What produces 0.5 *0.5 *4 ???

0i.5 *0.5 *4 =1m^3. better to say 5i *5i *40i =1000i^3.
Three factors produce Volume.

How Two Factors Produce m^2..
Alias 0.5y *4x = 2m^2..better to say: 5i *40i = 200i^2

Greetings T.n.p. The mat. that doesn't allow you to make mistakes-:)))
If => 5y *5x == 25m^2 ==> 5y *5x *5z == 125m^3.-:)))
Alias i50y *i50x =2500i^2==> i50^3 = 125'000i^3-:)))

What produces 0.5 *0.5 *4 ???

0i.5 *0.5 *4 =1m^3. better to say 5i *5i *40i =1000i^3.
Three factors produce Volume.

How Two Factors Produce m^2..
Alias 0.5y *4x = 2m^2..better to say: 5i *40i = 200i^2

Greetings T.n.p. The mat. that doesn't allow you to make mistakes-:)))
If => 5y *5x == 25m^2 ==> 5y *5x *5z == 125m^3-:)))
Alias i50y *i50x =2500i^2==> i50^3 = 125'000i^3-:)))

They don't understand the importance of volumes e.g.:
If 2 *6 =12m^2..0.5 *2 *6 =6m^3..between the two products..
there is an abyss since 12m^2 *100i^2 = 1200i^2
while 6m^3 *1000i^3 = 6,000i^3 = 6 Tons..
thanks to that miserable 5i that is on the Z axis.

Greetings T.n.p. 1m *2*6 = 12m^3 = 12 Tons-:)))
Alias 10i *20i *60i = 12'000i^3 = 12 Tons-:)))

Must have been Brave to do BTC today!
https://brave.com/bitcoin-wallet/

Lucrezio S schrieb:
> What produces 0.5 *0.5 *4 ???
>
> 0i.5 *0.5 *4 =1m^3. better to say 5i *5i *40i =1000i^3.
> Three factors produce Volume.
>
> How Two Factors Produce m^2..
> Alias 0.5y *4x = 2m^2..better to say: 5i *40i = 200i^2
>
> Greetings T.n.p. The mat. that doesn't allow you to make mistakes-:)))
> If => 5y *5x == 25m^2 ==> 5y *5x *5z == 125m^3.-:)))
> Alias i50y *i50x =2500i^2==> i50^3 = 125'000i^3-:)))
>

On 02/29/2024 05:34 PM, Mild Shock wrote:
>
> Must have been Brave to do BTC today!
> https://brave.com/bitcoin-wallet/
>
> Lucrezio S schrieb:
>> What produces 0.5 *0.5 *4 ???
>>
>> 0i.5 *0.5 *4 =1m^3. better to say 5i *5i *40i =1000i^3.
>> Three factors produce Volume.
>>
>> How Two Factors Produce m^2..
>> Alias 0.5y *4x = 2m^2..better to say: 5i *40i = 200i^2
>>
>> Greetings T.n.p. The mat. that doesn't allow you to make mistakes-:)))
>> If => 5y *5x == 25m^2 ==> 5y *5x *5z == 125m^3.-:)))
>> Alias i50y *i50x =2500i^2==> i50^3 = 125'000i^3-:)))
>>
>

So, let me see if I understand this, the idea of
a bitcoin, is a block of data, what SHA-256 run
on it twice, that is zeroes, except the low bytes,
the count of those, being the difficulty target.

Then, those are just sort of "uniform random",
in blocks of data, to be found, or Merkle-y,
Merkle-ish.

https://en.wikipedia.org/wiki/Merkle%E2%80%93Damg%C3%A5rd_construction
https://en.wikipedia.org/wiki/SHA-2

So, SHA-256, is this algorithm, and its seeds,
just happen to be the non-integer parts of some
computed roots of primes.

And you want to know, how to produce inputs,
that, result zeroes filling in the SHA-256(SHA-256(block))?

Where, how that works is just shuffling the input
data basically with a bridge shuffle?

And, it's just SHA-256 twice, not with a seed or
otherwise is some HKDF or this way.

That being printing money?

https://en.bitcoin.it/wiki/Protocol_documentation

And every now and then the difficulty target just
floats so there's "always constantly more"?

Then, it uses public-key asymmetric cryptography
to change hands? And that it settles until the ledger
is propagated?

And it's ECDSA? Elliptic Curve?

So the idea to get money is to either
find money that is new money and
sign that with the longest chain
supporting it, or, get money signed
to you, then that some people will
cash that out.

And it entirely depends on SHA(SHA)
not finding zeroes, and ECDSA being strong.
And everybody noticing.

What could go wrong?

"The main branch is defined as the branch with
highest total difficulty, summing the difficulties
for each block in the branch."

It really seems that it needs a sort of double-blind,
about that contention should have it so that
there's a settling then an unveiling and so on, ....

I.e. it should work out that the chain works itself
out that basically things get added encrypted into
the mix, then result irrepudiable as they are.

I mean, you figure that constant competition
constantly ups the bar and nobody holds back
"more difficult" things found to grease the ditches,
i.e. to steal work, when the work is "found money"
not just "ditto diminishing".

So, it's totally as strong as nobody finding
a pre-image attack on SHA-2, or breaking
Elliptic Curve with reference curve, or
finding pathologically difficult inputs.

Then the rest is just bartered scarcity
based on digital anonymity and side agreements?