Topic: Please run a full node - page 3. (Read 6669 times)

ask him
not to be biased on the leanest linear block... but a average block that has some quadratics and where UTXO cache delays things
and
not to be biased of FIBRE header only.. but a average full block relay or average where latency and other things are included, such as average connections
and
all the other non hashing functions, then come to a total

and guess what.. if they try to argue its all milliseconds of non hashing function...
then that debunks all the issues core extremists ever had against "big blocks"

P.S
im gonna laugh when he wants to knit pick '2min' difference.. but cannot explain himself out of the 50-60 min difference he thinks

not gonna argue with you Franky , cause i'd be simply repeating myself.

On a sidenote.... question for _ck:  -- how much time is actually spent validating, and is this typically done in parrellel?

actually there are a few things, which help.
in laymens.(simplified so dont knitpick literally)

say you had to go from "helloworld-0000001" to "helloworld-9999999" hashing each try where the solution is somewhere inbetween
solo mining takes 10mill attempts and each participent does this
"helloworld-0000001" to "helloworld-9999999" hashing each try (very inefficient)
however, pools gives participant
A: "helloworld-0000001" to "helloworld-2499999" hashing each try
B: "helloworld-2500000" to "helloworld-4999999" hashing each try
C: "helloworld-5000001" to "helloworld-7499999" hashing each try
D: "helloworld-7500000" to "helloworld-9999999" hashing each try

which is efficient...
which at 1-d makes people think that killing POOLS takes 4x longer...

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but here is the failure...
pool U does "helloWORLD-0000001" to "helloWORLD-9999999" hashing each try 20min to get to 10mill where a solve is somewhere inbetween (average 10min to win)
pool V does "HELLOworld-0000001" to "HELLOworld-9999999" hashing each try 20min to get to 10mill where a solve is somewhere inbetween (average 10min to win)
pool W does "helloworld-0000001" to "helloworld-9999999" hashing each try 20min to get to 10mill where a solve is somewhere inbetween (average 10min to win)
pool X does "HElloworld-0000001" to "HElloworld-9999999" hashing each try 20min to get to 10mill where a solve is somewhere inbetween (average 10min to win)
pool Y does "HelloWorld-0000001" to "HelloWorld-9999999" hashing each try 20min to get to 10mill where a solve is somewhere inbetween (average 10min to win)
pool Z does "HelLoWorLd-0000001" to "HelLoWorLd-9999999" hashing each try 20min to get to 10mill where a solve is somewhere inbetween (average 10min to win)
it takes each pool similar times to get to 9999999 and each would get a solution inbetween should they not give up
and if you take away pool W,X,Y guess what..
pool Z doing "HelLoWorLd-0000001" to "HelLoWorLd-9999999" hashing each try would NOT suddenly take 4x longer to get to 99999999
because Z is not working on a quarter of the nonce of other pools!!!!!!!!!!!!!!!!!

because the work pool Z is doing 'HelLoWorLd' is not linked to the other 3 pools.

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so 2 dimensionally
pool U does "helloWORLD-0000001" to "helloWORLD-9999999" 20min to get to 10mill (average 10min to win)
pool V does "HELLOworld-0000001" to "HELLOworld-9999999" 20min to get to 10mill (average 10min to win)
pool W does "helloworld-0000001" to "helloworld-9999999" 20min to get to 10mill (average 10min to win)
pool X does "HElloworld-0000001" to "HElloworld-9999999" 20min to get to 10mill (average 10min to win)
pool Y does "HelloWorld-0000001" to "HelloWorld-9999999" 20min to get to 10mill (average 10min to win)
pool Z does "HelLoWorLd-0000001" to "HelLoWorLd-9999999" 20min to get to 10mill (average 10min to win)

because they are not LOSING efficiency pool Z does "HelLoWorLd-0000001" to "HelLoWorLd-9999999" still takes 20min to get to 10mill (average 10min to win)

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now do you want to know the mind blowing part..
lets say we had 10minutes of time
you would think if pool W had 650peta and that if pool Z had 450peta
you would think pool Z =14 minutes due to hash difference

but
what if i told you out of the 10 minutes upto 2minutes is wasted on the propogation, latency, validation, utxo cache.. (note: not the hashing)
so
if pool W had 650peta
if pool Z had 450peta
pool Z =11min33 due to other factors because the calculating of hash is not based on 10 minutes.. but only ~8ish (not literally) of hashing occuring per new block to get from 0-9999999 (not literally)

now imagine Z done spv mining.. to save the seconds-2minutes of the non-hashing tasks- propogation, latency, validation, utxo cache.. (note: not the hashing))
Z averages under 11min:33sec

so if Z went alone his average would be UNDER 11:33sec average

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so while some are arguing that out of 6 blocks
U wins once, V wins once, W wins once, X wins once, Y wins once, Z wins once..
they want you to believe it take 60 minutes per pool to solve a block (facepalm) because they only see W having 1 block in an hour

if you actually asked each pool not to giveup/stale/orphan .. you would see the average is 10 minutes(spv:10min average or 11:33 if validate/propagate).. but only 1 out of 6 gets to win thus only 1 gets to be seen.

but if you peel away what gets to be seen and play scenarios on the pools that are not seen (scenarios of if they didnt give up).. you would see it not 60 minutes

Yeah, like he said, exponential. My first full run was using a low difficulty and the solve times were really low as I just wanted to get a feeling for how it would work out. For that run, the "average" solve time was 4 seconds. The longest was 28 seconds. The median was somewhere between 2 and 3 seconds. i.e. half the block were solved before that, the other half after that. For the last run I was doing I had increased the difficulty and tweaked some performance issues and was running a much longer test but the damn computer locked up so I have't bothered starting it up again. The interm data from that test was following the same sort of distribution anyway.

gotcha. thx.

Exponential.
(at least, the "inter-block times" are distributed according to an exponential distribution)

Perhaps he is mentally stuck on "proof of work" when a more descriptive term might be "proof of lottery participation"

(Hint:  A powerball lottery doesn't have a winner ever week necessarily.  If no one wins, the pot keeps growing until a winning ticket is found)

Interesting.  So what is the distribution then , if not gaussian?

I would like to recall to you, that our long digression into how mining works, was needed to try to make you see that you had an erroneous argument in favour of full nodes, because a miner with 10% of the hash power has NO INCENTIVE to step back from remaining in agreement with the other miners, simply because he's then hard-forking all by himself, and will make a 10 times shorter chain.

Your erroneous understanding of mining made (probably still makes) you think that that betraying miner is going to mine all by himself a fork of just the same length as the chain of the rest of the miners, and hence "reap in all the rewards, orphaning the 90% chain" because full nodes agree with him, and not with the miner consortium.  

But this is not the case: our dissident miner will make just as many blocks on his own little fork, than he would have made on the consortium chain (*), with just as many rewards: so there's no incentive for him to leave the consortium, and make his little hard fork.  There *may be* such an incentive, if our dissident miner thinks that he will have the majority of the USERS on his side, and that the majority of the USERS will dump their coins on the long chain, and will fight for the coins on HIS fork.  But this is "hard forking power games", and has nothing to do with full node power.
The irony in this is that in order for his hardforking to work, users need to be ABLE TO TRANSACT on the chain they would dump and which, in our gedanken experiment, no node wants to accept !

==> "the dissident miner" story has hence no relevance to the "full node power" story.  It is a different power game that is described.

You are in a kind of bleak position to use an argument of authority, after having blatantly shown your total misunderstanding of elementary mining dynamics.  Of course, any logically built up argument is independent of its author, but there isn't any such in your post.

I do recognise UTILITY (but no POWER) to non-mining full nodes.  The elements of utility are:

1) to the owner: to be informed of the eventual deviation of the actually used protocol on the block chain, and the protocol requirement of his node (in other words, to acknowledge that the miners are using a different protocol than the one he would like with his node)

2) to the owner: anonymity of sending transactions (deniability of IP address).

3) helping a P2P network of data propagation, in case that direct internet links to miner nodes have problems (technical/political/....)

4) as a proxy, relieving the miner nodes from user traffic and helping them to make bigger profits because they can spend less on their network infrastructure towards users (uh )

Edit (*): caveat: at least, if in his hard fork, he didn't mess with the difficulty of course and kept the difficulty of the parent chain.

Lol.

Well, 95% of all blocks being mined into existence are mined using my software so if he can't believe me telling him he's wrong then I'm not sure how much higher an authority you can appeal to?

Funny, me too !    I edited my post with a long reply concerning the "lottery" stuff, when I wanted to submit, your post was there, I also realized the futility and didn't submit after all

anyway back on topic..

nodes DO have a crucial role to play..
anyone saying other wise just wants people to turn off their node..

lol. I had this really long reply to him typed up and just deleted it as I knew it would go no where. I was thinking maybe there was some unique thing that happens when you stick a bunch of miners in a pool that doesn't happen if they were all solo mining. So I was writing a simulation of 100k miners. But after seeing the true distribution (which, despite knowing better I still was thinking of it as a normal one), I shelved it as I had proved enough to myself that what he's saying just can't be true. For me the whole "average" thing is sort of misleading as it immediately puts a normal distribution in the back of your mind. But when you actually see it, things get pretty clear. At least it did for me.

It IS hopeless.

by the way im not an ignoramus about cryptography.
nor am i ignoramus about the odds of finding a block with 1 leading 0 vs odds of finding 18 leading 0's (im not being literal, im talking laymens)
nor am i ignoramus about for instance antpools 650peta hash a second and how many hashes over ~10minutes that equates too

p.s its not 1:1800

but until you realise that each pool works on EVERY BLOCK win or lose, theres no point talking about the the 3rd dimension


wow.. finally seems now you admit the pools work on every block.

so now your seeing that it resets each height (thank god your now seeing that.. after 2 days)
now imagine if it didnt stop hashing 469998 even when C won.. it kept hashing 469998. you would see that it would have a solved 469998 in approx:

i hope you see now that you realise its not working on say 470,000 based on the time from 469994.. but 469999 you can see things more two dimensionally..

can you atleast see how just VIEWING 2 blocks in a blockchain in a 1 hour period. DOES NOT mean it takes 30 minutes to solve a block

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as for the new analogy of the lottery game

for 2 pages others were taking all the lottery tickets bought over 6 week of 6 games. and when winning only 2 lotteries. thinking it took all 6 weeks worth of tickets to win the 2 game .. but then double failed by saying it would still take 3 weeks to win..

rather than thinking that each week without anyone else buying tickets. then you could just yorself keep buying tickets until you win that weeks lottery. and it wont take 3-6x the amount of tickets

Think of it like this:  Imagine a ginormous fish tank, 1000 miles high , wide, and deep, filled with plastic balls.  Most of the balls are yellow, but a few select balls are red.  The tank is constantly churning and mixing the balls, so if you reach your hand down and pull one ball it, its completely random.  You can toss a ball back in after choosing one, and still have the same exact chance of choosing a red ball.  Let's same three miners: Alice, Bob, and Carol each can pull one ball a second , check if its red, and then toss it back before trying again, and lets say the ratio of yellow to red balls is 1:1800.  After 10 minutes, each of them has done 600 pulls, and together 1800.  There's no guarantee they will pull a red ball in 10 minutes but that's the average.  Sometimes Alice will find a red ball quickly after Bob finds one, sometimes no one will find one for an hour.  The key thing is that whenever they pull a ball, the chances are always almost exactly 1:1800, because the tank is so huge.  


The mistake is to think that after 30 minutes of pulling balls, Alice's next pull will be anything different than 1:1800 chance.

Now lets say Bob starts to get really quick on the draw and he starts pulling 2 balls a second, so that now instead of 1800 pulls in 10 minutes, the 3 miners can do 2400 pulls in 10 minutes.  After 2 weeks, the tank magically adjusts the difficultly and now there's only 1:2400 red balls to yellow balls.

Be careful.  They are working on SIMILAR blocks, but each pool has made his own block, with its own picking of transactions from the mem pool, and with its own specific order of course.  They are working on top of the same consensus block, 469997, of course, because this one has been published.

This is what tricked me in trying many times too. And yes, they warned me and I didn't want to believe I couldn't get him to see this simple fact.  franky1 is not a total ignoramus concerning crypto, so it is beyond comprehension that he has a misunderstanding on which he is locked in so hard that nobody can get him out of there.... but there's no hope I'm afraid.  

But I'm starting to see part of franky1's confusion.  Yes, of course, all miners work on average about 10 minutes on EACH BLOCK.  When they win a block, they were only working on it for about 10 minutes.   It seems that "all the rest was wasted".  Probably franky1's confusion comes from the fact that "if there hadn't been that bastard of a competitor, I could have continued working on the block instead of having to give up on it and start all over again with a fresh one". So if, each time I WIN, I only needed about 10 minutes on average, I would only need 10 minutes on those that I had to give up because a bastard of a competitor forced me.

I think that is the core of his reasoning.

But this is wrong, because, exactly, it is not a race, but a lottery, like jonald tried to explain.

There is statistically not a single difference between mining on the same block, or changing blocks.  Your probability of success doesn't change.  It is as if you  had to throw different dice.  You chances of throwing a number don't change because you use different dice.  In other words, the fact that a "bastard of a competitor" forced you to change the block doesn't alter anything to your chances of winning one.  But I can somehow understand how this can be counter-intuitive, because by far most tasks we know, are a cumulative effort towards a result.

Let us try once more (you see, I'm tricked again in trying !).

Let us say that the "block on which you are working" is the specific dice you throw.   If I give you 3 red dice, and you need to throw 3 times a 6, you will have to try several times.  I have 3 blue dice and I try too.  You have one chance out of 216 to get 3 times a six.  Now after you've thrown 50 times without success, I have also thrown 50 times, and hey, I have 3 six !  At this point, you have to leave your red dice, and you have to use my blue dice now, while I take grey dice.  And we continue playing.
Do you think that you were "close to having 3 six" with your 3 red dice, and because I was a bastard of a competitor, forcing you to abandon the red dice, you can start over again ?

At each "round" (each second, say), both of us throw our dice.  As each of us has 1 chance in 216 to have a triple six, there will be a triple six found every 108 seconds on average, do you agree ?  But you will only find a triple six, on average, every 216 seconds, and me too.  The fact that I am present, doesn't change your chances of winning the next time.
It is only in very rare circumstances, that both of us win at the same throw.  In that case, one of us gets orphaned.  But that only happens very rarely.  Most of the time, each time you win, I don't win, and vice versa.  And whether I win or not, doesn't change the rate at which you win.

While C was solving 469998, A is also working on the same block.  All miners are working on that block (unless there was some kind of latency and they didn't get 469997)


10 minutes.

your taking things too one dimensionally literally.
ok lets try this.. be warned there is a test

470000 A 10:00
469999 B 10:00
469998 C 10:00
469997 D 10:00
469996 C 10:00
469995 D 10:00
469994 A 10:00
469993 B 10:00

1. what height is A working on while C was solving 469998 (imagining current height was 469997 and 469998-4700000 have not yet been solved)
[] 469998      [] 469996      [] 470000    [] 469995

2. imagine A did win 469998 instead of C, how long would you think it took using approx (meaning give or take a little variance.. not literal/exact)
[] 10 mins     [] 60mins      [] 30 mins     [] 40 mins  

please think beyond 1 dimension of counting blocks