If you add an opponent with opposite values from you into a decision theory problem, you are adding a component that is approximately as hard to compute the behavior of as it is to solve the problem as a whole (since it depends on a computation by the opponent which is of roughly the same form).
My point is exactly that P and PSPACE are not the same. Addition of an opponent cannot turn NP into P, so “decision problem” vs “zero-sum game”cannot be the distinction which defines the problem space.
Oh my—I completely misread the post, thinking it was comparing P vs PSPACE, when in fact it’s comparing NP to PSPACE. My point that competitive decisions are just a special-case of decision theory stands, but it’s far less important.
If you add an opponent with opposite values from you into a decision theory problem, you are adding a component that is approximately as hard to compute the behavior of as it is to solve the problem as a whole (since it depends on a computation by the opponent which is of roughly the same form).
P and PSPACE are not the same.
My point is exactly that P and PSPACE are not the same. Addition of an opponent cannot turn NP into P, so “decision problem” vs “zero-sum game”cannot be the distinction which defines the problem space.
Why do you think zero-sum game theory is P? The quantified Boolean formula problem is PSPACE-complete.
Oh my—I completely misread the post, thinking it was comparing P vs PSPACE, when in fact it’s comparing NP to PSPACE. My point that competitive decisions are just a special-case of decision theory stands, but it’s far less important.
Nevermind!