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The miner needs to send a POST http request to the server containing the authentication info and a json getwork struct and get the result like described by Zagitta. The data field is a int[32] (big endian ) , the first 20 integers are the actual block header that needs to be hased. There are other 12 ints because it's the first step of the sha256 algo(pseudocode here http://en.wikipedia.org/wiki/SHA-2 ). The midstate contains the a,b,c,d,e,f,g,h , registers of the sha256 transform after processing the first 512 bits chunk of data. Notice that for processing the first chunk , the nonce isn't needed, because it is contained in data[19] (last int of the actual header) that's why is handful to have them , because you can calculate the hashes starting from that point (actually you can go some step further because the nonce is in position 19, the processing of the fist chunk finish on 15 , you can process positions 16 17 18 , and start each calculation from there) . Your goal is to find a nonce to be inserted in header[19] for which sha256(sha256(header ) ) has a specified (by the difficulty ) number of final zero bits. After finding the nonce you need to put it into the data[] field of the json sent by the server , put this data[] in a string , build a json getwork request , add to it a params array, put the string in the arra , send the json to the server. It will rely with an other json whose result field will be true if the solution was accepted. I just finished writing my miner in c#. I found really confusing the endianess of integers for the sha256 transform. The ints in data[] are represented as big endian hex strings. If you parse the string '000000ff'(big endia) it will return the int 0xff , that's written in memory as 0xff000000 (if you use windows or linux) BUT when you perform the arithmetical operations on that number , the operators work as that is a big endina number. So for instance if you have the number int num = 0x1 , it's written in memory as 0x80000000 , but if you write int res = num << 1 , you get res = 2, becase << operate on the big endian representation. This means that before doing your calculations you need to reverse the bytes of each int of the data array, because the sha256 transform requires to operate on little endian integers. |
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Mi sa che l'ha già venduta perchè io non ho avuto notizie
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Ok penso che ora è chiaro spezzando i blocchi si ha un input al problema diverso. In pratica invece di avere un un'unico blocco grosso 100 si hanno 10 blocchi grossi 10. Il vantaggio c'è perchè risolvere un blocco da 100 è più impegnativo che risolvere 10 volte un blocco da 10, perchè la complessità cresce più che linerarmente con la grandezza dei blocchi.  Comunque qua in ufficio non riesco a connettermi ne al client ufficiale (i ragionamenti che ho fatto prima sono dovuti a prove fatte ieri sera da casa) ne ai pool tramite GUIMiner. Il firewall aziendale non mi permette di uscire su certe porte immagino. Esistono interfacce web per partecipare ad un pool? Voglio dire una specie di sito dove tramite protocollo http ricevo l'input del mio problemino e il mio pc lo elabora e rimanda la soluzione sempre in http cosi si riesce a bypassare sto firewall? No la complessità è sempre la stessa. L'algoritmo calcola l'hash dell'header del blocco , il quale è sempre 640bit che sono 80bytes e quindi un array di 20 int. Insomma la grandezza dell'header è la stessa. I soldi guadagnati medi minando in pool sono gli stessi che guadagneresti minando in solitaria MA in pool hai un guadagno costante mentre in solitaria potresti non trovare un blocco per giorni ma quando lo trvi becchi 650 euro circa in un colpo solo. In termini tecnici il mining in pool diminiusci la varianza della produzione. Insomma se ti senti iperfortunato mina da solo e magari dopo 1 min di mining becchi un blocco  . |
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AFAIK declaring those variable as local will make theme shared for all the the threads in the workgroup so it's possible to have a performance increase because you are using less memory , but speed doesn't matter when the calculations are all fucked up! !
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Probably i got it , you use two queues so you don't have to call clFinish to wait for the current work to end before reading the buffer.
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In the part where you enqueue the work to the gpu and then read the output buffer , why do you alternate two buffers? ( buffer and output are two arrays of two buffers)
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Actually you can get robbed also using paypal ? And all other payment methods. Ok paypal can refound you taking the money from the account that robbed you , but do you think that after taking your money , the thief will leave the money in his account just for paypal to return them back to you ??
That's why there is reputation on ebay , because also using paypal you are not sure to receive what you buy.
There is also the possibility to be robbed when using cash actually.
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If this happened two times in a row , probably there is some problem?
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The problem with Shared Maximum Pay Per Share is that in short lucky runds you get payed per share , instead if you get a long rounds streak there is the possibility that the credit isn't enough to pay all the shares and it switches to proportional method !! This goes in favour to the pool hopping cheat!! Btw we are in a very unlucky round  18 hours and still no block.  |
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Ma è anche in garanzia?
Di dove sei ? (per eventuale ritiro a mano )
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Difficulty is doubled now , and the value of BTC has lowered , so the profit will be negative if you keep these prices.
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Is it possible tu use both gpus on the same system? Won't there be any driver conflict?
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Yes that was actually the only problem i was thinking about , differences in the grounds voltages. In theory there shouldn't be any difference , in reality it's possible. How can i connect the two grounds. Is the case of the psu connected to it's ground ?
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Is there something nice that can be said about me too ? |
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If i'm not wrong each card needs 2 (6PIN PCI-E) connectors but that psu has only 2 of those connectors , this mean that you can't have 2 GPUs.
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I'm thinking about buying a new GPU that i would add to my computer. The problem is that my PSU can't handle 2 GPUs so i thought about buying a new psu for the second GPU . Theorically there shouldn't be any problem , but has anyone tried that in practice? ?
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Compared to other FPGA's i've seen this is a pretty good price.
But potential ROI ?
Assuming we get an average price of 10$/BTC and average difficulty of 4mill... Would yield ~$328.70/mo, or 63 months to payoff. (thats still ~18% return per year, if this holds which is pretty good as an investment)
The problem is that the difficulty is increasing, in the next inrease will be almost doubled and so also the payback time , and also you can't forsee what will happen in 63 months. |
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I'm having a look at your code because i would like to understand how a miner works. I have a few questions. Where i can find how is stored the data contained in the work sent to the miner from the server ? Also could you explain a little bit in few words what happens in lines of code between 888 and 944? Here: bufferIndex = (bufferIndex == 0) ? 1 : 0;
workSizeTemp.put(0, workSize);
currentWork.update(workSizeTemp.get(0) * loops * vectors);
System.arraycopy(currentWork.midstate, 0, midstate2, 0, 8);
sharound(midstate2, 0, 1, 2, 3, 4, 5, 6, 7, currentWork.data[16], 0x428A2F98);
sharound(midstate2, 7, 0, 1, 2, 3, 4, 5, 6, currentWork.data[17], 0x71374491);
sharound(midstate2, 6, 7, 0, 1, 2, 3, 4, 5, currentWork.data[18], 0xB5C0FBCF);
int fW0 = currentWork.data[16] + (rot(currentWork.data[17], 7) ^ rot(currentWork.data[17], 18) ^
(currentWork.data[17] >>> 3));
int fW1 = currentWork.data[17] + (rot(currentWork.data[18], 7) ^ rot(currentWork.data[18], 18) ^
(currentWork.data[18] >>> 3)) + 0x01100000;
int fW2 = currentWork.data[18] + (rot(fW0, 17) ^ rot(fW0, 19) ^ (fW0 >>> 10));
int fW3 = 0x11002000 + (rot(fW1, 17) ^ rot(fW1, 19) ^ (fW1 >>> 10));
int fW15 = 0x00000280 + (rot(fW0, 7) ^ rot(fW0, 18) ^ (fW0 >>> 3));
int fW01r = fW0 + (rot(fW1, 7) ^ rot(fW1, 18) ^ (fW1 >>> 3));
int fcty_e = currentWork.midstate[4] + (rot(midstate2[1], 6) ^ rot(midstate2[1], 11) ^ rot(midstate2[1], 25)) +
(midstate2[3] ^ (midstate2[1] & (midstate2[2] ^ midstate2[3]))) + 0xe9b5dba5;
int fcty_e2 = (rot(midstate2[5], 2) ^ rot(midstate2[5], 13) ^ rot(midstate2[5], 22)) + ((midstate2[5] & midstate2[6]) |
(midstate2[7] & (midstate2[5] | midstate2[6])));
int fcty_e_plus_e2 = fcty_e + fcty_e2;
int fcty_e_plus_state0 = fcty_e + currentWork.midstate[0]; I also noticed that i get 'Spurious CL_INVALID_KERNEL_ARGS error, ignoring' when running with -d enabled, what could it be ? Thanks for the patience |
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Here you can find how much time in average it will take for you to generate a share http://www.alloscomp.com/bitcoin/old_calculator.phpThe common difficulty for shares is 1 , so put '1' in the difficulty field and 30000 in the speed field then hit calculate. |
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