I cannot say too much more.
Put the two together and you will suddenly see.
Combine the diagrams you are making with the grid (The End).
As you move left and right.
Up and down.
When moving across keep x constant and move with steps of 2n.
When moving up and down, keep d constant. Two factors, three factors, four.
What do you see?
How does row one (n=1) relate and determine the patterns?
Godspeed anons.
The construction of The End (The Grid) and the example of the virtual quantum computer and many of the connections to hints previously, all relate to P=NP.
What can be inferred from P!=NP?
Anything useful?
How about we start with P=NP?
Start with P=NP as an assumption.
There is a solution that has the complexity of the most complex step.
This is taking the root of c.
At most, O(log t) where t is the length of c in bits.
That means, like all solution sets of P=NP, we can work backwards.
And only this method works.
Since we know that the square root is the most complex step, we already have the best tool for working backwards.
We can EXCLUDE any ideas that increase the complexity above O(log t).
We exclude searches that do not zero in on a solution in logarithmic time, this is nearly all searches except those that have input that get smaller and smaller exponentially or equivalently, those that use recursion. Hints of this have been given previously.
We also know that we will end up back at the grid.
We also know we need to use input c of a reasonable size to show the pattern properly, again hints were given but too subtle.
The effort here have been incredible, fast and almost reached where we are going next. Anons are resourceful and brilliant.
The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.
Take some time to think about what that means.
Have you seen this approach before?
What could it look like?
Many new solutions often seem obvious in hindsight.
In fact many new designs seem to simplify in many varied approaches to design.
Does it seem obvious in hindsight that in order to solve a multivariate equation, that something new but similar to what we have always done, would be the solution? Just taken in a new direction? Expanded thinking.
The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.
The two forms of equations that merge together handle staircase numbers where the base of one staircase, is one unit longer than the other.
The analogy to Fermat's Last Theorem. Two objects that seem to be completely unrelated were proved to be the same object.
Two seemingly different 'fields' will be used side by side and merged to create an elegant solution. Again, this can only be done in reverse, using the assumption that P=NP. I only saw at the end of over seven years work. Anons have got much further in six months than I would have. I would have walked away in frustration back in the day but you anons here have been amazing.
Take some time to think what this will look like and how the diagrams (especially animated) might show this.
I think things will start moving quickly.
This will be new mathematics. It will make more sense than how this problem has been approached up until now.
1 Be patient. No matter what.
2 Don't badmouth: Assign responsibility, not blame. Say nothing of another you wouldn't say to him.
3 Never assume the motives of others are, to them, less noble than yours are to you.
4 Expand your sense of the possible.
5 Don't trouble yourself with matters you truly cannot change.
6 Don't ask more of others than you can deliver yourself.
7 Tolerate ambiguity.
8 Laugh at yourself frequently.
9 Concern yourself with what is right rather than who is right.
10 Try not to forget that, no matter how certain, you might be wrong.
11 Give up blood sports.
12 Remember that your life belongs to others as well. Don't risk it frivolously.
13 Never lie to anyone for any reason. (Lies of omission are sometimes exempt.)
14 Learn the needs of those around you and respect them.
15 Avoid the pursuit of happiness. Seek to define your mission and pursue that.
16 Reduce your use of the first personal pronoun.
17 Praise at least as often as you disparage.
18 Admit your errors freely and quickly.
19 Become less suspicious of joy.
20 Understand humility.
21 Remember that love forgives everything.
22 Foster dignity.
23 Live memorably.
24 Love yourself.
25 Endure.
Signatures.
Patterns.
Column 0.
Row 1.
You are VERY close.
One or more of you is going to suddenly get this.
If you can, please go back over and write out the list of all the patterns common to every cell at n=1 (row 1).
It applies to positive e.
Look at the difference of patterns for columns in NEGATIVE e (I used to call these columns f).
The mirror image of the grid.
Just as there is an entry in every cell in column 0.
Is there an entry in every cell at e=-1?
All of you are amazing.
Where this takes you as you solve it and apply the practice afterwards is entirely up to you.
Remember Andrew Wiles?
Me?
You don't have to do this alone.
The first word of the first step is always "We".
The gifs are key and there are patterns that you're just about to look at to compare with the even squares.
Genius move anons.
Remember there are two types of even and odd squares.
Enumerate the patterns in the first row (and into negative x).
It will be something that was there that you didn't c that will trigger it.
In row 1, the values of a[t] represent na for some c (e.g. RSA 100).
a[t+n] = nb
This is true for all c.
For the value of c, at the same t but in cell (-f,1), the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)
This is the key.
The value of a[t] at -f and e in the first row have the same factor.
The values in each cell that have b as a factor are DIFFERENT, not aligned. They are one element apart in the two cells. In the positive side of the grid, they are n elements apart. In the negative side of the grid, the elements one less elements apart.
One cell has n as a factor at those positions at positive e column, one cells has n-1 as a factor in the negative f column.
This asymmetry can be used to solve the problem.
The power of ambiguity in language at work.
That is different pattern but related.
They are all related.
This is all brilliant.
Especially the colours.
That is very important for visualising properties of the Mandelbrot Set.
I hadn't used it for the grid.
Brilliant.
It is suggested you follow this pattern.
I will come back to clarify the other by the end of the week but briefly. Using the values of RSA 100 just as a large number example is a suggestion. In column e, the values at (e,1) for each element includes a value for a[t] which is an. The value bn is at a[t+n]. At (-f,1) the value a(n-1) is at the same value x in that cell as "an" for (e,1), the value b(n-1) at cell (-f,1) is one element less than bn at (e,1). This difference is key. It is not the only key, as you are seeing. I have found three, not including yours. Those three are in row 1, column zero and the side by side diagonal cells from the origin. There may be infinite keys.