Difference between revisions of "2012 IMO Problems/Problem 3"

Latest revision as of 00:23, 19 November 2023

The liar’s guessing game is a game played between two players A and B. The rules of the game depend on two positive integers $k$ and $n$ which are known to both players.

At the start of the game the player A chooses integers $x$ and $N$ with $1 \le x \le N$ . Player A keeps $x$ secret, and truthfully tells $N$ to the player B. The player B now tries to obtain information about $x$ by asking player A questions as follows: each question consists of B specifying an arbitrary set $S$ of positive integers (possibly one specified in some previous question), and asking A whether $x$ belongs to $S$ . Player B may ask as many questions as he wishes. After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful.

After B has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x \in X$ , then B wins; otherwise, he loses. Prove that:

(a) If $n \ge 2^k$ then B has a winning strategy.

(b) There exists a positive integer $k_0$ such that for every $k \ge k_0$ there exists an integer $n \ge 1.99^k$ for which B cannot guarantee a victory.

Solution to Part (a)

It suffices to show that there is a winning strategy for $n = 2^k-1$ , as a winning strategy for any $n \ge 2^k$ easily gives a winning strategy for $n = 2^k-1$ .

Player B first divides all integers $1$ through $N$ into $2^k$ nonempty sets $T_s$ , where $s$ is some binary string of length $k$ . On the $i^{th}$ question for $1 \le i \le k$ , player B asks whether the set containing $x$ is labeled with a string that has a $1$ in the $i^{th}$ position. After asking these questions, there is one set that player A must indicate does not contain $x$ at $k$ times in a row; WLOG we may assume this set is $T_{00 \dots 0}$ .

On the next question, B asks if $x \in T_{00 \dots 0}$ . If A says 'no', then B can rule out $T_{00 \dots 0}$ . If A says 'yes', then B asks if $x \in T_{100 \dots 0}$ . If A says 'no', then B can rule out $T_{100 \dots 0}$ . If A says 'yes', then B asks if $x \in T_{0100 \dots 0}$ . If A says 'no', then B can rule out $T_{0100 \dots 0}$ , while if A says 'yes', B can rule out $T_{1100 \dots 0}$ .

In any case, B can ascertain that $x \not \in T_s$ for some string $s$ , thus reducing the number of possibilities for $x$ . Player B can repeat this procedure as long as at least $2^k$ elements remain, eventually reducing the number of possible values of $x$ to at most $2^k - 1$ .

Solution to Part (b)

Take $k$ sufficiently large so that there is some positive integer $n$ with $n \ge 1.99^k$ but $n+1 \le .001*1.999^{k+1}$ .

Player A sets $N=n+1$ . Let $f(i,m)$ denote the number of times in a row that A has just indicated that $i \neq x$ . In other words, if A indicates on the $m^{th}$ step that $x$ was in a set containing $i$ , then $f(i,m)=0$ , while if A indicates otherwise, we would have $f(i,m)=f(i,m-1)+1$ .

At the $m^{th}$ step, player A makes a decision so as to minimize the sum: $g(m)=\sum_{i=1}^N 1.999^{f(i,m)}$ I claim that, by using this strategy, A will preserve the invariant that $g(m) < 1.999^{k+1}$ . We can prove this by induction on $m$ . Note that initially, $g(0)=N<1.999^{k+1}$ .

Suppose that $g(m-1) < 1.999^{k+1}$ . Note that if B asks if $x \in S$ for some $S \subset \{1,2,\dots,N\}$ , then the two possibilities for $g(m)$ are: $g_\text{in}(m)=N-|S|+1.999\sum_{i \in S} 1.999^{f(i,m-1)}$ and $g_\text{out}(m)=|S|+1.999\sum_{i \not \in S} 1.999^{f(i,m-1)}$ Note that $g_\text{in}(m)+g_\text{out}(m)=N + 1.999g(m-1)$ . Therefore, we have that: $\min(g_{\text{in}}(m), g_{\text{out}}(m)) \le \frac{g_{\text{in}}(m) + g_{\text{out}}(m)}{2} = \frac{N + 1.999g(m-1)}{2} < \frac{.001*1.999^{k+1} +1.999^{k+2}}{2} = 1.999^{k+1}$

This completes the induction. In particular, since $g(m)<1.999^{k+1}$ at all times, A will never indicate that $x\neq i$ more that $k$ times in a row. So player B will never be able to rule out any element of $\{1,2,\dots,N\}$ and submit a set of $n$ integers that are guaranteed to contain $x$ .

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-The liar’s guessing game is a game played between two players A and B. The rules of the game depend on two positive integers k and n which are known to both players.
+The liar’s guessing game is a game played between two players A and B. The rules of the game depend on two positive integers <math>k</math> and <math>n</math> which are known to both players.
-At the start of the game the player A chooses integers x and N with 1≤x≤N. Player A keeps x secret, and truthfully tells N to the player B. The player B now tries to obtain information about x by asking player A questions as follows: each question consists of B specifying an arbitrary set S of positive integers (possibly one specified in some previous question), and asking A whether  x belongs to S. Player B may ask as many questions as he wishes. After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any k+1 consecutive answers, at least one answer must be truthful.
+At the start of the game the player A chooses integers <math>x</math> and <math>N</math> with <math>1 \le x \le N</math>. Player A keeps <math>x</math> secret, and truthfully tells <math>N</math> to the player B. The player B now tries to obtain information about <math>x</math> by asking player A questions as follows: each question consists of B specifying an arbitrary set <math>S</math> of positive integers (possibly one specified in some previous question), and asking A whether  <math>x</math> belongs to <math>S</math>. Player B may ask as many questions as he wishes. After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any <math>k+1</math> consecutive answers, at least one answer must be truthful.
-After B has asked as many questions as he wants, he must specify a set X of at most n positive integers. If x∈X, then B wins; otherwise, he loses. Prove that:
+After B has asked as many questions as he wants, he must specify a set <math>X</math> of at most <math>n</math> positive integers. If <math>x \in X</math>, then B wins; otherwise, he loses. Prove that:
-(a) If n≥2k then B has a winning strategy.
+(a) If <math>n \ge 2^k</math> then B has a winning strategy.
-(b) There exists a positive integer k0 such that for every k≥k0 there exists an integer n≥1.99^k for which B cannot guarantee a victory.
+(b) There exists a positive integer <math>k_0</math> such that for every <math>k \ge k_0</math> there exists an integer <math>n \ge 1.99^k</math> for which B cannot guarantee a victory.
+==Solution to Part (a)==
+It suffices to show that there is a winning strategy for <math>n = 2^k-1</math>, as a winning strategy for any <math>n \ge 2^k</math> easily gives a winning strategy for <math>n = 2^k-1</math>.
+Player B first divides all integers <math>1</math> through <math>N</math> into <math>2^k</math> nonempty sets <math>T_s</math>, where <math>s</math> is some binary string of length <math>k</math>. On the <math>i^{th}</math> question for <math>1 \le i \le k</math>, player B asks whether the set containing <math>x</math> is labeled with a string that has a <math>1</math> in the <math>i^{th}</math> position. After asking these questions, there is one set that player A must indicate does not contain <math>x</math> at <math>k</math> times in a row; WLOG we may assume this set is <math>T_{00 \dots 0}</math>.
+On the next question, B asks if <math>x \in T_{00 \dots 0}</math>. If A says 'no', then B can rule out <math>T_{00 \dots 0}</math>. If A says 'yes', then B asks if <math>x \in T_{100 \dots 0}</math>. If A says 'no', then B can rule out <math>T_{100 \dots 0}</math>. If A says 'yes', then B asks if <math>x \in T_{0100 \dots 0}</math>. If A says 'no', then B can rule out <math>T_{0100 \dots 0}</math>, while if A says 'yes', B can rule out <math>T_{1100 \dots 0}</math>.
+In any case, B can ascertain that <math>x \not \in T_s</math> for some string <math>s</math>, thus reducing the number of possibilities for <math>x</math>. Player B can repeat this procedure as long as at least <math>2^k</math> elements remain, eventually reducing the number of possible values of <math>x</math> to at most <math>2^k - 1</math>.
+==Solution to Part (b)==
+Take <math>k</math> sufficiently large so that there is some positive integer <math>n</math> with <math>n \ge 1.99^k</math> but <math>n+1 \le .001*1.999^{k+1}</math>.
+Player A sets <math>N=n+1</math>. Let <math>f(i,m)</math> denote the number of times in a row that A has just indicated that <math>i \neq x</math>. In other words, if A indicates on the <math>m^{th}</math> step that <math>x</math> was in a set containing <math>i</math>, then <math>f(i,m)=0</math>, while if A indicates otherwise, we would have <math>f(i,m)=f(i,m-1)+1</math>.
+At the <math>m^{th}</math> step, player A makes a decision so as to minimize the sum: <cmath>g(m)=\sum_{i=1}^N 1.999^{f(i,m)}</cmath> I claim that, by using this strategy, A will preserve the invariant that <math>g(m) < 1.999^{k+1}</math>. We can prove this by induction on <math>m</math>.
+Note that initially, <math>g(0)=N<1.999^{k+1}</math>.
+Suppose that <math>g(m-1) < 1.999^{k+1}</math>. Note that if B asks if <math>x \in S</math> for some <math>S \subset \{1,2,\dots,N\}</math>, then the two possibilities for <math>g(m)</math> are: <cmath>g_\text{in}(m)=N-|S|+1.999\sum_{i \in S} 1.999^{f(i,m-1)}</cmath> and <cmath>g_\text{out}(m)=|S|+1.999\sum_{i \not \in S} 1.999^{f(i,m-1)}</cmath> Note that <math>g_\text{in}(m)+g_\text{out}(m)=N + 1.999g(m-1)</math>. Therefore, we have that:<cmath>\min(g_{\text{in}}(m), g_{\text{out}}(m)) \le \frac{g_{\text{in}}(m) + g_{\text{out}}(m)}{2} = \frac{N + 1.999g(m-1)}{2} < \frac{.001*1.999^{k+1} +1.999^{k+2}}{2} = 1.999^{k+1}</cmath>
+This completes the induction. In particular, since <math>g(m)<1.999^{k+1}</math> at all times, A will never indicate that <math>x\neq i</math> more that <math>k</math> times in a row. So player B will never be able to rule out any element of <math>\{1,2,\dots,N\}</math> and submit a set of <math>n</math> integers that are guaranteed to contain <math>x</math>.
+==See Also==
+{{IMO box|year=2012|num-b=2|num-a=4}}

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Difference between revisions of "2012 IMO Problems/Problem 3"

Latest revision as of 00:23, 19 November 2023

Solution to Part (a)

Solution to Part (b)

See Also