12x12 Queens Puzzle #159
Difficulty: easy. Place 12 queens on this 12x12 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 36-step reasoning is written out below.
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Step-by-step solution for puzzle #159
What follows is a complete solve of puzzle #159, worked the way a person works it rather than the way a computer does. It runs to 36 deductions — 12 queen placements and 24 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: last square in a region, region shadow, region confined to one line, proof by contradiction, pairing argument on columns, pairing argument on rows.
Show the full walkthrough (30 steps) — contains the solution
Step 1. Only row 7, column 8 is still open in pale yellow, and since each colour must take a queen, that settles it. That clears row 7, column 8, the rest of the pale yellow region and every square adjacent to it — diagonals included.
Step 2. Blue is down to 5 candidates in rows 3–5, columns 1–3, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about blue itself.
Step 3. Orange cannot leave column 5 — every one of its open squares (2 open squares) sits there. That column belongs to orange, so every square in it owned by another colour is dead: 13 squares in all.
Step 4. Wherever teal 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 5. Rose cannot leave row 12 — every one of its open squares (4 open squares) sits there. That row belongs to rose, so every square on it owned by another colour is dead: 6 squares in all.
Step 6. The teal region is down to one open square, at row 11, column 12. Every colour holds exactly one queen, so that square is forced. Cross out the whole of row 11 and column 12, the remaining teal squares, and the ring of neighbours around row 11, column 12.
Step 7. Amber cannot leave column 11 — every one of its open squares (2 open squares) sits there. That column belongs to amber, so every square in it owned by another colour is dead: 6 squares in all.
Step 8. All of the sea green region's remaining squares (2 open squares) lie in row 10. Its queen therefore has to be somewhere on that row, which means no other colour may use it: 6 squares come off.
The board after step 8 — 2 of 12 queens placed. Step 9. Only row 9, column 5 is still open in orange, and since each colour must take a queen, that settles it. That clears row 9, column 5, the rest of the orange region and every square adjacent to it — diagonals included.
Step 10. The khaki region is down to one open square, at row 8, column 10. Every colour holds exactly one queen, so that square is forced. Cross out the whole of row 8 and column 10, the remaining khaki squares, and the ring of neighbours around row 8, column 10.
Step 11. Slate is down to 2 candidates in rows 5 and 6, columns 4 and 9, and they overlap enough to cover 2 squares elsewhere. Cross those out without deciding anything about slate itself.
Step 12. Now a short proof by contradiction. Suppose the indigo queen went to row 1, column 1. Follow the forced moves that would create and the red region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 13. Test row 1, column 2 for indigo: place a queen there, run the forced consequences through, and column 4 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 14. Now a short proof by contradiction. Suppose the indigo queen went to row 1, column 3. Follow the forced moves that would create and the red region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 15. Test row 2, column 1 for indigo: place a queen there, run the forced consequences through, and the red region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
The board after step 15 — 4 of 12 queens placed. Step 16. Now a short proof by contradiction. Suppose the indigo queen went to row 2, column 2. Follow the forced moves that would create and the red region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 17. Test row 2, column 3 for indigo: place a queen there, run the forced consequences through, and the red region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 18. Now a short proof by contradiction. Suppose the indigo queen went to row 2, column 4. Follow the forced moves that would create and the slate region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 19. Test row 2, column 6 for indigo: place a queen there, run the forced consequences through, and the red region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 20. Now a short proof by contradiction. Suppose the indigo queen went to row 2, column 7. Follow the forced moves that would create and the red region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 21. Test row 3, column 3 for indigo: place a queen there, run the forced consequences through, and the red region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 22. Here is the pairing argument. The indigo, mint, slate, and rose regions have no open squares outside columns 4, 6, 7, and 9 — 4 colours confined to 4 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 23. Red is down to 4 candidates in rows 5 and 6, columns 1–3, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about red itself.
The board after step 23 — 4 of 12 queens placed. Step 24. Here is the pairing argument. The slate and red regions have no open squares outside rows 5 and 6 — 2 colours confined to 2 rows. Between them they must take one row each, in some order, so every other colour is shut out of all of them: 4 squares come off in one stroke.
Step 25. Only row 2, column 11 is still open in amber, and since each colour must take a queen, that settles it. That clears row 2, column 11, the rest of the amber region and every square adjacent to it — diagonals included.
Step 26. 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 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 27. Test row 3, column 7 for indigo: place a queen there, run the forced consequences through, and column 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 28. Wherever indigo ends up among its 2 squares, the same squares fall within reach of it — 1 square in total, impossible regardless of how the colour resolves.
Step 29. Test row 4, column 6 for indigo: place a queen there, run the forced consequences through, and column 4 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 30. From here the board cascades. Keep alternating the region sweep and the line sweep and 7 queens fall out in order without any new idea being needed: indigo has one square left, so its queen takes row 1, column 4; slate has one square left, so its queen takes row 5, column 9; mint has one square left, so its queen takes row 4, column 7; blue has one square left, so its queen takes row 3, column 1; sea green has one square left, so its queen takes row 10, column 3; red has one square left, so its queen takes row 6, column 2; rose has one square left, so its queen takes row 12, column 6. 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 12, column 6.
The board after step 30 — 12 of 12 queens placed.
That completes the board: 12 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 12x12 Queens puzzle #159
12x12 puzzle #159 is rated easy. Twelve colored regions cover its 144 squares, and your job is to seat twelve queens so that every row, column and color contains exactly one — with no queen touching another, including diagonally.
Like every board on this site, #159 has been verified to have exactly one valid solution, and that solution is reachable by deduction alone. You never have to guess and then unwind; every queen can be justified by something already visible on the grid. What changes from puzzle to puzzle is where the first certainty appears and how long the chain of reasoning runs before the next one — and that is entirely a product of how these particular colors are shaped.
How the colors divide this board
From smallest to largest, the regions on puzzle #159 are: pale yellow (1 square), orange (2 squares), teal (3 squares), blue (5 squares), amber (7 squares), rose (8 squares), sea green (11 squares), mint (18 squares), red (19 squares), indigo (22 squares), slate (24 squares), khaki (24 squares). An even split would hand every color 12 squares, so the pale yellow region is running 11 squares under that average while the khaki region is the roomiest area on the grid with 24 squares.
That imbalance is the first thing to read, because every region takes exactly one queen regardless of how much space it owns. Pale yellow has only 1 place to put its queen; khaki has 24. The spread between them on this board is 23 squares, and the wider that spread runs, the more it pays to solve from the small end inward.
A region's outline can constrain it more than its size does. Here pale yellow is the most solid, filling 100% of its 1×1 box, and indigo the most diffuse at 46% of its 6×8 box. Solid blocks give clean eliminations; diffuse ones spread across the board and stay open longer.
Region by region
The complete breakdown of each region on this board follows, ordered from tightest to roomiest. Skimming the spans first tells you where the board is crowded and where it has slack.
Pale yellow: 1 square across rows 7–7, columns 8–8. It never leaves row 7, so its queen is locked to that row.
Orange covers 2 squares, spanning 2 rows and 1 column (rows 9–10, columns 5–5). It sits wholly inside column 5, so its queen must be in that column.
Teal occupies 3 squares within rows 11–12, columns 11–12.
Blue: 5 squares across rows 3–5, columns 1–3.
Amber covers 7 squares, spanning 5 rows and 2 columns (rows 1–5, columns 11–12).
Rose occupies 8 squares within rows 11–12, columns 5–10.
Sea green: 11 squares across rows 10–12, columns 1–4.
Mint covers 18 squares, spanning 5 rows and 6 columns (rows 1–5, columns 6–11).
Red occupies 19 squares within rows 3–10, columns 1–4.
Indigo: 22 squares across rows 1–6, columns 1–8.
Slate covers 24 squares, spanning 5 rows and 8 columns (rows 5–9, columns 4–11).
Khaki occupies 24 squares within rows 6–11, columns 6–12.
Where to start on this board
Begin where the choices are fewest. On this board that means pale yellow, which owns just 1 square inside a 1×1 window running from row 7 to row 7 and column 8 to column 8. A queen is guaranteed somewhere in there, so every line that window crosses is under pressure from the very first move.
Because pale yellow never leaves row 7, its queen must be in that row — and that instantly settles row 7 for everyone else. Sweep along it and X out every square that is not pale yellow. 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.
Work outward to the other flat regions next. Orange, rose and teal are no more than two rows tall, so their queens are confined to a narrow band. When two flat regions overlap the same pair of rows, those rows are spoken for and every other color is locked out of both — the single most productive pattern in the game, and it is live on this board.
The tightest lines on puzzle #159
A quick way to find the next move is to count how many different colors touch each line. On this board, rows 1, 2, 8 and 12 are the narrowest, touched by only 3 colors. Column 12 is the tightest column, with 3 colors.
The fewer colors a line touches, the sooner it surrenders its queen. A row meeting just 3 colors has to be claimed by one of them, which pushes every remaining region into the other 11 rows. Row 5, meeting 6 colors, is the opposite case — the busiest line on #159 and the last one you should expect to crack.
Working through the middle game
Once two or three queens are down, the board changes character. Each queen removes an entire row, an entire column and the ring of up to eight squares around it, so the grid shrinks faster than most people expect. The habit that pays here is marking rather than remembering: after every placement, cross out the full row, the full column and all eight neighbours before hunting for the next move.
With twelve queens to seat and twelve regions competing over 144 squares, keep two questions in rotation. Does any region now have exactly one open square left? That queen is forced. Does any row or column now have exactly one open square left? That queen is forced too. Alternating between the region view and the row-and-column view is what keeps a solve moving, and it is the clearest difference between players who finish 12x12 boards quickly and players who stall halfway.
Resist the plausible-looking queen. On #159 every correct placement has a stated reason behind it, and any move you cannot justify is a guess wearing a disguise. Backing out of a wrong guess on a single-solution board takes far longer than finding the real deduction would have.
Why puzzle #159 is rated easy
Ratings track how generous the opening is. On #159 you get three regions pinned to a single row or column, a smallest color of 1 square, a tightest row crossed by 3 colors, and a 23-square gap between the largest and smallest regions. Those four numbers largely explain the easy label.
The scale has nothing to do with maths or luck — every puzzle here is solvable by deduction and has exactly one answer. It measures reasoning depth: how many eliminations must stack up before a placement becomes provable. That depth is what separates a easy board from the rest.
If #159 feels harder than its rating suggests, the usual culprit is incomplete marking rather than a difficult deduction. Check that every queen already on the grid has its full cross of eliminations recorded — including the diagonal neighbours, which are by far the easiest to miss. More stalled solves come from one forgotten diagonal X than from any genuinely deep chain of logic.
Techniques that work well at 12x12
At 12x12 you are managing 144 squares, and holding the whole board in your head will not work. Split it into quadrants and solve locally. Most useful deductions on a grid this large are regional — a color pinched between two placed queens, or a band of rows that two regions must share. Trust the marks on the board over your memory.
One pattern is worth learning at every size, and it is available here. When two regions can only fit inside the same two rows, they must take one row each — which locks every other region out of both. The same argument works for columns, and for any matching count: three regions sharing three rows, four sharing four. Spotting these pairings turns a slow grind into a quick cascade, and on #159 the place to look for one is around row 1.
Here is a shape check that gives nothing away: when #159 is solved, 4 queens sit on the outer edge of the grid and 8 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 12x12 Queens puzzle #159?
Puzzle #159 is rated easy. Its smallest color region holds 1 square, its tightest row is touched by 3 colors, and three of its regions are confined to a single row or column, which gives you a forced move early.
Does 12x12 puzzle #159 have only one solution?
Yes. Every puzzle in the 12x12 collection, including #159, has been verified to have exactly one valid arrangement of twelve queens, and it can always be reached by logic alone — no guessing or backtracking needed.
Where should I place my first queen on puzzle #159?
Start with the pale yellow region — the most constrained color on this board, with 1 square spanning rows 7–7 and columns 8–8. Narrowing the smallest region first removes the most squares elsewhere on the grid.