Math Detours · Grade 8 · Around the corner · 25

A cube in four dimensions

A point, a segment, a square, a cube. Keep going: the 4-cube has 16 corners, 32 edges and 8 cubes for faces, and you can count them.

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Tonight's idea

Start with a point. Drag it one unit to the right and it sweeps out a segment: 2 corners, 1 edge. Drag the segment one unit up and it sweeps out a square: 4 corners, 4 edges. Drag the square one unit out of the page and you get a cube: 8 corners, 12 edges, 6 square faces.

12 · 14 · 4 · 18 · 12 · 6 · 116 · 32 · 24 · 8
Corners · edges · squares · cubes. The 4-cube is drawn as a cube dragged "inwards": the start cube, the end cube, and the eight dashed edges swept out by the corners.

Nothing stops you dragging the cube in a fourth direction, perpendicular to the other three. You cannot see it, but you can count what the drag produces. Every piece of the dragged shape appears twice, at the start and at the end, and as it moves it sweeps out a new piece one dimension higher: a corner sweeps an edge, an edge a square, a square a cube.

new count = 2 × (old count of that piece) + (old count of the piece one dimension lower)

Check it on the cube: edges \(2 \times 4 + 4 = 12\), squares \(2 \times 1 + 4 = 6\). Now drag the cube. Corners \(2 \times 8 = 16\). Edges \(2 \times 12 + 8 = 32\). Squares \(2 \times 6 + 12 = 24\). Cubes \(2 \times 1 + 6 = 8\). The 4-cube, or tesseract, has eight cubes for faces, just as a cube has six squares.

Watch one: the 32 edges, counted a second way

Give the 4-cube coordinates. A corner is a list of four numbers, each 0 or 1, such as \((1, 0, 1, 1)\): \(2^4 = 16\) corners. An edge runs between two corners that differ in one coordinate, so an edge is "three coordinates fixed, one running from 0 to 1". There are 4 choices for the running coordinate and \(2^3 = 8\) ways to fix the other three: \(4 \times 8 = 32\) edges, agreeing with the dragging rule.

The same count works for every piece: a square has two running coordinates, a cube has three. For an \(n\)-cube, choose which \(k\) coordinates run, then fix the other \(n - k\) in \(2^{n-k}\) ways.

Your turn

  1. Use the dragging rule to find the counts for the 5-cube: corners, edges, squares, cubes, and 4-cubes.
  2. Use the coordinate count instead. (a) Edges of the 5-cube: 5 choices of running coordinate, times what? (b) Squares of the 6-cube: the number of pairs of running coordinates is \(6 \times 5 / 2 = 15\); finish the count. Check (b) against the dragging rule applied to your answer to question 1.
  3. The diagonal of a unit square is \(\sqrt2\). Use Pythagoras once more to show the long diagonal of a unit cube is \(\sqrt3\). What is the long diagonal of the unit 4-cube, and how does it compare with an edge? Which \(n\)-cube has a long diagonal of exactly 10?
  4. Each corner of the 4-cube has 4 edges, one for each direction. Multiplying, \(16 \times 4 = 64\), counts every edge twice, so there are 32. Each corner also lies on 6 squares and 4 cubes. Use the same idea to recover the number of squares and the number of cubes, and say what you divide by each time.

Stretch: the corners as binary strings

Write the 16 corners as 4-digit binary strings, 0000 to 1111. Two corners share an edge when the strings differ in exactly one digit. How many corners are exactly 2 edges away from 0000? Sort all 16 corners by their distance from 0000 and compare the counts with a row of Pascal's triangle.

Answers

  1. Corners \(2 \times 16 = 32\). Edges \(2 \times 32 + 16 = 80\). Squares \(2 \times 24 + 32 = 80\). Cubes \(2 \times 8 + 24 = 40\). 4-cubes \(2 \times 1 + 8 = 10\).
  2. (a) \(5 \times 2^4 = 80\), the same as question 1. (b) Two coordinates run, four are fixed: \(15 \times 2^4 = 240\) squares. Dragging rule from the 5-cube: \(2 \times 80 + 80 = 240\). They agree.
  3. A face diagonal \(\sqrt2\) and an edge 1 meet at a right angle, so the long diagonal is \(\sqrt{(\sqrt2)^2 + 1^2} = \sqrt3\). Repeating, the unit 4-cube has long diagonal \(\sqrt{3 + 1} = \sqrt4 = 2\), exactly twice an edge. In general the diagonal is \(\sqrt n\), so \(\sqrt n = 10\) at \(n = 100\).
  4. Squares: \(16 \times 6 = 96\), and each square has 4 corners, so \(96 / 4 = 24\). Cubes: \(16 \times 4 = 64\), and each cube has 8 corners, so \(64 / 8 = 8\). You divide by the number of corners each piece has, because that is how many times it was counted.

Stretch. Six corners are 2 edges away: the strings with exactly two 1s (0011, 0101, 0110, 1001, 1010, 1100), since each edge changes one digit. The distance from 0000 is just the number of 1s, so the counts by distance are 1, 4, 6, 4, 1, row 4 of Pascal's triangle (rows numbered from 0): the number of ways to choose which digits are 1. (The number of digits where two binary strings differ is called the Hamming distance; the mind-reader sheet uses a Hamming code built on the same idea.)

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