Re: DDD emulated by HHH diverges from DDD emulated by HHH1

On 05/06/2025 05:27, olcott wrote:

On 6/4/2025 10:55 PM, Mike Terry wrote:

On 05/06/2025 02:39, olcott wrote:

On 6/4/2025 8:28 PM, dbush wrote:

On 6/4/2025 9:08 PM, olcott wrote:

On 6/4/2025 7:41 PM, dbush wrote:

On 6/4/2025 8:32 PM, olcott wrote:

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Show me this side-by-side trace and I will point out your mistake.

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See below, which shows that the simulations performed by HHH and HHH1 are identical up to the point that HHH aborts, as you have agreed on the record.
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False.  The correct trace is the one I posted, which shows all levels of emulation performed by HHH and HHH1.  See the corrections I made to your comments

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It is not supposed to do that.

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Are you saying it's not supposed to include /nested/ emulations?  It is perfectly sensible to include nested emulations.
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 It can include nested simulations yet nested
simulations are in a hierarchy thus not side-by-side.
A side-by-side analysis must be side-by-side.
 

Hierarchies can be compared side-by-side.  In the case of these traces, the hierarchy can be "flattened" into one stream of nested simulations.  You do this yourself every time you present one of your nested simulation traces.  Such a trace should include a simulation depth (or equivalent) for each entry.
Two nested simulation traces can easily be presented side-by-side for comparisson.  You are just trying to divert attention from your own failings to properly understand the requirements.

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It is supposed to show
the emulation of DDD by HHH1 and
the emulation of DDD by HHH
side-by-side to show the point where these
emulations diverge.

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Just to be perfectly clear:
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It is supposed to show
the emulation of DDD by *OUTER* HHH1 and
the emulation of DDD by *OUTER* HHH
side-by-side to show the point where these
emulations diverge.
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 No, that is not the way I was doing it.

So you were doing it wrong.  No problem, but now you (hopefully) understand you can do it right.

I was
doing it all in the execution trace of HHH1(DDD).

I.e. You were doing it wrong.
The emulation of DDD by *OUTER* HHH1 is represented by the trace entries produced by HHH1 when called from main:  HHH1(DDD).
The emulation of DDD by *OUTER* HHH is represented by the trace entries produced by HHH when called from main:  HHH(DDD).
To compare them, generate those two traces and put them side by side.
In fact, since HHH1 does a full simulation, you /could/ extract the trace of HHH's emulation of DDD from the HHH1(DDD) trace, as long as you don't get confused.  Bear in mind that a given point in HHH1's emulation will be a /different/ point in HHH's emulation.  (This is what I did when I generated the two traces to compare.)
Since there is in reality no chance of you NOT becoming confused, you should just follow the obvious procedure of performing the two native computations HHH1(DDD) and HHH(DDD) and extracting their emulation traces.
[..snip remainder of post, which is just confused nonsense.  There is no need or point in you explaining /why/ you did it wrong, just do it right next time..]
Mike.