Answer :
N ≥ [tex]2^{n}[/tex] shows that our construction is optimal up to a multiplicative constant.
Suppose the number of switches is [tex]N=2^{n}[/tex]. Here is a construction using 6n+3 settings.
First, we take the three constant settings [tex]0^{N} , 1^{N},2^{N}[/tex]. Next, for any two different values a,b ∈ {0,1,2} and i ∈ {0,…,n−1}, we have the setting such that switch k gets value a or b depending on the value of the [tex]i^{th}[/tex] bit of k. Now suppose s, t are the two correct switches, and their correct settings are a, b.
If a= b then one of the constant settings unlocks the safe. Otherwise, suppose s and t differ on a bit i. Then either the setting (a, b, i) or the setting (b, a, i) unlocks the safe.
On the other hand, suppose {s1,…, sn} is a set of settings of N switches such that for all pairs i, j ∈ [N], there is a setting in which switch i is set to 0 and switch j is set to 1. Then N ≥ [tex]2^{n}[/tex]. This shows that our construction is optimal up to a multiplicative constant.
To know more about the multiplicative constant visit: brainly.com/question/11872107
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