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Cognitive Computing 2013 Consciousness and Computations 8. THE REDUCTION PRINCIPLE Prof. Mark Bishop

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01/04/2014(c) Bishop: Consciousness and computations2 Unsolvable decision problems (1) Syntax: If u is a decision problem… Then U is a Norma program to decide u … and U (x) is a Norma program to decide u with input (X = x). NormaP (a) Norma program P executing with input reg. (X = a). Some unsolvable decision problems: The Halting Problem (HP) Input:Requires two parameters {p, x} Action:To determine if NormaP (x) is defined for (X = x)? The Zero Input Halting Problem (ZIHP) Input:Requires one parameter {p} Action:To determine if NormaP (0) is defined for (X = 0)?

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01/04/2014(c) Bishop: Consciousness and computations3 Unsolvable decision problems (2) The Emptiness Problem ( P) Input:Requires one parameter {p} Action:To determine if ( (x): NormaP (x)) is defined? The Totality Problem ( P) Input:Requires one parameter {p} Action:To determine if ( (x): NormaP (x)) is defined? The Equivalence Problem (P Q) Input:Requires two parameters {p, q} Action:To determine if NormaP (x) = NormaQ (x)? Theorem : All Unsolvable Decision problems are reducible to SAP.

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01/04/2014(c) Bishop: Consciousness and computations4 The reduction principle (1) Suppose q & u are two decision problems And further that we know that u is undecidable … it cannot be constructed. We need to show how to modify Q … an algorithm for solving problem q.. into R an algorithm for solving problem u Of course there must be no doubt that the modifications to Q can be constructed.

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01/04/2014(c) Bishop: Consciousness and computations5 The reduction principle (2) Loosely speaking, the modifications to Q will take the form of a pre-processing algorithm P and there must be no doubt that P can be constructed. Our composite NORMA algorithm R … (R = P + Q).. for deciding u will have the form: R = Pre-process (x); Q; Hence: IF (algorithm P can be constructed) AND (algorithm Q can be constructed) THEN algorithm R, (R = P + Q), can also be constructed (and is decidable). Conversely IF the algorithm (P + Q) is cant exist (R is undecidable) THEN either P cant exist [p is undecidable] or Q cant exist [q is undecidable]. IF we know that R cant exist (problem u is undecidable) THEN we can conclude that Q cant exist and q is also undecidable)

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01/04/2014(c) Bishop: Consciousness and computations6 The emptiness problem (1) Input in X:(X = p)Where p is the nCode of Norma algorithm P. Output in Y:(Y = 0)Iff NormaP (x) is defined for at least one x. (Y = 1)Otherwise. Solution: Consider the following NORMA algorithm NormaA X := a;... where a = nCode (NormaA) A; Clearly: Given nCode (A) is a total and computable function, the algorithm NormaA can always be formally constructed from any algorithm for A. Hence the function f (a) a, (where a is nCode of NormaA), must be total & computable. But note, from the definition of A, NormaA (x) is defined for at least one x, iff NormaA (a) is defined As the first thing A' does is to force the content of (X) to be overwritten with (a).

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01/04/2014(c) Bishop: Consciousness and computations7 The emptiness problem (2) Now Consider the NORMA algorithm R: X = f (X); Q (X); Q (X) is a NORMA algorithm to decide the Emptiness property on the algorithm NormaX Where x = nCode (NormaX) Clearly: As pre-processing algorithm, P, is total and computable P = f (x) Clearly the algorithm for, R … (ie. P + Q) … can be constructed (and is decidable) iff Q can be constructed (and is decidable).

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01/04/2014(c) Bishop: Consciousness and computations8 The emptiness problem (3) However it is apparent that R (x) will decide SAP (x) Since given any input to R, (X = a) ….. the input to Q will always be of the form (X = a) Where a is the nCoding of NormaA'. But from the definition of A normaA(x) is defined for at least one x IFF NormaA (a) is defined. But we know NormaA (a) is not computable as this is SAP and cannot be constructed Hence R cannot be constructed. But R cannot be constructed iff Q cannot be constructed. Thus Q cannot be constructed and the Emptiness property is not NORMA decidable. NB.The essence of the above is that every computation of P' reduces after the first step to the computation of P with input p (ie. SAP).

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01/04/2014(c) Bishop: Consciousness and computations9 The halting problem Input in X:{p, x} Where {p} is the nCode of Norma algorithm P. {x} is the input to P. Hence need to code X register input in the form {2 p 3 x } Output in Y: (Y = 0) Iff NormaP (x) is defined. (Y = 1) Otherwise. Solution: Transform input X such that a HP solver will solve SAP for some program P. Consider the following NORMA pre-processing algorithm P: X := 6 x ;

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01/04/2014(c) Bishop: Consciousness and computations10 Solution (part 1) Clearly: P (X = 6 x ) is total and computable. Observation: (ab) x = a x b x Thus, for any input value, (X = x) the function of the pre-processor, P, will be to set the X register to (X = 2 x 3 x ). Now Consider the NORMA algorithm R: X := 6x; Q; Where, Q (x), is a NORMA algorithm to decide the Halting property on the algorithm NormaX Where x = nCode (NormaX).

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01/04/2014(c) Bishop: Consciousness and computations11 Solution (part 2) Clearly the pre-processing algorithm, P, can be constructed (is total and computable) P: (x = 6 x ) Hence R.. R: (P + Q).. can be constructed (and is decidable) Iff Q is decidable. However it is apparent that Q will decide SAP Since given any input to R, (x) the above transform will force the input to Q to be of form (2 x 3 x ).

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01/04/2014(c) Bishop: Consciousness and computations12 Solution (part 3) From the definition of R, normaR (x) is defined, iff NormaX (x) is defined. … but we know NormaX (x) is not computable (as this is SAP) and so cannot be constructed Hence R cannot be constructed. But R cannot be constructed iff Q cannot be constructed! Thus Q cannot be constructed and the Halting Problem is not NORMA decidable. The essence of the above is that every computation of R reduces after the first step to the computation of Q with input {X = 2 p 3 p } and this is SAP.

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