# Steinberg's conjecture (Solved)

**Conjecture**Every planar graph without 4-cycles and 5-cycles is 3-colourable.

Borodin et al. [BGRS] who proved that every planar graph without cycles of length in is 3-colourable.

Borodin and Raspaud [BR] proposed the following conjecture which implies Steinberg's Conjecture.

**Conjecture (Strong Bordeaux Conjecture)**Every planar graph without 5-cycles and without adjacent triangles is 3-colourable.

This conjecture is in turn implied by the following stronger one due to Borodin et al. [BGJR]

**Conjecture**Every planar graph without 3-cycles adjacent to cycles of length 3 or 5 is 3-colourable.

Borodin et al. [BGJR06] proved that every planar graph without triangles adjacent to cycles of length from 3 to 9 is 3-colourable.

Steinberg's Conjecture cannot been generalized to list colouring: Voigt [V] constructed a planar graph without 4-cycles and 5-cycles which is not 3-choosable.

## Bibliography

[BGJR] O.V. Borodin, A.N. Glebov, T.R. Jensen, and A.Raspaud. Planar graphs without triangles adjacent to cycles of length from 3 to 9 are 3-colorable. Sib. Elektron. Mat. Izv., 3:428--440, 2006.

[BGRS] O. V. Borodin, A. N. Glebov, A. Raspaud, and M. R. Salavatipour. Planar graphs without cycles of length from 4 to 7 are 3-colorable. J. Combin. Theory Ser. B, 93(2):303--311, 2005.

[BR] O. V. Borodin and A. Raspaud. A sufficient condition for planar graphs to be 3-colorable. J. Combin. Theory Ser. B, 88(1):17--27, 2003.

[V] M. Voigt. A non-3-choosable planar graph without cycles of length 4 and 5. Discrete Math., 307(7-8):1013--1015, 2007.

* indicates original appearance(s) of problem.

## Steinberg Conjecture disproved

All of the conjectures mentioned here were disproved by Vincent Cohen-Addad, Michael Hebdige, Daniel Kral, Zhentao Li and Esteban Salgado. See https://arxiv.org/abs/1604.05108 .