>Anyone validate this? ..
>The e looks to be 1979924809378250681468 too large?
Ok, validated the e was correct.
>>8725 ty, used image to corroborate validation.
My mistake, was using too low of a precision with BigInt (was at 256, upped it to 512 and all is fine). Lost digits with the d^2 which gets large.
precision(BigFloat) = 512
Works. Will check next against the 2048 numbers that AA, PA and others posted back in July:
>>6753 dejaVu…
>>6794 PA
Ready.
> - Works. Will check next against the 2048 numbers that AA, PA and others posted back in July.
Ok, all checking out. Had to set precision to:
setprecision(2048) to get a match.
setting to 4096 provides same results (just more significant digits out in the decimal floatS).
Is there significance in the sequence by which the grid elements are constructed in the original code loops with i and j?
Could this have something to do with TIMING?
[e, n, t] creation sequence:
[0, 0, 1] is 1st, then [2, 1, 1], then [0, 0, 3], then [1, 1, 1], [6, 1, 1], [3, 2, 1], etc.
>Lets say you take the first fifteen primes and multiple c to make c'.
Would you clarify which small primes are valid?
Do they need to come from the e or -f column associated with our c (e.g., from elements where a=1 and b=prime)?
Or, can we just use the same sequence of primes (2, 3, 5 .. p), until large enough that their product is sqrt(c)?
So, for RSA_2048, we would use the first 132 primes, generating a primeProd for the sequence of:
2903775511555279803023174864166140404845228761969778627101141839318630354131913148718157825308298335287985368473982611555148661663333023178291402119421484553385515765831512441803996425951513086447100980862026122164262081616021130587497853802885830115639327460909636718944783285770656342604598491113326147439710
>use odd primes.
Haha don't be daft! All primes are odd, Kek! Roger, will avoid consecutive primes and two too. ty.
>>495 (/pb) this was the comment was referring to:
>Choose a prime number that is a factor of any value of a in a cell in the first row (e,1).
> E.g. 5
> E.g. (1,1)
VA, take a BigFloat on over to
>>4379 and just holler.
Have been doing all this in a…
Notebook!
From planet Jupyter. Seems to be working nicly. You'd be off in no time.
>If you multiply c by small primes, the smoothness of BigN-n increase.
>Once the size of the product of small primes is larger than the root of c, when that product is multiplied by c, there is enough information to imply n.
>which small primes are valid?
>Avoid the use of two, use odd [alternating] primes.
Ok, think this is clear. Here is result for RSA2048:
1) take the first 233 primes (min number required while still providing enough 'information').
2) Remove alternating primes (so delete 3, 7, 13, 19, …). Result is this list of 116 primes:
BigInt[5, 11, 17, 23, 31, 41, 47, 59, 67, 73, 83, 97, 103, 109, 127, 137, 149, 157, 167, 179, 191, 197, 211, 227, 233, 241, 257, 269, 277, 283, 307, 313, 331, 347, 353, 367, 379, 389, 401, 419, 431, 439, 449, 461, 467, 487, 499, 509, 523, 547, 563, 571, 587, 599, 607, 617, 631, 643, 653, 661, 677, 691, 709, 727, 739, 751, 761, 773, 797, 811, 823, 829, 853, 859, 877, 883, 907, 919, 937, 947, 967, 977, 991, 1009, 1019, 1031, 1039, 1051, 1063, 1087, 1093, 1103, 1117, 1129, 1153, 1171, 1187, 1201, 1217, 1229, 1237, 1259, 1279, 1289, 1297, 1303, 1319, 1327, 1367, 1381, 1409, 1427, 1433, 1447, 1453, 1471]
3) Using BigInt/BigFloat, calculate product of these primes. Result is:
387690662375116154189306786300948040682152301112381332365947690358877016252291271579386805963882041509361136031463830160262212467528052601358967283927566989520797030933142943740776342583911079131737628443244072008838040734929847409534239275952452284284160980866268104515160752851645533924749226331947534836485
4) Test if product * sqrt_c c. If not, use more primes, if too large, trim list until min number of primes attained for list.
if sqrt_c < primeProd "OK, primeProd is larger than sqrt_c by", c - primeProd;
else "Add more primes, gap is", (c - primeProd);
end
Result for this list with RSA2048:
("OK, primeProd is larger than sqrt_c by", 25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987161491760058521104895835079161095536116271086072393115554792243877707807571989926584428204710928410151016170024737789515993921376615551232545447668706865200593860513521035480280148273931812271646970121373881700459124885651426525880377915555485715615600879464497755422866790625784918857287261170790874585883872)
>Lets say you take the first fifteen primes and multiple c to make c'.
>You would focus on the column with e',d' and c'.
Did you mean multiply the primeProd result by sqrt_c here to get c'? Otherwise very large like the D value.
Think you're right VA. Was reading it that way too and wrapping head around that. We're going to need to get good at searching those solution spaces! ps, doing this in the notebook.
One error with the listed gap in Step 4. Can see was showing Diff for c and primeProd, not sqrt_c. Here are values fixed. Will try not to post to many loooong numbers.
RSA2048 (c) =
25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357
primeProd of set:
2.2895847132472495001482427763988503310279383766757161657022106260529704617154132317510816270431394037668973397527380869550879298705523576071279870870493805484939129171966550420690585279287210596490355970701005164717322046794212049008088245181444490229836448724503506940231137980416138558565393918984970002864052+308
Ok, so now to build out the c' and respective cells for each prime in that list. Taking simple approach with single primes, vs 2 or more * c for a c'.
>Haha don't be daft!
reminded me of listening to Delta Notch. There's a new 12hr study set out last week… (embedded)
VA, I got it running but it was a nightmare. Weren't you running VS Code? VS and VS Code are different. Senpai uses VS (not Code, according to an earlier post last year, same drill we're doing now). After wrangling for a while, and reading online, found that the .DLL's for Forms are not part of VS Code / don't play well, but installing VS provides them and can then got place them where needed for VS Code to pick up (it's about attached Resources). Punted that and finally got the code in VS, which requited Numerics addition for BigInt. In VS, compiled you need to create the form, it will generate a couple more files, including a Forms.cs file (in addition to your primary Program.cs file).
This might not be quite accurate, am a newfag w/ C# and coding.
Did all this in Julia very easily. Created the primes list using code versus searching lists online and processing in Excel with things like =TRIM(CLEAN(SUBSTITUTE(B1,CHAR(160)," "))) to remove whitespace and such. It's night and day.
Only risk is if Senpai actually posts the code, would need to adapt it to Julia. I can live with that based on everything seen so far.
Oh, one more thing, Julia can actually call C code and run it as well. So, could probably use it without re-writing if needed.
2) Remove alternating primes (so delete 3, 7, 13, 19, …) and then delete 2, the first prime.
Julia code for this, operating on the array of the first 233 primes:
#remove every other prime
primeListAlt = primeList[1:2:end]
This creates the "primeListAlt" array, the alternating one. Next remove "2" from the list to get a new primeListAlt:
#remove 2, the first prime in the list
filter!(x->x≠2,primeListAlt)
How to find the product of all primes remaining in that list?
#cal product of all primes in list
primeProd = prod(primeListAlt)
Now, it's straightforward to create a new array, multiplying the trimmed array by c (produces an array of c'), with another one-liner of code.
>>8765 ty for coming by, with old Steele it's starting to get a bit Rusty toward the End. WWG1WGA (mostly).
>The excluded middle.
>That’s a clue right thar
As they say in the UK: "MIND the GAP!"
Chris Curtis may have passed, after breaking all records, but not to worry, our five eyes see a promise that must be kept, a cross that must be Bourne, by our Resurrected fisher of mEN.
Anyone catch these two posts by Anon - quite interesting, and highly related to our work here:
>Twitter. Looked for twitter/tweet as a keyword for shits and giggles. Not an obsessive freak who says that when u reverse all the words, and shit on your keyboard Q appears.
>Came across Q post 1221.
(I mean imagine that, shitting on your sweet mechanical keyboard, with Cherry KEYs, just to get Chris to appear? Topre be damned!)
got quite the (you) from Queue
>>>qresearch/5568629
>Something you should explore further.
>Impressive, most impressive.
>Q
Posts were FIRE!