Minimum chain to open an electronic lock

Question

An electronic lock can be opened by entering three numbers from $\{0,1,2\}$ and this independent from what we enter before. For example if the correct password is $012$, then $11012$ will open the lock. The question is how long the minimum sequence of numbers that open for sure the lock.\

I have tried to write the problem as a graph problem, in fact there are $27$ possibilities for the correct password. Let consider the oriented graph with vertices the set of possible passwords and $(xyz, tuv)$ is an edge in the graph if $xyz \neq tuv$ and $y = t,~ z = u$. I claim that to answer the problem it is enough to find the length of a shortest path which goes through every vertex. I tried to find an Hamiltonian path, but I did not find a good algorithm to check if one exist (the problem of finding the hamiltonian path is NP-complete). Does any one have a suggestion to solve this problem ?

https://math.stackexchange.com/questions/48808/shortest-sequence-containing-all-permutations — Austin Weaver, Oct 02 '17 at 16:52
Thanks for the link but, the password is not necessarily a permutation of $0,1,2$. It can be $101$ — Kroki, Oct 02 '17 at 16:59
You seem to be asking for the De Bruijn Sequence for the alphabet ${0,1,2}$ for words of length $3$. — JMoravitz, Oct 02 '17 at 17:02

score 2 · Accepted Answer · answered Oct 02 '17 at 17:04

Consider instead the nine vertex graph whose vertices are pairs of numbers $(x,y)$, and where two pairs $(x,y)$ and $(z,w)$ are connected by an edge whenever $y=z$. Show that this graph has an Eulerian circuit (cycle which traverses each edge exactly once) and that such a circuit translates to a strategy for opening the lock.

Minimum chain to open an electronic lock

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