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Statement

Theorem 7 (p. 13). "There exists a $2nm$-fold colouring with $2\lceil(b+1)2n\rceil\lceil(b+1)\frac{2m}{3}\rceil$ colours of the graph $G_{[1,b]}$ i.e. $\frac{\chi_{2nm}(G_{[1,b]})}{2nm}\le\frac{2\lceil\sqrt3(b+1)\frac{2n}{3}\rceil\lceil(b+1)\frac{2m}{3}\rceil}{2nm}$ [sic]."

The two expressions in the statement differ: the first factor is $\lceil(b+1)2n\rceil$ in the colour count and $\lceil\sqrt3(b+1)\frac{2n}{3}\rceil$ in the ratio. The proof (pp. 14-15) ends with $2\lceil(b+1)\frac{2m}{3}\rceil\cdot\lceil\sqrt3(b+1)\frac{2n}{\sqrt3}\rceil$ colours, which equals the first count, and Table 2 (p. 15) agrees with the first count (for $b=1$, $n=m=1$ it lists $16$ colours, where the ratio's form would give $12$). This page reads the theorem as the first count, the ratio's $\frac{2n}{3}$ standing for $\frac{2n}{\sqrt3}$; this is a filing observation, not a review verdict.

The statement does not quantify $n$, $m$ or $b$; the proof treats $n$ and $m$ as positive integers, and section 2 (p. 5) reduces the graphs studied to $G_{[1,b]}$ with $b\ge1$. Here $G_{[1,b]}$ joins two points of the plane whose distance lies in $[1,b]$, and $\chi_j$ is the $j$-fold chromatic number (Definition 2, p. 4).

For $G_{[1,1]}$ the paper tabulates (Table 2, p. 15) $k=16$, $24$, $32$, $48$, $56$, $64$ colours for $j=2nm=2$, $4$, $6$, $8$, $10$, $12$, and says (p. 16) that for $G_{[1,2]}$ the method of Theorem 7 does not give good results.

Source. J. Grytczuk, K. Junosza-Szaniawski, J. Sokół, K. Węsek, Fractional and $j$-fold coloring of the plane, Discrete Comput. Geom. 55 (2016), 594-609, doi:10.1007/s00454-016-9769-3; read in arXiv:1506.01887v2 (5 October 2015), Theorem 7 on p. 13 and its proof on pp. 13-15 of that version. The source card records the edition.

Read depth. Claims checked: the statement was read clause by clause on the print, and every Table 2 count for the $2nm$ method was recomputed from the first count. The proof was read for structure only.

Proof pointer

pp. 13-15. Start from the tiling by hexagons of side $1/2$ and form $nm$ translated grids, shifted down a column by multiples of $\frac1m[0,-3/2]$ and along a row by multiples of $\frac1n[\sqrt3/2,0]$. The next same-colour hexagon in a column lies $\lceil(b+1)\frac{2m}{3}\rceil$ rows away, at centre distance at least $b+1$, and in a row $\lceil(b+1)2n\rceil$ hexagons away, at centre distance at least $\sqrt3(b+1)$; these two are then at least $2b+2$ apart, leaving room for a second family of $nm$ grids shifted into the gaps, which doubles both the fold number and the row count.

Dependencies

The method combines the two-layer construction of Theorem 4 with that of Theorem 6 (p. 13); it does not use either as a lemma.

Bears on

• Problem 508: the problem asks for the chromatic number of $G_{[1,1]}$. At $b=1$ Theorem 7 bounds $2nm$-fold chromatic numbers of $G_{[1,1]}$ from above, at best the ratio $32/6\approx5.33$ in the paper's Table 2; it gives no bound on the chromatic number itself.