Toggling lights 3 at a time

Alice and Bob are playing a game. $n$ lights are put on a line with individual switches, initially all off. Then the two phases happen as follow:

1. Alice turns on some of these lights

2. Bob may do these actions: choose a contiguous block of 3 lights and toggle these 3 lights. He may repeat these actions as many times as he wants for different blocks.

And the scoring of the game is as follows:

In phase 1, Alice gets a -1 point for each light that she turns on. In phase 2, Alice gets a +1 point every time Bob performs an operation. Then at the end of the game, Alice gets $i$ points if light $i$ is on. Obviously Alice wants her score to be as high as possible, and Bob wants her score to be as low as possible. Determine the maximum points that Alice can get if both players are playing optimally.

Solution

Let $T_k$ denote the action by Bob of toggling a contiguous block of $k, k+1, k+2$, where $k \in [1, 98]$. If light $k+3$ is on and light $k$ is off, by performing $T_k + T_{k+1}$ he can turn off $k+3$ and turn on $k$. The net effect to the score is -1 so this move is beneficial to Bob. Therefore, he would keep doing it as long as there are lights on at $k > 3$. Furthermore, if there are 2 lights on $k > l$ such that $k \equiv l \mod 3$ then by repeating the operation above Bob can turn off both lights. The net effect to the score is $-k-l+2\frac{k-l}{3} < 0$. This means that it is not optimal for Alice to turn on two lights that have the same residue mod 3. So at most Alice should turn on 3 lights. But even then, if we ever arrive at the condition where lights $(1,2,3)$ are on, then Bob can perform $T_1$ to turn off all the lights. So Alice should turn on at most 2 lights.

The final condition that is most beneficial to Alice is either $(3)$ or $(1,2)$, both having the same value, and both are accessible from each other in phase 2. But obviously it's more beneficial for Alice to turn on only 1 light, and it should be the largest light divisible by 3, so that Bob has to spend a lot of moves moving it to the smaller number. So Alice should turn on $3\lfloor n/3 \rfloor$, and Bob will have to move it all the way to 3.

The total score is then $-1 + 2\lfloor n/3 \rfloor + 3 = 2\lfloor n/3 \rfloor +1$

Note: the number 3 can be replaced by a larger number $p$ but the principle is the same. Any light on above $p$ can be shifted down for a net gain for Bob. However, the final condition is tricky. Take $p=5$ for example. We don't want $(1,2,3,4,5)$ because then Bob can simply turn them all off in one fell swoop. We also don't want $(3,4,5)$ because Bob can still replace it by $(1,2)$ for a net gain. In fact, the optimal condition for Alice is $(3,5)$ because while Bob can replace it by $(1,2,4)$, it costs him 1 point to do so, so it's not a beneficial move for Bob. Likewise, for $p=6$, the best condition for Alice is $(5,6)$ because it's "just under" $(1+2+3+4+5+6)/2$. The number of lights that would be left on so that the sum is just under $n(n+1)/2$ is $\lfloor (n-1)/2 \rfloor$ which can be proved by induction. However, the exact configuration is harder to determine in closed form expression. This matters because in the course of shifting the lights down during phase 2, Bob may inadvertently have to perform the same $T_k$ twice in which case he would just not perform that action, so the calculation of final point count becomes more complicated.

Number of contiguous blocks containing black balls over all permutations

$k$ black balls and $n-k$ white balls are arranged on a straight line, where $n \geq 1$, $0 \leq k \leq n$. The score of the said arrangement is the number of contiguous subsequences that contain at least one black ball. For example, the arrangement $W W B W B$ has two subsequences of length 1, and three subsequences of length 2 that contain at least one black ball. Of course there are subsequences of length 3 or higher as well, so the total score for this arrangement is 11.

What is the sum of the score of all $\binom{n}{k}$ arrangements? You may express this quantity in a recursive formula computable in $O(n^2)$ time, $O(n \log n)$ time, or $O(n)$ time.

Solution 1

We give a recursion formula computable in $O(n^2)$ time with DP.

More to come...

Solution 2A

We prove that the number is $\binom{n+2}{k+2} . \binom{k+1}{2}$ which is computable in $O(n)$ time simply by calculating the factorial table

First let $T_n$ be the $n$-th triangular number, $T_n = n(n+1)/2 = \binom{n+1}{2}$. And we also will use the following well-known identity a lot: $$\sum_{i=1}^{n} \binom{i+p-1}{p} = \binom{n+p}{p+1}, p \geq 0$$ It can be proven many ways, either with induction on $n$ keeping $p$ fixed, induction on $p$ keeping $n$ fixed, or telescoping sum, or using combinatorial arguments (choosing $p+1$ items, and condition on the first item chosen). For $p = 0$ it is a trivial sum, and for $p = 1$ it is the definition of triangular numbers.

Note that if all balls were black, then every single contiguous blocks would contribute to the score. There are $1 + 2 + \dots + n = T_n$ such blocks and that is the score. But if there are white balls, then we want to calculate how many scores are lost due to these pockets of white balls, over the course of all different arrangements.

Now given a particular arrangement of $k$ black balls and $n-k$ white balls, we come with an associated sequence of $k+1$ integers denoting the number of white balls in between black balls, including before the first black ball and after the last black ball. For example, the sequence WWBWWWBBBW corresponds to the sequence $(2, 3, 0, 0, 1)$. Each element in this sequence must be non-negative and they must add up to $n-k$, and there must be exactly $k+1$ integers.

Lemma

Suppose $A$ is the set of all sequences of non-negative integers with length $m \geq 2$ that sum up to $M$, that is $$ A = \{ (a_1, \dots, a_m) | a_i \geq 0, a_1 + \dots + a_m = M \}$$ then:

1. There are $|A| = \binom{M+m-1}{m-1}$ possible sequences.

2. If $m \geq 2$, the number $t \in [0,M]$ occurs $m \binom{M-t+m-2}{m-2}$ times in $A$, in any position.

We prove by induction. For $m=2$, the possible sequences are $A = \{(M,0), (M-1,1,), \dots, (0,M)\}$ so $|A| = M+1$ and there are two occurrences of each number.

Now suppose it holds for $m$, now for $m+1$, consider the first element $a_1 \in [0,M]$. Once its value is chosen, then we have $m-1$ numbers that need to sum up to $M - a_1$. So there are $\binom{M-a_1+m-2}{m-2}$ ways to do this, by induction hypothesis. The total number of ways is the sum of that expression when $a_1$ ranges from $0$ to $M$. $$\sum_{a_1 = 0}^M \binom{M-a_1+m-2}{m-2} = \sum_{i=m-2}^{M+m-2} \binom{i}{m-2} = \sum_{j=1}^{M+1} \binom{j+(m-2)-1}{m-2} = \binom{M+m-1}{m-1}$$ The first two equalities are change of variables, but the last one utilizes the identity above. We have proved the first assertion. For the second assertion, consider the case where the first element $a_1 = t$. The rest of the $m-1$ elements must sum up to $M-t$, so there are $\binom{M-t+m-2}{m-2}$ ways to complete the sequence. However, the number $t$ may appear in any of the other positions as well, so in total there are $m$ times that amount.

One might ask that sometimes the number $t$ itself my appear as a duplicate in multiple elements. For example for $m=4$ and $M=11$, we have have a sequence $(2,2,5,2)$ which should only be counted once. Indeed, when we consider $a_1=1$, we contributed one count towards the total number of "2" appearing. Likewise, when we consider $a_2=2,a_4=2$ we also added one count each time, but the sequence itself contributes three counts towards the appearance of "2." So the total amount is exact and correct, and there is no need to compensate for double counting. Thus the lemma is proven

Alternatively, one could also prove the first assertion by using a combinatorial argument. Consider $M+m-1$ spaces and we are to choose $m-1$ spaces to serve as separators. The spaces between the separators will be the numbers in the sequence.

Now observe that each arrangement corresponds one-to-one with a sequence like we described above. The process to convert from an arrangement to a sequence is invertible. Therefore, the total number of contiguous blocks in all arrangements is $$ T_n \binom{(n-k)+(k+1)-1}{(k+1)-1} = T_n \binom{n}{k} $$ But a lot of these blocks would contribute zero towards the total score. That is, for each sequence $(a_1, \dots, a_{k+1})$, the total loss of scores is the number of contiguous blocks that are contained fully within those pockets of white balls, that is: $$T_{a_1} + \dots + T_{a_{k+1}}$$ So we have to add these over all the possible sequences whose sum is $n-k$. This is where we utilize the lemma. Each time the number $t$ appears, it contributes $T_t$ towards the total loss of score. But thanks to the lemma, we know that the number $t$ appears exactly $$(k+1) \binom{(n-k)-t+k-1}{k-1} = (k+1) \binom{n-t-1}{k-1}$$ times. So the total loss of scores is the sum of this quantity as $t$ ranges from one to $n-k$. (Technically $t=0$ is a valid scenario but it incurs no loss of score because $T_0 = 0$ so we ignore it). $$\sum_{t=1}^{n-k} (k+1) \binom{n-t-1}{k-1} T_t $$ At this point, this sum can be computed in $O(n \log n)$ times using FFT technique, as it's a sum of terms expressed in $t$ and $n-t$, but we can simply the terms further to arrive at a closed form solution, by repeated use of the identity above. Let $B_i = \binom{i+k-2}{k-1}$ (where $k > 0$ is fixed), and for any $m > 0$: $$ B_1 + \dots + B_m = \sum_{i=1}^{m} \binom{i+k-2}{k-1} = \binom{m+k-1}{k}$$ by direct application. Then: $$ \sum_{i=1}^{m} (m+1-i). \binom{i+k-2}{k-1} = m.B_1 + (m-1).B_2 + \dots + 1.B_{m} = B_1 + (B_1+B_2) + \dots + (B_1 + \dots + B_m) = \sum_{i=1}^{m} \binom{i+k-1}{k} = \binom{m+k}{k+1}$$ One more time: $$ \sum_{i=1}^{m} T_{m+1-i}. \binom{i+k-2}{k-1} = T_m.B_1 + T_{m-1}.B_2 + \dots + 3.B_{m-1} + 1.B_{m} $$ $$ = [1+2+ \dots + m]B_1 + [1+2+ \dots + (m-1)]B_2 + \dots + [1+2]B_{m-1} + 1.B_m$$ $$ = 1.B_1 + (2.B_1 + 1.B_2) + (3B_1 + 2B_2 + B_3) + \dots + (m.B_1 + \dots + 1.B_m)$$ $$ =\sum_{i=1}^m \binom{i+k}{k+1} = \binom{m+k+1}{k+2}$$ Now, the total loss of scores becomes: $$ \sum_{t=1}^{n-k} (k+1) \binom{n-t-1}{k-1} T_t = (k+1) \sum_{i=1}^{n-k} T_{n-k+1-i} \binom{i+k-2}{k-1}$$ due to change of variable. Now we apply the latest identity above, with $m=n-k$: $$ = (k+1) \sum_{i=1}^{n-k} T_{n-k+1-i} B_i = (k+1) \binom{n+1}{k+2}$$

So the total score is then: $$ T_n \binom{n}{k} - (k+1) \binom{n+1}{k+2} = \frac{n(n+1) n!}{2. k!(n-k)!} - \frac{(k+1)(n+1)!}{(k+2)!(n-k-1)!} = \frac{(n+1)!}{k!(n-k-1)!} \left(\frac{n}{2(n-k)} - \frac{(k+1)}{(k+2)(k+1)} \frac{n-k}{n-k} \right)$$ $$ \frac{(n+1)!}{k!(n-k)!} \left(\frac{n}{2} - \frac{n-k}{k+2} \right) = \frac{(n+1)!}{k!(n-k)!} \frac{kn+2n-2n+2k}{2(k+2)} = \frac{(n+1)!}{k!(n-k)!} \frac{k(n+2)}{2(k+2)}.\frac{k+1}{k+1} = \frac{(n+2)!}{(k+2)!(n-k)!} \frac{k(k+1)}{2} = \binom{n+2}{k+2} . \binom{k+1}{2}$$

Solution 2B

We also prove that the number is $\binom{n+2}{k+2} . \binom{k+1}{2}$ but using a different (simpler) combinatorial argument.

Given $n+2$ tokens in row, mark $k+2$ of these as blue and $n-k$ as red. There are $\binom{n+2}{k+2}$ ways to choose. And now, among the blue tokens, further mark 2 tokens as "start" and "end", with the condition that "start" must come before "end", and there must be at least one blue token between them. There are $\binom{k+2}{2} - (k+1) = \binom{k+1}{2}$ ways to choose. So the end result is an arrangement of red and blue tokens, and two tokens marked start and end containing at least a blue token in between. We will show that this arrangement corresponds to exactly one score-producing contiguous block in a particular arrangement of black and white balls, and that the mapping is invertible and thus we obtain a bijection.

Arrange the balls according to the $n$ unmarked tokens, where all $n-k$ red tokens correspond to white balls, and $k$ unmarked blue tokens correspond to black balls. The contiguous block is represented by the tokens between start and end, including all red and blue tokens. Because the start and end must flank at least one blue token, then this contiguous block contributes one point towards the score.

On the reverse, given an arrangement of balls, create the corresponding arrangement of red and blue tokens. And then insert the start token to the space right before the first ball in the contiguous block, and the end token to the space right after the last ball. Thus we have a bijection, and the number of total score is $\binom{n+2}{k+2} . \binom{k+1}{2}$.

Solution 3

We present a different combinatorial argument, this time in the permutation space, not a combination space.

First we label all the balls from $1,\dots,n$ such that all the black balls are labeled $1,\dots,k$ and the white balls are labeled $k+1, \dots, n$. Consider any given permutation of the balls, and define the score of this permutation the same way as the problem statement: the number of contiguous blocks that contain at least one black balls. Then we also consider the total number of scores of all $n!$ permutations.

Note that each arrangement of unlabeled balls corresponds to $k! (n-k)!$ permutations of labeled balls, all having the same score. So the total number of scores of all $n!$ permutations is $k! (n-k)!$ times the total score of all arrangements. We will show that the total number of scores of all permutations is $\frac{(n+2)!}{2} \frac{k}{k+2}$.

First we arrange the items $1,\dots,n, X, Y$ in a row, with the condition that $X$ happens before $Y$. There are $\frac{(n+2)!}{2}$ ways to do so. This permutation corresponds to permutation of balls, and the contiguous block is represented by the items between $X$ and $Y$. There may be some permutations where $X$ and $Y$ are next to each other, which means the contiguous block is empty, but that's okay for now. Because we have conditioned that $X$ happens before $Y$, the mapping is injective, meaning no two item permutation map to the same (ball permutation, contiguous block) tuple.

However, not all of those permutations will produce a score. In order for it to produce a score, there must be at least one of $1,\dots,k$ between $X$ and $Y$. So now we consider the permutations of these $k+2$ items. Each of the $n+2$-item permutations has a unique corresponding $k+2$-item permutation, but every $(n-k)!$ of the large permutations produces the same smaller permutation.

There are $\frac{(k+2)!}{2}$ smaller permutations because they still all have $X$ before $Y$. We want to exclude the ones where $X$ and $Y$ are together in this permutation (even if in the original permutations there may be numbers larger than $k$ between them). There are $(k+1)!$ such permutations. So the remaining permutations are: $$\frac{(k+2)!}{2} - (k+1)! = \frac{(k+2 - 2}{2} (k+1)! = \frac{k}{2} (k+1)! = \frac{k}{k+2} \frac{(k+2)!}{2} $$ In other words, the good permutations are only $k/(k+2)$ of the total number of permutations, when we consider all small permutations. But since all large permutations map to the small permutations evenly ($(n-k)!$ to one), then the number of good large permutations is $\frac{(n+2)!}{2} \frac{k}{k+2}$. Then, we map it back to the space of ball arrangement, where each permutation corresponds to $k!(n-k)$ arrangements, so the total number of score of arrangements is: $$\frac{(n+2)!}{2} \frac{k}{k+2} . \frac{1}{k!(n-k)!} .\frac{k+1}{k+1} = \frac{(n+2)!}{(n-k)!(k+2)!} \frac{(k+1)k}{2} = \binom{n+2}{k+2} \binom{k+1}{2}$$

Questions in an exam

A test is taken by 2019 students. The test consists of $n$ questions (where $n > 0$ is even), where each question is a true/false problem worth 1 point. It is known that:

1. Each student answered exactly half of the questions correctly.

2. Given two students, their answers agreed on exactly half of the questions.

Find the smallest $n$ such that this situation is possible.

Solution

If $n = 2^k m$ where $m$ is odd, then there are at most $2^k - 1$ students. Therefore, with 2019 students, we need $2048$ questions for this situation to be possible. Proof:

Suppose $m=1$, we will give a configuration of $2^k-1$ students that satisfy both conditions. First of all, given a number $b$, let $f(b)$ denote the number of 1s in its binary representation, modulo 2. That is $f(b)=0$ if $b$ contains an even number of 1s in its binary representation, and $f(b) = 1$ otherwise.

Label the students $1$ to $2^k-1$ and consider their binary representation (from $0...01$ to $1...1$, all having exactly $k$ digits). Also label the problems $0$ to $2^k-1$ and consider their binary representation (from $0...0$ to $1...1$, all having exactly $k$ digits). We set student $s$ answered problem $p$ correctly if $f(p \wedge s) = f(s)$, where $\wedge$ denotes bitwise AND operation between $p$'s binary representation and $s$' binary representation. In other words, if the number of digits where $p$ and $s$ are both 1 in their binary representation is the same parity as the number of 1s in $s$' representation, then student $s$ answered problem $p$ correctly.

For example, student $5=0...101$ answered question $2=0...010$ correctly because $f(5 \wedge 2) = f(0...000) = 0 = f(5)$. But student $7 = 0...111$ answered question $3 = 0...011$ incorrectly because $f(7 \wedge 3) = f(111 \wedge 011) = f(011) = 0 \neq 1 = f(7)$.

From there, it's easy to see that condition 1 is satisfied. A student's correct answers depends only on the digits of $s$ that are 1, and there are $2^{f(s)}$ possible combinations of those digits. Out of those, there will be half that has even number of 1s and half that has odd number of 1s, so that student answers half the questions correctly.

To prove condition 2, consider student $s$ and $t$. Consider $r = s \oplus t$ where $\oplus$ denotes the bitwise XOR operation. In other words, $r$ is a number whose binary representation is 1 at the digits where $s$ and $t$ are different, and 0 at the digits where $s$ and $t$ are the same. The problems where $s$ and $t$'s answers agree are exactly the problems where student $r$ answered correctly. To prove that assertion, note that $f$ is distributive with respect to $\oplus$. That is $f(a \oplus b) = f(a) \oplus f(b)$. This is because if $a$ has $d_a$ number of 1s and $b$ has $d_b$ number of 1s then $a \oplus b$ has $d_a + d_b \mod 2$ number of 1s, because the digits that were both 1s and turned into zero always come in pairs.

So suppose both $s$ and $t$ get problem $p$ right, then $p \wedge r = p \wedge (s \oplus t) = (p \wedge s) \oplus (p \wedge t)$ so $f(p \wedge r) = f(p \wedge s) \oplus f(p \wedge t) = f(s) \oplus f(t) = f(s \oplus t) = f(r)$ and similarly if they both get the problem wrong. We can show that if $s$ gets problem $p$ right and $t$ gets it wrong, then $r$ gets problem $p$ wrong as well. That way, we showed that student $s$ and $t$'s answers agree exactly on the problems that student $r$ gets right, and by condition 1, this is exactly half of all the problems. So condition 2 is proven.

Now for $m > 1$, let $l = 2^k$ so $n = lm$, we simply assign student $s$ answering problem $p$ correctly if and only if student $s$ answers problem $(p \mod l)$ correctly. In essence, we are duplicating problems $0,\dots, l-1$'s answers (which were constructed above) $m$ times. Condition 1 and 2 are still satisfied.

The second part of the proof is to show that this arrangement is optimal. That is, with $n = 2^km$ there is at most $2^k-1$ students.

For $n$ to be even then $k \geq 1$. For $k = 1$ we show that there is at most 1 student. That is, condition 1 can be fulfilled easily, but condition 2 cannot be fulfilled. Indeed, WLOG we may assume that student $A$ answered questions $1,\dots,m$ correctly and $m+1,\dots,2m=n$ incorrectly. If there is a second student $B$, then suppose above questions $1,\dots,m$ he answered $t$ questions correctly, then he must answer $m-t$ questions among the second half correctly. However, that means the number of questions that $A$ and $B$ agreed on is $t + (m- (m-t)) = 2t \neq m$, contradiction. So there is at most $2^1 - 1 = 1$ student for $k = 1$.

Now we use induction on $k$. Suppose that $k > 1$ and let $h = n/2 = 2^{k-1}m$. Take student $A$, and WLOG assume that he answered questions $1, \dots, h$ correctly (i.e. the first half). For any other student $B$, suppose he answers $t$ questions correctly on the first half, so $h-t$ questions correctly from the second half. The number of questions that $A$ and $B$ agreed on is $t + (h-(h-t)) = 2t = h$, so student $B$ answered $h/2$ questions correctly on the first half. This means all students other than $A$ must answer $h/2$ questions correctly on the first half.

Also note that if we replace $B$ with $B'$ whose answers are completely opposite of $B$, then condition 1 and 2 are both still preserved. $B'$ still answered half the questions, and the questions that $B$ agreed with other students are just "flipped." However, $B$ and $B'$ cannot both coexist because obviously condition 2 would be violated.

So now suppose that there are two students $B,C$ distinct from $A$. It's possible that $B$ and $C$ agree completely on the first half (and disagree completely on the second half). We see this a lot in the construction above. In that case, we call $B$ and $C$ "friends", and we put them under one "group." Also, if $B$ and $C'$ would have been friends, we still call $B$ and $C$ friends (as in, if $B$ and $C$ disagree completely on the first half but agree completely on the second half, we still call $B,C$ friends). Note that because $C$ and $C'$ cannot both coexist, we are not claiming that $B$ can be friends with two existing students. Indeed we shall show that $B$ is friends with at most one other person.

If $B$ and $C$ are friends, and $B$ and $D$ are friends. WLOG we may assume that $B,C,D$ agree completely on the first half (otherwise replace $C$ by $C'$ or $D$ by $D'$ or both). Suppose :
1. $X$ is the set of problems that $A$ got right, $B$ got right
2. $Y$ the set of problems $A$ got right, $B$ got wrong
3. $Z$ the set of problems $A$ got wrong, $B$ got right
4. $W$ the set of problems $A$ got wrong, $B$ got wrong

So $X \cup Y$ is the first half, and $Z \cup W$ is the second half. Because $C$ agree with $B$ on the first half, he must disagree completely on the second half. That means $C$ got $X \cup W$ completely right, and $Y \cup Z$ completely wrong. But we my apply the same argument to $D$, therefore $D$ got $X \cup W$ completely right and $Y \cup Z$ completely wrong. This violates condition 2 because now $C,D$ agree everywhere. Contradiction. Therefore, each group of friendship contains at most 2 students. (Some students may not have friends, so they are in a group each by himself).

Next, suppose $B$ and $C$ are not friends. Consider the answers of $C$ in $X,Y,Z,W$ that agree with $B$. Suppose there are $t_X, t_Y, t_Z, t_W$ answers that agree with $B$ respectively. Now by condition 2, the total number of agreement between $B$ and $C$ must be $h$, so: $$t_X + t_Y + t_Z + t_W = h$$ By condition 1, the total number of answers that $C$ got right must be $h$, so: $$t_X + (|Y| - t_Y) + t_Z + (|W| - t_W) = h$$ And note that $|Y| + |W|=h$ because those are the questions that $B$ got wrong. So the second equation can be written as: $$t_X + t_Z = t_Y + t_W$$ That means $t_X + t_Z = t_Y + t_W = h/2$. But also, $C$ must agree with $A$ for $h$ questions, so: $$t_X + |Y| - t_Y + |Z| - t_Z + t_W = h/2$$ But $|Y| + |Z| = h/2$ because those are the questions that $B$ disagreed with $A$, so $t_X + t_W = t_Y + t_Z = h/2$ Solving the two equations, we have $t_X = t_Y$. Next, because $C$ must answer $h/2$ questions correctly on the first half, then $t_X + |Y|-t_Y = h/2$, so $|Y| = h/2$, which means $|X| = |Y| = |Z| = |W| = h/2$. we get $t_X = t_W = t_Z = t_Y = h/4$. Because $k>1$ and $h = 2^{k-1}m$ then $h/4$ is still an integer. Therefore $t_X + t_Y = h/2$. So the conclusion here is that if $B$ and $C$ are not friends, then $C$ agreed with $B$ on exactly $h/2$ on the first half, and $h/4$ on the second half.

Suppose there are $2^k$ students, and because $A$ is one of them, then there are $2^k-1$ "non-A" students. Because each group contains at most 2 students, then there are at least $\lceil (2^k-1)/2 \rceil = 2^{k-1}$ groups. Take any student from each group to be the "representative" of that group. Now, consider their answers only on the first half of the exam. These answers satisfy the condition for the exams with $n/2 = h$ questions. Indeed, condition 1 is satisfied because all students other than $A$ must answer $h/2$ questions correctly. Condition 2 is also satisfied because every two students who are not friends must agree exactly $h/2$ on the first half. Therefore, now we have at least $2^{k-1}$ students (excluding student $A$) with $h = 2^{k-1}m$ problems that satisfy conditions 1 and 2, violating our induction hypothesis. Contradiction. So there must be at most $2^k-1$ students in an exam with $2^km$ problems.

Finally, because $2^{10} < 2019 < 2^{11}$, then in order for this situation to happen with 2019 students, $k \geq 11$. That means the least number of problems is $2^{11}.1 = 2048$.

Spy And Radio

A spy is being commissioned by the government to infiltrate a mafia rank in order to find out who the head honcho is. There are sixteen potential candidates and the government needs to know which one of them is the leader. However, the spy only has one chance to pass the information back to the government before she gets murdered.

On an previously-agreed date, the local radio station will announce the winning lottery number. This number will be from 0 to $2^{15}-1$ inclusive, and will be saved as a 15 binary digit data in that radio station's computer. The spy only has one chance to infiltrate the radio station's computer mainframe, change one bit (or none) in that number, and leave. The broadcaster will announce the number without knowing that the spy has changed it, and the government agents will listen to that changed number. The government and the spy don't know the winning number in advance and they have no further power to rig the lottery.

To make things more complicated, the spy does not have that much time in the radio station's computer room. Therefore, she would not be able to perform lengthy and complex calculations. Of course, lengthy and complex is a relative term, but she is a sufficiently accomplished mathematician and she needs to perform this task in less than one minute.

What strategy can be done by the spy and the government so that she can pass the mafia leader's identity accurately with that one chance?

Remark: The following weaker versions of the problem are due to David Rager:

1. Prove that if there are 3 mafia members and 3 bit number lottery winner, you can pass the information with only 1 flip.

2. Prove that if there are 27 mafia members and 27 bit number lottery winner, you can pass the information with only 3 flips.

Prisoners and hats

2047 prisoners are kept in a dungeon. One day, the King decided to do an experiment. They will each be given either a blue hat or a red hat and then collected in a room. Each prisoner can see the color of everyone else's hats but not his own. No form of communication is allowed. After a few minutes, the warden will come and ask each of them individually what he thinks the color of his hat is. They can answer "red" or "blue." Each prisoner must answer by whispering to the warden's ear, so that the other prisoners can't hear his answer. Those who can answer correctly will be freed, but those who answer incorrectly will be executed on the spot. The prisoners found out about this experiment the night before, so they had some time to devise a scheme to ensure they can save as many people as possible.

Question 1: What is the scheme that allows everyone to live in this case?

Suppose now the King decided to alter the experiment a little bit. The warden will come to the room immediately after everyone get their hats. Each prisoner is allowed to answer "red", "blue", or "I don't know." If everyone answers "I don't know", they will all be executed. If someone answers "red" or "blue" and he is wrong, they will all be executed. Otherwise, everyone is freed at the same time. In other words, in order for them to live, at least one person must answer "red" or "blue" and he must be right. As before, the prisoners can't hear everyone's answers, and the release or execution is delayed until everyone had given their answers.

Question 2: What is the scheme that allows everyone to live with the highest probability? What is that probability?