Sujet : Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 28. Nov 2024, 20:48:44
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <d8ba9d736f495798c05e3b28acba4b5481fad05a@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Thu, 28 Nov 2024 11:57:23 -0600 schrieb olcott:
On 11/28/2024 11:06 AM, Richard Damon wrote:
On 11/28/24 11:43 AM, olcott wrote:
On 11/28/2024 9:47 AM, Richard Damon wrote:
On 11/28/24 10:01 AM, olcott wrote:
DDD emulated by any HHH cannot possibly reach its "ret" instruction
final halt state.
But that DDD CAN'T be emulated more than 4 instructions by ANY pure
function, as you can't emulate past the call HHH instruction.
You just aren't paying any attention at all or are woefully inaccurate
in your word choice. HHH1 does emulate all of DDD.
HHH1 <is> a pure function.
Strawman. We are talking about HHH.
-- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:It is not guaranteed that n+1 exists for every n.