Shade Link strategy: use adjacent pairs, then protect the kept paths.
Use a different fixed board to force a third copy out, reject choices that isolate kept cells, and compare five checked deductions with reproducible wrong branches.
A repeated pair tells you that at least one cell must be shaded. It rarely tells you which one. This lesson turns that ambiguity into useful deductions: combine an adjacent equal pair with a third copy, then follow the consequences of a choice until a required unshaded cell loses its path.
The board below differs from the beginner example. Its clean solution has five shaded cells. The objective is still to satisfy all three conditions together, rather than minimize the duplicate count on the next click.
Load the strategy board and make a paper ledger
Use Practice this example beside this article, or Load strategy board in the game. Retry first. The starting check is 10 equal-number pairs, 0 shaded side pairs and 1 unshaded group. R and C count rows and columns from the top left.
1 3 1 1 3 2 2 1 1 2 3 3 4 1 4 3
On paper, write S for a cell proved shaded and K for a cell proved kept unshaded. The game has no K annotation mode: an untouched cell is unshaded and already counts. Record the reason for K on paper; do not click it just to mark it as decided. A shaded neighbour makes a K necessary because shaded cells cannot share a side.
An adjacent equal pair forces a third copy out
R1C3 and R1C4 are adjacent ones. They cannot both be shaded, so at least one of them must remain unshaded. R1C1 is another one in the same row. If R1C1 stayed unshaded, both of those other ones would have to be shaded, which is forbidden. Therefore shade R1C1. R1C2 and R2C1 must stay unshaded next to it.
This deduction requires an adjacent equal pair and another copy in the same line. Three equal values in unrelated rows and columns do not justify it. It also does not choose between the two cells of the adjacent pair.
A related paper deduction appears in row 4: the two fours at C1 and C3 have one cell between them. R4C2 must stay unshaded. Shading that middle cell would force both fours to remain unshaded, leaving a repeated four. This A–B–A argument needs the two copies to have exactly one cell between them; it is not a rule about arbitrary distant endpoints.
Reject a choice by following its forced consequences
After shading R1C1, compare the two remaining ones in row 1. Suppose R1C3 were shaded. Its neighbour R1C4 would have to remain unshaded. That kept one would force R2C4, also a one in column 4, to be shaded. R1C4 would then be trapped between two shaded neighbours: R1C3 on its left and R2C4 below. No later shade can open a path while those deductions remain in force.
So R1C3 must remain unshaded, and its equal neighbour R1C4 must be shaded. This is a short contradiction, not permission to assume a successful click is a correct deduction. The game allows wrong choices and only reports the rules for the current shading; it does not search for future completions.
Try the rejected top-row choice, then reveal its checks
Retry and shade R1C1, then R1C3. At Toggles 2 the check is 6 equal-number pairs, 0 shaded side pairs and 1 unshaded group. It looks locally permissible even though it has no valid completion. Shade the forced R2C4 next: Toggles 3, 5 equal-number pairs, 0 shaded side pairs and 2 unshaded groups. The isolated group is the single unshaded R1C4.
Undo once restores the two-toggle trial, not a safe solution prefix. Undo again removes the unsupported R1C3 choice and returns to the proved R1C1 shade at Toggles 1. Now shade R1C4 instead. The clean prefix has Toggles 2 and 5 equal-number pairs, with 0 shaded side pairs and 1 unshaded group.
Repeat the argument on the crossing pairs
- Shade R2C2. R2C2 and R2C3 are adjacent twos, so exactly one is shaded. If R2C3 were shaded, R3C3 would remain unshaded. That kept three would force R3C4 to be shaded. But R2C4, already kept next to shaded R1C4, would now have shaded neighbours above, left and below. It would be isolated. Therefore keep R2C3 and shade R2C2.
- Shade R3C4. Consider shading R3C3 instead. That forces neighbouring R3C4 to remain unshaded, which in turn forces the equal three at R4C4 to be shaded. Together with the proved shades R1C1, R1C4 and R2C2, this splits the unshaded cells into two groups. One group contains the required kept R1C2; the other contains required kept R2C1. Further shading cannot reconnect them or remove either required cell. Reject R3C3, keep it, and shade its row partner R3C4.
- Shade R4C1. One of the two fours in row 4 must be shaded. Shading R4C3 would trap the unshaded bottom-right R4C4 between R3C4 above and R4C3 to its left. Keep R4C3 and shade R4C1 instead.
The connectivity argument is stronger when you identify a cell that must stay unshaded in each separated group. Simply seeing two groups in an arbitrary trial is not a proof that every extension fails: in some settings a whole group might be removed. Here the kept witnesses rule out that escape.
Replay the complete deduction route
Retry and activate each cell below exactly once. Check after each step if you want to compare the counts. Each check adds no toggle. All listed checkpoints have 0 shaded side pairs and 1 unshaded group.
- Shade R1C1: the adjacent ones in C3 and C4 force the other one out. Toggles 1; 7 equal-number pairs.
- Shade R1C4: reject shading R1C3 because it would force an isolated top-right cell. Toggles 2; 5 pairs.
- Shade R2C2: keeping R2C3 avoids the forced isolation of R2C4. Toggles 3; 3 pairs.
- Shade R3C4: the alternative at R3C3 would split required unshaded cells into separate groups. Toggles 4; 1 pair.
- Shade R4C1: the other four cannot be shaded without trapping the bottom-right cell. Toggles 5; 0 pairs.
Press Check shading after the fifth toggle to register the win. A correct pattern alone does not trigger the checked finish. There are 11 unshaded cells; their paths may turn and need not follow a straight row or column.
Reveal the finished map and compare two later rejected choices
X 3 1 X 3 X 2 1 1 2 3 X X 1 4 3
The row lists are (3, 1), (3, 2, 1), (1, 2, 3) and (1, 4, 3). The column lists are (3, 1), (3, 2, 1), (1, 2, 3, 4) and (1, 3). None repeats a number. The five shades share no sides, and the eleven unshaded cells form one group. Independent exhaustive checking finds exactly one valid shading pattern for these printed numbers.
To inspect the R2C3 branch, Retry and shade R1C1, R1C4, R2C3 and R3C4. Check at Toggles 4: 2 equal-number pairs, 0 shaded side pairs and 2 unshaded groups; R2C4 is isolated. This reproduces the contradiction without silently counting any paper K marks as game edits.
For the R3C3 branch, Retry and shade R1C1, R1C4, R2C2, R3C3 and R4C4. Check at Toggles 5: 1 equal-number pair, 0 shaded side pairs and 2 unshaded groups. The groups are {R1C2, R1C3, R2C3, R2C4, R3C4} and {R2C1, R3C1, R3C2, R4C1, R4C2, R4C3}. The first contains kept R1C2; the second contains kept R2C1.
Self-check: does zero repeated pairs settle it?
Retry, reproduce the first four correct shades, then shade R4C3 instead of R4C1. Predict all three check results before checking. Also predict what Undo does after a genuine five-toggle win.
Reveal both predictions and the limits of this method
The wrong final four gives Toggles 5, 0 equal-number pairs, 0 shaded side pairs and 2 unshaded groups. R4C4 is isolated. Zero duplicate pairs is necessary but insufficient. Undo once returns to the four-toggle correct prefix; shade R4C1 and Check to finish.
After the genuine win, Undo unshades R4C1 and returns to Toggles 4. Checking reports 1 equal-number pair, 0 shaded side pairs and 1 unshaded group. The repeated pair is R4C1 with R4C3. Retry instead keeps the numbers and clears the whole attempt to Toggles 0.
These deductions solve this small board without an arbitrary final choice. They do not promise that every larger Hitori board can be solved with adjacent pairs and short connectivity arguments alone. A lower pair count is not a safety score; a paper deduction is useful only when its stated conditions hold.
Load strategy board restores this fixed example. Load beginner board returns to the first lesson's different numbers. New puzzle cycles the original seven prepared boards and removes the practice parameter; this strategy example names one of those boards rather than expanding the collection. Refreshing its practice address restores the printed numbers with no saved attempt. No login, payment, timer or advertisement action is required.