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# Find the chromatic number of the surface ofMn

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Problem 11208

(American Mathematical Monthly, Vol.113, March 2006) Proposed by Li Zhou (USA).

The stage-n Menger sponge Mn is generated recursively, starting with the unit cube M0. To drill a cube is to partition it into 27 congruent subcubes and remove the closed central cube along with the 6 remaining subcubes sharing a face with that cube, resulting in a solid that is the union of 20 congruent cubes. GivenMn−1, constructMn by drilling each of the20n−1subcubes of edge-length 31n inMn−1. Find the chromatic number of the surface ofMn.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We first find the number of vertices v(n), sides s(n), and faces f (n) of the tassellation given by the recursive construction of the surface Mn. Since M0 is the unit cube then v(0) = 8, s(0) = 12, and f (0) = 6. Morever, following the construction, for n ≥ 1

v(n) = v(n − 1) + 2s(n − 1) + 4f (n − 1) + 8 · 20n−1 s(n) = 3s(n − 1) + 12f (n − 1) + 36 · 20n−1

f (n) = 8f (n − 1) + 24 · 20n−1

In particular f (n) = 4 · 8n+ 2 · 20n and therefore the Euler characteristic is

χ(n) = v(n) − s(n) + f (n) = χ(n − 1) − f (n − 1) − 4 · 20n−1= χ(n − 1) − 4 · 8n−6 · 20n. Since the genus g(n) = 1 − χ(n)/2 and g(0) = 0 then

g(n) = g(n − 1) + 2 · 8n+ 3 · 20n= 2 · (8n−1)/7 + 3 · (20n−1)/19.

The first terms of the sequence g(n) are: 0, 5, 81, 1409, 26433.

Finally by Heawood’s formula (note that Mn is not a Klein bottle because Mnis orientable) we are able to compute the chromatic number of Mn:

γ(n) = 1 2

7 +p48g(n) + 1 .

The first terms of the sequence γ(n) are: 4, 11, 34, 133, 566.  Remark. Note that Heawood’s formula was proved by Ringel and Youngs (1968) with two exceptions:

the sphere, and the Klein bottle. When the four-color theorem was proved in 1976 then the formula was confirmed also for the sphere. As regards the only exception left, the Klein bottle, Heawood’s formula gives 7 but the correct is 6 as demonstrated by the Franklin graph. See for example Four Colors Suffice : How the Map Problem Was Solvedby Robin Wilson.

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