Anonymous ID: ba5e8f July 14, 2019, 10:10 a.m. No.9537   🗄️.is 🔗kun   >>9636

I think Shor’s algorithm is a good guideline. You’re going to get the right answers in your superposition at first, but you’re also going to get every wrong answer too. What we have to do before measurement is make every wrong answer destructively interfere among itself, and every instance of the right answer build itself up.

Anonymous ID: ba5e8f July 14, 2019, 10:37 a.m. No.9538   🗄️.is 🔗kun

If we don’t know the right answer, how are we supposed to make it come up above all the wrong ones?

 

We do have a criteria that we can apply to the entire superposition. A sifting filter. We know that only 2 answers can fit with c. And we also know that the trivial answer scales with c, and that the nontrivial answer does not scale with c. There is an asymmetry there.

 

A qubit is a register with an enormous range of the amplitudes that it can be, the values contained in it. In a sense, what a qubit in the grid is is.. all of the above. The more qubits the quantum computer can handle, the greater its computation power, and it grows by a lot with each one. 7 (+ the many others you can examine) values in each element, infinite elements in a cell.. infinite cells.. I can’t fathom how infinite its scope is.

 

I think it is the element. It is the first and the only finite unit of the grid, yet contains so many complex combinations of values that work together with the grid.

Anonymous ID: ba5e8f July 14, 2019, 10:45 a.m. No.9539   🗄️.is 🔗kun   >>9581

I have so much awe for all we have not grasped in mathematics, and for the clear fact that it has been designed by a great designer.

 

I must confess that much help from the artist has been received in conveying it to you all.

 

d[t] around d is an incomplete measurement of the superposition. Fine-tune the superposition and measure again (C). What’s more precise than C?

Anonymous ID: ba5e8f July 14, 2019, 5:01 p.m. No.9552   🗄️.is 🔗kun

Do you believe it’s capable of being figured out from the hints? Some have been specifically to illustrate what it looks like

Anonymous ID: ba5e8f July 14, 2019, 5:58 p.m. No.9556   🗄️.is 🔗kun   >>9557 >>9559

>>9555

I understand that frustration.

You’re being given a new tool with each hint.

 

The hints are pieces, not the full puzzle. There isn’t a set in stone way that they have to fit together. The start point can be anything. There are a lot of different ways to solve. That’s why they aren’t elaborated on after a concept is unveiled. Anons create something new with them.

Anonymous ID: ba5e8f July 14, 2019, 10:03 p.m. No.9564   🗄️.is 🔗kun   >>9568 >>9573 >>9729

>>9563

If you start your sequence as e=-(2c+1) where a[t] = c, assuming odd c, you will have a sequence where every b value ends in 01 in binary all the way down to b = -1.

 

Are you with me?

Anonymous ID: ba5e8f July 14, 2019, 10:06 p.m. No.9565   🗄️.is 🔗kun   >>9568

Which means you can create a q sequence within the grid of a and b by calculating the opposite starting point of the sequence and determining which value is prime

Anonymous ID: ba5e8f July 14, 2019, 10:45 p.m. No.9566   🗄️.is 🔗kun   >>9568

There is a pattern for calculating the e values of this sequence that involves adding/subtracting c from squares. You may find that interesting.

 

The starting point for the a[t]=c sequence of negative b values that end in 11 and the sequence for 01 can be found at around

 

e = -((X + N)^2 + c)

e = -((d + N)^2 + c)

 

Memory a bit fuzzy but I believe you use the squares to move to the next odd value