#### kubesolver

##### Member

- Joined
- Nov 15, 2019

- Messages
- 25

The idea

Let's celebrate the new Ao5 3x3x3 world record by Feliks Zemdegs with a competition dedicated to it!The $500 in prizes will be roughly equally divided between the best and random competitors.

How to enter

To enter competition please post in this thread a video of you solving 3 out of 5 solves from the world record attempt. The video can be edited (e..g the solutions don't have to happen immediately one after another). Preferably please provide a reconstruction or at least a short description of a method or something interesting about your solution.

Scramble 1: R2 F' U2 L D' L' D' R D2 U2 F2 L' D2 L' D' B' U'

Scramble 2: D' L D' U2 R2 U' B L U2 B2 R U R2 F2 D' U2 B'

Scramble 3: U' R2 B U' L' F' R D2 U L' F2 D2 R' F' D B' D' U'

Scramble 4: L B2 U2 B' R' D2 R' B L B F' U2 L D R2 B R

Scramble 5: U' R F U2 F' U2 F2 D' R D' L2 R' F D B2 L' B R' U2

Duration

The competition will last 1 month. It starts on Monday 18th November GMT 0:01 and ends on Tuesday 17th December GTM 0:01.

Prizes

The prize pool will be 500 USD and will be given out as vouchers to any speed cubing show which offer vouchers / gift cards online.

Everyone who participates will have a chance to win but the better your times the higher chance to win.

There are 5 100$ prizes to be won. One person can win only one prize.

Score = total time of 5 attemps.

1) 100$ for the best score at the end of competition

2) 100$ for the person with the most points, where the points are assigned according to the following formula:

1 points for the person with the best score on the day 1

2 points for the person with the best score after day 2

3 points......

3) 100$ to a random person. Randomly weighted by (1/score) (so the person with 2 times shorter time is 2 times more likely to win.

4) 100$ for a random competitor. Randomly weighted by the number of days the participant entry was present in a competition (1 point for someone who entered the competition on the last day, 2 points for someone who entered the second last day etc). So someone who entered a competition on the first day is 2 times more likely to win than the person who entered a competition in the middle of it.

5) 100$ for a random person. All participants equally likely to win this price.

Disclaimer

The fixed part is the world record celebration and the prize pool. The exact rules and prize structure might change later based on the feedback received from participants.

Edit: changed the rules to make it easier to enter competition

Let's celebrate the new Ao5 3x3x3 world record by Feliks Zemdegs with a competition dedicated to it!The $500 in prizes will be roughly equally divided between the best and random competitors.

How to enter

To enter competition please post in this thread a video of you solving 3 out of 5 solves from the world record attempt. The video can be edited (e..g the solutions don't have to happen immediately one after another). Preferably please provide a reconstruction or at least a short description of a method or something interesting about your solution.

Scramble 1: R2 F' U2 L D' L' D' R D2 U2 F2 L' D2 L' D' B' U'

Scramble 2: D' L D' U2 R2 U' B L U2 B2 R U R2 F2 D' U2 B'

Scramble 3: U' R2 B U' L' F' R D2 U L' F2 D2 R' F' D B' D' U'

Scramble 4: L B2 U2 B' R' D2 R' B L B F' U2 L D R2 B R

Scramble 5: U' R F U2 F' U2 F2 D' R D' L2 R' F D B2 L' B R' U2

Duration

The competition will last 1 month. It starts on Monday 18th November GMT 0:01 and ends on Tuesday 17th December GTM 0:01.

Prizes

The prize pool will be 500 USD and will be given out as vouchers to any speed cubing show which offer vouchers / gift cards online.

Everyone who participates will have a chance to win but the better your times the higher chance to win.

There are 5 100$ prizes to be won. One person can win only one prize.

Score = total time of 5 attemps.

1) 100$ for the best score at the end of competition

2) 100$ for the person with the most points, where the points are assigned according to the following formula:

1 points for the person with the best score on the day 1

2 points for the person with the best score after day 2

3 points......

3) 100$ to a random person. Randomly weighted by (1/score) (so the person with 2 times shorter time is 2 times more likely to win.

4) 100$ for a random competitor. Randomly weighted by the number of days the participant entry was present in a competition (1 point for someone who entered the competition on the last day, 2 points for someone who entered the second last day etc). So someone who entered a competition on the first day is 2 times more likely to win than the person who entered a competition in the middle of it.

5) 100$ for a random person. All participants equally likely to win this price.

Disclaimer

The fixed part is the world record celebration and the prize pool. The exact rules and prize structure might change later based on the feedback received from participants.

Edit: changed the rules to make it easier to enter competition

Last edited: Nov 21, 2019