Hard Riddle #1 • page 4/6 • Off-Topic Discussion • lichess.org

@polylogarithmique Its a/(1+t) sorry, just a basic proportion. Also, yeah I meant quadratic my bad.

@polylogarithmique So actually as I think about this more, my switch to a straight line has a small issue. The drag placed on the robot is greatest at the 1/2 way mark, and it decreases thereafter, whereas by going to a straight line, we create a linear drag all the way, which is not correct. However, once the robot has passed the halfway mark, the increase in the remaining distance to travel is less than .5 meters per second, so it can still reach the end anyway. I could create a more technical and complex function for the rate of the robot as a function of the distance, but that would involve going into some circle stuff and it isn't necessary. Basically, if the robot can reach past the halfway mark, it can finish.

InkyDarkBird

#33

t = time passed

Let us assume that the robot CAN reach back to its original point.
This would make the equation:
t + 1 - 0.5t = 0

t + 1 is the circumference of the circle
0.5t is the distance covered by the robot

Unfortunately, this is a circle, so the expansion would be equally spread across the entire circle.
For example, let's say that when 1 second passes, the robot moves 0.5 meters and then, the circle expands by 1 meter.
Before the circle expanded, it only had a circumference of 1, so the robot covered half of the distance.
Now, when the circle expands, only 50% of that 1 meter actually inhibit the robot from getting to its original point.
The other 50% is behind the robot and will not affect it.
(now you see why this question depends on which thing goes first)
This means that the robot only has (1 - 0.5) + 0.5 = 1 meter left, which means, after the circle expanded, it now has to travel 2/3 of the circle instead of 100% of the circle.

The new equation:
(t + 1) x (0.5t)/(t + 1) - 0.5t + 1 = 0
t + 1 x (0.5t)/(t + 1) is how much the distance from the robot's goal increases each second.
-0.5t is the robot's distance per second.
1 is the initial circle diameter.
I'm too lazy to solve the rest, but I hope that this is correct so far.

t = time passed Let us assume that the robot CAN reach back to its original point. This would make the equation: t + 1 - 0.5t = 0 t + 1 is the circumference of the circle 0.5t is the distance covered by the robot Unfortunately, this is a circle, so the expansion would be equally spread across the entire circle. For example, let's say that when 1 second passes, the robot moves 0.5 meters and then, the circle expands by 1 meter. Before the circle expanded, it only had a circumference of 1, so the robot covered half of the distance. Now, when the circle expands, only 50% of that 1 meter actually inhibit the robot from getting to its original point. The other 50% is behind the robot and will not affect it. (now you see why this question depends on which thing goes first) This means that the robot only has (1 - 0.5) + 0.5 = 1 meter left, which means, after the circle expanded, it now has to travel 2/3 of the circle instead of 100% of the circle. The new equation: (t + 1) x (0.5t)/(t + 1) - 0.5t + 1 = 0 t + 1 x (0.5t)/(t + 1) is how much the distance from the robot's goal increases each second. -0.5t is the robot's distance per second. 1 is the initial circle diameter. I'm too lazy to solve the rest, but I hope that this is correct so far.

ChessMathNerd

#34

@InkyDarkBird That doesn't work because as the circle expands, the robot is dragged forwards depending on where it is on the circle. The speed of the robot is slightly faster than .5. A correct solution requires continuous calculus to solve because everything is moving at the same time. It is like drawing a line on a balloon while someone else is blowing it up at the same time. You actually end up drawing a longer line than you intend to.

InkyDarkBird

#35

@ChessMathNerd said in #34:

@InkyDarkBird That doesn't work because as the circle expands, the robot is dragged forwards depending on where it is on the circle.
What if the circle expands from the robot, instead of the robot being dragged by the circle? This would cause the robot to not move and then my equation should technically work.

@ChessMathNerd said in #34: > @InkyDarkBird That doesn't work because as the circle expands, the robot is dragged forwards depending on where it is on the circle. What if the circle expands from the robot, instead of the robot being dragged by the circle? This would cause the robot to not move and then my equation should technically work.

kyanite111 edited

#36

Well it is apparent that many have solved it already, perhaps some incorrectly :P. But I am putting in my solution for the record.

Distance of destination from start: f(t) = t + 1
Distance of robot from start: g(t) = 0.5*t + f(t)*g(t-1)/f(t-1)

Note this takes into account the expansion of the circle at 1 m/s along its circumference (it was never stated - it could have been expansion of the radius, but I assume the simplest interpretation that the circle itself is elongating by 1 m/s)

Where f(0) = 1 and g(0) = 0.

Question: Can g(t) >= f(t) for some t?

It would be nice if we could derive some function for g(t). But alas, I think it can only be solved recursively.

f(0) = 1
f(1) = 2
f(2) = 3
f(3) = 4
f(4) = 5

g(0) = 0
g(1) = 0.5 + 20/1 = 0.5
g(2) = 1 + 30.5/2 = 1.75
g(3) = 1.5 + 41.75/3 = 1.5 + 7/3 = 1.5 + 2.33 = 3.8333...
g(4) = 2 + 5(1.5 + 7/3)/4 = 2 + (7.5 + 35/3)/4 = 2 + (3.75 + 35/6)/2 = 2 + (1.875 + 35/12) = 3.875 + 2 + 11/12 = 6.79166...

g(4) > f(4), so yes.

Well it is apparent that many have solved it already, perhaps some incorrectly :P. But I am putting in my solution for the record. Distance of destination from start: f(t) = t + 1 Distance of robot from start: g(t) = 0.5*t + f(t)*g(t-1)/f(t-1) Note this takes into account the expansion of the circle at 1 m/s along its circumference (it was never stated - it could have been expansion of the radius, but I assume the simplest interpretation that the circle itself is elongating by 1 m/s) Where f(0) = 1 and g(0) = 0. Question: Can g(t) >= f(t) for some t? It would be nice if we could derive some function for g(t). But alas, I think it can only be solved recursively. f(0) = 1 f(1) = 2 f(2) = 3 f(3) = 4 f(4) = 5 g(0) = 0 g(1) = 0.5 + 2*0/1 = 0.5 g(2) = 1 + 3*0.5/2 = 1.75 g(3) = 1.5 + 4*1.75/3 = 1.5 + 7/3 = 1.5 + 2.33 = 3.8333... g(4) = 2 + 5*(1.5 + 7/3)/4 = 2 + (7.5 + 35/3)/4 = 2 + (3.75 + 35/6)/2 = 2 + (1.875 + 35/12) = 3.875 + 2 + 11/12 = 6.79166... g(4) > f(4), so yes.

InkyDarkBird

#37

So what's the solution? @TakeThePawnOrLose

Nomen-Nonatur edited

#38

@Akbar2thegreat said in #30:

I thought this forum was meant to be fun and not for intellectuals, sorry, my bad!

If in your opinion "fun" is defined as "not for intellectuals" you have to come to grips with yourself pondering what that means. I'll happily leave you alone while you weave the rope onto which you hang yourself.

@TakeThePawnOrLose said in #15:

@Nomen-Nonatur The problem is well defined. A rubber band is a circle

This was not my point. The point was where the increase in length happens. Maybe you thought by specifying "rubber band" you implied that the increase in size is happening at all points evenly distributed and if you say so i'll rest my case. Then, all the calculations of @CalbernandHowbe, @polylogarithmique, @ChessMathNerd, etc. are correct, as i said already.

I'll explain what i meant by "evenly distributed increase of length": let us start by cutting the band at point x, which you mentioned in #1: if the increase in length is evenly distributed then x at t=0 will stay at x, x+0.5 m (the point opposite x in the circle) at t=0 will move to x+1 m at t=1 and x+1 m at t=0 (the point neighboring x, where we cut) will move to x+2 m. In other words, the point initially at half the distance from x will always stay at half the distance, etc.. And this will hold true for all points.

In this case the robot will not only be moved by its own movement but also by the part "behind" it increasing in length and the length it has still to travel will not increase at the full rate of 1m/s because the part it already traveled (the "behind" part) takes away from the increase of the rubber band before it.

Now, if the increase takes place at a certain point, x + some y, so that all points x + z (z<y) at t=0 remain at the same place and all points x+z (z>y) at t=0 move to x+z+1m at t=1 these calculations are not correct any more. In this case what i wrote beginning with "it depends" holds: either the robot reaches x after 2 seconds (because no increasing in length modifies the part of the way it still has to travel) or it never reaches x (because all the increase modifies part of the way it still has to travel and is more than its movement rate).

@Akbar2thegreat said in #30: > I thought this forum was meant to be fun and not for intellectuals, sorry, my bad! If in your opinion "fun" is defined as "not for intellectuals" you have to come to grips with yourself pondering what that means. I'll happily leave you alone while you weave the rope onto which you hang yourself. @TakeThePawnOrLose said in #15: > @Nomen-Nonatur The problem is well defined. A rubber band is a circle This was not my point. The point was *where* the increase in length happens. Maybe you thought by specifying "rubber band" you implied that the increase in size is happening at all points evenly distributed and if you say so i'll rest my case. Then, all the calculations of @CalbernandHowbe, @polylogarithmique, @ChessMathNerd, etc. are correct, as i said already. I'll explain what i meant by "evenly distributed increase of length": let us start by cutting the band at point x, which you mentioned in #1: if the increase in length is evenly distributed then x at t=0 will stay at x, x+0.5 m (the point opposite x in the circle) at t=0 will move to x+1 m at t=1 and x+1 m at t=0 (the point neighboring x, where we cut) will move to x+2 m. In other words, the point initially at half the distance from x will always stay at half the distance, etc.. And this will hold true for all points. In this case the robot will not only be moved by its own movement but also by the part "behind" it increasing in length and the length it has still to travel will not increase at the full rate of 1m/s because the part it already traveled (the "behind" part) takes away from the increase of the rubber band before it. Now, *if* the increase takes place at a certain point, x + some y, so that all points x + z (z<y) at t=0 remain at the same place and all points x+z (z>y) at t=0 move to x+z+1m at t=1 these calculations are not correct any more. In this case what i wrote beginning with "it depends" holds: either the robot reaches x after 2 seconds (because no increasing in length modifies the part of the way it still has to travel) or it never reaches x (because *all* the increase modifies part of the way it still has to travel and is more than its movement rate).

Akbar2thegreat

#39

@Nomen-Nonatur
That wasn't my definition of fun!
Those are your words.
And fun is not 'always' about solving something.

@Nomen-Nonatur That wasn't my definition of fun! Those are your words. And fun is not 'always' about solving something.

Nomen-Nonatur edited

#40

@Akbar2thegreat said in #39:

That wasn't my definition of fun!
Those are your words.

You know what a "quote" is, yes? I just repated your own words and if you don't know what you said you are just proving my point.

@Akbar2thegreat said in #39: > That wasn't my definition of fun! > Those are your words. You know what a "quote" is, yes? I just repated your own words and if you don't know what you said you are just proving my point.