Planar Graphs - Numberphile
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Date: 2022-04-09
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Comments and reviews: 9
BrunoJMR
Relating to the adjacent countries only needing 4 colours. Imagine there was a country that was a circle, and arround it were 10 countries adjacent to that countrie's borders. If we only use 4 colours, then the circle country will have to share colour with at least 2 of those countries, thus breaking the rule that no 2 adjacent countries should have the same colour, right?
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Relating to the adjacent countries only needing 4 colours. Imagine there was a country that was a circle, and arround it were 10 countries adjacent to that countrie's borders. If we only use 4 colours, then the circle country will have to share colour with at least 2 of those countries, thus breaking the rule that no 2 adjacent countries should have the same colour, right?
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totlyepic
Nitpick just so students watching this don't get lost looking at some other source:
around 12: 50 when it's said that V is the # of verts and E is the # of edges, this is not standard. V is the set of vertices, and E is the set of edges. -V- is then the cardinality (size) of the set V (i. e. the number of vertices) and -E- is the cardinality of the set E.
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Nitpick just so students watching this don't get lost looking at some other source:
around 12: 50 when it's said that V is the # of verts and E is the # of edges, this is not standard. V is the set of vertices, and E is the set of edges. -V- is then the cardinality (size) of the set V (i. e. the number of vertices) and -E- is the cardinality of the set E.
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Tom
I told my son when he was in high school that K_-3, 3- was impossible (specifically, the houses-and-utilities variant. He took it as a challenge, and came up with several solutions, such as running a line through a house to another house, or stacking the houses together into an apartment building. Lesson: you really have to look for unspoken assumptions.
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I told my son when he was in high school that K_-3, 3- was impossible (specifically, the houses-and-utilities variant. He took it as a challenge, and came up with several solutions, such as running a line through a house to another house, or stacking the houses together into an apartment building. Lesson: you really have to look for unspoken assumptions.
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JNCressey
3: 58 -anything you can embed with curved edges, you can with straight edges. -
Well. only if they're simple.
Two edges joining the same pair of vertices or an edge that uses the same vertex for both its ends can't be drawn straight.
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3: 58 -anything you can embed with curved edges, you can with straight edges. -
Well. only if they're simple.
Two edges joining the same pair of vertices or an edge that uses the same vertex for both its ends can't be drawn straight.
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jdub
Now, the K3, 3 graph, I remember this. My Question is: Why only take advantage on 1 side of the plane? If you use Both sides of the plane it is solvable (via a tunnel, or on circuit boards it's called a 'via'. The bridge to terabithia.
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Now, the K3, 3 graph, I remember this. My Question is: Why only take advantage on 1 side of the plane? If you use Both sides of the plane it is solvable (via a tunnel, or on circuit boards it's called a 'via'. The bridge to terabithia.
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asailijhijr
Incidentally, Alaska is a terrible example of a non-contiguous territory in terms of the 4-colour theorem. This is because Alaska has only one land border with a country that has land borders with only one country.
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Incidentally, Alaska is a terrible example of a non-contiguous territory in terms of the 4-colour theorem. This is because Alaska has only one land border with a country that has land borders with only one country.
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simplecoffee
So if you're given a tangled-up graph you know is planar, what's the quickest way to arrive at its planar embedding? Asking for science. Not because it's the basis of a puzzle game I sometimes play or anything. '
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So if you're given a tangled-up graph you know is planar, what's the quickest way to arrive at its planar embedding? Asking for science. Not because it's the basis of a puzzle game I sometimes play or anything. '
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Tiamat
I was working on transition-metal complexes that had a non-planar graph in its base structure a couple of years ago.
My former Prof. really wanted to coin the term Kuratowski-complexes for these kinds of complexes.
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I was working on transition-metal complexes that had a non-planar graph in its base structure a couple of years ago.
My former Prof. really wanted to coin the term Kuratowski-complexes for these kinds of complexes.
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jamespfp
3: 00 -- SO YEAH, this particular graph is NOT PLANAR, -BUT- the number of vertices has a numerical relationship with the number of points where multiple lines cross. So -- are some non-planar graphs fractal?
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3: 00 -- SO YEAH, this particular graph is NOT PLANAR, -BUT- the number of vertices has a numerical relationship with the number of points where multiple lines cross. So -- are some non-planar graphs fractal?
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