Writing each figure on squared paper, with a dot for an empty cell:
| Figure | Arrangement of the six squares | Net of a cube? |
| (i) | Two squares on top, four in a row below, the row starting under the second square | No |
| (ii) | A row of three, and a second row of three below it overlapping in one column | Yes |
| (iii) | A staircase — three rows of two, each row stepped one place to the right | Yes |
| (iv) | A row of four with one square above and one square below the third column | Yes |
| (v) | A row of four with two squares hanging one below the other from the second column | No |
| (vi) | A row of four with one square above the third column and one below the second | Yes |
So (ii), (iii), (iv) and (vi) fold into a cube; (i) and (v) do not.
Why (iv) and (vi) work. They are of the ‘1 – 4 – 1’ kind. Roll the row of four into a band — that band is the four side walls. One extra square is above the row and one below, so one closes the top and the other the bottom. It does not matter which column each extra square hangs from.
Why (v) fails. Its two extra squares are both below the row, stacked in the same column. Roll the row of four into a band; the first extra square becomes the bottom face. The second is attached to the far edge of that bottom face, and that edge already belongs to the wall opposite the one it came from. So the sixth square lands on a wall that is already there, and the top of the cube is left open.
Why (i) fails. Same trouble. Both extra squares are on one side of the row of four. Fold the row into a band and let the square directly above the row become the lid. The last square is joined to that lid along an edge which the lid already shares with one of the four walls — so it folds down on top of a wall, and the base is never covered.