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[Ilshat Shakirov <im.shakirov@xxxxxxxxx>]

>
> For me your reasonings sound fine. Good job!
>
thanks!

> Could be:
>
> \sum_{j=1}^{A} E(n-j)
>
> where E(x) = your formula (where n=x) and A = number of attackers?
>
Thanks again! To be honest, I thought the same variant 20 minutes later I
sent the previous message :D
So the final variant will be


f(n,T) = \sum_{i=1}^{n-T}i*\frac{\binom{n-i-1}{T-1}}{\binom{n-1}{T}}


g(n,T,A) = \sum_{i = 1}^{A}f(n-i+1, T)


Where g(n,T,A) computes the expected value of poisoned chunks in the team
with T trusted peers and A attackers. Of course T + A <= n

Thanks again. I will perform tests asap =)




2015-04-01 3:31 GMT+06:00 Vicente Gonzalez <vicente.gonzalez.ruiz@xxxxxxxxx>
:

>
>
> On Tue, Mar 31, 2015 at 8:29 PM Ilshat Shakirov <im.shakirov@xxxxxxxxx>
> wrote:
>
>> Hello!,
>>
>> Thanks for your feedback =)
>>
>
> You're welcome :-)
>
>
>>
>> I couldnt generalize my formula for different values of trusted peers and
>> attackers. But I generalized it for different number of trusted peers in
>> the team.
>> Here it is:
>>
>> ​where n is size of the team without splitter and T is number of trusted
>> peers.
>> The main idea is that the attacker will be detected when he will send
>> poisoned chunk to trusted peer. The time when he will send poisoned chunk
>> to trusted peer depends on position of trusted peer in his peer list.
>> So the denominator of fraction is just all the possible combinations of
>> trusted peers in peer list.
>> And the numerator is all possible combinations of T-1 trusted peers in
>> peer list. In more detail:
>> The attacker is able to send *#i* poisoned chunks when he has trusted
>> peer exactly in* #i* position in his peer list. In [1, #i) he can't have
>> trusted peers. So we have T-1 peers and we should place them in remaining
>> positions ((#i, n-1)) of peer list.
>>
>
> For me your reasonings sound fine. Good job!
>
> Tomorrow I will perform experiments on simulator in order to confirm that
>> formula.
>>
>
> OK.
>
>
>> Now I don't know how to generalize the formula to different number of
>> attackers, because we expel attacker after we detect him and team size
>> become lower. I think I need help with that problem =)
>> Thanks in advance for your feedback =)
>>
>
> Could be:
>
> \sum_{j=1}^{A} E(n-j)
>
> where E(x) = your formula (where n=x) and A = number of attackers?
>
> (The idea is that the size of the team will be reduced by 1 each time a
> attacker has been uncovered. So, the accumulated average number of poisoned
> chunks should be the sum the expected number of poisoned chunks, for each
> number of attackers).
>
> Hope this helps!
>
>
>>
>> ps Python impl is OK =)
>>
>
> OK, thanks.
>
> Best,
> Vi.
>