 Download Download Presentation Cognitive Computing 2013

# Cognitive Computing 2013

Télécharger la présentation ## Cognitive Computing 2013

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. Cognitive Computing2013 Consciousness and Computations 8. THE REDUCTION PRINCIPLE Prof. Mark Bishop

2. 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)? (c) Bishop: Consciousness and computations

3. 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. (c) Bishop: Consciousness and computations

4. The ‘reduction’ principle (1) • Suppose q & uare 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. (c) Bishop: Consciousness and computations

5. 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 can’t exist (R is undecidable) • THEN either P can’t exist [p is undecidable] or Q can’t exist [q is undecidable]. • IF we know that R can’t exist (problem u is undecidable) • THEN we can conclude that Q can’t exist and q is also undecidable) (c) Bishop: Consciousness and computations

6. 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). (c) Bishop: Consciousness and computations

7. 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) iffQ can be constructed (and is decidable). (c) Bishop: Consciousness and computations

8. 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 iffQ 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). (c) Bishop: Consciousness and computations

9. 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 {2p 3x} • 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 := 6x; (c) Bishop: Consciousness and computations

10. Solution (part 1) • Clearly: • P (X = 6x) is total and computable. • Observation: (ab)x = ax bx • Thus, for any input value, (X = x) • the function of the pre-processor, P, • will be to set the X register to (X = 2x 3x). • 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). (c) Bishop: Consciousness and computations

11. Solution (part 2) • Clearly the pre-processing algorithm, P, can be constructed (is total and computable) • P: (x = 6x) • Hence R .. • R: (P + Q) • .. can be constructed (and is decidable) • IffQ 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 (2x 3x). (c) Bishop: Consciousness and computations

12. 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 iffQ 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 = 2p 3p} and this is SAP. (c) Bishop: Consciousness and computations