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Suppose that there is a thin, circular ring of wire with circumference of 10 meters, and that we have some ants with the following properties,

1) All the ants when placed on the wire will travel at a fixed, constant speed of 1 meter/minute in the direction that they are heading.

2) When any two ants collide on the wire, each ant will instantaneously turn around and travel in the opposite direction at the same speed.

Suppose that we simultaneously place 14 ants at random locations on the ring. The orientation in which each ant is headed (clockwise or counter-clockwise) is determined by a fair, random coin flip. 13 of the ants are colored black, and 1 of the ants is colored red. We let the ants run around the ring as specified by the conditions above. After precisely 10 minutes, what is the exact probability that the red ant will end up at the spot it started in the beginning of the game?

Extra bonus:

The red ant would always return to the original position if and only if (1) the original orientation of all ants is the same, or (2) the number of ants with an initial clockwise orientation equals the number of ants with an initial counter-clockwise orientation.

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If all the ants were identical, it wouldn't make any difference if the ants simply passed through each other instead of colliding.So if we had fourteen ants on the wire that could pass through each other, in ten minutes, each ant would return to the exact spot it started.

Edited by rjsghk107
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Posted · Report post

is 1. If all the ants were black, the instantaneous turning around

would look precisely like the ants are going through one another.

So, in 10 minutes, it would appear that each ant went through

others in its way until it got to its initial position. So, each

ant gets to the initial position of some ant. But since they can't

actually go through each other, their initial order around the

circle remains unchanged. Therefore, after 10 minutes, all ants

get to their initial positions. If one, some or all of the ants

are colored differently, the above analysis doesn't change. So,

exact probability that the red ant returns to its starting position

is 1.

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Posted · Report post

I agree that each ant gets to the initial position of some ant after 10 minutes. And I agree that, since they can't actually go through each other, their initial order around the circle remains unchanged. But I don't think that requires each ant to return to its own original position -- there could be a shift of all ant positions that retains the original order.

Looking at the case of 3 ants, a shift seems quite possible, if not probable.

From examining a few smaller cases, it seems that all ants will return to their original position after 10 minutes if one of the following is true: (1) the original orientation of all ants is the same, or (2) the number of ants with an initial clockwise orientation equals the number of ants with an initial counter-clockwise orientation.

I've only taken a look at a few cases, so this might not always hold true. Can anyone prove this or come up with a counter-example?

If my previous hypothesis is true, then I think the probability in the case of 14 ants works out to 1717/8192, which is about 21%.

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Posted (edited) · Report post

I'm with exagorazo--if there were three ants, the R going CW and the two B going CCW, I think the R does not get home. I realize the OP has an even number of ants, but the Superprismatic argument does not seem to depend on the parity of the number of ants. So we wonder (exagorazo and I) whether the argument is really as strong as he suggests.

Edited by CaptainEd
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The bounce back effect of black ants would be the same as pass through, we can use the case of three to determine if shifted position is possible. Assuming position of red ant at 0, with the wire loop marked at meters clockwise. Place an ant at A at 8 and black ant B at 9. if both black ants start clockwise and red ant starts counter clockwise, red ant collides with A in 30 seconds at 9.5. At 1 minute A and B collide at 9 at which instant red ant is at 0. When red ant reaches 4.5, it collides with A and ant B has traveled to 3.5 and time is at 5.5 minutes. The next collision is at 6 minutes with red hitting B at 4, reversing direction of red ant to clockwise while A is still moving clockwise at 5. At 10 minutes, Red ant will be at 8, and A at 9 and ant B at 0. Ergo by demonstration, we see that shifting is possible under some circumstances. I have not worked out the probability yet but believe the probability of the red ant is greater than .5 and the more ants involved, the higher probability of all ants ending in original position.

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I'm with exagorazo--if there were three ants, the R going CW and the two B going CCW, I think the R does not get home. I realize the OP has an even number of ants, but the Superprismatic argument does not seem to depend on the parity of the number of ants. So we wonder (exagorazo and I) whether the argument is really as strong as he suggests.

If the ants are evenly spaced, I could not find and scenario of direction of movement or number of ants that would interfere with all ants reaching original positions in 1o minutes. The problem comes to how much variation from evenly spaced is allowed before a shift occurs. In my example above, I made the red ant having a small gap on one sife and large gap on the other but note it makes no differece which of the positions is the red ant for all must shift.

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What fun! I love the argument that the ants' positions move independent of the actual ant identities.

BUT...

Let's refer to a specific point O as an admittedly arbitrary Origin.

The locations have increasing labels, going clockwise.

let's refer to times with distance labels. That is, if an ant is at location x, facing O, we could say that, after time x, the ant has reached O.

And let "e" be "epsilon", a tiny amount.

Now, arrange the ants as follows:

R is at O - e facing CW. In other words, after e time units, R will be at O.

the Bs are located at O+e, O+3e, ..., O+25e, all facing CCW. We can think of them as 13 blind mice.

GO!

At time e, R and B1 meet at O, and reverse direction. Now R is the "image" of B1. After about 13e, R will be leading the other 12 blind mice CCW, with B13 acting as the image of R, moving CW at location 13e.

Since the ring is 10 meters in circumference, halfway is 5.

At time e + 5, R and B13 will meet at location O+5. R will bounce, and start moving CW again, but will hit B2 at e + 5 + e, and bounce again. R will arrive back at location O+5 at time 5 + 3e, moving CCW. To get back home (O - e), R has to travel more than 5, but he has less than 5 units of time remaining.

So, here is a layout that appears to fail--R does not get back home, someone else gets to R's home.

If this is right, then I think exagorazo's insight comes into play: the pattern of CW and CCW ants may affect whether R acts as his own image or not.

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Posted · Report post

For random placement of ants on ring.....

A)>> The no. of ants travelling in clockwise direction should be equal to the no. of ants travelling in counter clockwise direction.

So for 14 ants scenario, 7 ants should travel clockwise and remaining 7 counter clockwise.

the answer i reckon would be irrelevant of spacing between the ants....

I got 1936 possibilities i.e. 1934 possibilities for assumption A.

and +2 for all ants travelling in same direction..

All ants would end up in their original positions after 10 minutes....

Probability=121/1024..

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I agree that each ant gets to the initial position of some ant after 10 minutes. And I agree that, since they can't actually go through each other, their initial order around the circle remains unchanged. But I don't think that requires each ant to return to its own original position -- there could be a shift of all ant positions that retains the original order.

Looking at the case of 3 ants, a shift seems quite possible, if not probable.

From examining a few smaller cases, it seems that all ants will return to their original position after 10 minutes if one of the following is true: (1) the original orientation of all ants is the same, or (2) the number of ants with an initial clockwise orientation equals the number of ants with an initial counter-clockwise orientation.

I've only taken a look at a few cases, so this might not always hold true. Can anyone prove this or come up with a counter-example?

If my previous hypothesis is true, then I think the probability in the case of 14 ants works out to 1717/8192, which is about 21%.

This is indeed the correct probability. Congrats! (14.swapnil.14 made the same conjecture but made some mistake in the calculations). However, the solution to this puzzle is slightly incomplete (for me) because the probability computation relies on the unproven fact that

The red ant would return to the original position if (1) the original orientation of all ants is the same, or (2) the number of ants with an initial clockwise orientation equals the number of ants with an initial counter-clockwise orientation.

There is a neat proof for this fact, and I think the fine denizens of the den would appreciate the challenge of working why the above statement is true. Instead of posting that fact as a new puzzle, I think I'll declare this puzzle to be half-solved, and revise the OP to include this new challenge.

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There is a neat proof for this fact, and I think the fine denizens of the den would appreciate the challenge of working why the above statement is true. Instead of posting that fact as a new puzzle, I think I'll declare this puzzle to be half-solved, and revise the OP to include this new challenge.

This has been a fun puzzle, bushindo. I haven't yet come up with a proof, but I look forward to seeing what the denizens come up with!

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