Missionaries and cannibals

Pure logicLevel 3 · Intermediate · ●●●○○

Three missionaries and three cannibals must cross a river with a boat that only carries a maximum of two people. On no shore can there be a group where the cannibals are more than the missionaries (if there are missionaries present), because then they devour them.

How does everyone cross safely?

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Hints

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In all the intermediate states, when there are missionaries on a shore, they are never in the minority compared to the cannibals.
Notation: (M, C mid M, C) = (left | right).
Yes, and it can be done in 11 crossings.

Solution

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Answer: Yes, and it can be done in 11 intersections.
Notation: $(M, C \mid M, C)$ = (left | right).
Initial state: $(3,3 \mid 0,0)$

$CC \rightarrow$ $(3,1 \mid 0,2)$
$C \leftarrow$ $(3,2 \mid 0,1)$
$CC \rightarrow$ $(3,0 \mid 0,3)$
$C \leftarrow$ $(3,1 \mid 0,2)$
$MM \rightarrow$ $(1,1 \mid 2,2)$
$MC \leftarrow$ $(2,2 \mid 1,1)$
$MM \rightarrow$ $(0,2 \mid 3,1)$
$C \leftarrow$ $(0,3 \mid 3,0)$
$CC \rightarrow$ $(0,1 \mid 3,2)$
$C \leftarrow$ $(0,2 \mid 3,1)$
$CC \rightarrow$ $(0,0 \mid 3,3)$

In all the intermediate states, when there are missionaries on a shore, they are never in the minority compared to the cannibals.

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