The Recursive Chocolate Bar

Exactly m×n − 1 breaks are always required. The strategy is completely irrelevant — every sequence of breaks takes the same total count. A 4×3 bar always takes 11 breaks. A 10×10 bar always takes 99 breaks. Proof: each break adds exactly 1 piece; you need m×n − 1 additional pieces beyond the starting 1.

The trick is finding the right invariant: after each break, the number of pieces increases by exactly 1. This is true regardless of which piece you break or along which line. The total number of breaks is therefore completely determined by the start and end states. Step 1: Initial state: 1 piece (the whole bar). Target state: m×n pieces (all individual squares). Step 2: Each break takes 1 piece and splits it into 2 pieces. Net change: pieces increases by 1. Step 3: After k breaks: number of pieces = 1 + k. Step 4: When does the process end? When all pieces are individual squares: m×n pieces. Step 5: Set 1 + k = m×n → k = m×n − 1. Step 6: For a 4×3 bar: 12 − 1 = 11 breaks always. For a 2×5 bar: 10 − 1 = 9 breaks always. No strategy is better or worse — all require the same number of breaks. Key Insight: The invariant 'each break adds exactly one piece' is the key. Once you see it, the answer is immediate and requires no casework. This argument style — finding a quantity that changes by a fixed amount per step — is one of the most powerful proof techniques in combinatorics and the foundation of telescoping sum arguments.