Calculate the number of unique 5x5 chessboard colorings with specific black and white square constraints under rotational symmetry. | Step-by-Step Solution

Chess Coloring Challenge 🎯

What We're Solving:

We need to find how many unique ways we can color a 5-row chessboard where each row has exactly 5 squares, the first and last squares in each row are always black, and we can choose any color (black or white) for the middle 3 squares. The twist? We consider colorings that look the same after rotation as identical.

The Approach:

This is a classic application of Burnside's lemma (also called the Cauchy-Frobenius lemma)! When we have symmetries (like rotations), we can't just count all possibilities and divide by the number of symmetries. Instead, we need to count how many colorings remain unchanged under each specific rotation, then use Burnside's formula.

The key insight: The number of distinct colorings = (Average number of colorings fixed by each rotation)

Step-by-Step Solution:

Step 1: Identify what we can actually choose

• Each row has 5 squares: Black-?-?-?-Black
• We only choose colors for the middle 3 squares in each row
• With 5 rows, we make choices for 5 × 3 = 15 squares total
• Each choice is binary (black or white), so without symmetry, we'd have 2¹⁵ possibilities

Step 2: Determine the rotations of a square A square has 4 rotational symmetries:

• 0° (identity - no rotation)
• 90° clockwise
• 180°
• 270° clockwise

Step 3: Count fixed colorings for each rotation

For 0° rotation (identity): Every coloring is "fixed" by doing nothing, so we have 2¹⁵ = 32,768 colorings.

For 90° rotation: For a coloring to look the same after 90° rotation, the middle squares must form a pattern that maps onto itself. This creates very strict constraints - essentially, all middle squares in positions that rotate into each other must have the same color. This gives us 2³ = 8 fixed colorings.

For 180° rotation: Row 1 maps to row 5, row 2 maps to row 4, and row 3 stays as row 3. Within each row, the middle squares must be symmetric. This gives us more freedom than 90° rotation, resulting in 2⁹ = 512 fixed colorings.

For 270° rotation: This has the same constraints as 90° rotation (just in the opposite direction), so we get 2³ = 8 fixed colorings.

Step 4: Apply Burnside's lemma Number of distinct colorings = (32,768 + 8 + 512 + 8) ÷ 4 = 33,296 ÷ 4 = 8,324

The Answer:

There are 8,324 distinct ways to color the chessboard when considering rotational symmetry.

Memory Tip:

Remember Burnside's lemma as "Fix and Average":

• Fix: Count what stays the same under each symmetry
• Average: Add them up and divide by the number of symmetries

Think of it like taking a group photo with friends - if you take 4 identical shots, you don't have 4 different photos, you have 1 photo taken 4 times! Burnside's lemma helps us avoid this "overcounting" in combinatorics.

Great job working through this advanced combinatorics problem! The key was recognizing that symmetry creates dependencies between our choices. 🌟