Step 1
Definition: a 1‑cell protrusion.
A peninsula is a 1‑cell protrusion compared to the underlying rectangle of the cut.
Legend (quick reminder)
- Grey region = ODD dots. White region = EVEN dots.
- O = filled dot, X = empty.
- When a cut leaves only 1 (or 2) unknown cells, those cells are forced by ODD/EVEN.
1) Peninsula & Bay Cuts (Bay)
Now split into clean micro‑steps (no in‑image text).Core idea: use a “parity cut”. Grey regions are ODD, white regions are EVEN. If the cut leaves only one unknown cell, that cell is forced.
Peninsula
Step 2
Choose the cut + the single “?” cell.
Take the cut (red outline) so that it includes exactly one unknown cell (the peninsula).
Step 3
Compute ODD/EVEN on the cut.
Row/column totals give the cut parity. Fully enclosed regions contribute fixed parity (Grey=ODD, White=EVEN).
Step 4
The single cell is forced.
After accounting for enclosed regions, the only way to match the required parity forces the “?” to be O.
Bay
Step 1
Definition: a 1‑cell indentation.
A bay is a 1‑cell indentation compared to the rectangle of the cut above it.
Step 2
Choose the cut so it leaves one “?”.
Pick the cut (red outline) that creates a single unknown cell in the bay.
Step 3
Add fixed parity from enclosed regions.
Sum the fixed parity inside the cut. If parity already matches, the “?” must be empty.
Step 4
Conclusion.
Here the cut parity is already satisfied, so the “?” cell is forced to be X (empty).
2) Double Cuts (Peninsula/Bay)
Images are grid‑only; explanations are below.Core idea: a parity cut can leave two unknown cells (A and B). If the cut is ODD, the pair must be different (exactly one dot). If the cut is EVEN, the pair must be the same (0 or 2 dots).
Rule (two unknowns)
Step 1
Odd cut ⇒ A ≠ B.
When the cut is ODD, the two “?” cells contain an odd number of dots — therefore exactly one is a dot. So one is O and the other is X (A ≠ B).
Step 2
Even cut ⇒ A = B.
When the cut is EVEN, the two “?” cells contain an even number of dots — so they must match: OO or XX (A = B).
Example (how to use it)
Step 3
Identify the two unknowns (top/bottom parity).
Mark the two “?” cells as a linked pair. Here the top part is EVEN and the bottom part is ODD, so the pair must be different (one O, one X).
Step 4
Use any extra constraint to decide which is which.
Now apply row/column totals (or another cut). If one choice breaks a line total, discard it — the other placement is forced.
3) Region Counts (advanced)
Label regions and count remaining dots to unlock stuck positions.Idea: when progress stalls, label a few regions with their required dot counts (e.g., “one”, “two”, “three”). Those labels are part of the method: they tell you the exact number of dots each region must contain.
Step 1
Starting position.
We begin from a mid‑game position. Standard line‑by‑line filling doesn’t immediately break it open.
Step 2
First exact count.
Use line totals to determine that a specific region must contain exactly three dots.
Step 3
Add another count.
Now pin down a second region as exactly one dot. This reduces branching immediately.
Step 4
Keep labeling.
Continue: a top region becomes two, and another becomes one. Each label is an exact dot‑count constraint.
Step 5
All labels interact.
When multiple labeled regions share rows/columns, their counts interact. This forces at least one new cell (O or X), triggering propagation.
Step 6
Propagation begins.
Each forced placement updates several line totals, which forces more cells. This is the start of the chain reaction.
Step 7
More forced cells.
With updated totals, additional cells become forced. The position “collapses” quickly once the key counts are fixed.
Step 8
Finish.
After the chain reaction, the remainder is straightforward: complete the remaining rows/columns using standard deductions.
4) No Cross‑Swaps (uniqueness)
Avoid “swappable” patterns that would create two valid solutions.What this catches: sometimes two distant 2‑cell areas can be filled as OX / XO in two different ways, and both ways still satisfy every row total, column total, and parity cut. That means the puzzle would have two solutions — not allowed. So any branch that produces a cross‑swap must be rejected.
The forbidden pattern
Step 1
Assume a crossed placement.
Here two separate 2‑cell areas are set as XO and OX in a crossed way. Suppose this comes from a trial assumption.
Step 2
Swap them — still valid.
If the puzzle constraints do not distinguish the two, you can swap the pair (OX ↔ XO) and still satisfy all totals and parity. That would give a second solution.
How to apply it while solving
Step 3
Detect a potential cross‑swap.
During solving, you may reach a situation where two areas could end up in a crossed swap. Mark it (red) as “danger”.
Step 4
If the cross is possible, discard the branch.
If your current assumption allows this crossed configuration without breaking any totals, it means the branch does not guarantee uniqueness. Abandon it.
Step 5
Force the non‑crossed orientation.
Therefore the correct continuation is the one that does not create a cross‑swap: the two areas must align consistently, so the remaining placements become forced.