Here's a heuristic suggesting that maybe something like $2^N/N$ forbidden words could be enough to give 0 entropy:

Consider $N$-step transitions between blocks of length $N$. That is: if $L_N$ is the set of all words of length $N$, you're asking: for which $U,V$ in $L_N$ is the concatenation $UV$ legal.

Clearly in the full shift, every transition between $N$-blocks is legal, that is, there are $2^{2N}$ such legal transitions. When a word of length $N$ is forbidden, I want to count the number of $N$-block transitions that are excluded, if the word $W$ is excluded. That is, I want to count the number of pairs $(U,V)$ such that the concatenation $UV$ contains the block $W$. There are $N+1$ locations in a block of length $2N$ where a $W$ could appear; and each one of them rules out $2^N$ possible transitions, as the remaining symbols can be filled in in any way. That is: a forbidden $N$-word, $W$, gives rise to $(N+1)2^N$ forbidden transitions between $N$-blocks (of course if there are repetitions then it may be smaller than this). Now, forbidding something like $2^N/N$ words, assuming that these forbidden words can be chosen very carefully with minimal overlaps *might* be enough to rule out all of the $2^{2N}$ transitions.

For a bound in the other direction, notice that if you forbid $\alpha 2^N/N$ words with $\alpha<\frac 12$, then the crude bounds that I gave rigorously show that there are at least $(1-\alpha)2^{2N}$ transitions remaining. This is sufficient for positive entropy by the criterion in the answer to your previous post.

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