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Episode 10. Blind quantifiers. Unistructurality The blind universal and existential quantifiers DeMorgan’s laws for blind quantifiers The hierarchy of quantifiers How blind quantification affects game trees. 0. 10.1. Unistructurality.

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Blind quantifiers

Episode 10

Blind quantifiers

  • Unistructurality

  • The blind universal and existential quantifiers

  • DeMorgan’s laws for blind quantifiers

  • The hierarchy of quantifiers

  • How blind quantification affects game trees

0


Unistructurality

10.1

Unistructurality

A game A is said to be unistructural iff, for any valuations e1 and e2, we have

Lre1[A]=Lre2[A].

Intuitively, unistructural games are games all instances of which have the same

structure. All of the examples of games that we have seen so far are unistructural, and

so are all naturally emerging games. Also, all of our game operations preserve the

unistructural property of games. Hence, if you wish, it is safe to assume that “game”

always means “unistructural game”. For the sake of generality, however, we do not

officially impose this unnecessary restriction on games.

Our definition (next slide) of the blind quantifiersx (universal) and x (existential),

however, assumes that the game A(x) to which they are applied satisfies the weaker

condition of x-unistructurality.

We say that a game A(x) is x-unistructural iff, for any valuation e and any two

constants c and d, Lre[A(c)]=Lre[A(d)]. Intuitively, this means that the structure of

any given instance of A does not depend on the value of x (but may depend on the

values of some other variables).

Of course, every unistructural game is also x-unistructural for any variable x.


Formal definitions of the blind quantifiers

10.2

Formal definitions of the blind quantifiers

Definition 10.1. Let A(x)be an x-unistructural game.

(a) The gamexA(x)is defined by:

(b) The gamexA(x) is defined by:

  • LrexA(x)= LreA(x).

  • WnexA(x)  = ⊤ iff, for all constants c, WneA(c)  = ⊤.

  • LrexA(x)= LreA(x).

  • WnexA(x)  = ⊥ iff, for all constants c, WneA(c)  = ⊥.

Intuitively, playing xA(x) or xA(x) means just playing A(x) “blindly”, without

knowing the value of x. In xA(x), Machine wins iff the play it generates is successful

for every possible value of x, while in xA(x) being successful for just one value is

sufficient.

When applied to elementary games, the blind quantifiers, just like the parallel

quantifiers, act exactly as the quantifiers of classical logic.


Demorgan s laws for blind quantifiers

10.3

DeMorgan’s laws for blind quantifiers

From the definition one can see a perfect symmetry between  and ,

⊤and⊥.

Therefore, just as for the choice and parallel quantifiers, DeMorgan’s

laws hold:

xA= xA

xA= xA

xA= xA

xA= xA


X versus x

10.4

x versus ⊓x

Unlike xA(x) which is a game on infinitely many boards, both

⊓xA(x)and xA(x) are one-board games. Yet, they are very different

from each other.

To see this difference, compare the problems

⊓x(Even(x)⊔Even(x))

x(Even(x)⊔Even(x))

and

Depth=

2 (Environment selects a

number m, to which Machine

replies by 0 or 1)

Depth=

1 (only Machine has a

move, 0 or 1)

Winnable?

Of course. Evenness

is a decidable problem.

Winnable?

No. Machine cannot

select a ⊔-disjunct

that would be true

for all values of x.


The hierarchy of quantifiers

10.5

The hierarchy of quantifiers

Easiest

to win

xA(x)

xA(x)

incomparable

⊔xA(x)

⊓xA(x)

Hardest

to win

xA(x)

xA(x)

incomparable


How blind quantification affects game trees

10.6

How blind quantification affects game trees

Suppose A is a unistructural game. To turn the game tree for A into a

game tree for xA, just add the prefix “x” to each node of it.

Similarly for xA.

Even(x)⊔Odd(x)

x

(Even(x)⊔Odd(x))

x

0

1

0

1

Even(x)

Odd(x)

x

Even(x)

x

Odd(x)


How blind quantification affects game trees1

10.6

How blind quantification affects game trees

Suppose A is a unistructural game. To turn the game tree for A into a

game tree for xA, just add the prefix “x” to each node of it.

Similarly for xA.

Even(x)⊔Odd(x)

x

(Even(x)⊔Odd(x))

⊥⊔⊥

x

0

1

0

1

Even(x)

Odd(x)

x

Even(x)

x

Odd(x)


A computable problem

10.7

x(Even(x)⊔Odd(x)  ⊓y(Even(x+y)⊔Odd(x+y)))

A computable -problem

This problem is computable. The idea of a winning strategy is that, for any given y,

in order to tell whether x+y is even or odd, it is not really necessary to know the value

of x. Rather, just knowing whether x is even or odd is sufficient. And such knowledge

can be obtained from the antecedent.

In other words, for any known y and unknown x, the problem of telling whether

x+y is even or odd is reducible to the problem of telling whether x is even or odd.

Specifically, of both x and y are even or both are odd, then x+y is even; otherwise

x+y is odd.

Below is the evolution sequence induced by a run where Machine used a winning

strategy.

Game Move

x(Even(x)⊔Odd(x)  ⊓y(Even(x+y)⊔Odd(x+y)))

1.7

x(Even(x)⊔Odd(x)  Even(x+7)⊔Odd(x+7))

0.0

x(Even(x)  Even(x+7)⊔Odd(x+7))

1.1

x(Even(x)  Odd(x+7))

The play hits ⊤, and thus Machine wins.


Exercise

10.8

Exercise

Do the following problems appear to be (always) computable?

x(P(x)⊔P(x))

No

x(P(x)P(x))

Yes

x(P(x)P(x)P(x))

Yes

x(P(x)P(x)P(x))

No

xP(x)⊓xP(x)

Yes

⊓xP(x) xP(x)

No

⊓xP(x) xP(x)

Yes

x(P(x)Q(x))  (xP(x)xQ(x))

Yes

⊓x(P(x)Q(x))  (⊓xP(x)⊓xQ(x))

No

⊓xP(x)⊓xQ(x) ⊓x(P(x)Q(x))

Yes


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