Enigmatic Code

Programming Enigma Puzzles

Enigma 1376: Square facts and faces

From New Scientist #2536, 28th January 2006

I have just constructed a simple polyhedron, i.e. a three-dimensional solid figure with some vertices, straight edges, and faces. The number of vertices is a square; the number of edges is a square; the number of faces is a square; the total number of vertices and edges is a square; and over half of the faces are squares! How many faces are there, and how many of those are triangles?



2 responses to “Enigma 1376: Square facts and faces

  1. Jim Randell 3 November 2013 at 8:41 am

    By using the Euler Characteristic of simple polyhedra we can work out possible values for the number of vertices (V), faces (F) and edges (E) of the polyhedron. This Python program looks for polyhedra with up to 100 faces/vertices/edges. It runs in 34ms.

    from itertools import product
    from enigma import irange, printf
    # some squares
    squares = list(i * i for i in irange(2, 15))
    # for a simple polyhedron V + F - E = 2
    for (V, F) in product(squares, repeat=2):
      E = V + F - 2
      # also 2E >= 3F and 3V
      if 3 * max(V, F) > 2 * E: continue
      # E and V + E should be squares
      if not(E in squares and V + E in squares): continue
      printf("E={E} V={V} F={F}")

    The code only finds one candidate tuple: V=9, F=9, E=16.

    Looking at Wikipedia [ http://en.wikipedia.org/wiki/List_of_small_polyhedra_by_vertex_count ] for polyhedra with 9 vertices, only the elongated square pyramid and the octagonal pyramid have V=9, F=9, E=16. And only the elongated square pyramid has any quadrilateral faces, and they can all be made to be square giving 5 square faces out of the 9 faces, so more than half the faces are square, and the remaining 4 faces are triangular.

    Solution: There are 9 faces. 4 of the faces are triangles.

    Enigma 1376 - Solution

  2. Jim Randell 3 November 2013 at 8:46 am

    You can find larger (V, F, E) tuples that satisfy the conditions, by increasing the number of squares considered. The next one is V=2601, F=2025, E=4624, and on rec.puzzles Dean Hickerson showed how to construct a 2025-hedron with 2601 vertices that satisfies the conditions of the puzzle. [ https://groups.google.com/d/msg/rec.puzzles/RaoavExEw18/CwDAyNGzcUUJ ].

    So, technically the puzzle is flawed, as there is not a unique solution. But it can easily be salvaged by putting a reasonable limit on the maximum number of faces allowed in the polyhedron.

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