Difference between revisions of "2016 USAMO Problems/Problem 6"
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For which values of <math>n</math> and <math>k</math> is the game winnable? | For which values of <math>n</math> and <math>k</math> is the game winnable? | ||
==Solution== | ==Solution== | ||
− | + | === Case I: <math>n>k</math> === | |
We first prove that the game is winnable whenever <math>n > k</math> by demonstrating a winning strategy in this case. | We first prove that the game is winnable whenever <math>n > k</math> by demonstrating a winning strategy in this case. | ||
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We have created a strategy that is guaranteed to win in at most <math>m = 2n-k+2</math> moves. Thus, the game is winnable for all <math>n > k.</math> | We have created a strategy that is guaranteed to win in at most <math>m = 2n-k+2</math> moves. Thus, the game is winnable for all <math>n > k.</math> | ||
+ | |||
+ | === Case II: <math>n=k</math> === | ||
We now prove that the game is not winnable if <math>n=k.</math> We will say that the game is in a state <math>S</math> if your knowledge about the card labels is of the following form: | We now prove that the game is not winnable if <math>n=k.</math> We will say that the game is in a state <math>S</math> if your knowledge about the card labels is of the following form: | ||
− | There exists a group of <math>n</math> cards for which you know that those <math>n</math> cards have all of the labels <math>1, 2, ..., n</math> (i.e. you know that they have all distinct labels) in some order, but you know nothing about which of those <math>n</math> cards have which labels. (Call this group of cards Group A.) | + | There exists a group of <math>n</math> cards for which you know that those <math>n</math> cards have all of the labels <math>1, 2, ..., n</math> (i.e. you know that they have all distinct labels) in some order, but you know nothing about which of those <math>n</math> cards have which labels. (Call this group of cards Group <math>A.</math>) |
Suppose that the game is in such a state <math>S.</math> We will now show that, regardless of your next move, you cannot guarantee victory or an escape from state <math>S.</math> | Suppose that the game is in such a state <math>S.</math> We will now show that, regardless of your next move, you cannot guarantee victory or an escape from state <math>S.</math> | ||
− | Clearly, the <math>n</math> cards that are not in Group A must also have all of the labels <math>1, 2, ..., n.</math> (You might know something about which cards have which labels, or you might not.) Call this other collection of cards Group B. | + | Clearly, the <math>n</math> cards that are not in Group <math>A</math> must also have all of the labels <math>1, 2, ..., n.</math> (You might know something about which cards have which labels, or you might not.) Call this other collection of cards Group <math>B.</math> |
− | If, on the next move, you pick all of the cards from Group A or all of the cards from Group B, then you clearly will not get a matching pair. The wizard will then arbitrarily permute those cards. Thus, for those <math>n</math> chosen cards, you know their labels are all distinct, but you know nothing about which cards have which labels. Thus, you are back in state <math>S.</math> | + | If, on the next move, you pick all of the cards from Group <math>A</math> or all of the cards from Group <math>B,</math> then you clearly will not get a matching pair. The wizard will then arbitrarily permute those cards. Thus, for those <math>n</math> chosen cards, you know their labels are all distinct, but you know nothing about which cards have which labels. Thus, you are back in state <math>S.</math> |
− | Now, suppose you pick <math>x</math> cards from Group <math>A</math> and <math>n-x</math> cards from Group <math>B,</math> where <math>x</math> is an integer and <math>1\le x\le n-1.</math> Then, the cards chosen from Group B will form a set of labels <math>P\subset Z_n,</math> where <math>Z_n = \left\{ {1, 2, ..., n} \right\}.</math> However, you know nothing about which cards in Group A have which labels. Thus, there is no way for you to prevent the <math>x</math> cards from Group A to form the exact set of labels <math>Q = Z_n\setminus P.</math> In such a case, there will be no matching cards, so you will not win. Furthermore, the wizard will then arbitrarily permute these <math>n</math> cards, so you will know that they have all <math>n</math> distinct labels, but you will know nothing about which cards have which labels. Therefore, you are again in state <math>S.</math> | + | Now, suppose you pick <math>x</math> cards from Group <math>A</math> and <math>n-x</math> cards from Group <math>B,</math> where <math>x</math> is an integer and <math>1\le x\le n-1.</math> Then, the cards chosen from Group <math>B</math> will form a set of labels <math>P\subset Z_n,</math> where <math>Z_n = \left\{ {1, 2, ..., n} \right\}</math> and <math>|P| = n-x.</math> However, you know nothing about which cards in Group <math>A</math> have which labels. Thus, there is no way for you to prevent the <math>x</math> cards from Group <math>A</math> to form the exact set of labels <math>Q = Z_n\setminus P.</math> In such a case, there will be no matching cards, so you will not win. Furthermore, the wizard will then arbitrarily permute these <math>n</math> cards, so you will know that they have all <math>n</math> distinct labels, but you will know nothing about which cards have which labels. Therefore, you are again in state <math>S.</math> |
We have covered all cases, so it follows that, once you enter state <math>S,</math> you cannot guarantee escape from state <math>S</math> or victory. | We have covered all cases, so it follows that, once you enter state <math>S,</math> you cannot guarantee escape from state <math>S</math> or victory. | ||
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We therefore conclude that the game is not winnable if <math>n=k.</math> We proved earlier that the game is winnable if <math>n>k,</math> so the game is winnable if and only if <math>n>k\ge 2.</math> | We therefore conclude that the game is not winnable if <math>n=k.</math> We proved earlier that the game is winnable if <math>n>k,</math> so the game is winnable if and only if <math>n>k\ge 2.</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | We claim that the game is winnable if and only if <math>n>k</math>. Suppose after the first step, the cards <math>1</math> to <math>n</math> are shuffled around. Notice that we have <math>n</math> cards that we don't know the position of (which are all cards from <math>1</math> to <math>n</math>). Now, suppose we pick <math>p</math> known cards. Note that the <math>p</math> cards are all different(since the known cards are the cards from <math>1</math> to <math>n</math>), and there is still a possibility that the other cards from the unknown cards complement and cause <math>1</math> to <math>n</math>. Therefore, we are in the same state as before, and the game is unwinnable. | ||
+ | |||
+ | Now, suppose <math>n>k</math>. Denote the ith card counting from the left. We pick cards <math>1</math> to <math>k</math>, keeping track of the set of values of the cards. Then, we pick cards <math>2</math> to <math>k+1</math>, adding the value of the <math>k+1</math>th card into the set of value of cards. We keep doing this, until we pick cards <math>2n-k</math> to <math>2n-1</math>, at which point we know the exact number on the <math>2n</math>th card. Now, we go back to <math>1</math> through <math>k</math>, and repeat this process, until we reveal the <math>2n-1</math>th card(unless we win during the process). This process terminates only when there are less or equal to <math>k</math> cards that we don't know the exact numbers on or if we somehow win, clearly, as otherwise we're still revealing new information by picking cards from <math>1</math> through <math>k</math>. Note that we now know the exact values on <math>2n-k</math> of the cards. By the Pigeonhole Principle, since <math>k<n</math>, <math>2</math> of them are the same, and we pick those <math>2</math> cards and <math>k-2</math> other random cards and we win. | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:11, 8 April 2023
Problem
Integers and
are given, with
You play the following game against an evil wizard.
The wizard has cards; for each
there are two cards labeled
Initially, the wizard places all cards face down in a row, in unknown order.
You may repeatedly make moves of the following form: you point to any of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the
chosen cards and turns them back face-down. Then, it is your turn again.
We say this game is if there exist some positive integer
and some strategy that is guaranteed to win in at most
moves, no matter how the wizard responds.
For which values of and
is the game winnable?
Solution
Case I: ![$n>k$](//latex.artofproblemsolving.com/c/8/9/c897386eb7397046d7b260082487115a130fd463.png)
We first prove that the game is winnable whenever by demonstrating a winning strategy in this case.
On the th move, choose the
cards in positions
through
Assuming that you do not win on any earlier move, repeat this for
Assume that you did not win on any of the first moves, as described above. Let
be an integer such that
On the
th move, the wizard revealed the cards in positions
through
so you know the labels of all of these cards (just not necessarily in the right order). Then, on the
th move, the wizard revealed the cards in positions
through
which means that you get to see all of the cards that were moved to positions
through
This means that you can uniquely determine the label on card
since you knew all of the labels from
through
and the card in position
could not have moved anywhere else since your last move.
It follows that, after the sequence of moves described above, you know the labels on the first
cards. Since
we have
so there must be a pair of cards with matching labels in this group of
cards, by the Pigeonhole Principle. On your next move, you can pick a group of
cards that includes that pair of matching cards, and you win.
We have created a strategy that is guaranteed to win in at most moves. Thus, the game is winnable for all
Case II: ![$n=k$](//latex.artofproblemsolving.com/d/6/f/d6fbf6b1b2bc81af471428fe2dc93cf475d6cd05.png)
We now prove that the game is not winnable if We will say that the game is in a state
if your knowledge about the card labels is of the following form:
There exists a group of cards for which you know that those
cards have all of the labels
(i.e. you know that they have all distinct labels) in some order, but you know nothing about which of those
cards have which labels. (Call this group of cards Group
)
Suppose that the game is in such a state We will now show that, regardless of your next move, you cannot guarantee victory or an escape from state
Clearly, the cards that are not in Group
must also have all of the labels
(You might know something about which cards have which labels, or you might not.) Call this other collection of cards Group
If, on the next move, you pick all of the cards from Group or all of the cards from Group
then you clearly will not get a matching pair. The wizard will then arbitrarily permute those cards. Thus, for those
chosen cards, you know their labels are all distinct, but you know nothing about which cards have which labels. Thus, you are back in state
Now, suppose you pick cards from Group
and
cards from Group
where
is an integer and
Then, the cards chosen from Group
will form a set of labels
where
and
However, you know nothing about which cards in Group
have which labels. Thus, there is no way for you to prevent the
cards from Group
to form the exact set of labels
In such a case, there will be no matching cards, so you will not win. Furthermore, the wizard will then arbitrarily permute these
cards, so you will know that they have all
distinct labels, but you will know nothing about which cards have which labels. Therefore, you are again in state
We have covered all cases, so it follows that, once you enter state you cannot guarantee escape from state
or victory.
Now, look at the very first move you make. Obviously, you cannot guarantee victory on the first move, as you know nothing about which cards have which labels. Assuming that you do not win on the first move, the cards you chose have all distinct labels. The wizard then permutes the
cards you chose, so you now know that those
cards have all distinct labels but know nothing about which cards have which labels. Therefore, if you do not win on your first move, then the game enters state
and we have already proven that you cannot guarantee victory from this point.
We therefore conclude that the game is not winnable if We proved earlier that the game is winnable if
so the game is winnable if and only if
Solution 2
We claim that the game is winnable if and only if . Suppose after the first step, the cards
to
are shuffled around. Notice that we have
cards that we don't know the position of (which are all cards from
to
). Now, suppose we pick
known cards. Note that the
cards are all different(since the known cards are the cards from
to
), and there is still a possibility that the other cards from the unknown cards complement and cause
to
. Therefore, we are in the same state as before, and the game is unwinnable.
Now, suppose . Denote the ith card counting from the left. We pick cards
to
, keeping track of the set of values of the cards. Then, we pick cards
to
, adding the value of the
th card into the set of value of cards. We keep doing this, until we pick cards
to
, at which point we know the exact number on the
th card. Now, we go back to
through
, and repeat this process, until we reveal the
th card(unless we win during the process). This process terminates only when there are less or equal to
cards that we don't know the exact numbers on or if we somehow win, clearly, as otherwise we're still revealing new information by picking cards from
through
. Note that we now know the exact values on
of the cards. By the Pigeonhole Principle, since
,
of them are the same, and we pick those
cards and
other random cards and we win.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2016 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |