Baltic Way 2019

Let $18=n$, so that the initial message has length $4n$, with exactly $n$ symbols of each kind. We first establish an invariant. Assign $E=3$, $X=-3$, $I=2$, $T=-2$, and let $S$ denote the sum of the values of all symbols appearing in the message. Initially, $S=0$, and the sum stays invariant under all the legal transformations.

Our next observation is that we may always insert or delete the combination TETET, for $$\varnothing \mapsto \{TI\}\mapsto \{TEXI\}\mapsto \{TETIXI\}\mapsto \{TETEXIXI\}\mapsto \{TETETIXIXI\}\mapsto \{TETET\},$$ and this works also in reverse. Required are six operations.

To crack the system, first insert XE behind every I, and XE in front of every T: $I\mapsto IXE$, \quad $T\mapsto XET$. This requires at most $p_1=4n$ operations, after which the message has length at most $12n$. It may now be considered a sequence of the four possible strings $E$, $X$, $IX$, $ET$.

Second, expand any single $X$ (not preceded by an $I$) into $IXIXIX$, and any single $E$ (not succeeded by a $T$) into $ETETET$: $X\mapsto IXIXIX$ (one operation), $E\mapsto ETETET$ (six operations). This requires at most $p_2=12n\cdot 6=72n$ operations, and the message now has length at most $72n$. It is at present reduced to some binary combination of the two strings $IX$, $ET$.

Third, effectuate all possible reductions $$\{ETIX\}\mapsto \{EX\}\mapsto \varnothing \text{and}\{IXET\}\mapsto \{IT\}\mapsto \varnothing .$$ There can be at most $18n$ such reductions, totalling $p_3=18n\cdot 2=36n$ operations. The hacker will be left with a message of the types $ETET\dots ET$ or $IXIX\dots IX$ or an empty screen. But since the invariant $S=0$, the screen must now, in fact, be empty.

Fourth, insert EXIT, using $p_4=2$ more operations. The number of operations was at most $$p_1+p_2+p_3+p_4=4n+72n+36n+2=112n+2=112\cdot 18+2=2018<2019.$$