8x8 Queens Puzzle #289
Difficulty: easy. Place 8 queens on this 8x8 board so that no two share a row, column or color region and none touch — including diagonally. This puzzle has exactly one solution, and it has been verified solvable by deduction alone: the complete 25-step reasoning is written out below.
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Step-by-step solution for puzzle #289
What follows is a complete solve of puzzle #289, worked the way a person works it rather than the way a computer does. It runs to 25 deductions — 8 queen placements and 17 elimination steps that narrow the board between them. Each step names the rule that forces it, so nothing below is a guess, and the hardest idea required is the proof by contradiction.
Techniques this board needs: region shadow, last square in a region, region confined to one line, pairing argument on rows, pairing argument on columns, proof by contradiction, last square in a column, last square in a row.
Show the full walkthrough (22 steps) — contains the solution
Step 1. Wherever indigo ends up among its 6 squares, the same squares fall within reach of it — 1 square in total, impossible regardless of how the colour resolves.
Step 2. Red is down to 4 candidates in rows 6–8, columns 6 and 7, and they overlap enough to cover 2 squares elsewhere. Cross those out without deciding anything about red itself.
Step 3. Only row 8, column 5 is still open in khaki, and since each colour must take a queen, that settles it. Everything else in row 8, in column 5, in khaki and in the ring around that square is now dead.
Step 4. Wherever slate ends up among its 4 squares, the same squares fall within reach of it — 1 square in total, impossible regardless of how the colour resolves.
Step 5. Red is down to 2 candidates in rows 6 and 7, columns 6 and 7, and they overlap enough to cover 2 squares elsewhere. Cross those out without deciding anything about red itself.
Step 6. All of the slate region's remaining squares (2 open squares) lie in column 8. Its queen therefore has to be somewhere in that column, which means no other colour may use it: 3 squares come off.
The board after step 6 — 1 of 8 queens placed. Step 7. Wherever indigo ends up among its 3 squares, the same squares fall within reach of it — 2 squares in total, impossible regardless of how the colour resolves.
Step 8. Pale yellow is down to 5 candidates in rows 6 and 7, columns 1–3, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about pale yellow itself.
Step 9. Red and pale yellow are now boxed into rows 6 and 7 and nothing else. That is 2 colours needing 2 rows, so they fill those rows between them — which locks every remaining colour out and removes 2 squares at once.
Step 10. The slate region is down to one open square, at row 5, column 8. Every colour holds exactly one queen, so that square is forced. Cross out the whole of row 5 and column 8, the remaining slate squares, and the ring of neighbours around row 5, column 8.
Step 11. Blue is down to 3 candidates in rows 3 and 4, columns 1 and 2, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about blue itself.
The board after step 11 — 2 of 8 queens placed. Step 12. Here is the pairing argument. The indigo and red regions have no open squares outside columns 6 and 7 — 2 colours confined to 2 columns. Between them they must take one column each, in some order, so every other colour is shut out of all of them: 2 squares come off in one stroke.
Step 13. Wherever mint ends up among its 4 squares, the same squares fall within reach of it — 1 square in total, impossible regardless of how the colour resolves.
Step 14. Now a short proof by contradiction. Suppose the indigo queen went to row 3, column 6. Follow the forced moves that would create and column 4 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 15. Scan column 6 and only row 6, column 6 remains — a red square. Every column takes a queen, so it takes this one. Cross out the whole of row 6 and column 6, the remaining red squares, and the ring of neighbours around row 6, column 6.
Step 16. Now a short proof by contradiction. Suppose the indigo queen went to row 3, column 7. Follow the forced moves that would create and column 4 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 17. Only row 1, column 7 is still open in indigo, and since each colour must take a queen, that settles it. That clears row 1, column 7, the rest of the indigo region and every square adjacent to it — diagonals included.
The board after step 17 — 4 of 8 queens placed. Step 18. All of the amber region's remaining squares (3 open squares) lie in row 2. Its queen therefore has to be somewhere on that row, which means no other colour may use it: 1 square come off.
Step 19. Wherever mint ends up among its 3 squares, the same squares fall within reach of it — 1 square in total, impossible regardless of how the colour resolves.
Step 20. All of the blue region's remaining squares (2 open squares) lie in column 1. Its queen therefore has to be somewhere in that column, which means no other colour may use it: 2 squares come off.
Step 21. Amber cannot leave row 2 — every one of its open squares (2 open squares) sits there. That row belongs to amber, so every square on it owned by another colour is dead: 1 square in all.
Step 22. From here the board cascades. Keep alternating the region sweep and the line sweep and 4 queens fall out in order without any new idea being needed: row 3 is down to one square, putting a queen on row 3, column 1; amber has one square left, so its queen takes row 2, column 4; mint has one square left, so its queen takes row 4, column 3; pale yellow has one square left, so its queen takes row 7, column 2. After each one, clear that row, that column, the rest of the colour and the ring of neighbours before looking for the next. The last of them seats the queen on row 7, column 2.
The board after step 22 — 8 of 8 queens placed.
That completes the board: 8 queens seated, one in every row, one in every column and one in every colour, with no two of them touching. Notice that nothing in the sequence above required a guess — each placement followed from an elimination already on the grid, which is what “single solution, solvable by logic” actually means in practice.
About 8x8 Queens puzzle #289
Puzzle #289 is one of the easy-rated boards in the 8x8 set. Eight colors divide its 64 squares, and you win by placing eight queens so that each row, each column and each color holds exactly one, with no two queens adjacent in any direction.
As with every puzzle in this collection, #289 has a single verified solution and needs no guesswork. Each queen follows from something already visible. What the board controls is tempo — how soon the first forced move shows up and how far apart the rest are spaced — and that is a direct consequence of how these colors are cut.
How the colors divide this board
From smallest to largest, the regions on puzzle #289 are: khaki (2 squares), red (4 squares), indigo (6 squares), blue (6 squares), slate (6 squares), pale yellow (12 squares), amber (14 squares), mint (14 squares). An even split would hand every color 8 squares, so the khaki region is running 6 squares under that average while the mint region is the roomiest area on the grid with 14 squares.
Every color takes one queen and one only, whatever its size — which is why the 12 square difference between khaki and mint is worth noting before you touch the grid. The cramped colors resolve first and hand you information; the roomy ones absorb whatever is left over at the end.
Square count is only half the story — how a color is arranged counts too. Khaki is the tidiest shape on #289, packing 100% of its 1×2 bounding box, whereas mint straggles across its 5×7 box at just 40%. Tidy blocks are easy to work around; scattered ones touch more lines and resist resolution.
Region by region
Each region's exact extent on this grid is listed below, starting with the most constrained. Knowing which lines a color can and cannot reach is most of the work on a easy board.
Khaki covers 2 squares, spanning 1 row and 2 columns (rows 8–8, columns 5–6). It never leaves row 8, so its queen is locked to that row.
Red occupies 4 squares within rows 6–8, columns 6–7.
Indigo: 6 squares across rows 1–3, columns 6–8.
Blue covers 6 squares, spanning 4 rows and 2 columns (rows 3–6, columns 1–2).
Slate occupies 6 squares within rows 5–8, columns 7–8.
Pale yellow: 12 squares across rows 6–8, columns 1–5.
Amber covers 14 squares, spanning 3 rows and 7 columns (rows 1–3, columns 1–7).
Mint occupies 14 squares within rows 2–6, columns 2–8.
Where to start on this board
Open on the tightest color. Here that is khaki, confined to 2 squares between rows 8 and 8 and columns 5 and 6. Its queen has nowhere else to go, which means those 1 row and 2 columns carry more constraint than anywhere else on #289.
Because khaki never leaves row 8, its queen must be in that row — and that instantly settles row 8 for everyone else. Sweep along it and X out every square that is not khaki. Then look at which colors just lost ground: on a grid this size that one observation usually forces a second region within a move or two.
No region here is pinned to a shallow band of rows, so turn the same idea sideways: blue, slate, red and khaki span only two columns. Two narrow regions sharing the same two columns take one column each and lock everyone else out of both, which is usually enough to crack a board that offers no free opening.
The tightest lines on puzzle #289
A quick way to find the next move is to count how many different colors touch each line. On this board, rows 1 and 4 are the narrowest, touched by only 2 colors. Columns 1, 3, 4 and 8 are the tightest columns, with 3 colors.
The fewer colors a line touches, the sooner it surrenders its queen. A row meeting just 2 colors has to be claimed by one of them, which pushes every remaining region into the other 7 rows. Row 6, meeting 5 colors, is the opposite case — the busiest line on #289 and the last one you should expect to crack.
Working through the middle game
After the first few queens the puzzle becomes bookkeeping. A placed queen wipes its row, its column and every square it touches — as many as eight neighbours — so the space collapses quickly. Record those eliminations on the board instead of carrying them in your head, and the next forced move tends to announce itself.
The whole middle game reduces to two sweeps. Sweep the eight colors looking for one with a single square left; sweep the eight rows and eight columns looking for the same. Each sweep feeds the next, and switching between them deliberately — rather than drifting — is what separates a fast 8x8 solve from a stalled one.
A placement without a reason is a guess. Because this board has exactly one solution, a guess that happens to be wrong will not announce itself for several moves — and unwinding it is slower than simply finishing the eliminations first.
Why puzzle #289 is rated easy
The easy label comes out of four measurable things about #289: one of its regions is locked to one row or column, its smallest color spans 2 squares, its least crowded row meets 2 colors, and its largest and smallest regions differ by 12. Boards that score tight on those measures give up their first queen fast.
Nothing about the rating involves computation or chance. Each grid has a single solution reachable by pure logic; the rating simply measures how far ahead you have to reason. Easy boards resolve a queen every step or two, while hard ones can ask you to carry three unfinished lines of thought before any of them pays off.
A board that feels above its rating is nearly always a board with a missing X. Audit every queen on the grid: full row cleared, full column cleared, all eight touching squares cleared. Diagonals are the common omission, and missing one is far more often the cause of a stall than any genuine difficulty in the puzzle.
Techniques that work well at 8x8
At 8x8 the whole grid stays within a single glance, which makes systematic scanning practical. Read the rows in order and count the colors in each — a row touched by two colors is far more constrained than one touched by five, and that is usually where the next forced move hides. The board is small enough that re-scanning from the top after each placement costs almost nothing.
There is one technique that repays study at every grid size. Two regions restricted to the same two rows must occupy one row apiece, which excludes all other regions from those rows entirely; the identical logic applies to columns and to larger matched sets. It converts slow boards into fast ones, and the likeliest place to find it on #289 is row 1.
Here is a shape check that gives nothing away: when #289 is solved, 4 queens sit on the outer edge of the grid and 4 sit inside it, and no queen lands in a corner. A work-in-progress board that clusters its queens in one quadrant while leaving others bare is usually carrying a mistake.
How to play Queens
- The board is an N×N grid divided into N colored regions.
- Place exactly one queen (♛) in each row, each column, and each colored region.
- No two queens may touch each other — not horizontally, vertically, or diagonally.
- Tap once to mark a square with an X (where a queen can't go), tap again to place a queen, and once more to clear it.
- There is always exactly one valid solution. Use logic, never guessing, to find it.
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Questions about this puzzle
How hard is 8x8 Queens puzzle #289?
Puzzle #289 is rated easy. Its smallest color region holds 2 squares, its tightest row is touched by 2 colors, and one of its regions is confined to a single row or column, which gives you a forced move early.
Does 8x8 puzzle #289 have only one solution?
Yes. Every puzzle in the 8x8 collection, including #289, has been verified to have exactly one valid arrangement of eight queens, and it can always be reached by logic alone — no guessing or backtracking needed.
Where should I place my first queen on puzzle #289?
Start with the khaki region — the most constrained color on this board, with 2 squares spanning rows 8–8 and columns 5–6. Narrowing the smallest region first removes the most squares elsewhere on the grid.