Riddles

Pretty good riddle. I googled it but I’m not giving it away.

nothing

Nada, zip, zilch, efes,zero – an oldie, but goodie.

A rodef and a nirdaf (each viewed as a single point) are in a closed circle. They both move with the same maximum speed. Can the rodef catch the nirdaf?

If so, how does the rodef catch the nirdaf?

If not, how does the nirdaf avoid capture by the rodef?

Curiosity: the rodef and nirdaf are each a single point in a closed circle. A single point can’t carry a gun (since a single point has no dimension, and hence no mass). Alternatively, shooting the nirdaf would violate the given assumption that both move with the same maximum speed.

Here’s a hint for how to visualize the problem:

Let’s assume for a moment that we have a closed square with vertices at (0,0),(0,1), (1,0) and (1,1). This is a simple unit square. Now, assume the rodef begins at one corner of the square, say (1,1), and the nirdaf begins at the opposite corner (0,0). So the nirdaf is sqrt(2) units away from the rodef.

Now, as a conjecture, it would be a reasonable assumption that the rodef and nirdaf would have opposite strategies; namely, the rodef wants to minimize distance between the rodef and nirdaf while the nirdaf wants to maximize distance between the rodef and nirdaf.

In this example, the nirdaf actually has no incentive to move whatsoever. This is because (0,0) is as far as the nirdaf can get from (1,1) where the rodef is. So the nirdaf is maximizing distance from the rodef by remaining at (0,0). The rodef’s optimal strategy from (1,1) is to move in a straight line towards (0,0) because the shortest distance between two points is a straight line (extra credit to whoever proves that the shortest distance between two points is a straight line).

Since the nirdaf will never have an incentive to move given the rodef’s strategy (moving away from (0,0) at any time will shorten the distance with the rodef), and the rodef continually gets closer, then it is clear to see that the rodef will eventually “corner the nirdaf” and catch the nirdaf at (0,0).

So in this contrived example, the rodef catches the nirdaf regardless of what the nirdaf tries to do. The riddle, though, is what happens in a circle?

Another hint:

The optimal strategy for the rodef to minimize distance to the nirdaf is to move along the straight line connecting the rodef’s location on the circle with the nirdaf’s location on the circle.

557 – I was assuming you were going with the classical interpretation of rodef and nirdaf. How can one 2 dimensional point by a rodef? Furthermore, a nirdaf is allowed to kill a rodef. Assuming they were both people, moving at sprinting pace, a gun would have worked for either of them.

If they’re in a “closed” two dimensional circle (I guess this can be imagined like a circular pair of train tracks), and they both move at light speed, then they can’t catch eachother unless one of them breaks the circle…. Or if the rodef blocks the circle with some object and then corners the nirdaf on it.. your rules aren’t very clear.

Curiosity: Sorry for not being so clear. Perhaps this will clarify…

Consider a closed unit circle centered at the origin (0,0) in the Cartesian coordinate system in the Euclidean plane (i.e. your standard unit circle). Now, consider two different points in the closed unit circle (these points can be inside the circle or on the boundary of the circle, but since the circle is closed, they can never be outside the circle). Let’s call these points A and B, respectively.

Now, assume A and B move freely in the closed circle with equal maximum speed. A tries to “catch” B by getting to the same location as B (i.e. A catches B if both A and B have the same exact coordinates at any given time). B tries to “evade” A.

Assuming A and B play optimal strategies (i.e. A and B make the best possible moves at all times in accordance with their goals), does A “catch” B? If so, how can A be guaranteed to “catch” B? If not, how can B be guaranteed to “evade” A?

To make it more interesting for non-math-oriented people, I had called A the rodef and B the nirdaf since in a literal sense this is the case (i.e. A is the chaser and B is the chased). However, you correctly point out that from a halachic standpoint, the din of rodef and nirdaf is more complex. I apologize for the confusion.

Aside (if it’s confusing, just ignore): This is an example of a pursuit-evasion game. There are several interesting properties of these games in general, but there is something particularly interesting about this game.

One of the basic properties of these types of games is to note the path that the chaser takes to catch the chased. This path is called a pursuit curve, which is aptly named because the path of pursuit need not be straight. For example, if a mother is chasing a little child running straight in some direction, chances are that the mother is faster, and she can get onto the child’s path of travel and then follow the child’s path until she catches him. However, if she runs at the same speed as the child, then she cannot just follow the child’s path because she will never catch up. Instead, the mother runs to where the child is heading and intercepts him at some point. If the child changes direction during the chase in response to the mother’s chase, then the mother’s path of pursuit will be curved. Hence the term pursuit curve.

Ok you have a chicken,fox, and chicken feed on one side of the river you want to get them all to the other side eagerly but you can only take one item with you each time. If you leave the chicken with the chicken feed it will eat it, if you leave the fox with the chicken, you will come back to find a fox twice as big and no chicken….how do you do it?

I think you messed it up a bit. Don’t all three things have to eat one of the others?

Take the chicken,

Go back empty handed,

Take the fox,

On the way back to the seeds return the chicken with you,

take the seeds over to the fox, leaving the chicken at the starting side,

Go back empty handed, and bring the chicken.

Know that one!…curiosity that it correct! And the second riddle…well I screamed silently and decided whatever I needed in there was not important now…..

7 letters are in my name

If you subtract one 12 remains

What am I?

You are trapped in a room with only a mirror and table. There are no doors and windows, how do you escape?

557–

Unless Y can move away from X in a straight line, X can either cut the corner or turn inside the radius of the path Y is taking, thereby closing the gap and eventually catching Y.

Even if X and Y are both on the outer edge of the circle, X can still turn inside a smaller radius.

If there was an obstacle (such as an inner circle) within the circle or square, then Y could dodge X forever, since they would both circle that obstacle at the same speed.

Rough Rider–

Slice.

Shopping613–

It depends if the roach is the thing you went to get.

7 letters are in my name

What am I?

Twelves (or drytzen).

You look in the mirror, you see what you saw. You take the saw and cut the table in half, two halves make a hole and you climb out the hole! Very simple!

ICOT: This riddle became popular in the early 1930s, and if it were before 1952, you would not only have the correct answer, but you also noted within your answer the correct strategy X would use to catch Y.

Namely, X could be guaranteed to catch Y by staying on the same radius as Y. In other words, X moves at top speed in such a way that X always lies on the radius vector from the center to Y. If we assume without loss of generality that X stays on the boundary of the circle, then it is easy to check that X does now catch Y in

finite time. Indeed, in the time that it takes Y to run a quarter-circle at full speed, X (say starting at the center) performs a semicircle of half the radius of the disc, thus catching Y.

However, in 1952, it was proven that Y need not stay on the boundary and in fact can survive forever in a closed circle. The clever proof is as follows:

First, I establish a winning strategy for Y, and then I will establish that it is feasible, i.e. Y does not leave the closed circle.

Let T = {t_1, t_2, t_3, …} be a sequence of time periods of equal length. At the ith time period, Y runs in a straight line that is perpendicular to Y’s radius vector at the start of the period. Y runs into the half-plane that does not contain X (if X is on the radius then Y may choose either half-plane). Certainly, X does not catch Y in this time period.

Let Y repeat this procedure for all time periods. As long as T is an infinite sequence, Y is never caught.

Now, I must show that this strategy is feasible. Note that if R_i is the distance of Y from the center at the start of the ith time period, then (R_i+1)^2 = (R_i)^2 + (t_i)^2. (This is just a simple application of the famous Pythagorean Theorem.) If the sum of (t_i)^2 is finite, then the R_i’s are bounded so that (multiplying by a constant if necessary) Y does not leave the closed circle. Let t_i = 1/i.

Then, Y is never caught and never leaves the closed circle.

557-

Please let me know if the following are correct:

2) Do you mean half the diameter of the disc?

Please provide the following:

2) The values of t1, t2.

3) What direction (if any) is Y moving in during the non-nth time periods?

I realize the values are variable, but applying specific numbers may help me understand.

Thank you.

frummy in the tummy –

Let’s name the three people X, Y and Z.

557/ICOT –

Although it’s certainly an interesting problem as stated, the one problem with describing it as pursuer (X) and pursued (Y) is that it does not factor in reaction time. In order for Y to travel in the perpendicular direction of travel of X for any infinitesimal movement (during your time t_i), Y must know X’s direction of travel before X actually performs said movement. Assuming that this is impossible, X can randomly preempt the direction of travel of Y and cut him off, and after enough guesses should eventually catch him. I’m with ICOT in not understanding your entire logic (I’m an engineer, not a mathematician :)), so please correct me if I’m wrong.

ursketching: I think corn.

ICOT:

1) Yes.

2) Yes. Also, I meant to say that Y stays on the boundary of the circle.

Yes also to your square question – in fact, this strategy works in any compact metric space. (My square visualization hint at the beginning was correct given the assumption on X and Y’s optimal strategies. It turns out, though, that Y has an optimal strategy that does not necessarily maximize the distance from X.)

Radius vector – this is just the line drawn from the center of the circle to Y.

{t_1, t_2, t_3, …} – I don’t know if it’s possible to do math type here, but the “_number” is supposed to mean subscript. The number just identifies the time period, so t_1 is the first period of movement and so on.

ith – “i” just means that it refers to any number. It could refer to t_1 or t_2 or t_557 or anything you want. So using “i” instead of a number is a generalization. So if the ith time period happens to be t_557, then the (i+1)th time period is t_558. But “i” can be any number.

half-plane – in our example of a closed circle, we were using R^2 space, i.e. your standard 2-dimensional metric space. R^2 is the metric space that children do math in for all of grade school; when you draw a graph with an x-axis and y-axis, this is R^2 space. The whole R^2 space makes up a plane. If I only consider everything above the x-axis, then I am considering a “half-plane.”

So in the context of the problem, we could consider Y’s radius vector as dividing the “world” in half. By imagining that the vector actually is a line rather than a line segment, everything to one side of the line would constitute a half-plane whereas everything to the other side of the line would constitute the other half-plane.

The Pythagorean Theorem is actually quite easy; draw a line from the center of the circle to where Y is located. This is Y’s radius vector. Now Y is moving perpendicular to his radius vector, so draw a small line segment perpendicular to the radius vector that begins at Y’s current location. This line segment represents Y’s movement. Now connect the other end of the segment to the center of the circle. This will be Y’s new radius vector after moving. Note that since the original radius vector is perpendicular to the line segment of movement, we now have a right triangle, where the new radius vector after moving is the hypotenuse of the triangle.

The boundedness of the radii is a bit more difficult, but basically the idea is that in order for the strategy to work, Y cannot make “too big” of moves or else the strategy will require him to leave the circle, which he is not allowed to do.

frummy in the tummy: If I understand your kasha (which I might not), you are concerned that Y does not know where X is heading necessarily, so he cannot know which half-plane to choose.

Y does not need to know X’s direction; Y only needs to know X’s location.

ICOT that’s a creative solution! The original riddle is with a king who is watching over this scenario. He executes you if you walk through the wrong door, and it’s not some sort of mechanical or natural death that is physically connected by the door. Thus, your answer wouldn’t really work because the man you threw through the door wouldn’t be executed. Also, the question you are asking him doesn’t really guarantee anything, and he just kind of serves as a guinnea pig of sorts. Good effort, but try again!

Curiosity-

Thank you.

I already knew the “real” answer, but didn’t want to steal it from someone else who could solve it.

There are two possibilities. Let’s name the two people “X” and “Y” (such imaginative names, I know):

1) Ask X “which door would Y tell me is safe?”, then use the other door. If X is the honest one, he will point to the door that Y (the liar) would have indicated, while if X is the liar, he would dishonestly point to the door Y would not indicate.

2) Ask X “Hypothetically, if I asked you which door was safe, which door would you point to?”, then use the indicated door. If X was the truth-teller, he’d honestly point to the safe door that he’d point to if asked. If X is the liar, he’d dishonestly point to the safe door that he wouldn’t have pointed to if he’d actually (not hypothetically) been asked.

What do yeshivish’e cars and analogies have in common?

For the answer, see here:

http://www.theyeshivaworld.com/coffeeroom/topic/classic-emunah-question#post-503915

it

answer: A chair, a bed, and a toothbrush.