Update en.md for ioi prisoner

Author: jan876

ioi-2022-prison/en.md

@@ -1 +1,58 @@
-This problem is about generating a finite state machine to compare two different integers, A and B, with the minimum number of states.
+This problem is about generating a finite state machine to compare two different integers, $A$ and $B$, with the minimum number of states.
+
+### Subtask 1
+
+In this subtask, we can choose $x = N$. We can use the following strategy:
+
+* The first prisoner will always read 0 on the whiteboard, and then overwrites it with the number of coins in bag A.
+* The second prisoner will read a positive integer which represents the number of coins in bag A. So, the second prisoner has to check bag B and then answer which bag has fewer coins.
+
+### Subtask 2
+
+Let $S = {\lceil {\sqrt{N}} \rceil}$. In this subtask, we can do the similar thing in subtask 1 but with 2 phases instead of 1 phase. First, we can compare ${\lfloor {\frac{A}{S}} \rfloor}$ and ${\lfloor {\frac{B}{S}} \rfloor}$. If the values are the same, then we can continue to compare ($A$ mod $S$) and ($B$ mod $S$). This solution only needs $x = 2 \times S$.
+
+There’s also another solution that needs $x = 3 \times S$ using the fact that every non-negative number can be represented uniquely as $p^2 + q$ for non-negative numbers $p, q$ such that $q < 2 × p + 1$. First, we can compare ${\lfloor {\sqrt{A}} \rfloor}$ and ${\lfloor {\sqrt{B}} \rfloor}$. Then, if the values are the same, we can compare $A - {\lfloor {\sqrt{A}} \rfloor}^2$ and $B - {\lfloor {\sqrt{B}} \rfloor}^2$.
+
+### Subtask 3
+
+In this subtask, we can get some partial scores depending on how small the chosen $x$ is. The following are some possible solutions that could get some points from this subtask.
+
+#### 3.1 Solution with $x = 3 \times {\lceil {\log_2{N}} \rceil} − 1 = 38$
+
+We can simulate how to compare two binary numbers. By noting that it requires up to 13 bits to represent a number up to 5000, the value on the whiteboard represents the current state, i.e.
+
+* If the value is $3 \times d$, then we need to check the $(12 − d)$-th bit of $A$, and then overwrite it with $3 × d + 1$ or $3 × d + 2$ depending on the bit value.
+* If the value $is 3 × d + 1$ or $3 × d + 2$, then we know the $(12 − d)$-th bit of $A$ is either 0 or 1 from the previous prisoner. Then, the current prisoner has to check the $(12 − d)$-th of $B$ and compare the current bit value between $A$ and $B$. If the values are still the same, the prisoner can overwrite it with $3 \times (d + 1)$ to continue to the next bit. Otherwise, the prisoner can answers which bag has fewer coins.
+
+Therefore, this solution needs $x = 3 \times {\lceil {\log_2{N}} \rceil} − 1$ or $x = 38$ for $N = 5000$.
+
+#### 3.2 Solution with $x = 2 \times {\lceil {\log_2{N}} \rceil} = 26$
+
+Instead of storing the bit only from $A$ on the whiteboard, we can store the bit of either $A$ or $B$ alternately. We can use the parity of the current bit position $i$ to determine whether we are storing the $i$-th bit of $A$ or $B$. Therefore, we don't need an additional state for when we have not stored any value of $i$-th bit.
+
+#### 3.3 Solution with $x = 3 \times {\lceil {\log_3{N}} \rceil} = 24$
+
+We can compare them using base 3 instead of base 2 using a similar idea as the previous solution.
+
+#### 3.4 Solution with $x = 3 \times {\lceil {\log_3{N}} \rceil} - 2 = 22$
+
+We can prune two states when processing the last digit (in base 3 representation). If the last digit is 0 or 2, then we can be sure whether $A < B$ or $A > B$ without needing to check the last digit from the other bag since we know that $A \neq B$.
+
+#### 3.5 Solution with $x = 3 \times {\lceil {\log_3{(N + 1)}} \rceil} - 3 = 21$
+
+In the previous solution, we use the pruning idea on the last digit only. If we use the ternary number system like the previous solution, we can visualize it as a perfect ternary tree where we divide range [0, $3^k − 1$] into 3 parts recursively. And, the pruning in the previous solution is only applied for some leaf nodes (last digits).
+
+Instead of pruning only the last digit, we can apply it to any possible range. Let's say we know that both $A$ and $B$ are currently inside the range [$L, R$], and we are checking bag $A$. If we know that $A = L$, then we immediately know that $A < B$. If we know that $A = R$, then we know $A > B$. It means that we can "prune away" the leftmost and the rightmost integer from the range. After that, we can divide the range
+[$L + 1, R − 1$] into 3 parts and determine which sub-range $A$ belongs to. Then, the next prisoner can check bag $B$ and determine which sub-range it belongs to. If both bags still belong to the same sub-range, then we can continue the process with this smaller range. Otherwise, we can already determine which bag has fewer coins. We continue this process until we can determine the answer.
+
+When we divide a range into 3 sub-ranges, we need 3 additional states. If the base range is a range with 2 integers, then $3 \times K$ states can cover $3 × (3 × (. . . 3 × (2) + 2 . . .) + 2) + 2 = 3^{K+1} − 1$ numbers. So, we need approximately $3 \times K = 3 \times ({\lceil {\log_3{(N + 1)}} \rceil} - 1) = 21$ states for $N = 5000$.
+
+#### 3.6 Solution with $x = 20$
+
+We can optimize the previous solution. Dividing a range by 3 is not always optimal. We can use dynamic programming to determine what is the best divider for each level such that it can cover the largest range.
+
+$$
+dp[x] = \max_{1 \leq i \leq x} {(i \times dp[x - i] + 2)}
+$$
+
+where $dp[x]$ denotes the maximum range that can be covered using $x$ states and $dp[0]$ = 2. For $x = 20$, we can cover until $N = 5588$. Such $x$ is the minimum such that $N ≥ 5000$. It can be achieved by dividing the range [1, 5588] by 3, 3, 3, 3, 3, 2, 2, and 1 for each level recursively using the same process as the previous solution.