A ladder is a non-decreasing sequence $a_1, a_2, \dots, a_{2020}$ of non-negative integers. Diego and Pablo play by turns with the ladder $1, 2, \dots, 2020$, starting with Diego. In each turn, the player replaces an entry $a_i$ by $a_i'<a_i$, with the condition that the sequence remains a ladder. The player who gets $(0, 0, \dots, 0)$ wins. Who has a winning strategy? Proposed by Violeta Hernández
Problem
Source: Mexico National Olympiad Mock Exam (OMMock) P5
Tags: combinatorics, game strategy
plagueis
10.11.2020 21:29
Any solutions? @below thank you for your solution
Polipo2704
11.11.2020 04:10
I think the following strategy holds for B (Pablo) and for general $n$ divisible by $4$: Since $n$ is even we can pair the $2k$th term with the $2k-1$th term, call those pairs $P$ pairs, since $n$ is divisible by 4 we can pair the $4k-2$th term with the $4k$th term, call those pairs $Q$ pairs. If $A$ decrease a term of some member of a $Q$ pair exactly by $1$ at the first time, then $B$ decreases the other term of the $Q$ pair by exactly $1$, otherwise if $A$ decrease some term by $r$ then $B$ decrease by $r$ the other term corresponding to it's $P$ pair, it is easy to see that $B$ can always reply to $A$'s move since the difference between numbers of the same $P$ pair is at most $1$ at any time and the game must finish because the sum of the terms is strictly decreasing, then B wins.
R-sk
11.11.2020 09:19
I think pablo wins if he has the following strategy Since diago. Goes first then everytime when diago chooses an even number pablo replaces it with half of chosen number whereas when diago choses an odd number pablo replaces. It with half of (chosen number -1) then pablo wins