Can you fax it to me?
Include b[D2] * (4d), please
Maybe the next gap to look at is where c is between in i[t]
Far numerous too are the examples where the D and C (e, 1) element calculations give a solution i that is only off by a small amount
SeeIt 0.0.3 tested correcting this error value by checking below in the column for the correct combination based on factoring na recursively
SeeIt 0.0.4 will test a new error correction techniqu
Progressing it forward with real, programmatical examples
Recursive an factoring was a proof-of-concept, an idea to look for different d and n combinations that close the gap towards i.
The End’s qubits (elements) are all entangled. Each one contains the blueprints for the entire grid. One element contains all of it.
The application of this is that an infinite or unfeasibly large search space can be grasped with a tiny amount of data. All of a column can be represented in one cell, and all of a cell can be represented in one element. That’s what moving to (e, 1) from (e, n) and (e, N) is.
Since the factors of c exist in column e, but the n values of that column are too numerous to search, we can identify them by compressing the column together using n=1. We know that since the column has been compressed into a single cell, that the data for the factorisations of c exist in it. The route of identifying those factorisations efficiently must be informed by the nature of how the column has been compressed.
This nature is hidden in plain sight. The techniques of calculating the records c and d are between are the beginning of the unveiling of it, but with the stage finished being set, it will conclude anons discovering the solution.
Use the grid with this nature in mind.
Try EVERY idea that comes to mind.
Shor's algorithm consists of two parts:
A reduction, which can be done on a classical computer, of the factoring problem to the problem of order-finding [reduction to N -n, X -> x search space].
A quantum algorithm to solve the order-finding problem [VQC].
The quantum circuits [steps to factor based on family] used for this algorithm are custom designed for each choice of [c].
The input [d, e] and output [n, N] qubit registers need to hold superpositions of values [trivial and nontrivial elements].
Proceed as follows:
-
Initialize the registers [coordinates] to [n = 1]
This initial state is a superposition of [all n values in the column], and is easily obtained by generating [-f, 1; e, 1] each a superposition of [row N and row n] applying [e, 1] in parallel to [-f, 1].
Construct [the elements d is in the gap of]
[trivial and nontrivial elements] are now entangled, or not separable [patterns]
Apply the inverse Quantum Fourier transform [root and offspring of David transform] to the input register [gap elements]
I am here to accelerate.
Possibly into X and Y
If a variable can take you out of e, can one put you in e?
Would this make factoring e relevant?
The best place to learn about what multiplies to make a square is where you’re already headed.
A column filled with (in the valid, not shadow cells) factorisations of squares
Three different ways of achieving the same goal
Phone dying but will return soon.
I have flip flopped on whether I want to disclose it all.
If hints are explored and you still don’t see it.
Not Chris.
Just wanted to pick up the torch.
Convey what I’ve been shown.
Implications may not be correct.
Protected doesn’t equate to information required to solve being intentionally hidden.
I did not want to give away the code and it be attributed to a fleeting moment of genius.
I’m giving away how to do it from scratch with anything.
Well done on the summary.
I will go through it and find the best things to work on and put together.
You are right that linking concepts isn’t giving it away.
“We also find staircase numbers when we add the x values from an (e,1) element and an (f,1) element”
x is an index, BUT when something else is used as the index, it becomes more relevant.
Try the elements at x=d and x=d+1 (2d + 1).
ALSO, if the elements d is between are calculated with x ~= sqrt(f), what is at “x = f or x = f-1?”
It is much.
Let’s approach it from a different angle then
Yes
Keep going
It doesn’t have to be perfectly implemented to see (and use it as, but you will quickly find shortcuts) the solution. Same intent as the tree.
The correct way to view the shadow grid calculating from (e, n, t). Unity of method necessary also to work together with one another.
/ calculating new (e, 1) superposition /
Think of it as decreasing the search space until there’s nothing but the factors or a prime determination.
Image slowly becoming unblurred?
You’re unblurring the large square.
Try the other parts.
Other hints will suddenly find their true use.
A way to see why factorizing d is useful is to multiply d by its factors
It has to do with geometry - missing square piece of c.
Decompose d (thus dd) and e, and calculate the missing piece of c.
Take the sqrt(d).
Take the sqrt(i).
Apply to D, D2 scale search space estimate.
The problem can be solved with one of the squares in-between.
(2d+1)^2 - (2d)^2 = 4d+1
(4d+1)^2 - (4d)^2 = 8d+1
(8d+1)^2 - (8d)^2 = 16d+1
(16d+1)^2 - (16d)^2 = 32d+1
A 4096-bit private key can be calculated instantaneously.
Factorising numbers close to c can also result in factorising c.
33:00:11
Ironically solving it doesn’t require you to find some beautiful mathematical pattern that makes everything clear.
Just estimate and correction.
But there is one.
The root of a function gives you it’s “zero”
When you take away the root from c, you’re left with..
If e is a remainder and n is also a remainder and when roots are taken away remainders are left, are e and n the same type of thing? Does away that other “root” from c give you this nontrivial “remainder?”
But since n is to add to d+n, maybe the root n comes from is the root of david?
It was right on the mark.
Yes
There have been a lot of posts about where a solution value predictably scales to be close to.
Operations that take remainders.
If it was a snake it would bite you.
You can’t be a trailblazer if you’re following a trail
You are right.
I was wrong.
No one lives without God, admittedly or not, so there is no such thing as self-sufficiency. It’s God-sufficiency.
Will progress to full disclosure of the code.
Proverbs 12:1
The fact that e and -f are columns unique to c is used to adjust them in a manner like so:
Take the factor of na by subtracting (n-1)a from -f
Take the factor of na at t+n0
>either going from n0 to n1 to n2 etc changes the e and f values making them lower/closer to 0 logarithmically and ends up on a pair of elements where the n values is the same as the unknown n we're looking for,
Oo that’s brilliant
the En+d