This reminded me of yet another problem, but a more discreet one. It goes a bit like this.

A pair of lovers married yesterday, but in the craziness of the after party, they wake up in different houses (A house would be a point in the lattice grid Z*Z). The groom knows well that the wife is an anxious person, and will walk A streets North, B streets East and sleep there (A and B can be negative, and -4 north is 4 south for example), each day with constant A and B. "Woe is me!" He exclaims because he does not know where is she, and he does not know A nor B.

The groom is very fast so he can move from any house to another in a matter of seconds, but he can only check a house per day because he is a slow searcher. (That means, you can ask each day one house "Is the bride here?"). Can the groom, with complete certainty, catch the bride?

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## Comments

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TopNewestIs there any time limit on how long the groom can take to catch the bride (other than being finite)? If not:

Let $(x,y)$ be the initial house the bride is at, Then, on the $n$-th day, she is at $(x+An,y+Bn)$.

We know that $\mathbb{N}_0$ and $\mathbb{Z}^4$ have the same cardinality. So, let $f : \mathbb{N}_0 \to \mathbb{Z}^4$ be any bijection.

On the $n$-th day, compute $(x',y',A',B') = f(n)$. Then, have the groom search $(x'+A'n,y',B'n)$.

In other words, on the $n$-th day, the groom guesses that the bride started at the point $(x',y')$ and moves $(A',B')$ each day, and then searches the house the bride would be in based on that guess.

On the $f^{-1}(x,y,A,B)$-th day, the groom guessed correctly, and therefore found the bride.

Note that this could take an arbitrarily long, but finite time.

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This problem as well as the pirate ship problem have the following general strategy:

Find a formula for the position of the target at any time in terms of some set of initial parameters.

At each time, make a different guess for the values of those parameters, and go to the corresponding location.

Make sure that you will guess correctly after a finite amount of time.

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Seems about right to me

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Simple! Once the bride reaches a corner, she cannot go any further. So the groom waits Z days and searches all the corners

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When Diego wrote "Z*Z", I think he meant $\mathbb{Z} \times \mathbb{Z}$, i.e. the set of ordered pairs of integers.

This set extends infinitely in all directions and doesn't have "corners" for the bride to get stuck in.

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