VQC sez: "(You) are at The final lock and key construction steps. Happy to give it but it is the Eureka moment and anons are close. In hindsight, it shows exactly why this problem has existed for so long.You are solving two problems at once in this method of constructing the answer. Also in hindsight the steps give the pattern back in the grid (The End).
Previous Threads
RSA #0 —— https://archive.fo/XmD7P
RSA #1 —— https://archive.fo/RgVko
RSA #2 —— https://archive.fo/fyzAu
RSA #3 —— https://archive.fo/uEgOb
RSA #4 —— https://archive.fo/eihrQ
RSA #5 —— https://archive.fo/Lr9fP
RSA #6 —— https://archive.fo/ykKYN
RSA #7 —— https://archive.fo/v3aKD
RSA #8 —— https://archive.fo/geYFp
RSA #9 —— https://archive.fo/jog81
RSA #10 —— https://archive.fo/anGD7
RSA #11 ——http://archive.is/YERWc
Novice Baker here, dough was a little funky. To keep things clean and simple for this bread, I'm posting only links to previous breads. We should set up a new Pastebin for dough.
All comments and feedback appreciated!
Dang CA! Just finished studying your diagrams, and the numbers work! You and PMA are crushing it! Way to go, Lads!
PMA, you nailed this explanation. Also that is a beautful diagram for c6107!
>Per VQC,
>f is within each triangle but OUTSIDE of (n-1), where the (n-1) square should go in the middle.
>By taking one of the (n-1) portions out of the 2d(n-1) piece of the nn + 2d(n-1) + f - 1 equation, we are left with an equation that looks like:
>nn + (2d-1)(n-1) + (n-1) + f - 1
>This new (n-1) + f - 1 piece turns an "out of balance" f portion around the middle into something that is equally distributed among the 8 triangles.
>The revised full square pics attached for the same c6107 and c27707 tests show how this new layout comes together. Also attached are close ups of the (n-1) + f - 1 in the centers.
>The green squares in the middle represent the single (n-1) portion that has been removed from 2d(n-1).
>The yellow squares are the portions of (f-1) that have been "displaced" from their original positions.
>The red square is a single unit taken from (2d-1)(n-1).
>And this is most likely why (n-1) is so important as a factor to all of the transforms that we worked on previously. (d[t]-d)/(n-1), a(n-1) na transform into (e,1).
>(n-1) balances out the f center square.
Hello AA! Thanks for getting the dough sorted out. Thank you once more for being our BO, and doing a great job! Logos are for you. I love the red and white + eagle FYI.
Thanks Anon! Now we have a pastebin for our batter. Anyone who want to help add to it is welcome!
Want to work on building a "last bread notables" section? Do it.
It would be cool to have a summary of best posts for each bread. Do it.
Everyone can join in, this is a team effort.
Do it.
Everyone, this is awesome. I enjoy being here, very much. Energy of Gratitude is flowing outward from me to all of you. Can you feel it?