Video summary
L15. Sudoko Solver | Backtracking
Main summary
Key takeaways
Main ideas / concepts covered
- The video explains how to solve an incomplete Sudoku (a puzzle that has at least one valid solution, possibly more than one) using backtracking.
- It begins by defining Sudoku structure and rules, then outlines the solving strategy:
- Find the first empty cell.
- Try placing digits 1 through 9 in that cell if the placement is valid (respects Sudoku constraints).
- Recurse to solve the rest of the board.
- If a choice leads to a dead end, undo it (backtrack) and try the next digit.
- The approach is described as returning one valid Sudoku solution, not all possible solutions, and therefore stops searching once a valid completion is found.
Sudoku rules (constraints the solver must satisfy)
- The Sudoku board is 9×9, divided into nine 3×3 subgrids (boxes).
- For every placement of numbers 1–9, the following must hold:
- Row rule: Each digit 1–9 appears exactly once in every row.
- Column rule: Each digit 1–9 appears exactly once in every column.
- Box rule: Each digit 1–9 appears exactly once in every 3×3 subgrid.
Backtracking methodology (detailed step-by-step)
-
Data representation
- The Sudoku is stored as a 9×9 grid of characters (conceptually mentioned as C++
vector<vector<char>>).
- The Sudoku is stored as a 9×9 grid of characters (conceptually mentioned as C++
-
Core recursive function:
solve(board)(Boolean return)-
Step 1: Find the first empty cell
- Traverse the grid in row-major order.
- Select the first cell marked as “empty”.
-
Step 2: Try digits 1 through 9
- For each candidate digit
valfrom 1 to 9:- Check whether placing
valat the empty cell is valid with respect to:- the current row,
- the current column,
- the current 3×3 box.
- If valid:
- Place the digit.
- Recurse: call
solve(board)again.
- If recursion returns true:
- Propagate true upward (solution found).
- If recursion returns false:
- Backtrack by clearing/removing the digit and try the next candidate.
- Check whether placing
- For each candidate digit
-
Step 3: Dead-end handling
- If no digit 1–9 can be placed in the empty cell, return false.
-
Step 4: Completion handling
- If there are no empty cells left:
- The board is fully filled consistently, so return true.
- If there are no empty cells left:
-
-
Important stopping logic
- Once a recursive path returns true, the solver stops and does not try additional digits/branches.
- The video contrasts:
- solving for one valid Sudoku (stop early),
- vs. enumerating all solutions (more complex).
Validity checking (“isValid”) logic
-
To test whether a number can be placed:
- Ensure the candidate does not already appear in the row, column, or 3×3 box.
-
3×3 subgrid identification
- The subgrid location is computed using division and modulo by 3:
- Use division by 3 to find the block row/box row.
- Use modulo 3 to determine position within the block grid.
- The subgrid location is computed using division and modulo by 3:
-
The described approach uses single-iteration checks across row/column/box elements (rather than separate loops per structure).
Subgrid indexing approach (as described)
- For a cell at position
(i, j):- The subgrid boundaries are determined by grouping indices into blocks of size 3.
- The video mentions formulas like:
- block base row:
(i / 3) * 3 - block base column:
(j / 3) * 3
- block base row:
- It also discusses modulo 3 to handle offsets within the 3×3 span.
Note: The subtitles’ math/wording is sometimes garbled, but the core intent is to locate the 3×3 box using division/modulo by 3.
Speakers / sources featured
- Primary speaker: “Striver” (referred to by the narrator as “hey striver”)
- Company/source mentioned (promotional segment): Coding Ninjas (an external coding education provider, mentioned during an advertisement)
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