Edit: I misunderstood. Left for historical reasons, but not important. Please ignore.
Wait—may I ask for some remedial instruction on why zero-sum game-theory isn’t a strict subset of decision theory, just with pessimistic assumptions about some of the variables? I can turn any decision problem into a game-theory problem by adding an opponent with opposite values from me, right? I strongly suspect that _both_ are NP generally, and I can give many specific examples of both that can be solved in polynomial time.
I’m confused by your “cognitive modes” idea as well—it may be true, but I don’t see how it follows, if decision theory is more difficult (NP) than game theory (P), that
In decision theory, you plan with no regard to any opponents interfering with your plans, allowing you to plan on arbitrarily long time scales. In zero-sum game theory, you plan on the assumption that your opponent will interfere with your plans (your ∃s are interleaved with your opponent’s ∀s), so you can only plan as far as your opponent lacks the ability to interfere with these plans
This seems to lean in the opposite direction—in decision theory, you can easily answer as far out as you like, in zero-sum game-theory you’re far more limited. It is also more about certainty of game state than about motivation or existence of opponent. A plan with no opponents but also a bunch of randomness (or unknown starting state; same thing for these purposes) is MORE impossible to follow than a plan with known state and an active opponent.
I do somewhat agree with your final paragraph, but I don’t agree with your premise nor reasoning to get there.
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.
Edit: I misunderstood. Left for historical reasons, but not important. Please ignore.
Wait—may I ask for some remedial instruction on why zero-sum game-theory isn’t a strict subset of decision theory, just with pessimistic assumptions about some of the variables? I can turn any decision problem into a game-theory problem by adding an opponent with opposite values from me, right? I strongly suspect that _both_ are NP generally, and I can give many specific examples of both that can be solved in polynomial time.
I’m confused by your “cognitive modes” idea as well—it may be true, but I don’t see how it follows, if decision theory is more difficult (NP) than game theory (P), that
This seems to lean in the opposite direction—in decision theory, you can easily answer as far out as you like, in zero-sum game-theory you’re far more limited. It is also more about certainty of game state than about motivation or existence of opponent. A plan with no opponents but also a bunch of randomness (or unknown starting state; same thing for these purposes) is MORE impossible to follow than a plan with known state and an active opponent.
I do somewhat agree with your final paragraph, but I don’t agree with your premise nor reasoning to get there.
Where did this come from? Did you confuse PSPACE with P?
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!