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This makes me think of the 100 prisoners dilemma. It would be interesting to look at a generalization of the math in that, and actual practical implementation potential. It is such a huuuge improvement compared to intuition, and difficult to understand until the math is spelled out. You know, there are 100 prisoners individually uniquely numbered 1 to 100 on their backs. And there's a room with 100 boxes that each has a paper in it. Boxes and papers are all also uniquely numbered 1 to 100. The prisoners go in there one at a time and cannot communicate with anyone after that. The prisoner in that room has to find the box what contains the paper that has his prison number. He may look into only 50 out of the 100 boxes. So 50% chance, right? Thing is, that ALL the 100 prisoners have to find their own number in one out of those 50 boxes they are allowed to look into. Or they will be senteced badly (because the judge is on vacation and they cannot afford a lawyer). So that loos like 0.5^100 probability which never happens. BUT! There is a strategy, which they all together could agree upon before they enter the room individually, and cuts all feedback about anything that has to do with the boxes. And no prisoner is allowed to move any paper och mark any box or anything like that. Any prisoner could be the first to enter the room of doom. A strategy that results i about 1/3 or 30% probability that ALL 100 OF THEM FIND THEIR OWN NUMBER! By each one of them looking into only 50 out of 100 boxes. Isn't that fantastic? Let's all here come together and build some kind of AI RarearthMineral Com Startup Sick Valley company, or at least a casino, based upon this discovery!!!
Yes sure! Fire your torpedo on the number line horizontally (or in the same direction of the line)! Make sure your torpedo's velocity is much more faster than the submarine.
You forgot to say that the initial position is an integer. Without that constraint, it can be any real number and Cantor's diagonalization argument makes the sub impossible to guarantee a hit even if velocity is 0. This is such an obvious mistake that i feel you're engagement farming and ready to have a part 2 that's like 'pshych!'.
@WrathofMath, I have a probability problem that I think it would be interesting to solve. Say there are 100 people sitting at a round table for dinner. Each person has a napkin to the right and left of them, totaling 100 napkins in all. For this problem, the chance that an individual picks either the right napkin or the left napkin is 50%, the chance a person will pick a napkin for any given second, assuming they haven't already taken one, is 5%, the choice of a particular person is not influenced by the choices of anyone else, and a person cannot stand up to grab a napkin or take a napkin that is not adjacent to their seat. With all the assumptions aside, here are the questions. A, at the end of the experiment, what is the probability that all the people have a napkin? B, at the end of the experiment, what is the average amount of people that will not have a napkin? and C, is there any particular order that you could place people in their seats to minimize the number of people who don't get a napkin? I hope you see this, and if you do, I hope it makes a good video! P.S. upon further research, I did find that someone in 2003 came up with a similar problem, but I used that source only for a third question. Here is that article describing the problem: akc.is/papers/007-Conways-napkin-problem.pdf
@@thierrypauwels thanks for this scenario! There are some complicated ways I could expand on that issue, but for simplicities sake, let's say it's a 50% chance person 1 gets the napkin and likewise for person 2.
When i first saw the video, I thought this was a puzzle where both the torpedo and the submarine are described by linear three dimensional equations and you have to determine which of the intersections between the lines can actually happen at the same time
is this a pseudo collaboration with Independent Quark? or did both of you stumble upon the same textbook. always good to have different angles to the same solution. very unique take on it. thinking about it was Pk(k) helps me see the link to cantors diagonalization just that lil bit better
I like a lot of your videos, but I don't understand the solution of this one at all. Highest math level I got to was Calculus, so maybe that has something to do with it.
I think the think to keep in mind is that on the graph, you aren't plotting positions on the line directly. You're plotting possible positions of the submarine. Instead of saying "I'm going to shoot at position 0 then 1 then 2 etc" you're saying that you're going to shoot at where a submarine would be if it started at 0 with a velocity of 0 , then where it would be if it started at 0 with a velocity of 1, then a position of 0 with a velocity of -1, and so on. Since the number of possible submarines is "countable", you will (eventually, given an infinite amount of time) pick the right submarine and shoot it.
i'm going to use p to denote the initial position, then using the methodology given in the video... at t=0 we have p=0 and v=0, so we'd guess 0+0*0 = 0 at t=1 (following the spiral shown in the vid) we have p=1 and v=0 so we'd guess 1 + 1*0 = 1 at t=2 we have p=1 and v=1 so we'd guess 1 + 1*2 = 3 at t=3 we have p=0 and v=1 .... and so on
Every possible v and P0 combination is hit with the spiral eventually so neither the starting position nor the speed need to be known to be intercepted. The t is the same for the shooter and for the submarine.
@ actually not because in the line graph example the submarine moved from turn to turn. but since the coordinates in the graph with the spiral don't correspond to the current sub position but instead to all possible submarines, the target submarine is 'stationary' in that coordinate system. So with the line it's "we can hit every possible point, but we are probably outrun" and with the 2D graph it's "Its one of these and it's not changing to another, we just have to put the current time in the each equasion and shoot at the place that the sub we are cheking for would be by now. that way we will hit all possible subs."
Infinity screws with my mind. I came up with the spiral solution also. It's almost paradoxical to me that even if a spiral segment takes longer each time it goes around, it will eventually catch up to the submarine. I think the confusing part for me is that the submarine position could be written as P₀(start,velocity,time) = start + velocity × time with 3 variables. But for us, the only variable we get is time, and we're mapping that to an (x,y) If we have X(time) = x associated with that time Y(time) = y associated with that time Our torpedo position could be P₁(time) = P₀(X(time),Y(time),time) I guess I just have an infantesimal understanding of Infinity (pun intended)
Ah, I hoped this would be projectile calculation applicable to video games. I'll give you more interesting math problem (military and for video games). Imagine tower defense archer shooting at constantly moving infantry target. Projectile has to counteract gravity to hit target. Archer position, target position, target velocity and gravity is known. Easier version has known projectile X speed, but more interesting and harder version has known projectile velocity magnitude.
Hmm. We don't know how far away we are from P0, and we don't know the torpedo velocity. Whatever those numbers are, I'm not convinced we can hit every point on the spiral.
every point on the spiral is a combination of starting point and starting velocity. so somewhere on that point is the sub. and since the sub doesn't change its starting point or velocity mid way, our spiral will hit that combination eventually. where will the sub be by that time? we know the time t that has elapsed. we can put it into every combination and fire at the solution to. startingpoint + time x velocity. The spiral isn't catching up with the subs velocity, its eliminating every starting combination one by one.
If you hit (0,0) it will remove every possibility along the line where P0 = -v. You could then skip those options as you spiral. If you do it again, you remove every option along P0 = -2v. However, nothing is special about (0,0), so you could do this for every point you hit as you spiral - marking other solutions you already ruled out to skip over as you spiral, letting you catch the submarine "faster". However, as the majority of the options ruled out are getting further away as time goes on, it doesn't seem to significantly speed up the spiral as I try to do it mentally. It also can't efficiently rule out options in the first and third quadrents.
you hit every strategy ((P0,v) pair) that lies on lile with slope -t that goes throught the point you fired on I also wonder if you could optimize the hitting pattern by listing strategies by changing the way you create the "strategy space" like using P=P10 + v*(t-10) so you could actualy eliminate points in first and third quadrents but with infinity it probably wouldn't matter
@josefuher6117 the second and fourth quadrant represent strategies that move towards 0, while the first and third represent ones that move away from 0. Any move that is made can only remove a finite amount of possibilities in the first and third. This is best understood when you try to think of a point you could choose that would hit possibilities in both the first and third quadrant and why you can't.
Also, should not be infinite, since no submarine can move faster than the speed of light. There is a finite number of integer velocities below the speed of light. Thinking about that; if the submarine were moving very above 50% of speed of light and you start launching torpedos from 0, by the time to get close to 50% the speed of light you would have to move close to the speed of light back and forwars, or launch torpedos that move close to the speed of light; since the speed of light is the limit, a submarine movien above 50% the speed of light would not be hit with such method without breaking the laws of phisicas as we know them. I'd suggest, get 1 more torpedo to chech 2 possitions each time instead of 1 at the time so the speed of light would not be a limit.
But if you reach the heat death of the universe before getting to the correct position function, is your argument still valid? Also, constants should be capital letters and should come before variables, so P(t) = P0 + Vt, not P0 + tv.
This is a really cool theoretical puzzle. But in practice submarines aren’t really constrained to integer speeds and locations. Even if they were there is no guarantee the strategy would hit the submarine before the sub would do the damage we feared.
And then fire the next torpedo from negative infinity towards zero. If nothing else, you'll hit yourself! Several submarines actually have hit themselves when the rudder of the torpedo they launched was stuck. And thus made a big circle to come back home to Daddy. Boom! That was actually an important problem (for guys in submerged coffins) during the first half of the 20th century. It was launch and then get outta here: Full speed ahead, dive dive dive, help help help!
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This makes me think of the 100 prisoners dilemma. It would be interesting to look at a generalization of the math in that, and actual practical implementation potential. It is such a huuuge improvement compared to intuition, and difficult to understand until the math is spelled out.
You know, there are 100 prisoners individually uniquely numbered 1 to 100 on their backs. And there's a room with 100 boxes that each has a paper in it. Boxes and papers are all also uniquely numbered 1 to 100.
The prisoners go in there one at a time and cannot communicate with anyone after that. The prisoner in that room has to find the box what contains the paper that has his prison number. He may look into only 50 out of the 100 boxes. So 50% chance, right?
Thing is, that ALL the 100 prisoners have to find their own number in one out of those 50 boxes they are allowed to look into. Or they will be senteced badly (because the judge is on vacation and they cannot afford a lawyer). So that loos like 0.5^100 probability which never happens.
BUT! There is a strategy, which they all together could agree upon before they enter the room individually, and cuts all feedback about anything that has to do with the boxes. And no prisoner is allowed to move any paper och mark any box or anything like that. Any prisoner could be the first to enter the room of doom. A strategy that results i about 1/3 or 30% probability that ALL 100 OF THEM FIND THEIR OWN NUMBER! By each one of them looking into only 50 out of 100 boxes. Isn't that fantastic?
Let's all here come together and build some kind of AI RarearthMineral Com Startup Sick Valley company, or at least a casino, based upon this discovery!!!
This submarine is unhittable!
Cantor: Hold my Countable Infinity
Yes sure! Fire your torpedo on the number line horizontally (or in the same direction of the line)! Make sure your torpedo's velocity is much more faster than the submarine.
Pew Pew
Just bribe the commander to beach the submarine.
This is cantor diagonalization, isn't it? I'm mad at myself for not noticing that and getting the answer easily lol!
yup, it's definitely a fun surprise!
As soon as you drew and labeled the number line, I saw the answer.
You forgot to say that the initial position is an integer. Without that constraint, it can be any real number and Cantor's diagonalization argument makes the sub impossible to guarantee a hit even if velocity is 0. This is such an obvious mistake that i feel you're engagement farming and ready to have a part 2 that's like 'pshych!'.
No I didn't, it's one of the very first things I say in the video
@@WrathofMath You're right sorry. I need more sleep.
@mtardibu we all do haha!
You're like the cool teacher i never had thanks dude your explanations are amazing
@WrathofMath, I have a probability problem that I think it would be interesting to solve. Say there are 100 people sitting at a round table for dinner. Each person has a napkin to the right and left of them, totaling 100 napkins in all. For this problem, the chance that an individual picks either the right napkin or the left napkin is 50%, the chance a person will pick a napkin for any given second, assuming they haven't already taken one, is 5%, the choice of a particular person is not influenced by the choices of anyone else, and a person cannot stand up to grab a napkin or take a napkin that is not adjacent to their seat. With all the assumptions aside, here are the questions. A, at the end of the experiment, what is the probability that all the people have a napkin? B, at the end of the experiment, what is the average amount of people that will not have a napkin? and C, is there any particular order that you could place people in their seats to minimize the number of people who don't get a napkin? I hope you see this, and if you do, I hope it makes a good video! P.S. upon further research, I did find that someone in 2003 came up with a similar problem, but I used that source only for a third question. Here is that article describing the problem: akc.is/papers/007-Conways-napkin-problem.pdf
What if two adjacent people try at the same second to grab the same napkin ?
@@thierrypauwels thanks for this scenario! There are some complicated ways I could expand on that issue, but for simplicities sake, let's say it's a 50% chance person 1 gets the napkin and likewise for person 2.
It should be called "Enumerable" infinity
When i first saw the video, I thought this was a puzzle where both the torpedo and the submarine are described by linear three dimensional equations and you have to determine which of the intersections between the lines can actually happen at the same time
that's more like chapter 1 Calc 3 and less fun math puzzle accessible to anyone haha!
@WrathofMath I mean fair, but I don't really know how hard certain things would be for the average person anymore xD
is this a pseudo collaboration with Independent Quark? or did both of you stumble upon the same textbook.
always good to have different angles to the same solution.
very unique take on it. thinking about it was Pk(k) helps me see the link to cantors diagonalization just that lil bit better
I had never heard of him - just looked at his channel, odd coincidence! Will have to check out his videos on the sub!
Today I know the process of submarine
thank you for your video
I like a lot of your videos, but I don't understand the solution of this one at all. Highest math level I got to was Calculus, so maybe that has something to do with it.
I think the think to keep in mind is that on the graph, you aren't plotting positions on the line directly. You're plotting possible positions of the submarine. Instead of saying "I'm going to shoot at position 0 then 1 then 2 etc" you're saying that you're going to shoot at where a submarine would be if it started at 0 with a velocity of 0 , then where it would be if it started at 0 with a velocity of 1, then a position of 0 with a velocity of -1, and so on. Since the number of possible submarines is "countable", you will (eventually, given an infinite amount of time) pick the right submarine and shoot it.
I don't get it, what are the actual numbers you guess? Can someone give me the first few numbers so that I can work it out?
i'm going to use p to denote the initial position, then using the methodology given in the video...
at t=0 we have p=0 and v=0, so we'd guess 0+0*0 = 0
at t=1 (following the spiral shown in the vid) we have p=1 and v=0 so we'd guess 1 + 1*0 = 1
at t=2 we have p=1 and v=1 so we'd guess 1 + 1*2 = 3
at t=3 we have p=0 and v=1 ....
and so on
But the Pt of t function requires knowledge of the v in order to figure it out?
Every possible v and P0 combination is hit with the spiral eventually so neither the starting position nor the speed need to be known to be intercepted. The t is the same for the shooter and for the submarine.
@Culpride I mean, yes, every possible point is but by the spiral, but doesn't that run into the same problem as the line example? 0 1 -1 2 -2 one
@ actually not because in the line graph example the submarine moved from turn to turn. but since the coordinates in the graph with the spiral don't correspond to the current sub position but instead to all possible submarines, the target submarine is 'stationary' in that coordinate system.
So with the line it's "we can hit every possible point, but we are probably outrun"
and with the 2D graph it's "Its one of these and it's not changing to another, we just have to put the current time in the each equasion and shoot at the place that the sub we are cheking for would be by now. that way we will hit all possible subs."
Infinity screws with my mind.
I came up with the spiral solution also.
It's almost paradoxical to me that even if a spiral segment takes longer each time it goes around, it will eventually catch up to the submarine.
I think the confusing part for me is that the submarine position could be written as
P₀(start,velocity,time) = start + velocity × time
with 3 variables.
But for us, the only variable we get is time, and we're mapping that to an (x,y)
If we have
X(time) = x associated with that time
Y(time) = y associated with that time
Our torpedo position could be
P₁(time) = P₀(X(time),Y(time),time)
I guess I just have an infantesimal understanding of Infinity (pun intended)
Ah, I hoped this would be projectile calculation applicable to video games.
I'll give you more interesting math problem (military and for video games).
Imagine tower defense archer shooting at constantly moving infantry target. Projectile has to counteract gravity to hit target.
Archer position, target position, target velocity and gravity is known.
Easier version has known projectile X speed, but more interesting and harder version has known projectile velocity magnitude.
Hmm. We don't know how far away we are from P0, and we don't know the torpedo velocity. Whatever those numbers are, I'm not convinced we can hit every point on the spiral.
every point on the spiral is a combination of starting point and starting velocity. so somewhere on that point is the sub. and since the sub doesn't change its starting point or velocity mid way, our spiral will hit that combination eventually. where will the sub be by that time? we know the time t that has elapsed. we can put it into every combination and fire at the solution to. startingpoint + time x velocity.
The spiral isn't catching up with the subs velocity, its eliminating every starting combination one by one.
If you hit (0,0) it will remove every possibility along the line where P0 = -v. You could then skip those options as you spiral. If you do it again, you remove every option along P0 = -2v. However, nothing is special about (0,0), so you could do this for every point you hit as you spiral - marking other solutions you already ruled out to skip over as you spiral, letting you catch the submarine "faster".
However, as the majority of the options ruled out are getting further away as time goes on, it doesn't seem to significantly speed up the spiral as I try to do it mentally. It also can't efficiently rule out options in the first and third quadrents.
you hit every strategy ((P0,v) pair) that lies on lile with slope -t that goes throught the point you fired on
I also wonder if you could optimize the hitting pattern by listing strategies by changing the way you create the "strategy space" like using P=P10 + v*(t-10) so you could actualy eliminate points in first and third quadrents
but with infinity it probably wouldn't matter
@josefuher6117 the second and fourth quadrant represent strategies that move towards 0, while the first and third represent ones that move away from 0. Any move that is made can only remove a finite amount of possibilities in the first and third. This is best understood when you try to think of a point you could choose that would hit possibilities in both the first and third quadrant and why you can't.
Also, should not be infinite, since no submarine can move faster than the speed of light. There is a finite number of integer velocities below the speed of light.
Thinking about that; if the submarine were moving very above 50% of speed of light and you start launching torpedos from 0, by the time to get close to 50% the speed of light you would have to move close to the speed of light back and forwars, or launch torpedos that move close to the speed of light; since the speed of light is the limit, a submarine movien above 50% the speed of light would not be hit with such method without breaking the laws of phisicas as we know them.
I'd suggest, get 1 more torpedo to chech 2 possitions each time instead of 1 at the time so the speed of light would not be a limit.
By a certain speed the hydrodynamic friction creates more heat than the explosion of the torpedo.
But if you reach the heat death of the universe before getting to the correct position function, is your argument still valid?
Also, constants should be capital letters and should come before variables, so P(t) = P0 + Vt, not P0 + tv.
But in physics v == velocity but V == volume
This is a really cool theoretical puzzle. But in practice submarines aren’t really constrained to integer speeds and locations. Even if they were there is no guarantee the strategy would hit the submarine before the sub would do the damage we feared.
Start at positive infinity and fire the torpedo along the number line towards zero!
even easier: Shoot the ocean until all the water evaporates ...
And then fire the next torpedo from negative infinity towards zero. If nothing else, you'll hit yourself! Several submarines actually have hit themselves when the rudder of the torpedo they launched was stuck. And thus made a big circle to come back home to Daddy. Boom!
That was actually an important problem (for guys in submerged coffins) during the first half of the 20th century. It was launch and then get outta here: Full speed ahead, dive dive dive, help help help!