Re: Flibble’s Leap: Why Behavioral Divergence Implies a Type Distinction in the Halting Problem

On 5/11/2025 1:44 PM, olcott wrote:

On 5/11/2025 12:34 PM, dbush wrote:

On 5/11/2025 1:14 PM, olcott wrote:

On 5/11/2025 11:54 AM, dbush wrote:

On 5/11/2025 12:49 PM, olcott wrote:

The category error is actually the fact that everyone
here expects a termination analyzer to report on behavior
other than the behavior that its input finite string
actually specifies.

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That's because it's whether or not the algorithm described by the input halts when executed directly.
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No one cares what "the behavior that its input finite string specifies" because that's not what we asked about.

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int sum(int x, int y) { return x + y ; }
when you ask about the sum of 5 + 7 using sum(3,2)
you are asking the wrong question.
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HHH reports on the behavior that DDD specifies.
sum reports on the sum that its inputs specify.
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Category error.  (2,3) is not (5,7), but (DDD) is (DDD).
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 DDD correctly emulated by HHH SPECIFIES RECURSIVE EMULATION
DDD correctly emulated by HHH1 DOES NOT SPECIFY RECURSIVE EMULATION
  

What DDD "specifies" is irrelevent.  What matters is what algorithm is described by DDD, and that algorithm halts when executed directly.  And it is that behavior that we're asking about.
No one cares what "the behavior that its input finite string
specifies" because that's not what we asked about.

The finite string DDD is a description the algorithm consisting of the fixed code of the function DDD, the >>fixed code of the function HHH, and the fixed code of everything that HHH calls down to the OS level.
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That UTM(DDD) exactly replicates the behavior of algorithm DDD proves this is a correct description.
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So to meet the requirements, HHH is required to report the behavior of the algorithm described by its input:
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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly