Homework four quick answers
Brief Answers
1. 2.1 - 2 (a) - any odd n; (b) K2 (c) both r,s even
2. 2.1 - 4 lift twice - 6 odd degree vertices. connect 2 of them, then lift (1);
then connect 2 more, then lift (2); then connect the last two.
3. 2.1 - 10 Not possible: cells on board are vertices,
edges to those other cells where a knight can move. More
than 2 have odd degree.
4. 2.1 - 16 If all vertices have even degree, then
each line in the line graph has even degree. K4 has
no Euler cycle, but its line graph does.
5. 2.2 - 4 (d,e) d can’t work; nor can e.
6. 2.2 - 16 looking for a directed HAM circuit in the
graph where seat positions are nodes. Circuit has 25
vertices – so either north / south or east / west
pairs has a different count, and thus you can’t
return to start.
7. 2.3 - 6 a) 3-color the triangles, starting at a.
Then the a-f arc should create no problems. This is
equitable. 1-c is obviously 2-colorable and
equitable
8. 2.3 - 10 a-draw the n+1st circle. Inside it’s
internally consistent and properly colored, by
induction, but it’s wrong at every boundary. Flip
all colors inside this new circle. B- use color A
for all regions with even overlap, color B for odd
overlap. Must have different colors at every
boundary, since each boundary changes by 1 the
overlap count.
9. 2.3 - 12 verts = animals; edges = animals that
can’t get along; a color = open space (animals in
that open space get same color).
10. 2.4 - 2 Must be 7 regions. If all are bounded by 4
or more edges, then since 2E = region degree sum,
there must be 14 edges. But there are only 13, so
there must be a triangle, and thus a 3-cycle.
11. 2.4 - 6 a. color classes are independent sets, and
they cover the graph; b. If min degree is d, then no
independent set can contain more than n-d vertices.
12. 2.4 - 14 Do the standard proof, by induction: color
every vertex but the deg 4 vertex using 4 colors, then
color that last vertex using the argument in the book
(the big blob lemma).
--
RobbieMoll - 05 Oct 2006