It's just an explanation of the triangles in the (x+n)(x+n) square that we'd already done. It's not really any different, except I used odd n even x. It also isn't finished, by the way, mostly because of time constraints at the moment. There is another reason it isn't done, though, which is explained here >>7812
The following, in regards to using f/8 as a test triangle base to calculate n0, and extensions of this idea, is based around VQC's posts in RSA #10. Just so anyone who reads this knows, as I type this out, I don't necessarily know where this math leads. All I'm doing is following the logic of each post and expanding each one to reiterate and potentially decrypt what VQC is saying. That's part of the point behind this thread (having all the information in one place). Whether or not these examples use the related concepts for the correct reasons is up for debate, since this is the part when we started getting into diagrams we never seemed to fully understand.
I'll be using the following two semiprime cells for this post:
(26,3,8) = {26:3:51:14:37:71}, f=-77, c=2627
(46,5,7) = {46:5:31:12:19:53}, f=-17, c=1007
As explained previously, you can break up odd (x+n)(x+n) squares into eight triangle numbers plus one, and you can also break them up into nn+2d(n-1)+f-1. When n is odd, nn is another square made of eight triangle numbers plus one (since this is a trait of all odd squares). So when we have odd n and even x, we can break up a square into nn-1+2d(n-1)+f (notice the -1 has moved). This is shown in the first image (the x+n squares for the above cells), and was explained several posts ago.
An idea that VQC introduced in RSA #10 was a new variable called n0 ( >>4343 ). n0 is a guess version of n. To find it, we use the division of f among the x+n square to create a guess triangle base, and we use a function he posted to figure out what n would be if this value calculated from f was the correct base (knowing that it is not going to be correct because it's just an arbitrary division of f). First I'll explain the ways in which you can use f to find a guess triangle base, and then I'll (attempt to) explain the specific things they're used for.
We've been using u as the variable name for the base of each triangle in an x+n square (8T(u)+1). In the case of both these example cells, x+n is 17, so the base u=8. Let's forget about the distribution of 2d(n-1) and nn in these triangles for now and just focus on f. f is being divided by 8 and distributed among the triangles. It won't necessarily be divisible by 8, so there will be some left over. When we divide f by 8 and place it in the triangles with the nn and 2d(n-1) portions, as in the previous diagrams and explanations, we're just filling the triangle from the top row by row. f/8 (minus the remainder) can also be expressed as a line. If we were to divide f by 8 in each of these cases and treat them as an undivided line, we could put them into each triangle as an entire row instead. This is shown in the second image. Notice how in the first example, f/8 is too big to fit in the triangles (it's on the outside of the square, in case you can't tell). There are two cases in this sense: sometimes f/8 is larger than u (this case), and sometimes f/8 is smaller than u (the second x+n square in the image). In these images, the remainders of f/8 are placed underneath. In the first case, f/8 is equal to 9r5. In the second case, f/8 is equal to 2r1. If we were to treat these as triangle bases, they would create triangles with bases of 9 and 2 (which equal 45 and 3) respectively.
Another way in which VQC explained that we could create a guess triangle base out of f was to divide f further. In VQC's example (which is far too large to make accurate diagrams out of since it's based on RSA100) he found that f was divisible by 5. From there, instead of dividing f by 8, he divided it by 40 (85). This allowed him to create another guess base out of f/40, and to turn the way f is represented within the triangles to a polite number (which is any number that is equal to the sum of consecutive integers; for example, 3+4+5=12 is polite, 2+3+4+5+6=20 is polite, every triangle number (1+2+3+4+5+…) is polite). The guess base was placed in the diagram as the row that was the same length, and the other parts of the division were placed in the adjacent rows (if that doesn't make sense it will when you read the rest of this paragraph hopefully). We can't really divide the base of 2 in the second example further, so we'll look at this in the context of the first example x+n square, where f=77=72+5 (5 being the remainder). The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. So let's take 72=324 as an example. f is divisible by 3 and 8, so it's divisible by 24. 72/24=3. In the previous example, we divided 72 by 8 and put a row of 9 into each triangle. In this case, since we divided f by an extra 3, and the result of that division was 3, we're putting a polite number into the triangle with a center length of 3 which has one row either side (to create 3 rows). In other words, we're putting 9=2+3+4 into the triangles (as shown in the next image - hopefully that explains what I said might have been confusing above). VQC also mentioned that you can create a guess base by dividing f (the entire f; in this case, 77) by 8 and something else that doesn't divide it with no remainder. It would change the remainder created by f/8. There aren't any situations that work for this particular example (it's very difficult to find examples for this stuff that both meet all the criteria and aren't so big that they're impossible to create diagrams for). Think of it this way. If you had 1001/8=125r1, you could represent 125 in each triangle as 23+24+25+26+27 (dividing 125 by 5). But what if you wanted to divide 125 by 3? Then you would be dividing 1001 by 24 (3*8), which is 41r17. Here you would be inserting 20+21 into the related x+n square. Where in the pictured example we have 5 extra units dispersed unevenly throughout the triangles, with this 1001/24 example, we would have 17 extra units dispersed throughout the triangles (this still leaves only one triangle with an extra unit, which is what it had before - this isn't necessarily always the case but I don't know; we'd have to test it).
I'll be continuing with (26,3,8) = {26:3:51:14:37:71}, f=-77, c=2627 from the two previous posts.
VQC posted a function here >>4335 which he says you're meant to put the guess triangle base into. This function calculates what n would be if the divided f value we put into it actually happened to be the correct triangle base for our x+n. If you put the correct u into it (continuing with our example with a u of 8), it creates eight triangles out of that base and adds one (creating 8T(u)+1, which is our x+n square), then it adds c (since (d+n)(d+n)-(x+n)(x+n)=c, c+(x+n)(x+n)=(d+n)(d+n), so this step calculates (d+n)(d+n)), then it takes the square root (the correct d+n), and then it subtracts d (leaving us with n). This is shown in the first picture. As mentioned >>4343 here, when we put our guess base based on a division of f, the resulting return value is n0. Since we can divide f arbitrarily for this, if we create a new variable (we don't need to use this outside of this sentence for the sake of understanding n0), maybe w, which is the extra number you're dividing f by (if we just divide f by 8, w=1; if we divide f by 24 (8*3), w=3, and so on), then we can rearrange n0 with algebra.
u' = floor(f/8w)
n0 = (sqrt(8T(u')+1 + c)) - d
Continuing with the example cell above, we have an f of 77. If w is 1 (as explained above, this means f is being divided by 81), we get a guess base of 9. If w=2 (f/(82)), guess base = 4. If w=3 (f/(8*3)), guess base = 3. Etc. There is a range of possible guess base values for any given f, calculated from f/8 to f/f. This is shown in the second picture (with the current example cell).
VQC then posted another function a while later >>4594 which takes the guess base, n0, d and f as parameters. For the sake of explanation, continuing with the same example cell (26,3,8) = {26:3:51:14:37:71}, f=-77, c=2627, let's set the guess base to f/8. As shown previously, this guess base is equal to 9. It creates an n0 value of 3. This function, Get_Remainder_2dnm1, uses the guess base to calculate 8T(u') (so it does the same thing that the beginning of the previous function does) and uses n0 to calculate the n0 equivalent of nn+2d(n-1)+f-1 (using the correct f and d values). It subtracts (nn+2d(n-1)+f-1) from (x+n)(x+n). So if we input the correct u base and n values, it will return 0. If, however, we input each pair of potential guess base and n0 values, we won't necessarily get a gap between 8T(u) and nn+2d(n-1)+f-1 of 0. Using the same example as we've been looking at for this stuff, pic related is the output with every potential pair of guess bases and n0s. You'll notice none of the remainders is equal to 0. If any of them was equal to 0, we would have the correct square. We're not going to unless n is a factor of f and we happen to stumble upon an n0 value that equals our n value (which hasn't happened once throughout my testing, so I don't think it's probably something that can happen).
After going through these functions, VQC didn't really seem to say much else about this process. He went on to discuss other things that I'll cover eventually. It's not all stuff I entirely understand, but I didn't understand the relationship between the guess base and n0 until I went through this for the sake of this thread, so I'm sure by the time I post about it I'll understand it enough to not waste everyone's time with something that isn't true. Either way, my point here is that this is a process that is meant to be important, but it hasn't been completely explained. I have been going through RSA #10 for the past few hours to attempt to shed some light on it but it's quite a bit past midnight now.
Another way to visualize the (x+n)(x+n) square, this time in relation to (d+n)(d+n), is to look at it as analogous to an L shape on a square of d sides which incorporates the remainder e and the gap made by f between dd and (d+1)(d+1) in the L shape. This is shown in the attached diagrams. According to VQC, "the colours in the MAP will be the KEY to GUIDE you through the process of how the algorithm works", but nobody seems to have been able to figure out whatever that was supposed to mean.
For odd (x+n)(x+n), there is a finite set of cells where n!=1, and an infinite set where n==1.
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Since the d-value increase when you move right starting from (0, 1) to (2, 1) by one and (1, 1) to (3, 1) we can simply remove e/2 (for even e) and (e - 1)/2 for odd e from d. That should allow us to use the D[t] in (0, 1) or (1, 1) since the difference in d from (0, 1) to (e, 1) (for even e) is e/2 and the difference in d from (1, 1) to (e, 1) (for odd e) is (e - 1)/2.
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If a cell has values at (e,n) then there will be values in a cell at (e+2n,n). This is how you get the horizontal pattern of grid cells.
One thing VQC has brought up several times is this question: "Why are there gaps?". A simple answer to this question is that in these places, the pair of e and n values is not valid. A more complicated way to look at it is to take a look at individual patterns.
Horizontally, starting from both (0,n) and (-1,n), there will be a cell 2n to the right. So for example where n=5, there's a cell at (0,5), (10,5), (15,5), etc, and at (-1,5), (4,5), (9,5), etc. The thing about this pattern is that it can have multiple starting places. So also where n=5, it seems to start from (-1,5), (0,5) and (1,5). That's why when you go 5 cells across in this row, there are always groups of three. Where n=1, the starting places are only (-1,1) and (0,1), and they move 2n (which is just 2) across, so this is why row 1 has no gaps. The starting places seem a little weird to predict, though. Columns (-1,n) and (0,n) are always starting places, but some cells have extras in weird places. Where n=3, for example, (-1,3) and (0,3) are valid, but so are (2,3) and (3,3).
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c from (e,n,t) = a at (2c,1,1) and (2c-1,1.1)
The following posts are a collection of all of the hints that I personally didn’t understand or that nobody seemed to work on or find the significance behind. Since this thread is meant to be a place to collect all relevant material once it is understood, even though the following information isn’t understood or its point hasn’t necessarily been discovered (at least I didn’t see anything), it seems like a good idea to me to at least have any loose ends in the same place, just segregated off between some red text line breaks.
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Think about the distribution of n and (n-1) in cell (e,1) in the a[t] and d[t] values respectively. This is the quick way to utilising f.
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The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.
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.
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The remainder of f when divided by 8 is a key to what values of d and (n-1) can be used.
Essentially, as I was showing with RSA100, you take a stab at the number by creating a triangle base out of f div 8. The remainder (f mod 8) is divided among the eight triangles. This restricts the multiples of 2d that can be used because they must always be in a number that supplies the gap in f and satisfy another property. That property can be broken down.
It is easier to break down the less factors there are as there are less solutions.
The answer is easier to come to the more factors there are as there are more solutions.
Starting with the smallest odd squares.
(x+n) = 1,3,5,..,
For each odd square what are the permutations of possible values of f,d and (n-1)?
What are the patterns?
What do these look like?
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If you colo(u)r the values of f in the grids that show the triangles, patterns will be added, ESPECIALLY if you divide f up into 8. The remainder of f when divided by 8 is a key to what values of d and (n-1) can be used.
Playing with these patterns will organically lead you exactly where we will be going.
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Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)
This will REALLY help if you visualise the smaller (x+n) squares and throw in a few much larger (x+n) squares. Focus on the patterns in the way f is distributed in each of the 8 triangles. Look at symmetries and how to construct the triangles when you know you need the two different portion sizes of f. This affects the values of 2d(n-1). There is a pattern to these that matches the grid which is why later you'll look back at the grid (The End) and understand all the puns and analogies of using that term.
If you draw out some odd (x+n) squares, relate them graphically side by side with the grid and see any patterns that f make and each 8 triangle, especially if portions of f are outside the (n-1) square in the middle.
(in regards to (f-1)/8) To clear this up slightly, the base of a triangle WITHIN the triangle.
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What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.
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As stated at the start, the square root of (D)avid and what remains are the key to unsealing the The End, the grid.
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When does c first appear at a[t]?
When is the second time it appears?
When is the first time a squared appears?
What is the factor it is multiplied by?
What is the first time b squared appears?
The second?
When a appears as "an" it appears another time.
When does n first appear?
What are the rules?
In column -f too?
Any one of these patterns well understood between -f and e is the step you are looking for.
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Extend the cells at (e,1) into negative x.
Cell (1,1) is a mirror, what about cells to the right?
What additional information does this show?
What pattern does this show?
How about negative x in cells below (e,1)?
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Here are two points made by VQC that aren’t exactly patterns but are useful to think about in the broad scope of the problem:
(in regards to things being “key”) 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.
If there are more than two prime factors, d and e are the same, you can just find the largest two first, one will just be the product of further primes, divide and go again.
I'm pretty sure that's everything. I most likely missed something, since I've been doing this alone. If at any point in the future someone notices that I've missed an entire concept or something small, let me know. Otherwise, from here, I'll eventually go through and organize all of this information into relevant categories. As stated by VQC >>6506
>Patterns that apply to all c in the grid.
>Patterns that apply to all cells in a row.
>Patterns that apply to all cells in a column.
>Special rows.
>Special columns.
Categories. So I'll go through everything and organize it as best I can.
Someone had a Twitter conversation with VQC back when he had a Twitter account, and everything between quotation marks was in one of VQC's messages. After whoever was having that Twitter conversation with him posted that, no mention of the root of d came up until a few months later, when VQC mentioned "the root of d" being important (he was very ambiguous, as always). We all thought that meant sqrt(d) for a while (for all we know it actually does), and then we thought it was sqrt(2d) since that was mentioned in a code comment he posted, until I was putting this thread together and I found that root of d (0,n) cell that everyone had forgotten about. 3DAnon found some interesting patterns relevant to this cell here >>7446 but we still have no idea what to do with it.
>>8284
I don't know if it was that far back. It was certainly a long time ago. So is that a hint that you wanted to talk on Discord or was it just a pun? Topol was doing that the other day so now I just automatically assume.
Did you have a look at the code I posted on Discord? All you'd have to do is add one line of code and you could test for the factors of n-1. I don't know if it really counts having 1 as a factor so maybe 145 wasn't the best example.
Also shouldn't this be in RSA general? I suppose you must have posted this here because it was likely to get my attention?
Right, but if it isn't necessarily what he meant, and we need to study the thing and figure out if it is what he meant and if so what specific patterns are in play, shouldn't we do that in RSA general? This thread is for things we understand. and can explain. So far I don't even completely understand where you're going with this (since before you were looking at the triangle bases being a factor of (x+n)(x+n), and now it's n-1).