Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: dbush Newsgroups: comp.theory Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1 Date: Sat, 29 Mar 2025 11:23:58 -0400 Organization: A noiseless patient Spider Lines: 40 Message-ID: References: MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 7bit Injection-Date: Sat, 29 Mar 2025 16:23:59 +0100 (CET) Injection-Info: dont-email.me; posting-host="d82829ff2684f0f25de37249bda61e80"; logging-data="1714114"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1/zScw/Zd/1pRhIHVDNCQhB" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:296+CBdNKYMccW1DJ0VJHmSTS10= Content-Language: en-US In-Reply-To: Bytes: 3348 On 3/29/2025 11:12 AM, olcott wrote: > On 3/28/2025 11:00 PM, dbush wrote: >> On 3/28/2025 11:45 PM, olcott wrote: >>> >>> It defines that it must compute the mapping from >>> the direct execution of a Turing Machine >> >> Which does not require tracing an actual running TM, only mapping >> properties of the TM described. > > The key fact that you continue to dishonestly ignore > is the concrete counter-example that I provided that > conclusively proves that the finite string of machine > code input is not always a valid proxy for the behavior > of the underlying virtual machine. In other words, you deny the concept of a UTM, which can take a description of any Turing machine and exactly reproduce the behavior of the direct execution. All you've proved is that HHH(DDD) == 0 is incorrect for the same reason H0(DDD) == 0 below is incorrect: int H0(uintptr_t *p) { return 0; } Namely that it doesn't return 1 as per the requirements: Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as with input Y: A solution to the halting problem is an algorithm H that computes the following mapping: (,Y) maps to 1 if and only if X(Y) halts when executed directly (,Y) maps to 0 if and only if X(Y) does not halt when executed directly