Is it buggy for at least 2^(n)^2? It gives 4 for any n, but surely for example 2^^2 = 2^(2^2) != 4?
2^^2 != 2^(2^2). Instead, 2^^2 = 2^2.
This will make more sense if you look at how the inputs a,b,n in the toy (2,2,3) and (2,3,3) present differently.
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2^^2 != 2^(2^2). Instead, 2^^2 = 2^2.
This will make more sense if you look at how the inputs a,b,n in the toy (2,2,3) and (2,3,3) present differently.