Author

Topic: DefaultTrust changes - page 116. (Read 85473 times)

legendary
Activity: 2674
Merit: 2965
Terminated.
January 14, 2019, 11:26:18 AM
I'm honored to be DT1 too! I'll re-check my trust list again.

I'd like to propose a change to discourage DT1 members from adding users to DT2 just to improve their own trust:
-snip-
That doesn't discourage it, it completely eliminates it!

DT1 has been reconstructed with all of the criteria.
Does this account for exclusions?
I don't think exclusion behavior has changed.
legendary
Activity: 3654
Merit: 8909
https://bpip.org
January 14, 2019, 11:25:34 AM
DT1 has been reconstructed with all of the criteria.

Does this account for exclusions?
legendary
Activity: 1876
Merit: 1475
January 14, 2019, 11:24:55 AM
I'm honored to be DT1 too! I'll re-check my trust list again.

I'd like to propose a change to discourage DT1 members from adding users to DT2 just to improve their own trust:
If a DT2 member is added only by one DT1 user and the former left positive trust to the later then it should not appear as trusted on the DT1's profile.

So in practice the DT1's default trust should not be affected by changes made by that DT1 user to their trust list.
legendary
Activity: 1511
Merit: 1072
quack
January 14, 2019, 11:24:16 AM
Try to not take that DT position too heavily.

You can't truly judge people properly based on a couple of messages in the Internet. I am not too fond of placing strong ratings, positive or negative, especially based on nothing but messages. Something like person's judgement or trustworthiness cannot be properly evaluated based on a couple of messages here and there. People may talk on different context levels or have other kind of communication problems, etc., misunderstandings happen all the time. Cultural differences, language barriers, different values, different moods... etc..
legendary
Activity: 2674
Merit: 2965
Terminated.
January 14, 2019, 11:18:47 AM
Code:
achow101
actmyname
asche
BitcoinPenny
Blazed
coinlocket$
Coolcryptovator
cryptodevil
Cyrus
DarkStar_
dooglus
EcuaMobi
gmaxwell
greenplastic
Halab
Hhampuz
hilariousandco
iasenko
ibminer
ICOEthics
Jet Cash
JohnUser
krogothmanhattan
Lafu
Lauda
Lesbian Cow
LoyceV
marlboroza
minerjones
Mitchell
monkeynuts
mprep
OgNasty
owlcatz
qwk
SebastianJu
suchmoon
TheFuzzStone
The Pharmacist
theymos
TMAN
tmfp
TookDk
vizique
Vod
Welsh
xtraelv
yahoo62278
yogg
zazarb

I plan to do it again early every month from now on, with adjustments as needed. But if we assume that this general system continues forever, the long-term goal is to keep the critera basically stable and then let the list actually grow over time: the goal is not to constantly select the "top 50 forum users" or similar, but to select a wide swath of users who are basically acceptable.
Excellent work although there are some names that make no sense (e.g. this guy). Time for Cult of Lauda[1] to do your bidding master.

[1] This isn't even a joke anymore.
legendary
Activity: 2240
Merit: 3150
₿uy / $ell ..oeleo ;(
January 14, 2019, 11:16:04 AM
I'm honoured and surprised that I have become a DT1. Thank you for the promotion.

You are not the only one surprised here, now I ended up on DT1 if that's not a mistake, of course.
This is more than unexpected. I'll be very careful with my further trust/feedback judgements.
legendary
Activity: 2814
Merit: 2472
https://JetCash.com
January 14, 2019, 10:59:46 AM
I'm honoured and surprised that I have become a DT1. Thank you for the promotion.
administrator
Activity: 5222
Merit: 13032
January 14, 2019, 10:58:51 AM
M (number of V vertices) is not given right?

M is the number of vertices in the input graph, and is known. The number that should be in the optimal output is not known.

Typo on your end in the 2nd optimal solution? I figure you meant either v1 or v3 for one of those vectors?

Right, thanks.
administrator
Activity: 5222
Merit: 13032
January 14, 2019, 10:52:43 AM
DT1 has been reconstructed with all of the criteria. Old:
Code:
achow101
actmyname
BitcoinPenny
Blazed
Cyrus
dooglus
gmaxwell
greenplastic
HostFat
JohnUser
krogothmanhattan
Lauda
Mitchell
monkeynuts
OgNasty
owlcatz
qwk
SebastianJu
suchmoon
The Pharmacist
theymos
TMAN
zazarb
New:
Code:
achow101
actmyname
asche
BitcoinPenny
Blazed
coinlocket$
Coolcryptovator
cryptodevil
Cyrus
DarkStar_
dooglus
EcuaMobi
gmaxwell
greenplastic
Halab
Hhampuz
hilariousandco
iasenko
ibminer
ICOEthics
Jet Cash
JohnUser
krogothmanhattan
Lafu
Lauda
Lesbian Cow
LoyceV
marlboroza
minerjones
Mitchell
monkeynuts
mprep
OgNasty
owlcatz
qwk
SebastianJu
suchmoon
TheFuzzStone
The Pharmacist
theymos
TMAN
tmfp
TookDk
vizique
Vod
Welsh
xtraelv
yahoo62278
yogg
zazarb

I plan to do it again early every month from now on, with adjustments as needed. But if we assume that this general system continues forever, the long-term goal is to keep the critera basically stable and then let the list actually grow over time: the goal is not to constantly select the "top 50 forum users" or similar, but to select a wide swath of users who are basically acceptable.
legendary
Activity: 3654
Merit: 8909
https://bpip.org
January 14, 2019, 10:19:51 AM
This looks like quite a step backwards in terms of decentralization.

I wonder how it would look like if the voting power was capped at 10 and 2 respectively, e.g.

72 merits - has 7 votes
105 merits - has 10 votes
560 merits - still has only 10 votes

Same with 250s, i.e. anyone between 250-499 merits has 1 vote and anyone >=500 has 2.
sr. member
Activity: 742
Merit: 395
I am alive but in hibernation.
January 14, 2019, 08:52:20 AM
Code:
~snip list~
At least 6 users are banned:
Code:
979216: RuSS512
925926: gawlea
227450: identifyuser
915311: mstfprcn
944905: saulzaents
1331637: zentdex
Some of them are still included, only one still includes another user:
Code:
mstfprcn->Kubra Dam

@theymos: should the custom trust list of a banned user be deleted? If I understand correctly, their existing inclusions still count towards voting.

Look Like LoyceV forget ATMcoin  in list who is also banned.

I do not think satoshi will be coming to vote.

These 2 should be blacklisted from DT eligible list as well as voting too.
Apja94
pitipawn

legendary
Activity: 2240
Merit: 3150
₿uy / $ell ..oeleo ;(
January 14, 2019, 08:43:16 AM
Relatively new users put in their trust list those who have interactions with.
I just started my list and it's far from being ready, so I've added those whom I know well and I feel OK to add them. Looking through the other people's (DTs) lists, I see users I never heard of, or never saw ( or at least remembered with something) so they are quite unknown to me. I'll share some free time to check them, but this is a long term process.

I'm sure I'm not the only one in this situation. Regarding SaltySpitoon I've crossed some of his posts, but I'm not so familiar with the user.
 
legendary
Activity: 3290
Merit: 16489
Thick-Skinned Gang Leader and Golden Feather 2021
January 14, 2019, 08:31:08 AM
I like LoyceV just fine but 6X as many votes as SaltySpitoon? lol no..
If you put it like this, it is weird indeed. For what it's worth, I have SaltySpitoon in my trust list.

Quote
I think the DT1 list would be nice with around 50 members on it anyway..
Just a thought: what if theymos pickes a number for DT1, and adjusts the requirements on the fly in order to reach that amount?

Another thought:
At the moment a relatively new user can be voted into DT1. Isn't it better to require at least some time on DT2 (or even DT3) first, so that others can judge how the user is doing?
legendary
Activity: 2296
Merit: 2262
BTC or BUST
January 14, 2019, 08:18:21 AM
Nothing about them. If you're excluded, you're excluded.

As it was before a person needed more inclusions than exclusions by 10 [or 9? to have (0)].

Now any single 1 exclusion is the final answer? Whose exclusion counts first?


legendary
Activity: 2674
Merit: 2965
Terminated.
January 14, 2019, 08:13:47 AM
I think the DT1 list would be nice with around 50 members on it anyway..
At that point abuse would be somewhat noticeable and keeping track of every DT2 will be difficult.

I like LoyceV just fine but 6X as many votes as SaltySpitoon? lol no..
Is LoyceV 6X more trusted to make good votes than SaltySpitoon?
These votes have nothing to do with how trusted someone is. Stop confusing concepts.

Does the algorithm decide "most useful" before taking into account how many exclusions a member must overcome to stay on DT1 or after?
Exclusions are processed at the very end.
That is no good IMO..
What about all the extra votes someone needs to overcome rivals exclusions?
Nothing about them. If you're excluded, you're excluded.
legendary
Activity: 2296
Merit: 2262
BTC or BUST
January 14, 2019, 08:09:15 AM
This loos like quite a step backwards in terms of decentralization.
Given the previous size of the DT list, it really isn't.

To reduce the size of the DT1 list it could easily have been changed to 30 10 merit votes and 5 100 merit votes or something rather than greatly centralizing voting power.

I get the 10 merit standard, 100 merit standard, even the 250 merit standard, but anything over 500 merits is rediculous and depends too much on where you post, what you post about, and who your friends are..

I like LoyceV just fine but 6X as many votes as SaltySpitoon? lol no..
Is LoyceV 6X more trusted to make good votes than SaltySpitoon?

I think the DT1 list would be nice with around 50 members on it anyway..

Does the algorithm decide "most useful" before taking into account how many exclusions a member must overcome to stay on DT1 or after?
Exclusions are processed at the very end.

That is no good IMO..
What about all the extra votes someone needs to overcome rival exclusions?
legendary
Activity: 2674
Merit: 2965
Terminated.
January 14, 2019, 07:53:24 AM
This loos like quite a step backwards in terms of decentralization.
Given the previous size of the DT list, it really isn't.

Does the algorithm decide "most useful" before taking into account how many exclusions a member must overcome to stay on DT1 or after?
Exclusions are processed at the very end.
legendary
Activity: 2296
Merit: 2262
BTC or BUST
January 14, 2019, 07:45:46 AM
legendary
Activity: 3290
Merit: 16489
Thick-Skinned Gang Leader and Golden Feather 2021
January 14, 2019, 07:29:47 AM
Yes, true, and where the user Veleor?
Piggy uses theymos' Merit data dump, which updates on Friday. Veleor didn't have 250 Merit earned yet last Friday. By now he would have 1 vote.
sr. member
Activity: 938
Merit: 452
Check your coin privilege
January 14, 2019, 07:29:27 AM

G = (U, V, E) is a bipartite graph with edges from the left side, U, to the right side, V. There are N vertices in U and M vertices in V. A "capacity" function c: U -> Z is defined for every vertex in U. A constant integer "target value" T exists.

Candidate solutions are subgraphs S = (US, VS, ES) of G satisfing the following requirements:
 1. For each vertex u in US, the number of edges attached to u must be less than or equal to c(u).
 2. For each vertex v in VS, the number of edges attached to v must be greater than or equal to T.
 
Find an S such that the number of vertices in VS is maximal.

Example:


Note that this graph is depicted as directed, but that doesn't actually matter.

Note:

 - To satisfy requirement #1, you must exclude at least two of (u1, v1) or (u1, v3) or (u1, v4).
 - To satisfy requirement #1, you must exclude at least one of (u2, v1) or (u2, v2).
 - To satisfy requirement #1, you must exclude at least one of (u3, v1), (u3, v2) or (u3, v3)
 - To satisfy requirement #2, you must exclude at least v4 (since it cannot possibly get T=2 edges), and possibly more depending on the rest of S.

In a very naïve greedy algorithm, you might just fill up vertices on the right from top to bottom until you can't do anything else. That'd give a candidate solution of:
(u1, v1)
(u2, v1)
This is a valid candidate solution, but it's non-optimal because it includes only 1 vertex in V whereas 2 are possible. In order to achieve an optimal solution, you need some backtracking, at least. In this case there are two equally-good optimal solutions:
(u1, v3)
(u3, v3)
(u2, v2)
(u3, v2)
or:
(u1, v1)
(u3, v3)
(u2, v2)
(u3, v2)

It becomes more complicated as the graph gets bigger.

Writing it down in this way reminds me a lot of the stable marriage problem, which gives me hope that it can be solved exactly.

Two more questions :

M (number of V vertices) is not given right? We have the number of u vertices, their capacity, and T the constant, and our solution is finding the vectors (u,v) that satisfy the requirements with the maximum number possible of v vertices?

Typo on your end in the 2nd optimal solution? I figure you meant either v1 or v3 for one of those vectors?

----------

And where the user chimk?

Yes, true, and where the user Veleor?

Can you stop?
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