11x11 Queens Puzzle #102

Difficulty: easy. Place 11 queens on this 11x11 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 66-step reasoning is written out below.

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Step-by-step solution for puzzle #102

What follows is a complete solve of puzzle #102, worked the way a person works it rather than the way a computer does. It runs to 66 deductions — 11 queen placements and 55 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, pairing argument on columns, proof by contradiction, region confined to one line, last square in a row, last square in a region.

Show the full walkthrough (57 steps) — contains the solution

Step 1. Wherever amber ends up among its 5 squares, the same squares fall within reach of it — 1 square in total, impossible regardless of how the colour resolves.
Step 2. Slate is down to 6 candidates in rows 5–9, columns 1 and 2, and they overlap enough to cover 2 squares elsewhere. Cross those out without deciding anything about slate itself.
Step 3. The pale yellow region has 4 squares left, spread across rows 7 and 8, columns 6–8. You do not need to know which of them takes the queen: every one of those options attacks the same 2 squares, so they can be crossed out now.
Step 4. Wherever orange 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 5. Teal is down to 4 candidates in rows 9–11, columns 1 and 2, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about teal itself.
Step 6. Here is the pairing argument. The slate and teal regions have no open squares outside columns 1 and 2 — 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: 9 squares come off in one stroke.
Step 7. Test row 1, column 6 for indigo: place a queen there, run the forced consequences through, and the sea green region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 8. Now a short proof by contradiction. Suppose the indigo queen went to row 1, column 7. Follow the forced moves that would create and the sea green region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 9. Test row 2, column 3 for indigo: place a queen there, run the forced consequences through, and row 4 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 10. Now a short proof by contradiction. Suppose the indigo queen went to row 3, column 3. Follow the forced moves that would create and row 4 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 11. Test row 3, column 4 for indigo: place a queen there, run the forced consequences through, and row 4 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 12. Now a short proof by contradiction. Suppose the indigo queen went to row 4, column 4. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 13. Test row 4, column 5 for indigo: place a queen there, run the forced consequences through, and row 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 14. Now a short proof by contradiction. Suppose the amber queen went to row 4, column 6. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
The board after step 14 — 0 of 11 queens placed.
Step 15. Test row 1, column 8 for blue: place a queen there, run the forced consequences through, and column 10 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 16. Now a short proof by contradiction. Suppose the blue queen went to row 2, column 8. Follow the forced moves that would create and row 1 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 17. Test row 3, column 8 for blue: place a queen there, run the forced consequences through, and the khaki 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 blue queen went to row 4, column 8. Follow the forced moves that would create and row 1 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 19. Test row 4, column 9 for blue: place a queen there, run the forced consequences through, and row 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 20. Now a short proof by contradiction. Suppose the blue queen went to row 4, column 10. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 21. Test row 4, column 11 for blue: place a queen there, run the forced consequences through, and row 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 22. Now a short proof by contradiction. Suppose the indigo queen went to row 1, column 3. Follow the forced moves that would create and row 3 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 23. Test row 5, column 8 for blue: place a queen there, run the forced consequences through, and row 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 24. Now a short proof by contradiction. Suppose the blue queen went to row 5, column 9. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 25. Test row 5, column 10 for blue: place a queen there, run the forced consequences through, and row 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 26. Now a short proof by contradiction. Suppose the blue queen went to row 5, column 11. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 27. Test row 6, column 9 for blue: place a queen there, run the forced consequences through, and row 2 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 28. Now a short proof by contradiction. Suppose the blue queen went to row 6, column 10. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 29. Test row 5, column 4 for mint: place a queen there, run the forced consequences through, and the teal region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
The board after step 29 — 0 of 11 queens placed.
Step 30. Now a short proof by contradiction. Suppose the mint queen went to row 5, column 5. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 31. 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 32. Now a short proof by contradiction. Suppose the mint queen went to row 6, column 4. 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 33. Test row 6, column 5 for mint: 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 34. All of the mint region's remaining squares (2 open squares) lie in column 3. Its queen therefore has to be somewhere in that column, which means no other colour may use it: 5 squares come off.
Step 35. Test row 6, column 1 for slate: place a queen there, run the forced consequences through, and column 5 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 36. Now a short proof by contradiction. Suppose the blue queen went to row 1, column 11. Follow the forced moves that would create and row 2 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 37. Test row 2, column 11 for blue: place a queen there, run the forced consequences through, and the khaki region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 38. Now a short proof by contradiction. Suppose the blue queen went to row 3, column 11. Follow the forced moves that would create and column 5 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 39. Test row 7, column 1 for slate: place a queen there, run the forced consequences through, and the orange region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 40. Now a short proof by contradiction. Suppose the slate queen went to row 7, column 2. Follow the forced moves that would create and column 10 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 41. Slate is down to 3 candidates in rows 5, 8, and 9, columns 1 and 2, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about slate itself.
Step 42. Now a short proof by contradiction. Suppose the amber queen went to row 4, column 7. Follow the forced moves that would create and the khaki region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 43. Look along row 4: one square survives, row 4, column 3, which belongs to mint. A row cannot be left empty, so the queen goes there. That clears row 4, column 3, the rest of the mint region and every square adjacent to it — diagonals included.
The board after step 43 — 1 of 11 queens placed.
Step 44. Now a short proof by contradiction. Suppose the blue queen went to row 1, column 9. Follow the forced moves that would create and row 10 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 45. Test row 2, column 4 for indigo: place a queen there, run the forced consequences through, and the khaki region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 46. Now a short proof by contradiction. Suppose the indigo queen went to row 2, column 5. Follow the forced moves that would create and the khaki region is left with nowhere to put its queen — impossible. So that square is ruled out for good.
Step 47. Test row 3, column 5 for indigo: place a queen there, run the forced consequences through, and the khaki region is left with nowhere to put its queen. The assumption breaks, so the square is eliminated.
Step 48. All of the indigo region's remaining squares (2 open squares) lie in row 1. Its queen therefore has to be somewhere on that row, which means no other colour may use it: 1 square come off.
Step 49. Test row 2, column 9 for blue: place a queen there, run the forced consequences through, and row 10 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 50. Now a short proof by contradiction. Suppose the amber queen went to row 3, column 7. Follow the forced moves that would create and row 10 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 51. 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 52. Now a short proof by contradiction. Suppose the slate queen went to row 8, column 2. Follow the forced moves that would create and row 9 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 53. Slate is down to 2 candidates in rows 5 and 9, columns 1 and 2, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about slate itself.
Step 54. All of the teal region's remaining squares (2 open squares) lie in row 11. Its queen therefore has to be somewhere on that row, which means no other colour may use it: 7 squares come off.
Step 55. Wherever orange 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 56. Here is the pairing argument. The amber and orange 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: 7 squares come off in one stroke.
Step 57. From here the board cascades. Keep alternating the region sweep and the line sweep and 10 queens fall out in order without any new idea being needed: pale yellow has one square left, so its queen takes row 7, column 8; row 5 is down to one square, putting a queen on row 5, column 1; teal has one square left, so its queen takes row 11, column 2; row 6 is down to one square, putting a queen on row 6, column 11; sea green has one square left, so its queen takes row 10, column 9; blue has one square left, so its queen takes row 3, column 10; orange has one square left, so its queen takes row 9, column 7; amber has one square left, so its queen takes row 2, column 6; indigo has one square left, so its queen takes row 1, column 4; khaki has one square left, so its queen takes row 8, column 5. 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 8, column 5.
The board after step 57 — 11 of 11 queens placed.

That completes the board: 11 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 11x11 Queens puzzle #102

Board #102 of the 11x11 bank carries a easy rating. Its 121 squares are shared between eleven colored regions, and the solution places eleven queens: one per row, one per column, one per region, none of them side by side or corner to corner.

Like every board on this site, #102 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 #102 are: pale yellow (4 squares), amber (5 squares), teal (5 squares), slate (6 squares), orange (6 squares), red (8 squares), sea green (13 squares), indigo (14 squares), mint (15 squares), blue (22 squares), khaki (23 squares). An even split would hand every color 11 squares, so the pale yellow region is running 7 squares under that average while the khaki region is the roomiest area on the grid with 23 squares.

Why does the split matter? Because a region's square count has no bearing on how many queens it gets — the answer is always one. So pale yellow, with 4 squares, is under far more pressure than khaki with 23. A 19-square gap like this one rewards players who start where the choices are fewest.

Square count is only half the story — how a color is arranged counts too. Blue is the tidiest shape on #102, packing 92% of its 6×4 bounding box, whereas khaki straggles across its 7×7 box at just 47%. Tidy blocks are easy to work around; scattered ones touch more lines and resist resolution.

Region by region

Here is the full footprint of every color on puzzle #102, smallest first. Reading these spans before you place anything is the quickest way to see which parts of the grid are genuinely contested.

Pale yellow: 4 squares across rows 7–8, columns 6–8.

Amber covers 5 squares, spanning 3 rows and 2 columns (rows 2–4, columns 6–7).

Teal occupies 5 squares within rows 8–11, columns 1–2.

Slate: 6 squares across rows 5–9, columns 1–2.

Orange covers 6 squares, spanning 3 rows and 4 columns (rows 9–11, columns 4–7).

Red occupies 8 squares within rows 6–10, columns 9–11.

Sea green: 13 squares across rows 7–11, columns 7–11.

Indigo covers 14 squares, spanning 4 rows and 5 columns (rows 1–4, columns 3–7).

Mint occupies 15 squares within rows 1–6, columns 1–5.

Blue: 22 squares across rows 1–6, columns 8–11.

Khaki covers 23 squares, spanning 7 rows and 7 columns (rows 5–11, columns 2–8).

Where to start on this board

Begin where the choices are fewest. On this board that means pale yellow, which owns just 4 squares inside a 2×3 window running from row 7 to row 8 and column 6 to column 8. A queen is guaranteed somewhere in there, so every line that window crosses is under pressure from the very first move.

Pale yellow spans 2 rows, so it stays open for now — but it is still the shortest list on the board at 4 squares. Work through those squares first and mark the ones that would leave another region homeless. That pruning is what makes the rest of the grid legible.

No region here is pinned to a shallow band of rows, so turn the same idea sideways: amber, slate and teal 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 #102

A quick way to find the next move is to count how many different colors touch each line. On this board, row 1 is the narrowest, touched by only 3 colors. Columns 1, 3, 9, 10 and 11 are the tightest columns, 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 10 rows. Row 9, meeting 6 colors, is the opposite case — the busiest line on #102 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.

With eleven queens to seat and eleven regions competing over 121 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 11x11 boards quickly and players who stall halfway.

Say the reason out loud: "this region has one square left", "this column has one square left". If no such sentence is available, the eliminating is not finished. One more pass of X marks is almost always quicker than committing to a placement you cannot defend.

Why puzzle #102 is rated easy

The rating reflects how much the board concedes up front. Puzzle #102 has no region confined to a single row or column, its smallest color holds 4 squares, its tightest row meets just 3 colors, and the gap between its largest and smallest regions is 19 squares. Flat, lopsided boards open quickly; even, sprawling ones make you earn the first placement.

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.

If #102 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 11x11

At 11x11 you are managing 121 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.

Learn the pairing argument and you will use it on every board, this one included. If two colors can only sit inside the same two rows, between them they consume both rows and every other color is shut out. It generalises to any matching count — three colors across three rows, four across four. On #102, start hunting for it near row 1.

Here is a shape check that gives nothing away: when #102 is solved, 4 queens sit on the outer edge of the grid and 7 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 11x11 Queens puzzle #102?

Puzzle #102 is rated easy. Its smallest color region holds 4 squares, its tightest row is touched by 3 colors, and no region is confined to a single row or column, so the opening move takes some hunting.

Does 11x11 puzzle #102 have only one solution?

Yes. Every puzzle in the 11x11 collection, including #102, has been verified to have exactly one valid arrangement of eleven queens, and it can always be reached by logic alone — no guessing or backtracking needed.

Where should I place my first queen on puzzle #102?

Start with the pale yellow region — the most constrained color on this board, with 4 squares spanning rows 7–8 and columns 6–8. Narrowing the smallest region first removes the most squares elsewhere on the grid.