saved_chats/#4

USER: solve this: https://usaco.org/index.php?page=viewproblem2&cpid=1400

BOT: There was something wrong with the link. Could you paste it again?



USER: https://usaco.org/index.php?page=viewproblem2&cpid=1400

BOT: Thanks! Now could you provide a link to the editorial.



USER: https://usaco.org/current/data/sol_prob3_silver_feb24.html

BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 



USER: editorial

BOT: Subtask 1: $M \leq 16$:
For this subtask, it suffices to iterate through all of Elsie's $2^M$ candidate
turn sequences in lexicographical order. If with Elsie's current candidate turn
sequence, she is guaranteed to not lose, output the current sequence, and exit. 
It remains to describe how to check whether Elsie is guaranteed to not lose with
a given candidate turn sequence. Given Elsie's guess $p$ for a turn $t$, the
best response for Bessie is the one that causes Elsie to lose the most marbles
on that turn (or gain the least marbles if none of Bessie's responses can cause
Elsie to lose marbles on that turn). Define $\texttt{changes}[t][p]$ to be the
amount $N$ will change on turn $t$ if Elsie guesses parity $p$ and Bessie moves
optimally. After precomputing all entries of $\texttt{changes}$ in $O(MK)$ time,
we can check whether a given turn  sequence causes Elsie to not lose in $O(M)$
time. 
The overall time complexity is
$O(MK+M\cdot 2^M)$.
Implementation (Python):

Subtask 2: $M\le 1000$
To remove the $O(M\cdot 2^M)$ from the time complexity, we can make the
following optimization to the function
$\texttt{find_best_move_seq_starting_with}(seq)$ from the above solution code:
if there is no sequence of moves starting with $seq$ that allows Elsie to not
lose, then immediately return. To check whether such a sequence exists or not,
we assume that for every turn after $|seq|$, Elsie guesses the parity that
causes her to gain the most marbles, assuming Bessie responds optimally.  If
this allows Elsie to not lose, then a sequence exists. Otherwise, no sequence
exists, since Elsie lost despite always making the best move after $seq$. This
can be checked in $O(M)$ time.
After adding the optimization, the function
$\texttt{find_best_move_seq_starting_with}$  is called $O(M)$ times rather than
$O(2^M)$ times, because if $\texttt{find_best_move_seq_starting_with}(seq)$ does
not immediately return, then
If there is a move sequence starting with $seq + [0]$, then the answer will
be found after calling
$\texttt{find_best_move_seq_starting_with}(seq + [0])$.Otherwise, $\texttt{find_best_move_seq_starting_with}(seq + [0])$
immediately returns and the answer will be found after calling
$\texttt{find_best_move_seq_starting_with}(seq + [1])$.
The overall time complexity is $O(MK+M^2)$.
Implementation with recursion:

Implementation without recursion:

Full solution:
To reduce $O(M^2)$ to $O(M)$ time, we need to speed up the function
$\texttt{cannot_start_with}(seq)$ from the above solution code. This function
checks the following two conditions:
$N+\sum_{t=0}^{l-1}\texttt{changes}[t][seq[t]]>0$ for all $l < |seq|$.$N+\sum_{t=0}^{|seq|-1}\texttt{changes}[t][seq[t]] + \sum_{t=|seq|}^{l-1}\max(\texttt{changes}[t][0], \texttt{changes}[t][1])>0$
for all $|seq|\le l\le M$.
Suppose that the first condition already holds; then the second condition holds
if and only if the minimum prefix sum (including the empty prefix) of the array 

is greater than some threshold. Let $\texttt{min_psum}[t]$ denote the minimum
prefix sum when $|seq| = t$; then we have
$\texttt{min_psum}[t]=\min(0, \max(\texttt{changes}[t]) + \texttt{min_psum}[t+1])$
for $t<M$. So $\texttt{min_psum}[t]$ can be computed for all $t$ in reverse
order in $O(M)$ time.
The full solution is now clear; it is the same as the non-recursive solution to
the previous subtask but replacing the call to $\texttt{cannot_start_with}(seq)$
with a comparison involving the current value of $N$ and
$\texttt{min_psum}$.
The overall time complexity is $O(MK)$.
My Implementation:

Danny Mittal's Implementation (Java):


Let me know if you have any questions.