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Coloring
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Flight Gates
flights need gates, but times overlap. how many gates needed?
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Airline Schedule
122145 67257306 99
Flights
time
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Conflicts Among 3 Flights
99
145
306
Needs gate at same time
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Model all Conflicts with a Graph
257
67
99
145
306
122
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Color vertices so that adjacent
vertices have different colors.
min # distinct colors needed =
min # gates needed
Color the vertices
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Coloring the Vertices
257, 67122,14599306
4 colors4 gates
assigngates:
257
67
99
145
306
122
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Better coloring
3 colors3 gates
257
67
99
145
306
122
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Final Exams
Courses conflict if student takes both, so need different time slots.
How short an exam period?
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Harvard’s SolutionDifferent “exam group” for every teaching hour. Exams for different groups at different times.
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But This May be Suboptimal
• Suppose course A and course B meet at different times
• If no student in course A is also in course B, then their exams could be simultaneous
• Maybe exam period can be compressed!
• (Assuming no simultaneous enrollment)
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Model as a Graph
CS 20
Psych 1201
Celtic 101
Music 127r
AM 21b
M 9amM 2pmT 9amT 2pm4 time slots
(best possible)
A B
Means A and B have at least one student in common
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Map Coloring
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Planar Four Coloring
any planar map is 4-colorable.1850’s: false proof published (was correct for 5 colors).
1970’s: proof with computer1990’s: much improved
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Chromatic Number
min #colors for G is
chromatic number, χ(G)
lemma:
χ(tree) = 23/16/12
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Pick any vertex as “root.”if (unique) path from root iseven length: odd length:
Trees are 2-colorable
root
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Simple Cycles
χ(Ceven) = 2
χ(Codd) = 3
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Bounded Degree
all degrees ≤ k, implies
very simple algorithm…
χ(G) ≤ k+1
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“Greedy” Coloring…color vertices in any order. next vertex gets a color different from its neighbors. ≤ k neighbors, so k+1 colors always work3/16/12
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coloring arbitrary graphs
2-colorable? --easy to check3-colorable? --hard to check (even if planar)find χ(G)? --theoretically no harder than 3-color, but harder in practice
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Finis
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