Once you know the rules of Skyscrapers, the real challenge is turning a few edge numbers into a complete grid without guessing. This guide collects the deductions experienced solvers use, from simple edge clues to the techniques that crack hard puzzles. Every rule holds for any grid size, and each comes with the reason it works.
A one-paragraph recap
In an N×N grid, each row and each column contains the heights 1 to N exactly once. A clue outside the grid says how many buildings are visible when you look along that row or column from that side; a taller building hides every shorter building behind it. Throughout this article, N is the grid size, a clue is called c, and "distance d" means the d-th cell counted from the clue's edge (d = 1 is the cell right next to the clue). If you want a fresh puzzle to practise on while reading, open Skyscrapers in another tab.
1. Start with the extreme clues: 1 and N
Two clues give away information immediately:
- Clue 1: only one building is visible, so the first building must hide all the others. That means the tallest building, N, stands in the cell directly next to the clue.
- Clue N: every building is visible, which is only possible if each one is taller than the one before. The whole line reads 1, 2, 3, …, N going away from the clue.
Fill these in first: each clue 1 places a tallest building, which removes N from the rest of its row and column.
2. The distance-from-edge rule
This is the single most useful rule in Skyscrapers, and it applies to every clue, not just the extreme ones:
With clue c, the cell at distance d from that edge cannot hold a height greater than N − c + d.
Why? Suppose the cell at distance d holds height h. In front of it there are only d − 1 cells, so at most d − 1 buildings can be seen before it. After it, every additional visible building must be taller than h, and there are only N − h heights taller than h. So the total you can see is at most (d − 1) + 1 + (N − h). For that to reach c, h must be at most N − c + d.
For a 5×5 grid the rule gives these upper limits (a dash means no restriction beyond 5):
| Clue | d = 1 | d = 2 | d = 3 | d = 4 |
|---|---|---|---|---|
| 2 | 4 | – | – | – |
| 3 | 3 | 4 | – | – |
| 4 | 2 | 3 | 4 | – |
| 5 | 1 | 2 | 3 | 4 |
A practical way to remember it: a clue of c means the tallest building cannot be in the first c − 1 cells, the second tallest cannot be in the first c − 2 cells, and so on. Mark these exclusions as pencil notes at the start; they often leave a single possible spot for N in a row.
3. Reading clue 2 and clue N − 1
Clue 2
Exactly two buildings are visible: the first building, and the tallest one, N. Nothing else can be seen. Two consequences follow:
- The first cell is not N (otherwise you would see only one building).
- Every building between the first cell and N is shorter than the first building. If the first cell is 3 and N is three cells further in, the two cells in between must both be lower than 3.
Clue N − 1
Here all buildings but one are visible. Checking every arrangement shows that the first cell must be 1 or 2, and the tallest building N must stand in one of the last two cells (the far end of the line).
4. Pair the opposite clues
Looking at a row from both ends at once is often more powerful than looking from either side alone. Call the left clue a and the right clue b, and let p be the position of N counted from the left.
- From the left you can see at most p buildings (nothing behind N is visible), so a ≤ p.
- From the right you can see at most N − p + 1 buildings, so b ≤ N − p + 1.
Together: a ≤ p ≤ N + 1 − b. This window tells you exactly where the tallest building may stand. It also proves that a + b can never exceed N + 1.
When a + b = N + 1 exactly, the window shrinks to a single cell: N sits at position a. Moreover, every building on the left side must be visible from the left, so they increase toward N, and every building on the right side increases toward N from the right. For example, in a 4×4 row with clues 2 (left) and 3 (right), the 4 must be in the second cell and the last two cells must decrease toward the right edge. Only three rows fit: 1 4 3 2, 2 4 3 1 and 3 4 2 1.
5. Use Sudoku-style elimination constantly
Skyscrapers is a Latin square, so the elimination habits from Sudoku carry over directly: each height appears once per row and once per column. Whenever you place a building, remove that height from the candidates of its row and column. Look for:
- Single candidates: a cell where only one height is still possible.
- Hidden singles: a height that has only one possible cell left in a row or column.
- Pairs: two cells in a line that can only hold the same two heights; those heights can be removed from the rest of the line.
Clue rules and elimination feed each other: a clue limits a cell, elimination places a height, and that height tightens the crossing line.
6. The visibility check: test a candidate before committing
When a cell is down to two or three options, try each one mentally and count. Place the candidate, fill the rest of the line with the most favourable remaining heights, and ask: can the clue still be met exactly? Check both the maximum and the minimum you could see. If even the most favourable arrangement gives too few visible buildings, or every arrangement gives too many, the candidate is impossible. This is not guessing; it is a quick proof by contradiction on one line.
7. A fully worked example
Take one row of a 5×5 puzzle with clue 2 on the left and 3 on the right. From the columns we already know that cell 2's column contains a 5 and a 1, and cell 4's column contains a 4 and a 2. Without any column information, 18 different rows would satisfy these two clues, so the clues alone are not enough. Here is how to get to the answer step by step.
- Distance rule. Left clue 2: cell 1 is at most 4. Right clue 3: cell 5 is at most 3 and cell 4 is at most 4.
- Locate the 5. The pairing window gives 2 ≤ p ≤ 5 + 1 − 3 = 3. Cell 2's column already has a 5, so the 5 goes in cell 3.
- Read the clues again. From the left, cell 1 and the 5 are the two visible buildings, so cell 2 must be hidden: cell 2 < cell 1. From the right, three buildings must be seen before and including the 5, and there are only cells 5, 4 and 3, so all three are visible: cell 5 < cell 4.
- Place the 4. It cannot go in cell 2 (it would need a 5 in cell 1), nor in cell 5 (at most 3), nor in cell 4 (its column already has a 4). So cell 1 = 4.
- Visibility check on cell 4. The remaining heights are 1, 2, 3. Cell 4 cannot be 2 (column). Try 1: then cell 5 would have to be smaller than 1, impossible, and only two buildings would be visible from the right. So cell 4 = 3.
- Finish. Cell 2 cannot be 1 (column), so cell 2 = 2 and cell 5 = 1.
| Clue | Cell 1 | Cell 2 | Cell 3 | Cell 4 | Cell 5 | Clue |
|---|---|---|---|---|---|---|
| 2 | 4 | 2 | 5 | 3 | 1 | 3 |
Final count: from the left you see 4 and 5 (two buildings); from the right you see 1, 3 and 5 (three buildings). Both clues are satisfied, and no step involved guessing.
A solving routine that works
- Fill every clue 1 and clue N.
- Apply the distance rule to all clues and write down the pencil marks.
- Pair opposite clues to narrow the position of N in each row and column.
- Eliminate as in Sudoku after every placement.
- When progress stalls, pick the most constrained line and run visibility checks on its candidates.
Return to step 2 whenever a new building appears. To make the routine automatic, start on a small grid in our free Skyscrapers game and move up in size once the routine feels natural.