A solution to Hoffman's packing puzzle with 4×5×6 cuboids coloured by orientation (1), and exploded to show each layer (2). In the SVG file, hover over the cuboids for their dimensions.
Hoffman's packing puzzle, disassembled

Hoffman's packing puzzle is an assembly puzzle named after Dean G. Hoffman, who described it in 1978.[1] The puzzle consists of 27 identical rectangular cuboids, each of whose edges have three different lengths. Its goal is to assemble them all to fit within a cube whose edge length is the sum of the three lengths.[2][3]

Hoffman (1981) writes that the first person to solve the puzzle was David A. Klarner, and that typical solution times can range from 20 minutes to multiple hours.[2]

Construction

The puzzle itself consists only of 27 identical rectangular cuboid-shaped blocks, although physical realizations of the puzzle also typically supply a cubical box to fit the blocks into. If the three lengths of the block edges are x, y, and z, then the cube should have edge length x + y + z. Although the puzzle can be constructed with any three different edge lengths, it is most difficult when the three edge lengths of the blocks are close enough together that x + y + z < 4 min(x,y,z), as this prevents alternative solutions in which four blocks of the minimum width are packed next to each other. Additionally, having the three lengths form an arithmetic progression can make it more confusing, because in this case placing three blocks of the middle width next to each other produces a row of the correct total width but one that cannot lead to a valid solution to the whole puzzle.[2]

Mathematical analysis

Each valid solution to the puzzle arranges the blocks in an approximate 3 × 3 × 3 grid of blocks, with the sides of the blocks all parallel to the sides of the outer cube, and with one block of each width along each axis-parallel line of three blocks. Counting reflections and rotations as being the same solution as each other, the puzzle has 21 combinatorially distinct solutions.[2][4]

The total volume of the pieces, 27xyz, is less than the volume (x + y + z)3 of the cube that they pack into. If one takes the cube root of both volumes, and divides by three, then the number obtained in this way from the total volume of the pieces is the geometric mean of x, y, and z, while the number obtained in the same way from the volume of the cube is their arithmetic mean. The fact that the pieces have less total volume than the cube follows from the inequality of arithmetic and geometric means.[2][3]

Table of solutions

The 21 distinct solutions are tabulated here as described by the references cited above [2][4].

All boxes below are entered in the format (east-west length) x (north-south length) x (up-down length), denoting the size of each box with the dimensions A, B, and C, where A < B < C. (In the above example, A = 4, B = 5, and C = 6).

All 3x3 matrices describe a set of 9 boxes, with east-west neighbors along each row and north-south neighbors down each column, with the three stacked layers being listed in sequence for each solution.

Solution Bottom layer Middle layer Top layer
Solution 01 CxBxA AxCxB BxAxC
BxCxA CxAxB AxBxC
AxCxB BxAxC CxBxA
CxAxB AxBxC BxCxA
AxBxC BxAxC AxCxB
BxCxA CxBxA CxAxB
BxAxC CxBxA AxCxB
CxAxB BxCxA CxBxA
AxBxC AxCxB BxAxC
Solution 02 CxBxA AxCxB BxAxC
BxCxA CxAxB AxBxC
AxCxB BxAxC CxBxA
CxAxB AxBxC AxCxB
AxBxC BxAxC BxCxA
BxCxA CxBxA CxAxB
BxAxC BxCxA CxBxA
CxAxB CxBxA AxCxB
AxBxC AxCxB BxAxC
Solution 03 CxBxA AxCxB BxAxC
BxCxA CxAxB AxBxC
AxCxB BxAxC CxBxA
AxBxC BxAxC BxCxA
CxAxB AxBxC AxCxB
BxCxA CxBxA CxAxB
CxAxB CxBxA AxCxB
BxAxC BxCxA CxBxA
AxBxC AxCxB BxAxC
Solution 04 CxBxA AxCxB BxAxC
BxCxA CxAxB AxBxC
AxCxB BxAxC CxBxA
AxBxC BxAxC AxCxB
CxAxB AxBxC BxCxA
BxCxA CxBxA CxAxB
CxAxB BxCxA CxBxA
BxAxC CxBxA AxCxB
AxBxC AxCxB BxAxC
Solution 05 CxBxA AxCxB CxAxB
BxCxA CxBxA BxAxC
AxCxB BxAxC AxBxC
AxBxC BxCxA BxAxC
BxAxC AxCxB CxBxA
CxBxA CxAxB AxCxB
AxCxB BxAxC CxBxA
CxAxB AxBxC AxCxB
BxAxC CxBxA BxCxA
Solution 06 CxBxA AxCxB CxAxB
BxCxA CxBxA BxAxC
AxCxB BxAxC AxBxC
AxBxC BxCxA BxAxC
CxAxB AxCxB CxBxA
BxAxC CxBxA AxCxB
AxCxB BxAxC CxBxA
BxAxC AxBxC AxCxB
CxBxA CxAxB BxCxA
Solution 07 CxBxA AxCxB BxAxC
BxCxA CxBxA CxAxB
AxCxB BxAxC AxBxC
CxAxB CxBxA BxCxA
BxAxC AxCxB AxBxC
AxBxC BxCxA CxAxB
AxBxC BxAxC AxCxB
CxAxB AxBxC BxCxA
BxCxA CxAxB CxBxA
Solution 08 CxBxA AxCxB BxAxC
BxCxA CxBxA CxAxB
AxCxB BxAxC AxBxC
BxAxC CxBxA BxCxA
CxAxB AxCxB AxBxC
AxBxC BxCxA CxAxB
CxAxB AxBxC AxCxB
AxBxC BxAxC BxCxA
BxCxA CxAxB CxBxA
Solution 09 CxBxA AxCxB BxAxC
BxCxA CxBxA CxAxB
AxCxB BxAxC AxBxC
AxBxC BxCxA CxAxB
BxAxC AxCxB AxBxC
CxBxA CxAxB BxCxA
AxCxB BxAxC CxBxA
CxAxB AxBxC BxCxA
BxAxC CxBxA AxCxB
Solution 10 CxBxA AxCxB BxAxC
BxCxA CxBxA CxAxB
AxCxB BxAxC AxBxC
AxBxC BxCxA CxAxB
CxAxB AxCxB AxBxC
BxAxC CxBxA BxCxA
AxCxB BxAxC CxBxA
BxAxC AxBxC BxCxA
CxBxA CxAxB AxCxB
Solution 11 CxBxA AxCxB BxAxC
BxCxA BxAxC AxBxC
AxCxB CxBxA CxAxB
CxAxB AxBxC BxCxA
AxBxC CxAxB AxCxB
BxCxA BxAxC CxBxA
BxAxC CxBxA AxCxB
CxAxB BxCxA CxBxA
AxBxC AxCxB BxAxC
Solution 12 CxBxA AxCxB BxAxC
BxCxA BxAxC AxBxC
AxCxB CxBxA CxAxB
CxAxB AxBxC AxCxB
AxBxC CxAxB BxCxA
BxCxA BxAxC CxBxA
BxAxC BxCxA CxBxA
CxAxB CxBxA AxCxB
AxBxC AxCxB BxAxC
Solution 13 CxBxA AxCxB BxAxC
BxCxA BxAxC AxBxC
AxCxB CxAxB CxBxA
CxAxB AxBxC AxCxB
AxBxC BxCxA CxAxB
BxCxA CxBxA BxAxC
BxAxC CxBxA BxCxA
CxAxB AxCxB CxBxA
AxBxC BxAxC AxCxB
Solution 14 CxBxA AxCxB BxAxC
BxCxA BxAxC AxBxC
AxCxB CxAxB CxBxA
BxAxC AxBxC AxCxB
AxCxB CxBxA CxAxB
CxBxA BxCxA BxAxC
CxAxB CxBxA BxCxA
BxAxC AxCxB CxBxA
AxBxC BxAxC AxCxB
Solution 15 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
CxAxB AxBxC BxCxA
AxBxC BxAxC AxCxB
BxCxA CxBxA CxAxB
BxAxC AxCxB AxBxC
CxAxB BxCxA CxBxA
AxBxC CxAxB BxCxA
Solution 16 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
CxAxB AxBxC BxCxA
AxBxC BxAxC AxCxB
BxCxA CxAxB CxBxA
BxAxC AxCxB AxBxC
CxAxB CxBxA BxCxA
AxBxC BxCxA CxAxB
Solution 17 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
CxAxB AxCxB AxBxC
AxBxC BxAxC BxCxA
BxCxA CxAxB CxBxA
BxAxC AxBxC BxCxA
CxAxB CxBxA AxCxB
AxBxC BxCxA CxAxB
Solution 18 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
BxAxC AxCxB AxBxC
CxAxB BxCxA CxBxA
AxBxC CxAxB BxCxA
CxAxB AxBxC BxCxA
AxBxC BxAxC AxCxB
BxCxA CxBxA CxAxB
Solution 19 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
AxCxB BxAxC CxBxA
BxAxC AxBxC BxCxA
CxBxA CxAxB AxCxB
AxBxC AxCxB BxAxC
CxAxB CxBxA AxCxB
BxAxC BxCxA CxBxA
Solution 20 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
AxCxB BxAxC CxBxA
BxAxC AxBxC AxCxB
CxBxA CxAxB BxCxA
AxBxC AxCxB BxAxC
CxAxB BxCxA CxBxA
BxAxC CxBxA AxCxB
Solution 21 CxBxA BxCxA CxAxB
BxCxA CxAxB AxBxC
AxCxB AxBxC BxAxC
AxBxC AxCxB BxAxC
BxAxC BxCxA CxBxA
CxBxA CxAxB AxCxB
AxCxB BxAxC CxBxA
CxAxB AxBxC AxCxB
BxAxC CxBxA BxCxA

Higher dimensions

Solution to the 2d puzzle

A two-dimensional analogue of the puzzle asks to pack four identical rectangles of side lengths x and y into a square of side length x + y; as the figure shows, this is always possible. In d dimensions the puzzle asks to pack dd identical blocks into a hypercube. By a result of Raphael M. Robinson this is again solvable whenever d = d1 × d2 for two numbers d1 and d2 such that the d1- and d2-dimensional cases are themselves solvable. For instance, according to this result, it is solvable for dimensions 4, 6, 8, 9, and other 3-smooth numbers. In all dimensions, the inequality of arithmetic and geometric means shows that the volume of the pieces is less than the volume of the hypercube into which they should be packed. However, it is unknown whether the puzzle can be solved in five dimensions, or in higher prime number dimensions.[2][5]

References

  1. Rausch, John, "Put-Together – Hoffman's Packing Puzzle", Puzzle World, archived from the original on 2019-11-17, retrieved 2019-11-16
  2. 1 2 3 4 5 6 7 Hoffman, D. G. (1981), "Packing problems and inequalities", in Klarner, David A. (ed.), The Mathematical Gardner, Springer, pp. 212–225, doi:10.1007/978-1-4684-6686-7_19
  3. 1 2 Alsina, Claudi; Nelsen, Roger B. (2015), "Problem 3.10", A Mathematical Space Odyssey: Solid Geometry in the 21st Century, Dolciani Mathematical Expositions, vol. 50, Mathematical Association of America, p. 63, ISBN 9780883853580
  4. 1 2 Berlekamp, Elwyn R.; Conway, John H.; Guy, Richard K. (2004), Winning Ways for Your Mathematical Plays, vol. IV, A K Peters, pp. 913–915
  5. von Holck, Nikolaj Ingemann (January 2018), An Experimental Approach to Packing Problems (PDF), Bachelor's thesis, supervised by Søren Eilers, University of Copenhagen, archived (PDF) from the original on 2019-11-17, retrieved 2019-11-17
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