# Understanding Backtracking Through a Tetris Optimizer in Go

When I first heard the term backtracking , it sounded like a complicated algorithm reserved for computer scientists. After spending the last couple of weeks learning it and implementing it in a Tetris Optimizer project, I realized something surprising: Backtracking is simply the art of making a decision, checking whether it works, and if it doesn't, undoing it and trying something else. This article explains backtracking using a practical project instead of abstract examples. The Problem Imagine you have several Tetris pieces (tetrominoes), and your goal is to fit all of them into the smallest possible square . It might look something like this: A A A A B B B B C C C C D D D D The challenge is to arrange every piece so that: No pieces overlap. No piece extends outside the board. Every piece is used exactly once. The board is as small as possible. This is much harder than it looks. My First Thought Initially, I thought I could simply place one piece after another. Place A Place B Place C Place D Done! Unfortunately, programming isn't always that kind. Sometimes the first position you choose for piece A makes it impossible to place D later. The mistake wasn't with D . The mistake happened much earlier. Enter Backtracking Backtracking works like this: Place a piece. Try placing the next one. If you get stuck... Remove the last piece. Try a different position. Repeat until every piece fits. It's essentially saying: "If this path doesn't work, let's go back and explore another one." Visualizing the Search Suppose we have four tetrominoes. Start ├── Put A at (0,0) │ ├── Put B │ │ ├── Put C │ │ │ ├── D fits ✅ │ │ │ └── D fails ❌ │ │ └── Try another position │ └── Move A elsewhere └── Try another position for A Every branch represents another possibility. Backtracking explores these branches until it finds one that works. How It Looks in Go The heart of the algorithm is surprisingly small. func solve ( index int ) bool { if index == len ( pieces ) { return true } for every