Bitcoin Forum
September 15, 2019, 07:01:24 AM *
News: If you like a topic and you see an orange "bump" link, click it. More info.
 
   Home   Help Search Login Register More  
Pages: [1]
  Print  
Author Topic: How much Terabyte needed to get 51 characters all possible combination?  (Read 220 times)
Honey9
Newbie
*
Offline Offline

Activity: 2
Merit: 0


View Profile
August 22, 2018, 05:20:42 PM
 #1

How much Terabyte needed to get 51 characters all possible combination, While starter character will always 5?
1568530884
Hero Member
*
Offline Offline

Posts: 1568530884

View Profile Personal Message (Offline)

Ignore
1568530884
Reply with quote  #2

1568530884
Report to moderator
1568530884
Hero Member
*
Offline Offline

Posts: 1568530884

View Profile Personal Message (Offline)

Ignore
1568530884
Reply with quote  #2

1568530884
Report to moderator
PLAY NOW
Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction. Advertise here.
1568530884
Hero Member
*
Offline Offline

Posts: 1568530884

View Profile Personal Message (Offline)

Ignore
1568530884
Reply with quote  #2

1568530884
Report to moderator
markj113
Legendary
*
Offline Offline

Activity: 2212
Merit: 1041



View Profile
August 22, 2018, 05:28:57 PM
Merited by HeRetiK (1)
 #2

2 X chinese kids, an abacus and 30 minutes.
aliashraf
Hero Member
*****
Offline Offline

Activity: 896
Merit: 656


View Profile
August 22, 2018, 05:40:15 PM
 #3

Combination is not the term, I guess you mean permutations and it is 50! = 3.04140932*10^64 which you can't store it anywhere not just because of capacity problems but also because of the time required for writing them, it would take billions times the galaxy's age for a multiple penta byte per second write operation, I suppose.
mfyilmaz
Jr. Member
*
Offline Offline

Activity: 82
Merit: 2


View Profile
August 22, 2018, 06:42:03 PM
Merited by DarkStar_ (2)
 #4

Combination is not the term, I guess you mean permutations and it is 50! = 3.04140932*10^64 which you can't store it anywhere not just because of capacity problems but also because of the time required for writing them, it would take billions times the galaxy's age for a multiple penta byte per second write operation, I suppose.

you are wrong... permutations are only for cases where the sequence is not a factor like when you draw 5 cards from a 52 cards deck... it doesnt matter if you draw spade ace first or last... so As-2s-3s-4s-5s is the exact same as 5s-3s-2s-4s-As....

while in this case the sequence is important because AAAA5 and 5AAAA or AA5AA is NOT the same... so here you have to actually do like #character pow 50... like when you use 26 alphabet then its 26^50... if you use captions then its 52^50... if you want to use alphanumeric (26 alphabet + 26 ALPHABET + 10 Numbers) then its 62^50

this is such a huge number thats its not even possible to write all this different combinations down...

https://www.reddit.com/r/Bitcoin/comments/8ss3dp/bitcoin_is_money_secured_by_law_of_universe/

and btw

2^256 = 1,15 * 10^77
62^50 = 4,16 * 10^89

so this number is 3 trillion times bigger as the number of sequences described in that picture... so pretty much impossible to do this. Smiley

★ PRiVCY ➢ Own Your Privacy! ➢ Best privacy crypto-market! ★
✈✈✈[PoW/PoS]✅[Tor]✅[Airdrop]✈✈✈ (https://privcy.io/)
TheArchaeologist
Member
**
Offline Offline

Activity: 84
Merit: 87

Learn from the Past!


View Profile WWW
August 22, 2018, 07:16:51 PM
Last edit: August 23, 2018, 07:10:07 AM by TheArchaeologist
Merited by suchmoon (4), achow101 (2)
 #5

How much Terabyte needed to get 51 characters all possible combination, While starter character will always 5?
You mean the WIF's for all uncompressed private keys? Wink
aliashraf
Hero Member
*****
Offline Offline

Activity: 896
Merit: 656


View Profile
August 22, 2018, 10:01:44 PM
Last edit: August 23, 2018, 06:20:26 AM by aliashraf
Merited by DarkStar_ (3), ETFbitcoin (1)
 #6

Combination is not the term, I guess you mean permutations and it is 50! = 3.04140932*10^64 which you can't store it anywhere not just because of capacity problems but also because of the time required for writing them, it would take billions times the galaxy's age for a multiple penta byte per second write operation, I suppose.

you are wrong... permutations are only for cases where the sequence is not a factor like when you draw 5 cards from a 52 cards deck... it doesnt matter if you draw spade ace first or last... so As-2s-3s-4s-5s is the exact same as 5s-3s-2s-4s-As....

No, you are wrong.

Permutation IS about the cases sensitive to order.  We say: "What permutations we could have for 5 cards"?
This is order sensitive and is 5!= 120 i.e. we got 5 cards and we can arrange them in 120 different ways.

Combination is used for a very different scenarios, e.g. how many different hands(each with 5 cards) we can draw from a 52 cards deck? Here the answer is 52!/47!/5! = 52*51*50*49*48/5/4/3/2=2,598,960 different hands.

So, when you ask about combinations usually you have to give 2 numbers like 5 and 52 in my example, i.e.  5 of 52.

Now I understand you are asking about permutation with repetition for 50 placements of 51 characters. Which is obviously 51^50, but you should never use the term combination for expressing it because in math (and I'm not a mathematician by the way) we use this term a bit different with colloquial language.

bob123
Legendary
*
Offline Offline

Activity: 1022
Merit: 1506



View Profile WWW
August 23, 2018, 10:00:17 AM
 #7

How much Terabyte needed to get 51 characters all possible combination, While starter character will always 5?

I guess you are trying to bruteforce a private key (known first char = 5 and length of 51 chars) ?

Basically you would have to guess the correct key out of 2^256 possible ones (which is practically not possible).

Since you are looking for a private key in WIF format, you'll have to add 1 byte at the front. Then put the first 4 bytes of the double sha256 hash at the end (this is the checksum).
And after converting it into base58, you will have your private key in WIF format (starting with 5.. ).


Fortunately, there is not enough energy on this planet to fuel the machines needed to bruteforce this search space.

NeuroticFish
Legendary
*
Offline Offline

Activity: 1974
Merit: 1310


There are no mistakes. Only opportunities wasted.


View Profile
August 23, 2018, 11:07:27 AM
 #8

Fortunately, there is not enough energy on this planet to fuel the machines needed to bruteforce this search space.

You should have also added the very well made image about all this. It makes OP understand the problem better.
Here it is:


HeRetiK
Legendary
*
Offline Offline

Activity: 1232
Merit: 1118


the forkings will continue until morale improves


View Profile
August 23, 2018, 04:38:37 PM
Last edit: August 24, 2018, 12:31:21 AM by HeRetiK
Merited by xhomerx10 (1)
 #9

Combination is not the term, I guess you mean permutations and it is 50! = 3.04140932*10^64 which you can't store it anywhere not just because of capacity problems but also because of the time required for writing them, it would take billions times the galaxy's age for a multiple penta byte per second write operation, I suppose.

That's assuming you only have 50 characters available in total (over a string with the length of 50) of which each character can only be used once.

I'm afraid mfyilmaz is a bit closer to the truth.

Still none of you guys have answered OP's question Sad

Using WIF (as requested by OP) would add unnecessary overhead to storing the private keys, so we're gonna save some storage and will just go with barebone private keys.

So taking the numbers as calculated by mfyilmaz and bob123:

Quote
2^256 = 1.15 * 10^77

You get 1.15 * 10^77 possible keys with a size of 2^256 bits (= 32 bytes) each so you get 32 * 1.15 * 10^77 bytes for storing all possible combinations of a 256 bit key.

32 * 1.15 * 10^77 equals 48 * 10^77 bytes which is 48 * 10 ^ 68 65 Terabytes, ie. 48,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 Terabytes or roughly the amount of data you could store using 2 X chinese kids, an abacus and 30 minutes.

(edit: Thirdspace's post reminded me that Tera = 10^12 not 10^9)

Thirdspace
Hero Member
*****
Offline Offline

Activity: 1106
Merit: 717


Mixing reinvented for your privacy | chipmixer.com


View Profile
August 23, 2018, 11:38:01 PM
Merited by xhomerx10 (1)
 #10

How much Terabyte needed to get 51 characters all possible combination, While starter character will always 5?
You mean the WIF's for all uncompressed private keys? Wink
that's pretty much what he's asking for Cheesy
Base58 WIF uncompressed means 58 possible characters in a 51-character private key
but remember not all 51-char length private keys are valid, because the checksum at the end of it  
more reasonable to calculate based on possible private key hex format (64 chars length) 16^64 = 2^256
= possible private key hex format * private key for uncompressed address
= 16^64 * 51
= 1.15792... * 10^77 * 51
= 59.05392... * 10^77

since 1 TB = 10^12, so:
= ~5.9 * 10^66 Terabytes
imagine having a cluster hard drive, 1066 units of a 6 TB hard drive Shocked

Pages: [1]
  Print  
 
Jump to:  

Sponsored by , a Bitcoin-accepting VPN.
Powered by MySQL Powered by PHP Powered by SMF 1.1.19 | SMF © 2006-2009, Simple Machines Valid XHTML 1.0! Valid CSS!