Hello,

More philosophy about super artificial intelligence..

I am a white arab and i think i am smart since i have also invented many
scalable algorithms and algorithms..

I think i am smart and i say that super artificial intelligence will
not be much more smart than humans, since you have to look at the
the constraints on expressiveness of super artificial intelligence,
for example when we look at the power of expressiveness of what is
it to be Turing-complete, you have to know that a Turing-complete system
can be proven mathematically to be capable of performing any possible
calculation or computer program, but science is the most important tool
of our smartness, but science is constrained by the following way:

Let me give an example about how to measure IQs or smartness..

So if you are really smart you will give a smart example so that people
can understand, so here it is:

If i say: 2 + 2 = 4

So you will notice that this equality is also constrained by constraints
of reality, since for example you are noticing that it is not so
mathematically expressive, so this not mathematically expressive is also
constraining human IQ or human smartness, since if you understand and
learn this mathematical equality, another person will quickly do the
same, so the other person will adapt quickly to this level of smartness,
so now you are noticing the smart idea, it is that even science and
technology are constrained the same way, and this constraints on
science and technology constrain or limit the expressiveness of high
human IQs or high level of smartness so that other lower level human
IQs or smartness can attain the level of adaptability of high human
IQs, this is what is happening in our today world, and if you are smart
you will notice that there is something else that is happening and it
is that abstraction of complexity that reduce the complexity is making
others not understanding the complexity behind the abstraction and this
is not so efficient.

Here is more about the constraints on science and technology:

Is Science Going To End?

Read more here:

https://philosophynow.org/issues/68/Is_Science_Going_To_End

And read also the following

The Industrial Era Ended, and So Will the Digital Era

So by my same logical reasoning above i can say that we have to be
positive about super artificial intelligence, since it will not be much
more smart than human because of the above constraints on science.

Read more here:

https://hbr.org/2018/07/the-industrial-era-ended-and-so-will-the-digital-era

Read more here in my following thoughts about human smartness:

https://groups.google.com/g/alt.culture.morocco/c/Wzf6AOl41xs

More philosophy about the Rice theorem and about the halting problem and
more..

I think that Rice theorem is a "generalisation" of the Halting problem,
so i think that understanding the Rice theorem is really important,
Rice theorem states that any non-trivial semantic property of a language
which is recognized by a Turing machine is undecidable. A property, P,
is the language of all Turing machines that satisfy that property.

You can see the logical proof of it here:

https://www.tutorialspoint.com/automata_theory/rice_theorem.htm

More philosophy about Markov chains and invariants and Petri nets..

I am also working with Petri Nets and Markov chains in mathematics and
structural analysis of Petri nets in mathematics, so i will
talk about it now:

About Markov chains in mathematics and more..

In mathematics, many Markov chains automatically find their own way to
an equilibrium distribution as the chain wanders through time. This
happens for many Markov chains, but not all. I will talk about the
conditions required for the chain to find its way to an equilibrium
distribution.

If in mathematics we give a Markov chain on a finite state space and
asks if it converges to an equilibrium distribution as t goes to
infinity. An equilibrium distribution will always exist for a finite
state space. But you need to check whether the chain is irreducible and
aperiodic. If so, it will converge to equilibrium. If the chain is
irreducible but periodic, it cannot converge to an equilibrium
distribution that is independent of start state. If the chain is
reducible, it may or may not converge.

So i will give an example:

Suppose that for the course you are currently taking there are two
volumes on the market and represent them by A and B. Suppose further
that the probability that a teacher using volume A keeps the same volume
next year is 0.4 and the probability that it will change for volume B
is 0.6. Furthermore the probability that a professor using B this
year changes to next year for A is 0.2 and the probability that it
again uses volume B is 0.8. We notice that the matrix of transition is:

| 0.4 0.6 |
| |
| 0.2 0.8 |

The interesting question for any businessman is whether his
market share will stabilize over time. In other words, does it exist
a probability vector (t1, t2) such that:

(t1, t2) * (transition matrix above) = (t1, t2) [1]

So notice that the transition matrix above is irreducible and aperiodic,
so it will converge to an equilibrum distribution that is (t1, t2) that
i will mathematically find, so the system of equations of [1] above is:

0.4 * t1 + 0.2 * t2 = t1
0.6 * t1 + 0.8 * t2 = t2

this gives:

-0.6 * t1 + 0.2 * t2 = 0
0.6 * t1 - 0.2 * t2 = 0

But we know that (t1, t2) is a vector of probability, so we have:

t1 + t2 = 1

So we have to solve the following system of equations:

t1 + t2 = 1
0.6 * t1 - 0.2 * t2 = 0

So i have just solved it with R, and this gives the vector:

(0.25,0.75)

Which means that in the long term, volume A will grab 25% of the market
while volume B will grab 75% of the market unless the advertising
campaign does change the probabilities of transition.

So notice that in my previous post i have talked about invariants of the
system, here is my previous post read it carefully:

About Turing completeness and parallel programming..

You have to know that a Turing-complete system can be proven
mathematically to be capable of performing any possible calculation or
computer program.

So now you are understanding what is the power of "expressiveness" that
is Turing-complete.

For example i am working with the Tool that is called "Tina"(read about
it below), it is a powerful tool that permits to work on Petri nets and
be able to know about the boundedness and liveness of Petri nets, for
example Tina supports Timed Petri nets that are Turing-complete , so the
power of there expressiveness is Turing-complete, but i think this level
of expressiveness is good for parallel programming and such, but
it is not an efficient high level expressiveness. But still Petri nets
are good for parallel programming.

About deadlocks and race conditions in parallel programming..

I have just read the following paper:

Deadlock Avoidance in Parallel Programs with Futures

https://cogumbreiro.github.io/assets/cogumbreiro-gorn.pdf

So as you are noticing you can have deadlocks in parallel programming
by introducing circular dependencies among tasks waiting on future
values or you can have deadlocks by introducing circular dependencies
among tasks waiting on windows event objects or such synchronisation
objects, so you have to have a general tool that detects deadlocks, but
if you are noticing that the tool called Valgrind for C++ can detect
deadlocks only happening from Pthread locks , read the following to
notice it:

http://valgrind.org/docs/manual/hg-manual.html#hg-manual.lock-orders

So this is not good, so you have to have a general way that permits
to detect deadlocks on locks , mutexes, and deadlocks from introducing
circular dependencies among tasks waiting on future values or deadlocks
you may have deadlocks by introducing circular dependencies among tasks
waiting on windows event objects or such synchronisation objects etc.
this is why i have talked before about this general way that detects
deadlocks, and here it is, read about it in my following thoughts:

Yet more precision about the invariants of a system..

I was just thinking about Petri nets , and i have studied more Petri
nets, they are useful for parallel programming, and what i have noticed
by studying them, is that there is two methods to prove that there is no
deadlock in the system, there is the structural analysis with place
invariants that you have to mathematically find, or you can use the
reachability tree, but we have to notice that the structural analysis of
Petri nets learns you more, because it permits you to prove that there
is no deadlock in the system, and the place invariants are
mathematically calculated by the following system of the given
Petri net:

Transpose(vector) * Incidence matrix = 0

So you apply the Gaussian Elimination or the Farkas algorithm to the
incidence matrix to find the Place invariants, and as you will notice
those place invariants calculations of the Petri nets look like Markov
chains in mathematics, with there vector of probabilities and there
transition matrix of probabilities, and you can, using Markov chains
mathematically calculate where the vector of probabilities
will "stabilize", and it gives you a very important information, and you
can do it by solving the following mathematical system: