One-Answer WCA Competition and Regulations Question Thread

I have no idea why nobody started this thread yet.
anyway, my MGC 4x4, it broke at the bottom of the stem where the cubie stays attached to the cube and greatly impacts performance.
So I did what any logical person would do and super glued the corner piece back together.
image.jpg

As you can see here, the cube is fixed, but that corner that broke is slightly tilted. Would this break the rules of the cube showing any sign of elevated pieces? The puzzle is fully turnable, and no breaks have happened after I used Krazy Glue.
 
Would this break the rules of the cube showing any sign of elevated pieces?
Yes, it would. Depending on delegate, you might still be allowed to use this cube for sighted solves, but don't count on it. (For 4BLD, having one piece that can be relatively quickly be distinguished by touch would be too much of an advantage.)
 
Yes, it would. Depending on delegate, you might still be allowed to use this cube for sighted solves, but don't count on it. (For 4BLD, having one piece that can be relatively quickly be distinguished by touch would be too much of an advantage.)
Okay, thanks for clarification. I won’t be using this cube for blind solving anytime soon, so that should mean I will be fine.
 
Is it legal if some of your 3x3 center pieces stick out a bit from the rest of the cube (oh no what did I do this time)
That can happen if your cube is loose but that may not be the case since that would be pretty easy to diagnose. The YJMoyuMFJSRS3M2020 has a very interesting cap design so it should actually be easy to fix without effecting the cubes performance, i.e. if you get to a situation in which you must remove plastic.

Edit: I should elaborate on the first sentence. If your cube is loose it will cause the screws to push on the center caps, preventing the caps from going down.
 
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Is it legal if some of your 3x3 center pieces stick out a bit from the rest of the cube (oh no what did I do this time)
Remember Cube solve hero and his Mei GV yeet? The centers pop out a bit, yet the cube is still legal because:
1. All the plastic is spray-painted black.
2. Corners are from 1 cube, edges are from 1 cube, and the core is from 1 cube. Meaning the corners are not a mix from two cubes, for example.
 
Is it legal if some of your 3x3 center pieces stick out a bit from the rest of the cube (oh no what did I do this time)
This isn’t legal because the sides are sticking out differently. This is similar to CSH’s Mei GC Yeet, as mentioned above, but every piece of the cube was identical to each other within that piece type (corner, edge, center).

I think that regulations 3h and 3j cover this.

From 3h: “Modified versions of puzzles are permitted only if the modification does not make any additional information available to the competitor (e.g. orientation or identity of pieces), compared to an unmodified version of the same puzzle.”

If some centers were raised and others weren’t, then you would possibly be able to more quickly distinguish the identity of some center pieces.

From 3j: “Puzzles must … not have any … elevated pieces … that significantly distinguish any piece from a similar piece.”

Going along with 3h, this also explains that you can’t have an elevated center without elevating all of them so that they are identical.

No, make all of them stick out
That actually isn’t a terrible solution. You would just have to make sure that they all stick out equally.

Sorry for the lengthy post to answer a relatively simple question.
 
I was watching a video from "The Cubing Historian" based on the investigation of the Worlds 2007 cheating incident (
). He was able to get the official document of the investigation (https://drive.google.com/file/d/1Nw_R1EPiCDYoebkU8dNf1p4Rgua3eOY7/view) through some delegate connections. I've been wanting to see what happened in the Martin Telesforo 4.41 incident for a long time now and was wondering if there is a way for me to obtain a similar report document like the one from Worlds 2007? Does the WCA have some kind of FOIA like request I could fill out to reach this document?
 
This is super interesting, thanks for sharing. Never seen this document before.

Regarding Martin's incident,
Dear WCA Community, This notice is in regards to the Perry Open 2013 competition held in Distrito Federal in Mexico on May 25, 2013 with Uriel Gayosso Ruiz acting as WCA delegate. During the first round of the 3x3x3 speed solve competition, competitor Julio Martín Gómez Telésforo (2011TELA01) recorded the following average of five solves: (4.41), (18.90), 14.71, 18.68, 16.53. The IAC completed its own independent investigation of this incident and the competitor (2011TELA01) confirmed to the IAC that his solution for the 4.41 second solve was: x2 U' R D F2 R L B' D2 L U' B2 U F' L' U2 D' B' U' U B' + J-permutation where a J-permutation is a case that can be solved with R U2 R' U' R U2 L' U R' U' L. As reported to the WCA by the IAC on May 31, 2013, the competitor insisted that he solved the cube in this manner. The scramble of the solve in question as reported by the WCA delegate is: F2 D2 B2 F2 R F2 L' D2 B U B2 R' F2 U' F' D' L D2 The competitor reports his solution to the above scramble to be: x2 + [moves improvised by competitor] + J-permutation By applying a J-permutation before the scrambling sequence, we have the following sequence: (R U2 R' U' R U2 L' U R' U' L) x2 (F2 D2 B2 F2 R F2 L' D2 B U B2 R' F2 U' F' D' L D2) x2 The solution to this state produces a sequence of moves that solves the original scramble into a J-permutation. The optimal solution found by Cube Explorer for this sequence is: U' R D F2 R L B' D2 L U' B2 U F' L' U2 D' B2 (17f*) and the improvised moves reported by the competitor are: U' R D F2 R L B' D2 L U' B2 U F' L' U2 D' B' U' U B' which is identical to the optimal solution found with some minor adjustments made at the end. The solution presented here is suspicious in several regards. First, the solve does not match previous descriptions of the solution made by the competitor after the competition. Second, the solve matches output from an optimal solving software, does not resemble any known solving method, and does not have any logical flow. Finally, the solve proposed does not seem amenable to executing in under 4.41 seconds, as claimed by the competitor, due to multiple turns on each face of the cube. Given this information, we find that the competitor did not solve the WCA generated scramble for the round in question in 4.41 seconds by either the solution presented here or by any other solution. The WCA finds that competitor Julio Martín Gómez Telésforo (2011TELA01) has maliciously exploited the WCA and the organizers of Perry Open 2013 and declares that the 3x3x3 speed solve in 4.41 seconds is invalid and will be recorded as a DNF. Due to the dishonesty of the competitor, and the attempted exploitation of Perry Open 2013 and the WCA by the competitor, the WCA prohibits the competitor Julio Martín Gómez Telésforo (2011TELA01) from participating in any WCA events through December 31, 2015. The WCA urges all WCA delegates and organizers to remain vigilant in their tournament operations. The WCA thanks the IAC for its diligent work, and to the community members who supported the IAC's investigation. Regards, Tyson Mao World Cube Association Board Member Supported by: Ron van Bruchem, Sébastien Auroux, Timothy Reynolds, and the WCA Delegates
 
When did you make the purchase, what competition is it, and when is the deadline for the final purchase.
It is the March 12 Greeley Colorado comp, and I registered as soon as it opened, but my internet was not loading so i registered after it was completely filled, the deadline when it stopped was 35 minutes after it opened.
Also what do you mean by purchase?
 
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