7x7 Queens Puzzle #167
Difficulty: hard. Place 7 queens on this 7x7 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 21-step reasoning is written out below.
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Step-by-step solution for puzzle #167
What follows is a complete solve of puzzle #167, worked the way a person works it rather than the way a computer does. It runs to 21 deductions — 7 queen placements and 14 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 confined to one line, region shadow, pairing argument on rows, pairing argument on columns, proof by contradiction, last square in a column, last square in a region, line confined to one region, last square in a row.
Show the full walkthrough (18 steps) — contains the solution
Step 1. Indigo cannot leave row 1 — every one of its open squares (3 open squares) sits there. That row belongs to indigo, so every square on it owned by another colour is dead: 5 squares in all.
Step 2. Blue is down to 4 candidates in rows 2–4, columns 5 and 7, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about blue itself.
Step 3. The mint region has 4 squares left, spread across rows 2–4, columns 2 and 3. 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. All of the pale yellow region's remaining squares (2 open squares) lie in row 7. Its queen therefore has to be somewhere on that row, which means no other colour may use it: 7 squares come off.
Step 5. Amber, blue, and mint are now boxed into rows 2–4 and nothing else. That is 3 colours needing 3 rows, so they fill those rows between them — which locks every remaining colour out and removes 5 squares at once.
The board after step 5 — 0 of 7 queens placed. Step 6. Here is the pairing argument. The indigo, blue, and pale yellow regions have no open squares outside columns 5–7 — 3 colours confined to 3 columns. Between them they must take one column each, in some order, so every other colour is shut out of all of them: 4 squares come off in one stroke.
Step 7. Wherever slate 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 8. Now a short proof by contradiction. Suppose the indigo queen went to row 1, column 6. 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 9. Scan column 6 and only row 7, column 6 remains — a pale yellow square. Every column takes a queen, so it takes this one. Row 7 and column 6 are now spoken for, along with the squares touching that queen.
The board after step 9 — 1 of 7 queens placed. Step 10. Now a short proof by contradiction. Suppose the indigo queen went to row 1, column 7. Follow the forced moves that would create and row 5 runs out of open squares entirely — impossible. So that square is ruled out for good.
Step 11. Only row 1, column 5 is still open in indigo, and since each colour must take a queen, that settles it. Everything else in row 1, in column 5, in indigo and in the ring around that square is now dead.
Step 12. The amber region has 4 squares left, spread across rows 2–4, columns 1 and 3. You do not need to know which of them takes the queen: every one of those options attacks the same 1 square, so they can be crossed out now.
Step 13. Column 4 has only slate squares left (2 open squares). The column needs a queen and only slate can provide one, which pins that colour to this column and kills its squares elsewhere: 1 square.
Step 14. Now a short proof by contradiction. Suppose the amber queen went to row 2, column 1. 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.
The board after step 14 — 2 of 7 queens placed. Step 15. Test row 3, column 1 for amber: place a queen there, run the forced consequences through, and row 5 runs out of open squares entirely. The assumption breaks, so the square is eliminated.
Step 16. Row 3 has a single open square left, at column 7 in the blue region. Every row needs a queen, so this one is forced. Cross out the whole of row 3 and column 7, the remaining blue squares, and the ring of neighbours around row 3, column 7.
Step 17. Amber is down to 2 candidates in rows 2 and 4, columns 1 and 3, and they overlap enough to cover 1 square elsewhere. Cross those out without deciding anything about amber itself.
Step 18. 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: column 3 is down to one square, putting a queen on row 2, column 3; mint has one square left, so its queen takes row 4, column 2; red has one square left, so its queen takes row 6, column 1; slate has one square left, so its queen takes row 5, column 4. 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 5, column 4.
The board after step 18 — 7 of 7 queens placed.
That completes the board: 7 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 7x7 Queens puzzle #167
7x7 puzzle #167 is rated hard. Seven colored regions cover its 49 squares, and your job is to seat seven queens so that every row, column and color contains exactly one — with no queen touching another, including diagonally.
There is exactly one way to finish #167, and it has been verified. More importantly, you can reach it without ever gambling: at every stage some square is provably right or provably wrong. Boards differ only in how deep you must dig before that proof appears, and the geometry of these seven regions is what sets the depth.
How the colors divide this board
From smallest to largest, the regions on puzzle #167 are: pale yellow (2 squares), indigo (3 squares), mint (4 squares), blue (5 squares), red (7 squares), amber (10 squares), slate (18 squares). An even split would hand every color 7 squares, so the pale yellow region is running 5 squares under that average while the slate region is the roomiest area on the grid with 18 squares.
One queen per color, regardless of acreage: that is what turns this uneven split into a solving order. Pale yellow is working with 2 squares while slate has 18 to spare, a gap of 16. Attack the cramped end and the roomy regions tend to resolve themselves.
Look at arrangement as well as area. Indigo is the densest region on this grid at 100% of a 1×3 frame; blue is the loosest, occupying only 56% of its 3×3 frame. The loose one reaches into more rows and columns, which is exactly why it usually settles last.
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 hard board.
Pale yellow occupies 2 squares within rows 7–7, columns 6–7. It never leaves row 7, so its queen is locked to that row.
Indigo: 3 squares across rows 1–1, columns 5–7. It never leaves row 1, so its queen is locked to that row.
Mint covers 4 squares, spanning 3 rows and 2 columns (rows 2–4, columns 2–3).
Blue occupies 5 squares within rows 2–4, columns 5–7.
Red: 7 squares across rows 5–7, columns 1–3.
Amber covers 10 squares, spanning 4 rows and 4 columns (rows 1–4, columns 1–4).
Slate occupies 18 squares within rows 3–7, columns 3–7.
Where to start on this board
The most constrained color is where a solve should begin, and on #167 that is pale yellow with 2 squares. It reaches from row 7 to row 7 and from column 6 to column 7 — 1 row by 2 columns. One queen has to live inside that footprint no matter what else happens, which makes those lines the busiest real estate on the board.
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. Indigo is no more than two rows tall, so its queen is 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 #167
A quick way to find the next move is to count how many different colors touch each line. On this board, rows 1, 5 and 6 are the narrowest, touched by only 2 colors. Columns 1 and 4 are the tightest columns, with 2 colors.
A line with few colors is a line close to resolving. Where only 2 colors appear, one of them owns the row outright and the others must fit into the remaining 6 rows. Row 4 is the crowded extreme at 4 colors — worth revisiting later, not now.
Working through the middle game
The middle of #167 is where careless play costs time. Every queen you place kills a row, a column and the eight surrounding squares; if you do not mark all of that immediately, you will later "discover" a move that was available several turns ago. Mark first, then look.
With seven queens to seat and seven regions competing over 49 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 7x7 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 #167 is rated hard
The hard label comes out of four measurable things about #167: two of its regions are 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 16. Boards that score tight on those measures give up their first queen fast.
Rating is a measure of chain length, not of chance. With one guaranteed solution and no guessing required, the only thing that changes across the collection is how many eliminations you must link together before a queen is provably forced.
Stuck on #167 despite the hard rating? The odds strongly favour a bookkeeping error over a hard chain of logic. Re-check each placed queen for its complete set of eliminations, paying particular attention to the four diagonal squares — those are the ones that get overlooked, and one of them is usually sitting on the answer.
Techniques that work well at 7x7
At 7x7 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.
Whatever the grid size, the pairing rule is the technique to internalise. When a pair of regions is boxed into the same pair of rows, they split those rows between them and nobody else may enter — and the same holds for columns, or for any n regions confined to n lines. Look for an instance of it around row 1 on this board.
A non-spoiling distribution check: solved, #167 carries 4 queens on the perimeter against 3 queens inside, with the corners empty. Because no two queens may touch, real solutions always look fairly evenly spread — a lopsided board in progress is a warning sign.
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 7x7 Queens puzzle #167?
Puzzle #167 is rated hard. Its smallest color region holds 2 squares, its tightest row is touched by 2 colors, and two of its regions are confined to a single row or column, which gives you a forced move early.
Does 7x7 puzzle #167 have only one solution?
Yes. Every puzzle in the 7x7 collection, including #167, has been verified to have exactly one valid arrangement of seven queens, and it can always be reached by logic alone — no guessing or backtracking needed.
Where should I place my first queen on puzzle #167?
Start with the pale yellow region — the most constrained color on this board, with 2 squares spanning rows 7–7 and columns 6–7. Narrowing the smallest region first removes the most squares elsewhere on the grid.