I was just thinking the other day about being heavy dependent of the colour scheme of our cubes, so I felt that it takes away a little of the speedcubing magic (let's call it that way) we all try to learn and perform. Hypothetically speaking, what would happen if you were given a random coloured cube each and every time? You'd have to preinspect the colours too, and make them stick to you during the solve, which is cubing skill in its purest form. So, I just want to hear what you think of this nondeterministic aspect of cubing.
MHO of the colour scheme dependency...
- Thread starter mt40
- Start date
qqwref
Legend
- Joined
- Dec 18, 2007
- Messages
- 7,834
- Location
- a <script> tag near you
- WCA ID
- 2006GOTT01
- YouTube
- Visit Channel
I think with practice you can learn to get used to pretty much any color scheme. But the more time you've spent on one color scheme, the harder it will be.
I made a simulator that randomizes the color scheme after each solve. It's pretty tricky to solve it quickly on 3x3. (And 4x4 is really tough
)
I made a simulator that randomizes the color scheme after each solve. It's pretty tricky to solve it quickly on 3x3. (And 4x4 is really tough
)
freshcuber
Veteran
How many different combinations can there be?
cmhardw
Premium Member
Try to reason through it. With a given set of 6 colors, in how many ways can you place the first color? Keep in mind that you can rotate the cube around after placing a color. So as a hint for doing this calculation, if I place red on the cube first I can place it anywhere. But, no matter where I place red I can rotate the cube so that the red face is on U. Now place the next color(s) until the cube is covered.
joey
Member
6! = 720
right?
right?
cmhardw
Premium Member
Much smaller, less than 100 actually.
The first color can go anywhere, then we put one of five colors opposite it.
The third color also can go in any of the remaining four spots, and one of three goes opposite that one.
The fifth color can go in either of the remaining two spots, and the sixth goes opposite it.
We have 5, 3, and 2 options at each step, for 30 total possible color schemes from a set of 6 colors.
When we multiply this by the 24 possible orientations of a cube, we get 6! = 720.
The third color also can go in any of the remaining four spots, and one of three goes opposite that one.
The fifth color can go in either of the remaining two spots, and the sixth goes opposite it.
We have 5, 3, and 2 options at each step, for 30 total possible color schemes from a set of 6 colors.
When we multiply this by the 24 possible orientations of a cube, we get 6! = 720.
Erzz
Premium Member
who said we needed to stick to the 6 standard colours?
Nobody said 6 standard colors, but if we say we can choose from as many colors as we want, there could be infinitely many combinations and trying to calculate a number becomes futile. However, if we want the number of colorings possible using n>6 colors, we just multiply 30 by (n choose 6) - multiplying the number of combinations with each set of 6 colors by the number of sets of 6 colors in our pool of n colors.
Given white on top, yellow on botton, red on front, we can have, in order of R-B-L faces:
blue orange green (normal scheme)
blue green orange
green orange blue
green blue orange
orange blue green
orange green blue
all of which are uniquely identifiable. 30 is correct, I have already accounted for symmetry and different possible orientations.
cyoubx
Veteran
I got 30 as well
EDIT: I just reread your post, disregard what I said, haha.
Heh. I think you may've been confused because I tried to make some sense of joey's 720 number, being the number of uniquely identifiable color schemes (30) times the number of uniquely identifiable orientations (24)
But, where are the 4 options at the 3rd step? Isn't that 5! then?
EDIT: Oh, duh!.. It's the symmetry.. Sorry
Last edited:
freshcuber
Veteran
Multi BLD with different color schemes would be so hard.
joey
Member
Oh damn, I'm dumb.
Similar threads
