago in playlist More videos from singingbanana
When two ants collide they both switch their direction. This is equivalent to both ants just passing by each other, so we can ignore the collision factor. The longest it would take for the ants to walk off the ends of the cane is 1 minute - in case the furthermost ant(or two ants) on one side is facing the other direction.
The shortest amount of time it would take was already correctly answered.
romantz1 5 months ago
@romantz1
Knew I should have come up with a solution for maximum possible time before reading any of the other comments.... no point in me trying to solve it now since your explanation makes perfect sense.
CharBroiled04 5 months ago
@romantz1 good way to think about it! Cool!!! :)
vedhasp 5 months ago in playlist More videos from singingbanana
you are awesme btw!!!! I am very pleased to see somebody talking so passionately about maths, without any fear of getting any tag (nerd/geek etc etc etc). Reason why I called you awesome!
vedhasp 5 months ago in playlist More videos from singingbanana
the shortest amount of time for them to fall off would involve every any facing the closest edge... so it'd be slightly less than half a minute
I imagine the ants will not all initially lie on the integer amount of centimetres since otherwise there would be 101 ants
velocityeleven 5 months ago
Continuing from previous:
This will also take 0.3 seconds.
With this information we can compute that it will take 15 seconds for the ants at the very end to 'Walk Off'.
By now, basically every ant is now looking in its original direction, and initial position.
The time required for the final ants to fall off now will increase by 0.3 seconds as 1 cm of distance is increased for them to cover. This will form an AP with with 15.3 difference. Total time: 318.625 Minutes MAX
PandeyNisheeth 5 months ago
right???
vedhasp 5 months ago in playlist More videos from singingbanana
shortest 29.7 sec :)
vedhasp 5 months ago in playlist More videos from singingbanana
Well assuming the 50 on the left are facing left and the 50 on the right are facing right, (i.e. there are no collisions), it would take 30 seconds for all of them (specifically, for the innermost and final ants) to walk off their respective ends. That should be the minimum time unless I'm missing something obvious... I'll get back to you on the maximum.
CharBroiled04 5 months ago
@CharBroiled04
Actually I take that back. The ants being evenly spaced is a little ambiguous. For the case of n=100 ants, the ants could be on the cm markers for 0, 1, 2, 3,... 99, with an empty cm on one end, or they could be on 1, 2, 3,... 99, 100, with an empty cm at the other. But the minimum time actually occurs when the ants are on 0.5, 1.5, 2.5, 3.5,... 99.5. The farthest any ant has to travel, then, is 49.5cm (from 49.5cm to 0 and from 50.5cm to 100). This actually gives min=29.7sec.
CharBroiled04 5 months ago
If there are fifty ants in each direction, all towards the right on the left, and all towards the left on the right, then:
Each and has a velocity of 0.6 cm/s. So for two ants coming towards each other, their relative velocity will be 1.2 cm/s.
From that we can calculate that in 0.3 s, it will cover a distance of 1m, hence causing a collision and resulting in the ant turning backwards. This ant will now cover some more distance before colliding with the ant that was behind it.
(in next post)
PandeyNisheeth 5 months ago
Shortest 27.3 secs
mbcx9ts2 5 months ago
Walk off? So, if (or "when" because of a collision) an ant at an extreme end is walking outwards (i.e. in the direction of nearest end of the stick), it falls off right? So that suicidal mathematical ant will no longer be hindrance to others right?? PLEASE LET ME KNOW IF MY UNDERSTANDING IS RIGHT!
vedhasp 5 months ago
@vedhasp That's right, suicidal 1-dimensional mathematical point ants.
singingbanana 5 months ago 28
@singingbanana Forget the puzzle, why are the mathematical point ants so SAD! Come on Ants, life is worth living!
robertmetaDOTcom 5 months ago
@robertmetaDOTcom Life is worth living when there is something to look forward to, there exist choices to experiment with. These ants dont even have 'choice' for direction, let alone the dimension!!! I would have been highly surprised had they not become suicidal. ;)
vedhasp 5 months ago in playlist More videos from singingbanana
one minute
mkaatr 5 months ago
Longest 100 minutes, shortest 50 minutes.
n4v32x 5 months ago
Shortest: when 50 ants are on right side facing opposite to other 50 ants on the left side.
longest: when those two groups of 50 ants face the other group of 50.
hassanfasl 5 months ago
@hassanfasl
I agree with your answer for shortest, intuitively. Having said that, this is gonna sound stupid but... while your solution for "longest" seems (intuitively) that it'd make sense... I don't see any reason why it really would. I mean, can you prove that?
Also, can you give a specific amount of time for each of your answers?
CharBroiled04 5 months ago
@CharBroiled04
For shortest time, no ant should collide with other ants, and since we will spread all ants every 1 cm, for last pair of ants it will take 30seconds.
For longest, I think I made mistake. it is 1min @iamdansin is right about longest time, it will take for the ant to go fro one end to other, that is all of them facing one end. If they face 50-50 to each other, it is same as shortest (given that changing direction takes zero seconds). All of them will face one side then will go out
hassanfasl 5 months ago
@hassanfasl
For shortest time though, the last ants don't take 30 seconds to walk off the end. Consider the starting positions of each ant at 0.5cm, 1.5cm, 2.5cm,..., 49.5cm, 50.5cm,..., 98.5cm, 99.5cm. With this setup, all ants are equally spread across the cane (1cm apart), and the two ants that are furthest from their respective ends are at 49.5cm and 50.5cm. Both of those ants are 49.5cm away from the nearest end, meaning they have to travel slightly less than 1/2m... 49.5cm specifically.
CharBroiled04 5 months ago
@CharBroiled04 (continued)
This means it should take slightly less than 1/2 minutes to walk to their doom. The math happens to work out to be 29.7seconds, not 30.
As far as the longest goes... I believe 1 minute to be correct. Since 2 ants colliding with each other is mathematically equivalent to them just passing each other by, we can more or less ignore the whole collision aspect of the problem.
If we set the ants at 0, 1, 2,..., 99, all that matters is that the ant on the left faces right.
CharBroiled04 5 months ago
@CharBroiled04 (continued)
Or similarly we could set the ants @ 1, 2, 3,..., 100, and make sure that the ant @ 100cm is facing left. As long as either of those initial conditions is set, the longest it could take is 1 minute exactly (the time it'd take the final ant to travel all the way across the cane).
CharBroiled04 5 months ago
@CharBroiled04
Well explained. I don't see a reason why not. :)
hassanfasl 5 months ago
wow it is difficult
1dexlab 5 months ago
Hint: A collision is mathematically equivalent to the ants passing through each other ;)
Vellwander 5 months ago 7
@Vellwander This "hint" made it all too easy.
And here I was getting ready to simulate a few hundred trials. :(
iambored404 5 months ago
i think i have the same stick as u :)
scott2e4c 5 months ago
Are the ants moving in one single line? Or in multiple lines (meaning that are they going to collide if they are at the same distance from one end of the stick)?
And around how many ants would be moving in each direction at a time?
PandeyNisheeth 5 months ago
@PandeyNisheeth They are all in one line - so they are 1-dimensional ants as well! The ants are randomly facing left and right, but try it with 50 and 50 if you prefer.
singingbanana 5 months ago
@singingbanana then it will be half a minute for shortest amount of time.
hassanfasl 5 months ago
@singingbanana actually since they are point ants, they are 0 dimensional ants
bscutajar 5 months ago