Timeline for Generalizations of the four-color theorem
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2017 at 3:04 | review | First posts | |||
| Jan 23, 2017 at 4:10 | |||||
| Jan 10, 2017 at 12:30 | comment | added | patrick labarque | Yes, 4CT is equivalent to finding for any two given TPPs with v vertices a proper boundary edge-coloring which works for both. Each polygon with a fixed base color has 2^(v-2) different edge-colorings. The number of boundary edges (included the base) with the same color has the same parity as v. | |
| Jan 10, 2017 at 11:42 | comment | added | მამუკა ჯიბლაძე | Sorry I cannot get it again. What is the final statement? 4CT is equivalent to finding for any two given TPPs a proper boundary edge-coloring which works for both? You mean polygons with the same number of boundary edges? | |
| Jan 10, 2017 at 11:42 | history | edited | patrick labarque | CC BY-SA 3.0 |
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| Jan 10, 2017 at 11:34 | comment | added | patrick labarque | Three-edge-coloring of hamiltonian maximal planar graphs (hMPG) is equivalent to the 4CT. Then recombination of all triangulated planar polygons (TPP) gives all hMPG (with many doubles, also a lot with double edges, but this does n't matter). And if any TPP has a proper boundary edge-coloring in common with any other one then this is also equivalent to the 4CT. | |
| Jan 10, 2017 at 11:23 | comment | added | მამუკა ჯიბლაძე | Thank you! Not that I understand everything but it is certainly more than before :D I've made it visible, if you don't mind (if you do, just tell me and I will revert it back) | |
| Jan 10, 2017 at 11:22 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
made image visible
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| Jan 10, 2017 at 11:18 | history | edited | patrick labarque | CC BY-SA 3.0 |
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| Jan 10, 2017 at 11:16 | comment | added | patrick labarque | See the image I added (but can't see it). | |
| Jan 10, 2017 at 11:13 | history | edited | patrick labarque | CC BY-SA 3.0 |
added 81 characters in body
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| Jan 9, 2017 at 21:42 | comment | added | მამუკა ჯიბლაძე | Sorry too convoluted for me. Could not you add at least a sketchy explanation of why exactly is this equivalent to the four color problem? | |
| Jan 9, 2017 at 8:49 | comment | added | patrick labarque | Add a 0 to the right of every number in every table. Then each table gives the different edge colorings (dual Tait coloring) for the boundary of a triangulated planar polygon with color 0 as fixed base color. The first table is for a triangle and an ear is added with each operation. Combination of two idem colored polygons gives a proper edge colored hamiltonian maximal planar graph. A much stronger generalization is the conjecture in mathoverflow.net/questions/258830/… | |
| Jan 8, 2017 at 8:55 | comment | added | მამუკა ჯიბლაძე | Fantastic! Do you have a reference to read more about this? | |
| Jan 8, 2017 at 8:31 | history | edited | patrick labarque | CC BY-SA 3.0 |
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| Jan 7, 2017 at 22:55 | review | Late answers | |||
| Jan 7, 2017 at 23:06 | |||||
| S Jan 7, 2017 at 22:39 | history | answered | patrick labarque | CC BY-SA 3.0 | |
| S Jan 7, 2017 at 22:39 | history | made wiki | Post Made Community Wiki by patrick labarque |