Very simple strategy: #1 says the number equal to it's ID. Squid #2 knows it's own ID and then compares them and says which one is larger. The sum of the numbers is simply the ID of Squid #1 which can be up to $2023.$ A better way would be to write these numbers in base-2: Squid #1 expresses it's number in base-2, using $2$s instead of $0$s. In this time, Squid #2 juts spams saying $1$ to reduce the sum. When Squid #1 is done, it says $3$ as it's number, signaling that it is done, and Squid #2 compares the numbers as before.

StrahdVonZarovich wrote: Very simple strategy: #1 says the number equal to it's ID. Squid #2 knows it's own ID and then compares them and says which one is larger. The sum of the numbers is simply the ID of Squid #1 which can be up to $2023.$ A better way would be to write these numbers in base-2: Squid #1 expresses it's number in base-2, using $2$s instead of $0$s. In this time, Squid #2 juts spams saying $1$ to reduce the sum. When Squid #1 is done, it says $3$ as it's number, signaling that it is done, and Squid #2 compares the numbers as before. This method isn't optimal. We can assume both squid get an $11$ digit number (in base $2$), put leading zeros if they get a $10$ digit or less. Then they take turns shouting the digits from the highest place. (squid 1 shout his first, squid 2 shout his second, etc.). This way we can cut the number to a $20$ instead of your $33$. However, I heard someone said there is a strategy with $10$.

CrazyInMath wrote: StrahdVonZarovich wrote: Very simple strategy: #1 says the number equal to it's ID. Squid #2 knows it's own ID and then compares them and says which one is larger. The sum of the numbers is simply the ID of Squid #1 which can be up to $2023.$ A better way would be to write these numbers in base-2: Squid #1 expresses it's number in base-2, using $2$s instead of $0$s. In this time, Squid #2 juts spams saying $1$ to reduce the sum. When Squid #1 is done, it says $3$ as it's number, signaling that it is done, and Squid #2 compares the numbers as before. This method isn't optimal. We can assume both squid get an $11$ digit number (in base $2$), put leading zeros if they get a $10$ digit or less. Then they take turns shouting the digits from the highest place. (squid 1 shout his first, squid 2 shout his second, etc.). This way we can cut the number to a $20$ instead of your $33$. However, I heard someone said there is a strategy with $10$. You can cut the number to $18$ because $11111111110$ and $11111111111$ in binary are greater than $2023$ in base ten. However, you can't cut it any more using this strategy because $10111111110$ and $10111111111$ in binary are less than or equal to $2023$ in base ten.

I have a strategy that I think is either optimal or very close to optimal, but I have no proof of that. I also don't know what is the maximal sum of announced numbers in this strategy.

The answer is $\log_2(n)$. Just induct on both bound and construction.

gghx wrote: The answer is $\log_2(n)$. Just induct on both bound and construction. You are welcome to show us more details of your surely ingeniously short solution.

Tintarn wrote: gghx wrote: The answer is $\log_2(n)$. Just induct on both bound and construction. You are welcome to show us more details of your surely ingeniously short solution. Neither ingenious nor short, but here goes... Construction goes as follows (it should be $\log_2(n)-1$, btw): $n=2$ can be done with a sum of $0$, since the numbers are distinct. We now induct to show that $2^k$ can be done with $k-1$. Consider the first player, who has some number $A$. If $A=1$ or $2^k$ he is done. Else assign the following ranges: $[2,2^{k-1}+1]\rightarrow 1$, $[2^{k-1}+2,2^{k-1}+2^{k-2}+1]\rightarrow 2$, etc Basically, $i$ is assigned a range of size $2^{k-i}$, for $1\le i\le {k-1}$. Based on which range $A$ falls into, the first player says the assigned number. By the induction hypothesis, we are done with maximum $(k-i-1)+i=k-1$ sum. Bound: We induct to show that $n=2^k+1$ requires at least $k$. This is obviously true for $k=1$. Any strategy goes as follows: all numbers from $[1,n]$ are assigned some value, and whichever number the first player receives he says the assigned value. Let the number of numbers with value $i$ be $f(i)$. Suppose that $2^k+1$ can be done with $k-1$. Then in this strategy we have $f(1)+f(2)+\cdots+f(k-1)=2^k+1$. It is easy to see that there is some $i$ for which $f(i)\ge 2^{k-i}+1$. This means that $n=2^{k-i}+1$ can be done with $k-i-1$, a contradiction.