Timeline for Bound on queries to a tree with unusual probabilties -- follow-up
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2017 at 12:39 | vote | accept | Michael Jarret | ||
| Sep 7, 2017 at 9:52 | answer | added | fedja | timeline score: 1 | |
| Sep 7, 2017 at 1:06 | history | edited | Michael Jarret | CC BY-SA 3.0 |
edited body
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| Sep 7, 2017 at 0:51 | comment | added | Michael Jarret | Let us continue this discussion in chat. | |
| Sep 7, 2017 at 0:45 | history | edited | Michael Jarret | CC BY-SA 3.0 |
eliminated condition 4
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| Sep 7, 2017 at 0:43 | comment | added | Michael Jarret | @fedja very good point, yet again. I had to reference my notes on the original problem. It was a condition on another function. You are correct that it should not be a condition. | |
| Sep 7, 2017 at 0:36 | comment | added | fedja | Alas, you still declare equality of all sums for $b=r$ in $(6)$. Are you sure you want to keep $(4)$? The problem gets much neater without it with just $(1,2,3,5)$: the weight is determined uniquely and just conditioned upon the remaining sub-tree every time, and several other features get way more attractive. From purely aesthetic point of view, $(4)$ is a total disaster. Of course, if it holds in the model, nothing can be done about it, but does it? | |
| Sep 6, 2017 at 22:44 | history | edited | Michael Jarret | CC BY-SA 3.0 |
corrected condition again
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| Sep 6, 2017 at 22:42 | comment | added | Michael Jarret | @fedja you are correct as always. I will correct it. | |
| Sep 6, 2017 at 22:40 | comment | added | fedja | Erm... That completely destroys the phrase "By (5), this is independent of our choice of $l0\in L(r)$." so something is still fishy as written... | |
| Sep 6, 2017 at 22:30 | history | edited | Michael Jarret | CC BY-SA 3.0 |
clarified condition (5)
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| Sep 6, 2017 at 22:29 | comment | added | Michael Jarret | @fedja I should have included $b\neq r$ in the sums of condition (5), which should avoid your mentioned problem. Thanks for pointing this out. | |
| Sep 6, 2017 at 22:21 | comment | added | fedja | Interesting... It looks like the conditions $(1)-(5)$ actually determine $k_r$ uniquely, don't they? Moreover they may become incompatible for certain trees (consider the tree with just two path branches, one of length 2 and another of length 1: L0-V-R-L1. Are you sure that nothing is screwed up now? | |
| Sep 6, 2017 at 13:51 | history | asked | Michael Jarret | CC BY-SA 3.0 |