[Math Puzzle] Day[9]

I'd say you would start with 1, then -2, then 4, then -8 and so on. That way no matter what speed or direction it goes in you'll eventually kill it.

EDIT: Realised that this doesn't work after 3 steps. It has to be 1, 6, 15, 28, on the positive side, and -2, -8, -18 on the negative. So 1, -2, 6, -8, 15, -18, 28, etc.

Also, shadowdrg got there before me.

For positive integers only, you'd want to zap at t^2 where t = time in seconds. So you'd shoot at 1, 4, 9, 16, etc. With the negatives at play, there's the problem of "missing" the flea by shooting at +4 when he's at -4 so you need to shoot at both possibilities.

1, -2, 6, -8, 15, -18, etc.

I'll edit this if I can figure out an expression for that really fast, otherwise screw it.

Let's say that the flea is traveling at the speed of x. On turn one, it can either be at x or -x depending on which direction it chose to travel in. On turn two, it can be at 2x or -2x. Since we don't know x, but we do know that its domain is the natural number, we can take turns shooting at each one of those natural number. IE, one turn one, let's guess that x is 1 in the positive direction, and since it's turn one, it would have gone 1x1 in the positive direction. Second turn, let's change our assumptions and guess that x is still 1 but in the negative direction. Since it's now turn two, we will have to fire 2x1 in the negative direction, or -2. With this method, you will eventually reach x high enough that you have guessed what was the speed of the flea, therefore hitting him/it.

Edit: ShadowDrgn's got it, the expression is on turn n, you shoot at (-1)^(n+1) x ceiling of (n/2) x n

That works. I didn't think of using the ceiling function, but it's a lot cleaner than the mess I was putting together. Your pre-edit version shot at -1, 2, -6, etc. but that worked just as well.

Haha, yeah same thing since the line's symmetrical.

This is a direct result of the integers being countable.
Take any sequence of counting the integers, A[t], and multiply by t. B[t]=t*A[t] will work. Should work for any other countable system with (integral) scalar multiplication.

If it could be going at any constant speed that means at second 1 it could have travelled 5 points or 100 or 1064 or anything, and it could be anywhere on an infinitely long number line. Forgetting the fact that how many times or how fast you can shoot your laser isn't even strictly defined.. I'll just assume that you can shoot once every second for as long as it takes. The flea would have an infinite number of potential locations at any given second. How can you come up with a pattern to definitively shoot that? In fact, i don't even think it matters how many times you can shoot or how often.

As has been posted, alternate between positive and negative speeds starting from 1 and going up and shoot where he would be with that speed at that time. P(t) = t * S, where P(t) is position at time t and S is his speed. Then you just keep shooting using t = {1, 2, 3, ...} and S = {1, -1, 2, -2, 3, -3, ...} or any other sequence that covers all integer values.

Definition: natural numbers are 1, 2, 3, 4, ....
integers are natural numbers, 0, and the negatives of the natural numbers.
iterations of a set is a bijective function between this set and the natural numbers (if possible).


Let t be a natural number that represents time, and let S(t) be an iteration of the integers. Then we can define a new function that tells us where to fire at time t, we can call this function F:

F(t) = t * S(t)

We claim that if we follow F(t) as our firing scheme we will eventually hit this flee. To prove this, fix the speed of the flee, call it n (n is an integer). Then at time t the flee will be at position nt. Since S(t) is an iteration of integers, there exists t_0 such that n(t_0)=F(t_0). This completes the proof.