Calculating the storage that would be needed to store all addresses.

[BlackHatCoiner]

All private keys are 2²⁵⁶, but all addresses that can exist are 2¹⁶⁰, which is an enormous difference. This means that there must be some address collisions.

If I wanted to calculate the storage that would be needed for 1M of addresses I would do that on a csv file:

Code:

PRIVATE_KEY1,ADDRESS1_LEGACY,ADDRESS1_BECH32,ADDRESS1_P2SH
PRIVATE_KEY2,ADDRESS2_LEGACY,ADDRESS2_BECH32,ADDRESS2_P2SH
...
PRIVATE_KEY10000000,ADDRESS10000000_LEGACY,ADDRESS10000000_BECH32,ADDRESS10000000_P2SH

One line is 165 bytes:

Code:

L1CaDFTppeWGoz7pstPR1B8GUCTa2vw5C2fHBGp1G33vQd9PepBM,1BUp7h85TuZxPFarjD2zjChPpr5Upwnj1A,bc1qwtearx8uglkvf2ehv37na5k7myl9sll9uyjqx0,3PE7q99B22vj3qHYesF8YSioTpwC1EYCtk

So, 1 million of addresses would be 165 million bytes which is equal with 165 megabytes. Pretty heavy csv file, my computer would crash if I opened it.

_______________________________________________

But how about 2²⁵⁶ lines? Well, that's:

115​792​089​237​316​195​423​570​985​008​687​907​853​269​984​665​640​564​039​457​584​007​913​129​639​936 times 165.

= 1910569472415717224488921252643350479578954746983069306651050136130566639058944 0 bytes

We can't pretend to understand that number, but let's try. Based on this quora link, the 2018's 3D flash drives contain 1600 bits per cubic nanometer. This means that we will need:

9,552,847,362,078,586,122,444,606,263,216,752,397,894,773,734,915,346,533,255,250,680,652 cubic meters of pure hardware.

Universe has been calculated that it's 1.8 x 10^28 cubic meters. That's 1/530713742337699229024700347956486244327 of the needed hardware space. Have I made any mistakes? Because it seems too huge to me.

[jackg]

That's only 700 square light years though if it's right (you should've used a factor of 10^-27 for converting metres to nanometers).

[BlackHatCoiner]

That's only 700 square light years though if it's right (you should've used a factor of 10^-27 for converting metres to nanometers).

So, will I get an even bigger number? Isn't it going crazy?

[jackg]

That's only 700 square light years though if it's right (you should've used a factor of 10^-27 for converting metres to nanometers).

So, will I get an even bigger number? Isn't it going crazy?

Yeah if you didn't cube the 10^-9 then it'll be out by a factor of 10^18...

If you need an explanation, a decimetre square is 10cmx10cmx10cm. This is 1000cm2 but is also 0.1x0.1x0.1=0.001m2.

[odolvlobo]

All private keys are 2²⁵⁶, but all addresses that can exist are 2¹⁶⁰, which is an enormous difference. This means that there must be some address collisions.

An address is an encoding of a 1-byte network value plus a 20-byte hash, plus a checksum. So, the space needed to store an address is 21 bytes. Since there are 2⁸ possible network values and 2¹⁶⁰ possible hashes, the amount of space needed to store all of the addresses is 21 x 2¹⁶⁸ bytes, or approximately 8x10⁵¹ bytes. That's equivalent to 8,000,000,000,000,000,000,000,000,000,000,000,000,000 1 TB drives.

However, it raises the question, why would you want to store every address, when any address can easily be generated?

As for collisions... Yes. Each possible address can be derived from an average of 2⁹⁷ private keys, which is an extraordinarily small number compared to the 2²⁵⁶ possible private keys.

[odolvlobo]

So, 1 million of addresses would be 165 million bytes which is equal with 165 megabytes. Pretty heavy csv file, my computer would crash if I opened it.

I see. You are really asking about how much space is needed to store a private key for every address.

Since a private key is 256 bits (plus a checksum and an 8-bit prefix that we can assume is always 0x80), it requires 32 bytes. The space required to store a private key for every address is 32 x 2¹⁶⁰ bytes, or approximately 4.7x10⁴⁹ bytes.

Assuming that you can store 200 bytes (1600 bites) in a nm³, that means that the volume required is 235,000,000,000,000,000,000 m³ ( 4.7x10⁴⁹ / 200 / 10²⁷ ).