1
$\begingroup$

This puzzle is a follow-up of Coins on a table (warning: the link contains the answer to the original question).

You are playing a game against your friend Alice:

  • first Alice places one coin on a rectangular table (she can choose where to put it).
  • Then it's your turn: you choose where to place the second coin on the table.
  • Then Alice places another coin.
  • Then you place another coin.
  • and so on, until there is no space left to place any other coins. The player who cannot place its coin loses.

You have an unlimited supply of identical coins -although you can use them only for this game, I'm sorry. The coins cannot overlap nor lie outside the boundary of the table.

If Alice plays first, who has a strategy to win this game? (this was the original question)

Original part of the puzzle:

It is well known that Alice is a bit distracted. Even if she has a strategy to win this game (i.e. even if she can choose whether to go first or second depending on the answer of the previous question) she always get her first move wrong. That means that she does not play the move that allows her to win: she plays another move of their choice. Then she realizes her mistake and she plays correctly for the rest of the match.

In this scenario can you always win this game?

In other words: do you always have a strategy after Alice's first move or does it depend on "how wrong" is Alice's first move?

Improve this question
$\endgroup$
6
  • $\begingroup$ "she plays another move of their choice" - whose choice? (question consistently uses 'she' and 'her' etc. when referring to Alice, and 'you' to refer to the other player, so what third party is 'their' referring to?) $\endgroup$
    – Steve
    Mar 10, 2020 at 20:50
  • $\begingroup$ Their = Alice's $\endgroup$
    – melfnt
    Mar 10, 2020 at 20:54
  • 1
    $\begingroup$ The issue with this question is phrasing Alice always got the first move wrong. Note that, the strategy on the original question IS NOT THE ONLY winning strategy. If you mean Alice's first move is not any winning move, by definition it's trivial that Alice will always lose. If you mean Alice's first move is not that particular strategy on the original question, then this becomes non trivial -- and I suppose this is what your question is intended to. (In that case tho, please explicitly state that Alice won't rot13(ba gur pragre bs gur gnoyr)) $\endgroup$
    – athin
    Mar 10, 2020 at 23:44
  • $\begingroup$ Just noticed on "Related" section, that this exact question (with different phrasing) was asked 3 years ago: puzzling.stackexchange.com/questions/33694/… but not sure how to mark as a duplicate? $\endgroup$
    – Steve
    Mar 11, 2020 at 9:31
  • 1
    $\begingroup$ Does this answer your question? Perfect strategy for the 2nd player in 'The coins on a table game' (@Steve since you don't have 3k reputation yet, you can't vote to close but only flag for closure, which it seems you've done) $\endgroup$
    – Rand al'Thor
    Mar 11, 2020 at 9:57

3 Answers 3

Reset to default
2
$\begingroup$

By definition, Alice's first move is one that does not guarantee her a win. Therefore, the other player must have a viable path to victory. If the other player does not have a way to win, then the premise that Alice made the "wrong" move at the start is false. If Alice can guarantee a win from her second move regardless of what the other player does, then her first move was irrelevant, in which case there is no wrong first move, also defying the premise of the question.

There is no concept of "how wrong" is Alice's first move - it either guarantees her victory, or it does not. From the original question's answer, we see that the "right" move is to

place the first coin in the center of the table.

But even if the coin is placed slightly differently, it can still guarantee a win - a somewhat misaligned coin can still be the "right" move. A move isn't "wrong" if it's not exacly on the specified point, it's only "wrong" if it doesn't guarantee a win.

Improve this answer
$\endgroup$
2
$\begingroup$

non-constructive answer and proof

It is stated that she always get her first move wrong. That means that she does not play the move that allows her to win.

We also know that the game has a winner - there are no draws.

Therefore if Alice does not play a move that guarantees her the win there exists a strategy that will guarantee you the win from that point.

Specifically, if you do not have a winning strategy from this position, Alice did not in fact play an incorrect first move, but chose a correct opening move that guarantees a win for her.

Partial "constructive" answer

First, the easy case:

If Alice places the coin with its centre

at least half a coin diameter, and less than a cos(60°) of a full coin diameter from the exact centre of the table - i.e. sufficiently close to the centre that it does not allow a coin to be played that covers the centre after you mirror it, but not actually covering the centre itself.

then you can

place a coin "opposite" so that the table is rotationally symmetrical, but the coins are now blocking the centre.
For every subsequent move you just play directly opposite the space where Alice plays, so are guaranteed the win.

If Alice places her first coin

with its centre at least a full coin diameter from the centre of the table

then there are two moves you definitely should NOT do, as they trivially lead to a winning strategy for Alice:

1. Play in the centre => Alice can play directly oppposite her first coin.
2. Play directly opposite Alice's first coin => Alice can play in the centre.

I've not yet worked out what move you should do, but one possibility could be:

play close to, but not covering the centre. If Alice remembers she's supposed to mirror all your moves from then on, and mirrors this move, this leaves you free to mirror Alice's first move with your second move, and it carries in the simple case thereafter.
However, if Alice does NOT mirror your second move the complexity of ongoing analysis greatly increases - some coins may already be blocking a mirroring strategy w.r.t. other coins.

The third and fourth possible cases are

Alice places her first coin covering the centre but not properly aligned with it, or between cos(30°) of a full diameter and 1 full diameter, so that a mirroring move still allows Alice to place her second coin so it covers the centre, but not to fully centralise that second coin.

For which I've not yet determined any "obviously wrong" or "obviously right" moves in the general case. However, it seems likely that

a winning strategy will exploit and enhance any asymmetry. e.g. if you place a coin close to the centre, such that the spot "opposite" your coin would overlap Alice's first coin, then Alice cannot directly "mirror" your first move.

Improve this answer
$\endgroup$
5
  • $\begingroup$ but alice can place it where she wants/ $\endgroup$
    – Jason V
    Mar 10, 2020 at 20:30
  • 2
    $\begingroup$ @JasonV "she always get her first move wrong. That means that she does not play the move that allows her to win" $\endgroup$
    – Steve
    Mar 10, 2020 at 20:32
  • 1
    $\begingroup$ In the first "easy case", your distances are a bit too generous. You have to make sure that the third move cannot cover the central point at all (not even with the coin's edge). $\endgroup$
    – Jaap Scherphuis
    Mar 10, 2020 at 23:10
  • $\begingroup$ @JaapScherphuis edited to fix that. Thanks. $\endgroup$
    – Steve
    Mar 11, 2020 at 9:17
  • $\begingroup$ Just noticed that the "easy case" is a duplicate of an answer given to the previous instance of this question - see puzzling.stackexchange.com/a/33702/12582 - another answer to that question gives a bit more insight into the general case. $\endgroup$
    – Steve
    Mar 11, 2020 at 9:34
1
$\begingroup$

No, you do not always have a strategy to beat a distracted Alice.

Imagine a rectangular table that is 1.5 coins by 3 coins. Alice does not have to place the first coin directly in the center: any placement in the middle of the long axis, no matter where it is along the short axis, gives exactly two more moves (one on each side of the first placed coin).

Improve this answer
$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.