Which zeros are evidence?

[[1,2,3],[4,0,6]]

Erase axes containing an original zero.

If you zero the second row immediately, what happens at the newly written zero in column 2?

1 / 9

Preserve only needed facts

A copied board is a simple correct model.

For zero propagation, row and column flags summarize the necessary original facts.

Could those same summaries drive Game of Life?

2 / 9

Valid Sudoku — predict before coding

Validation is three simultaneous uniqueness checks. A board can satisfy these checks without being solvable; this function does not search for a completion.

Decision: Cells (0,0) and (1,1) both contain 7; all other cells are empty. Which rule fails?

Presenter notes

Model answer: The top-left 3-by-3 box contains two 7s. Their rows and columns differ, so row-only and column-only checks would miss the violation.

Correctness argument: For every processed nonempty cell, its digit bit is present in exactly the associated row, column, and box masks. Before inserting the next digit, a set bit in any corresponding mask witnesses an earlier equal digit in that unit and therefore a violation. If all three bits are absent, adding them preserves uniqueness in every processed unit. The box index (r/3)*3+c/3 identifies each of the nine disjoint 3-by-3 boxes. After 81 cells, every possible row, column, or box duplicate would have been detected when its second occurrence was processed; absence of a detection is exactly current-state validity.

Debugging prompt: Test row-only, column-only, and box-only duplicates separately. Dots never set a digit bit. A fixed 9-by-9 input has constant work; do not describe it as a solver.

3 / 9

Spiral Matrix — predict before coding

The unvisited cells remain a rectangle. Boundary variables describe that rectangle, not the original matrix after some edges have already been consumed.

Decision: What is the spiral order of [[1,2,3],[4,5,6]]?

Presenter notes

Model answer: [1,2,3,6,5,4]. There is no inner rectangle, and 4 must not be visited twice.

Correctness argument: At the start of a round, exactly the cells inside the current bounds remain unvisited. Emit the top edge and remove it, then the right edge and remove it. Emit the bottom edge only if a row remains, and the left edge only if a column remains; updated bounds exclude corners already emitted in this round. These sweeps visit disjoint parts of the current perimeter in clockwise order and leave the smaller interior rectangle. Induction peels every layer exactly once. Empty bounds terminate; in a single remaining row or column the extra guards prevent traversing it again in reverse.

Debugging prompt: Use rectangular shapes, one row, one column, and zero columns. Filling each cell with a unique ID catches duplicated or missing corners better than an all-equal matrix.

4 / 9

Rotate Image — predict before coding

Compose two easy coordinate transformations instead of memorizing a four-way cycle. A clockwise rotation maps original (r,c) to (c,n-1-r).

Decision: In a 3-by-3 matrix, where does the original cell (2,0) move after a clockwise rotation?

Presenter notes

Model answer: It moves to (0,0): (r,c) becomes (c,n-1-r). The original bottom-left cell becomes the new top-left cell.

Correctness argument: Transposition sends the value at (r,c) to (c,r). Swapping only positions with c>r performs each off-diagonal exchange exactly once and leaves the diagonal unchanged. Reversing each transposed row then maps (c,r) to (c,n-1-r), which is the clockwise-rotation formula. Both transformations are permutations, so no value is lost or duplicated. Their composition therefore places every original cell at exactly its required destination. All exchanges occur in the original matrix with constant auxiliary state.

Debugging prompt: Test an asymmetric numbered square. Four rotations returning to the original is useful but not enough: a no-op also passes that property. Check the coordinate mapping after one rotation.

5 / 9

Set Matrix Zeroes — predict before coding

The first row and column become metadata storage. Their original zero status must be saved before the algorithm writes its own markers.

Decision: Why can matrix[0][0] alone not describe both the original first-row and first-column zero conditions?

Presenter notes

Model answer: There are four possible combinations of those two conditions, but a single zero/nonzero marker carries only two states. Separate flags prevent one axis from falsely zeroing the other.

Correctness argument: First record whether the original first row or first column contains zero. Scanning only interior cells then sets row and column markers exactly for original interior zeros; marker writes cannot affect an unscanned interior value. In the second interior pass, a marker in either axis is equivalent to an original zero somewhere in that cell's row or column, so zero exactly the required cells. The first-row and first-column flags handle their independent original conditions after their marker role is finished. Deferring these final writes avoids erasing metadata prematurely. Every required row and column is covered, and a nonzero cell outside all marked axes is preserved.

Debugging prompt: Test [[1,0],[1,1]] and [[1,1],[0,1]] separately. Zeroing an entire row during discovery creates artificial zeros that would incorrectly propagate into unrelated columns.

6 / 9

Game of Life — predict before coding

This is a synchronous update. Two bits in each cell let old state and next state coexist until every next-state decision is complete.

Decision: For the 3-by-3 board with only the middle row live, what is the next board?

Presenter notes

Model answer: Only the middle column is live: [[0,1,0],[0,1,0],[0,1,0]]. The blinker tests that earlier writes do not contaminate later neighbor counts.

Correctness argument: During the first pass, bit 0 of every cell remains its original state. Setting bit 1 for a future live cell cannot change any neighbor count because all neighbor reads mask with &1. Thus each next-state bit is calculated from exactly the original eight-neighbor configuration and the stated birth/survival rules. Once all cells have next-state bits, shifting right discards the old bit and installs the computed new state. The original board contains only 0 or 1, so no unrelated higher bits remain. Every cell consequently receives the same value as an independent two-board synchronous simulation.

Debugging prompt: Compare against a copied-board oracle. Read neighbors with cell&1, not cell==1, because an encoded live-to-live cell has value 3. Count all eight neighbors but never the center.

7 / 9

Diagonal Traverse — predict before coding

The sum r+c identifies a diagonal. Its parity determines traversal direction, separating geometry from boundary special cases.

Decision: What diagonal order comes from [[1,2,3],[4,5,6]]?

Presenter notes

Model answer: [1,2,4,5,3,6]. Diagonal d=2 contains coordinates (1,1) then (0,2), giving 5 then 3.

Correctness argument: Every cell belongs to exactly one diagonal d=r+c between 0 and m+n-2. For even d, start at the largest valid row on that diagonal and repeatedly decrease row while increasing column; this visits it upward-right. For odd d, start at the largest valid column and walk downward-left. The chosen starting coordinate satisfies the bounds, and each step preserves r+c=d until the opposite boundary is reached. No cell can repeat across diagonals because its coordinate sum is unique, and the monotone walk visits every valid coordinate within a diagonal. Concatenating diagonals in increasing d yields the required zigzag order.

Debugging prompt: Use a wide rectangle and its tall counterpart. Mixing n with m in the starting coordinate can pass square cases. Check that output length is exactly rows*columns.

8 / 9

Reconstruct and transfer

Derive clockwise rotation from coordinates.

Explain the two first-axis flags.

Show why life updates need old and next state simultaneously.

9 / 9