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Nov 15, 2023 at 15:56 comment added Mark Chimes Thank you, that works!
Nov 14, 2023 at 21:56 comment added Florian Lehner Let's denote the vertices of the Wagner graph by $0,\dots, 7$ with edges from $i$ to $i+1 \mod 8$ and from $i$ to $i+4 \mod 8$. If i didn't overlook something, then $(0,1), (1,2), \dots, (6,7),(7,0)$ together with $(0,4)$ and $(1,5)$ form a bramble. The only way to hit the first 8 edges with 4 vertices is to take all odd or all even numbers, but then we'd miss at least one of the last two. Hence a minimal hitting set must contain at least 5 vertices.
Nov 14, 2023 at 18:00 history edited YCor CC BY-SA 4.0
removed capitals from title
S Nov 14, 2023 at 17:18 review First questions
Nov 14, 2023 at 18:05
S Nov 14, 2023 at 17:18 history asked Mark Chimes CC BY-SA 4.0