Timeline for Variant of lion and 100 zebras
Current License: CC BY-SA 3.0
40 events
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| Mar 11, 2015 at 18:40 | history | rollback | Abr001am |
Rollback to Revision 17
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| Mar 11, 2015 at 18:39 | history | rollback | Abr001am |
Rollback to Revision 16
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| Mar 11, 2015 at 18:39 | history | rollback | Abr001am |
Rollback to Revision 15
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| Mar 11, 2015 at 18:24 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 10, 2015 at 16:59 | comment | added | Abr001am | but i cant hide many flaws which my code could contain , like immediacy of reaction . such reaction is a fruit of many variants including the narrowest and largest gap between two lions , distance and position of each lion . this is what is called heuristics in artificial intelligence and game theory | |
| Mar 10, 2015 at 16:55 | comment | added | Abr001am | As i have said before a sucessfull encirlement is when the group of lions form a radius=16 u circle arround a zebra which i couldnt do using my simulation cuz zebra is always avoiding lion's center of gathering | |
| Mar 10, 2015 at 11:39 | comment | added | Veedrac | I don't think I agree your conclusion is particularly strong, but the simulation is definitely helpful! | |
| Mar 10, 2015 at 10:40 | comment | added | Abr001am | yes @Veedrac its hard to interpret lions behavior when they perform stochastic moves , but i have considered consistent demarche toward that zebra . it would be too hard when lions act slyly | |
| Mar 10, 2015 at 10:32 | history | edited | Abr001am | CC BY-SA 3.0 |
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| S Mar 10, 2015 at 0:37 | history | suggested | Veedrac | CC BY-SA 3.0 |
Tried to make the language more idiomatic; I was having some difficulty reading it before.
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| Mar 9, 2015 at 23:17 | review | Suggested edits | |||
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| Mar 9, 2015 at 22:48 | comment | added | Veedrac | @Agawa001 Also, I don't think you can assume that the lions all move uniformly inwards. They might do better if they some make more specific moves. Also, the strategy may be heavily dependent on trapping more than one zebra. | |
| Mar 9, 2015 at 22:45 | comment | added | Veedrac | @Agawa001 "When a zebra is doomed to be caught , rayon equals $\frac{(k-1)}{k}r$" → Why? | |
| Mar 9, 2015 at 22:34 | comment | added | Veedrac | @Agawa001 What is a rayon? Do you mean radius? | |
| Mar 9, 2015 at 14:07 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 9, 2015 at 14:05 | comment | added | Abr001am | is it clearer now? | |
| Mar 9, 2015 at 13:33 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 9, 2015 at 11:58 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 7, 2015 at 21:35 | comment | added | Abr001am | @Veedrac im already trying , thanks for encouragement , but i think this condition leads to advantageous results since zebras are randomly placed that helps the escaping vector V to be continuously variant and not nil | |
| Mar 7, 2015 at 21:23 | comment | added | Veedrac | @Abidare001 This property seems to be invariant of distance; if the zebra is centred inside a triangle then it gets caught. It would help if you gave a version with three lions. | |
| Mar 7, 2015 at 20:44 | comment | added | Abr001am | well of course u would be able to catch a zebra when you position it few steps from a lions claws , i have said before zebras must be divergent enough | |
| Mar 7, 2015 at 20:37 | comment | added | Veedrac | @Abidare001 I don't really get your proof. What invariant are you using to show this? Using your simulation and treating one of the zebras as a lion (since each zebra doesn't distinguish between who is where) I am able to capture the other zebra if it is surrounded by a triangle. It seems this would disprove your claim, but I don't get the claim well enough to say. | |
| Mar 7, 2015 at 20:26 | comment | added | Abr001am | @Veedrac maybe they can have a zebra between but not sufficiently near to be caught. | |
| Mar 7, 2015 at 20:01 | comment | added | Veedrac | "this solution can prevent any zebra to be stranded by two lions both sides" → What exactly do you mean by this? It is provably possible for three lions to form a triangle with which a single zebra is contained. | |
| Mar 7, 2015 at 13:00 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 7, 2015 at 11:26 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 23:58 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 23:16 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 14:15 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 14:10 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 13:59 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 13:25 | comment | added | Abr001am | @Ryan Durrant thx for noticing it , iv changed the content | |
| Mar 6, 2015 at 13:23 | history | undeleted | Abr001am | ||
| Mar 6, 2015 at 13:23 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 12:44 | history | deleted | Abr001am | via Vote | |
| Mar 6, 2015 at 12:15 | history | undeleted | Abr001am | ||
| Mar 6, 2015 at 12:15 | history | edited | Abr001am | CC BY-SA 3.0 |
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| Mar 6, 2015 at 11:32 | history | deleted | Abr001am | via Vote | |
| Mar 6, 2015 at 11:14 | comment | added | Ryan Durrant | But the lion in the second column could chase his zebra for a number of moves, until he is significantly below the zebra in column 4. If the zebras then stick to their strategy, and assume that the one closest to a zebra moves first, then the lion from col 2 could then move across into col 4 and turn around and it would then be really easy. The additional lions break this solution for the zebras and it actually becomes one of the easiest for the lion. They could last a while moving side to side but this would allow the lions to move alternately and trap the zebra. | |
| Mar 6, 2015 at 11:05 | history | answered | Abr001am | CC BY-SA 3.0 |