worldinacoin


April 11, 2012, 04:41:45 AM 

How do I measure difficulty? For eg I see Difficulty 1626553 , what does 1626553 means? How many ghash does it translate to?






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Foxpup
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April 11, 2012, 05:10:42 AM 

Difficulty is the maximum possible hash target divided by the current hash target. The average time to find a block (in seconds) is equal to the current difficulty times 2^32 (4,294,967,296) divided by your hashrate (in hashes per second) (eg, if your hashrate is 1 Ghash/s = 1,000,000,000 hashes per second then the average time it will take you to find a block is (1,626,553 x 4,294,967,296 / 1,000,000,000 = 6,985,992 seconds = 80 days). Since the difficulty is adjusted so that miners find blocks on average once every 10 minutes (600 seconds), the total hashpower of the network is equal to the difficulty times 2^32 divided by 600 (eg, 1,626,553 x 4,294,967,296 / 600 = 11,643,319,900,351 hashes per second = 11.6 terahashes per second). That's how much hashpower you need to pull off a 51% attack. Good luck!




Revalin


April 11, 2012, 05:20:00 AM 

tl;dr: it takes (difficulty * 2^{32}) hashes to solve a block on average.

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Red Emerald


April 11, 2012, 06:42:50 AM 

tl;dr: it takes (difficulty * 2^{32}) hashes to solve a block on average.
2 paragraphs needed a tl;dr?




Pieter Wuille


April 11, 2012, 10:36:33 AM 

difficulty * 2^48 / 65535, actually

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Foxpup
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April 11, 2012, 11:10:04 AM 

difficulty * 2^48 / 65535, actually Yes, that 0.0015% makes all the difference.




worldinacoin


April 14, 2012, 09:53:06 AM 

Difficulty is the maximum possible hash target divided by the current hash target. The average time to find a block (in seconds) is equal to the current difficulty times 2^32 (4,294,967,296) divided by your hashrate (in hashes per second) (eg, if your hashrate is 1 Ghash/s = 1,000,000,000 hashes per second then the average time it will take you to find a block is (1,626,553 x 4,294,967,296 / 1,000,000,000 = 6,985,992 seconds = 80 days). Since the difficulty is adjusted so that miners find blocks on average once every 10 minutes (600 seconds), the total hashpower of the network is equal to the difficulty times 2^32 divided by 600 (eg, 1,626,553 x 4,294,967,296 / 600 = 11,643,319,900,351 hashes per second = 11.6 terahashes per second). That's how much hashpower you need to pull off a 51% attack. Good luck! Thanks a million, I still do not get you on this part Since the difficulty is adjusted so that miners find blocks on average once every 10 minutes (600 seconds), the total hashpower of the network is equal to the difficulty times 2^32 divided by 600 (eg, 1,626,553 x 4,294,967,296 / 600 = 11,643,319,900,351 hashes per second = 11.6 terahashes per second). That's how much hashpower you need to pull off a 51% attack. Good luck! Grin Why is it 11 tera hashes for 51%, shouldn't it be 100% already?




Revalin


April 14, 2012, 10:06:15 AM 

If the rest of the network has 11.6 TH, and you have 11.6 TH, you have 50%. If you go up to 23.2 TH, you'd have 67%.

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wachtwoord
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April 14, 2012, 10:10:52 AM 

difficulty * 2^48 / 65535, actually Yes, that 0.0015% makes all the difference. What 0.0015% difference? 65535=2^16 so: 2^48 / 65535 = 2^48 / 2^16 = 2^(4816)= 2^32 Showing the division does show better where the numbers come from though




Revalin


April 14, 2012, 10:15:34 AM 

65535=2^16 Nope. 2^16 = 65536 .

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Sukrim
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April 14, 2012, 12:19:26 PM 

What 0.0015% difference? 65535=2^16
A multiple of 2 that ends in 5? That's not the kind of math I learned in school...

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worldinacoin


April 14, 2012, 01:09:20 PM 

I think he means 0 to 65535




Foxpup
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April 14, 2012, 01:51:27 PM 

I think he means 0 to 65535
2^16 is always 65536, regardless of whether you count from 0 or from 1. The number is exactly 2^48/(2^161). I thought I could simplify things by saying it was 2^32, but that plan didn't seem to work as intended.




wachtwoord
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April 14, 2012, 03:20:29 PM 

Yes I see I have been number blind. I actually read 65536 so scrap that 2^48/((2^16)1) is indeed the exact number




