Calculate the number of Orientation of Last Layer (OLL) cases on a 4x4 Rubik's cube after solving the first 3 layers without pairing edges | Step-by-Step Solution

Understanding 4x4 OLL Cases (Unpaired Edges)

What We're Solving:

We need to find how many different Orientation of Last Layer (OLL) cases exist on a 4x4 Rubik's cube when the first 3 layers are solved but the edges aren't paired yet. This is a combinatorics problem about counting possible orientations!

The Approach:

On a 4x4 cube's top layer, we have different types of pieces that can be oriented in different ways. We need to:

• 1. Identify what pieces we're working with
• 2. Determine how many ways each piece type can be oriented
• 3. Account for any constraints (some orientations might be impossible)
• 4. Multiply the possibilities together

Step-by-Step Solution:

Step 1: Identify the pieces on the last layer On a 4x4's top face after solving the first 3 layers, we have:

• 4 corner pieces
• 8 edge pieces (4 pairs, but we're treating them as unpaired individual edges)

Step 2: Count corner orientations Each corner can be oriented in 3 ways (the sticker that's on top can be the actual top color, or one of the two side colors). However, there's a constraint: if 3 corners are fixed, the 4th corner's orientation is determined.

• Corners: 3^3 = 27 possible orientations

Step 3: Count edge orientations Each edge piece can be oriented in 2 ways (correct orientation or flipped). Again, there's a constraint: if 7 edges are fixed, the 8th edge's orientation is determined.

• Edges: 2^7 = 128 possible orientations

Step 4: Apply the multiplication principle Since corner and edge orientations are independent of each other:

• Total OLL cases = 27 × 128 = 3,456

The Answer:

There are 3,456 theoretical OLL cases on a 4x4 Rubik's cube when edges aren't paired after solving the first 3 layers.

Memory Tip:

Remember "3-7" for the exponents! You always have one less than the total number of pieces due to the orientation constraint. So it's 3^(4-1) × 2^(8-1) = 3^3 × 2^7. The constraint exists because the cube's structure means once you fix most pieces' orientations, the last piece's orientation is forced!

Great question! This combines Rubik's cube knowledge with combinatorics principles beautifully. Keep exploring these mathematical connections in puzzles! 🧩