commit 1bcb4bfa744d0fe9c341724405b8344a97a402de
parent c6cd23bb8bcfb50da6075765a905d1c07425ba78
Author: Sebastiano Tronto <sebastiano@tronto.net>
Date: Mon, 3 Jun 2024 16:37:19 +0200
Backed up no cube solving
Diffstat:
2 files changed, 69 insertions(+), 1 deletion(-)
diff --git a/src/speedcubing/no-cube/no-cube.md b/src/speedcubing/no-cube/no-cube.md
@@ -0,0 +1,68 @@
+# Solving a cube... without a cube
+
+*In this page you can find the full text of a message I posted on the
+[speedsolving forum](https://www.speedsolving.com/forum/threads/the-3x3x3-example-solve-thread.14345/page-273#post-1173067)
+in 2016. It is the explanation of how I managed to write down a solution
+for a 3x3x3 Rubik's Cube scramble using just pen an paper, without
+a cube.*
+
+**Scramble**: B R2 B2 R' U2 R2 D2 L B2 F2 R2 D R' U' R U' L2 R D2 B'
+
+For this scramble, I have tried something I have had in my mind for
+quite a long time. I have found this solution without using any cube
+(physical nor simulator). As I would do in a blind solve, I solved edges
+and corners separately with commutators. First, I tracked every edge
+to see the permutation cycle (and check if there were any flipped). This
+was easier than I had thought. I have found this.
+
+(UR BU DR FL LB FR BD DF UF UL)(DL RB)
+
+Now I use commutators to "solve" these cycles. Actually I "copy" these
+cycle, so I am matching the scramble instead of solving it. No big deal,
+I just have to invert the final solution to get one for this scramble.
+The easiest way would be to just do a blind solve, but I struggled
+I little bit to save moves, in order to see if I could get under the
+80HTM limit for a legal FMC solve. I could break the first big cycle
+into 3-cycles in any order, and I choose this one. There are probably
+better ways to solve edges, but I didn't want to think to much about it.
+
+(A) (UF UL UR) = [F2 U: M', U2] = F2 U R' L F2 R L' U F2
+(B) (UF BU DR) = [M', B' R B] = R' L U' R' U R L' B' R B
+\(C\) (UF FL LB) = [L' U L, E] = L' U L U D' B' U' B U' D
+(D) (UF FR BD) = [U2: M, U R U'] = U2 R L' B R B' R' L U R' U
+
+Then I solve the 2-2 cycle left.
+
+(E) (UF DF)(DL RB) = [U B U': (L2 F2)3] = U B U' L2 F2 L2 F2 L2 F2 U B' U'
+
+Same thing for corners. One was twisted, so I use a different notation
+for the permutation.
+
+UBL->BLU
+UFL->LFD->DBR->UFR->RFD->LBD->URB->FUL
+
+(1) (UBL UFL LFD) = [U', L D L] = U' L D L' U L D' L'
+(2) (URB FUL UBL) = [F R' F', L] = F R' F' L F R F' L'
+(3) (UFR RFD LBD) = [R U R', D2] = R U R' D2 R U' R' D2
+(4) (DBR UFR URB) = [D: B2, D F2 D'] = D B2 D F2 D' B2 D F2 D2
+
+Now I can use this commutators in any order, as long as the relative
+order of the pieces of the same kind remains the same (i.e. I cannot use
+(2) before (1), but I can do either (A) (B) (1), (A) (1) (B), (1) (A)
+(B)). To cancel more moves, I used this order:
+
+(1) (A) (2) (B) \(C\) (D) (E) (3) (4)
+
+At this point I checked the solution on my cube and it was extremely
+satisfactory to see everything going as I expected :) Inverting and
+cancelling gives the following solution:
+
+D2 F2 D' B2 D F2 D' B2 D R U R' D2 R U' R' U B U' F2
+L2 F2 L2 F2 L2 U B' U2 R U' L' R B R' B' L R' U' D' B'
+U B D U' L' U' L B' R' B L R' U' R U R F R' F' L'
+F R F U' L R' F2 L' R U' F2 L D L' U' L D' L' U
+
+alg.cubing.net
+
+79 HTM, just under the limit :) There are many ways to improve this
+"method", but I just wanted a success, and I am satisfied with the result.
diff --git a/src/speedcubing/speedcubing.md b/src/speedcubing/speedcubing.md
@@ -65,4 +65,4 @@ They are outdated and not very good, I may upload a better version at some point
* [My WCA profile](https://www.worldcubeassociation.org/persons/2011TRON02)
* [My YouTube channel](https://www.youtube.com/c/SebastianoTronto)
-* [Solving a cube... without a cube](https://www.speedsolving.com/forum/threads/the-3x3x3-example-solve-thread.14345/page-273#post-1173067)
+* [Solving a cube... without a cube](no-cube)
