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Encouraged by the positive solutions to my question, Tiling with one of each shape, I'd like to pose the $\mathbb{R}^3$ equivalent:

Q. Is there a tiling of $\mathbb{R}^3$ by (bounded) polyhedra, one each of polyhedra of $4$ faces, $5$ faces, $\ldots$ ?

The two main versions of Q are (1) allowing nonconvex polyhedra, or (2) insisting on convex polyhedra. (Another variant counts vertices rather than faces.) For (1), Noam Elkies' 2D idea may apply, say using a tiling by congruent cubes, and partitioning each cube. For (2), it may be that Martin Tancer's idea can extend to a spherical design rather than his 2D circular construction.

If both variants are resolved, it would then be natural to generalize to $\mathbb{R}^d$ for $d > 3$.

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    $\begingroup$ Martin Tancer's construction works for any integer sequence, not just 3,4,5,6 etc. So one approach is to just thicken it to a 1x∞x∞ layer and stack such layers to fill 3-space. A tetrahderon cannot be a prism but we can start with a triangular prism and dissect it to a 4- and 5-hedron. $\endgroup$
    – Noam D. Elkies
    Commented Nov 20 at 23:52
  • $\begingroup$ @NoamD.Elkies: Thanks! But I should stipulate that each tile should be a bounded poyhedron. Will add that now. (Or: I might be misinterpreting $1 \times \infty \times \infty$.) $\endgroup$
    – Joseph O'Rourke
    Commented Nov 21 at 0:28
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    $\begingroup$ Partition {5,6,7,8,...} into countably many infinite subsets S_i. For each one, make a Tancer tiling of the plane by polygons of orders {n-2 : n in S_i}. Replace each n-gon with a width-1 prism. That gives a tiling of a 1x∞x∞ layer by polyhedra with face counts in S_i. Stack them to get a tiling of R^3 by prisms of width 1 which are polyhedra with each face count other than 4 appearing once. One of them is a triangular prism. Cut off a corner (=tetrahedron), leaving a pentahedron. Q.E.F. $\endgroup$
    – Noam D. Elkies
    Commented Nov 21 at 1:44
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    $\begingroup$ @NoamD.Elkies: Thanks for clarifying (I was misinterpreting "1x∞x∞".) Very nice construction! $\endgroup$
    – Joseph O'Rourke
    Commented Nov 21 at 1:53

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