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On 6/4/2025 1:54 AM, Mikko wrote:Which is nothing. However, it is quite obvious from the C codeOn 2025-06-03 19:57:09 +0000, olcott said:The code proves what it proves.
On 6/3/2025 2:37 AM, Mikko wrote:If the input string does not unambiguously specify one and only oneOn 2025-06-02 15:52:53 +0000, olcott said:No they do not. When DDD calls its own emulator its behavior
DDD correctly emulated by HHH diverges from DDDThe DDD emulated (correctly or otherwise) by HHH is the same DDD as
correctly emulated by HHH1 as soon as HHH begins
emulating itself emulating DDD, marked below.
the one emulated (correctly or otherwise) so both specify the same
behaviour.
is different than when DDD calls another different emulator.
behaviour it is incorrectly encoded and not a valid input string.
The halting problem of Truing machines requires that every pair of
a Turing macnine and input is descibed so that the behaviour to
be decided about is the only behaviour that meets to the description.
_DDD()--
[00002183] 55 push ebp
[00002184] 8bec mov ebp,esp
[00002186] 6883210000 push 00002183 ; push DDD
[0000218b] e833f4ffff call 000015c3 ; call HHH
[00002190] 83c404 add esp,+04
[00002193] 5d pop ebp
[00002194] c3 ret
Size in bytes:(0018) [00002194]
_main()
[000021a3] 55 push ebp
[000021a4] 8bec mov ebp,esp
[000021a6] 6883210000 push 00002183 ; push DDD
[000021ab] e843f3ffff call 000014f3 ; call HHH1
[000021b0] 83c404 add esp,+04
[000021b3] 33c0 xor eax,eax
[000021b5] 5d pop ebp
[000021b6] c3 ret
Size in bytes:(0020) [000021b6]
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