The Paradox of cube 1x1

[Deleted member 27249]

The 1x1 cube is impossible to solve because it is already solved! This is a paradox!

 

[Michael Wimberly]

That's why everyone gets confused when I show them my force cube

 

[Martyt7777]

Cubehead would like to have a word with you (0:56)

 

[cranberryjuice]

That's why everyone gets confused when I show them my force cube

What's a force cube?

 

[Martyt7777]

What's a force cube?

Cube where each face is the same colour. You need 6 regular cubes to make one, disassembling each one for enough pieces of the colour

 

[Sub20Soon]

I wonder what the 1x1 wr is

 

[cranberryjuice]

Cube where each face is the same colour. You need 6 regular cubes to make one, disassembling each one for enough pieces of the colour

Ohhh, those really clean looking solid color cubes! I really want to make one, but I have other priorities to spend my money on for now.

 

[Caleb Valenzuela]

I wonder what the 1x1 wr is

*cough cough
AND I PRESENT TO YOU… THE ONE AND ONLY… @1x1 wr holder!

 

[Sub20Soon]

*cough cough
AND I PRESENT TO YOU… THE ONE AND ONLY… @1x1 wr holder!

Him being one of the biggest active forum members and keeping the default pfp is so tuff

 

[Michael Wimberly]

Cube where each face is the same colour. You need 6 regular cubes to make one, disassembling each one for enough pieces of the colour

Ohhh, those really clean looking solid color cubes! I really want to make one, but I have other priorities to spend my money on for now.

I didn't make one myself, it's just a limited edition cube. But it's still really cool.

 

[Renslay]

The 1x1 cube is impossible to solve because it is already solved! This is a paradox!

I don't think so.
If we define the word "solve" as "doing an act that brings the cube into a solved state", then the act of "doing nothing", which is a mathematically valid move, also known as the "identity element" (https://en.wikipedia.org/wiki/Identity_element) fits into the description, and therefore: doing nothing solves the cube.

Or, more precisely:

Let G be the group of cube moves acting on the set C of cube configurations, and let e ∈ G denote the identity element. Let s ∈ C be the solved state. Define a move g ∈ G to solve a configuration x ∈ C if g⋅x=s.

Then, in particular, for the solved configuration s, we have e⋅s=s.

Therefore, since e maps s to itself, the identity element satisfies the definition of a solving move when applied to an already solved cube. Q.e.d.

 

[Deleted member 27249]

I don't think so.
If we define the word "solve" as "doing an act that brings the cube into a solved state", then the act of "doing nothing", which is a mathematically valid move, also known as the "identity element" (https://en.wikipedia.org/wiki/Identity_element) fits into the description, and therefore: doing nothing solves the cube.

Or, more precisely:

Let G be the group of cube moves acting on the set C of cube configurations, and let e ∈ G denote the identity element. Let s ∈ C be the solved state. Define a move g ∈ G to solve a configuration x ∈ C if g⋅x=s.

Then, in particular, for the solved configuration s, we have e⋅s=s.

Therefore, since e maps s to itself, the identity element satisfies the definition of a solving move when applied to an already solved cube. Q.e.d.

I was ironic but this for me Is difficult to understand.

 

[1x1 wr holder!]

I don't think so.
If we define the word "solve" as "doing an act that brings the cube into a solved state", then the act of "doing nothing", which is a mathematically valid move, also known as the "identity element" (https://en.wikipedia.org/wiki/Identity_element) fits into the description, and therefore: doing nothing solves the cube.

Or, more precisely:

Let G be the group of cube moves acting on the set C of cube configurations, and let e ∈ G denote the identity element. Let s ∈ C be the solved state. Define a move g ∈ G to solve a configuration x ∈ C if g⋅x=s.

Then, in particular, for the solved configuration s, we have e⋅s=s.

Therefore, since e maps s to itself, the identity element satisfies the definition of a solving move when applied to an already solved cube. Q.e.d.

From the math ive taken so far, I'm not extremely familiar with this stuff yet, but I think I understand. So, with the act of nothing being something like zero, isn't there still the idea that different things can be done, but the end result is the same: zero. This is called dependency I believe. In the case of the 1x1, where we can't do anything, we have independency where there is only trivial solution?

 

[Deleted member 27249]

From the math ive taken so far, I'm not extremely familiar with this stuff yet, but I think I understand. So, with the act of nothing being something like zero, isn't there still the idea that different things can be done, but the end result is the same: zero. This is called dependency I believe. In the case of the 1x1, where we can't do anything, we have independency where there is only trivial solution?

Make 1x1 Its similar meditation