4 ms·

Open to correction, but isn't 128 bits a little less than 40 decimal digits?

prezjordan 6y ago
3.4 x 10^38 So we'd be off by about <30 hydrogen atoms instead of <1 :)
mixedmath 6y ago
Yes, just slightly. It would take 133 bits. So I suppose a packet would only hold the width of the universe up to an error of maybe the width of 32 hydrogen atoms. Clearly unacceptable. I'm sure this be fixed in IPv7.
- tinus_hn 6y ago
  Also you’d need to account for the 3 to the left of the decimal point.
- neilwilson 6y ago
  The next question is what is that IPv6 address? c90f:daa2:2168:c234:c4c6:628b:80dc:1cd1 I think
- segfaultbuserr 6y ago
  There's no IPv7 beyond a proposal, but it was already fixed in IPv9 and deployed in production systems. RFC 1606 - A Historical Perspective On The Usage Of IP Version 9 https://tools.ietf.org/html/rfc1606 https://tools.ietf.org/html/rfc1606 > Whilst there are still many addresses unallocated the available space has been sharply decreased. The discovery of intelligent life on other solar systems with the parallel discovery of a faster-than-light transport stack is the main cause. This enables real time communication with them, and has made the allocation of world-size address spaces necessary, at the level 3 routing hierarchy. There is still only 1 global (spatial) level 2 galaxy wide network required for this galaxy, although the establishment of permanent space stations in deep space may start to exhaust this. This allows level 1 to be used for inter-galaxy routing. The most pressing problem now is the case of parallel universes. Of course there is the danger of assuming that there is no higher extrapolation than parallel universes...
chmod775 6y ago
The highest number in base 2 you can cram in 40bits is roughly 0.34 * 10^39. I shifted it so the exponent is equal to the number of digits we can fit. 128 bits could indeed only fit the first 39 digits of PI. And even then you haven't really stored PI, because you haven't stored where the comma is supposed to go.

3.4 x 10^38 So we'd be off by about <30 hydrogen atoms instead of <1 :)

Yes, just slightly. It would take 133 bits. So I suppose a packet would only hold the width of the universe up to an error of maybe the width of 32 hydrogen atoms. Clearly unacceptable. I'm sure this be fixed in IPv7.

tinus_hn 6y ago
Also you’d need to account for the 3 to the left of the decimal point.
neilwilson 6y ago
The next question is what is that IPv6 address? c90f:daa2:2168:c234:c4c6:628b:80dc:1cd1 I think
segfaultbuserr 6y ago
There's no IPv7 beyond a proposal, but it was already fixed in IPv9 and deployed in production systems. RFC 1606 - A Historical Perspective On The Usage Of IP Version 9 https://tools.ietf.org/html/rfc1606 https://tools.ietf.org/html/rfc1606 > Whilst there are still many addresses unallocated the available space has been sharply decreased. The discovery of intelligent life on other solar systems with the parallel discovery of a faster-than-light transport stack is the main cause. This enables real time communication with them, and has made the allocation of world-size address spaces necessary, at the level 3 routing hierarchy. There is still only 1 global (spatial) level 2 galaxy wide network required for this galaxy, although the establishment of permanent space stations in deep space may start to exhaust this. This allows level 1 to be used for inter-galaxy routing. The most pressing problem now is the case of parallel universes. Of course there is the danger of assuming that there is no higher extrapolation than parallel universes...

chmod775 6y ago

The highest number in base 2 you can cram in 40bits is roughly 0.34 * 10^39. I shifted it so the exponent is equal to the number of digits we can fit. 128 bits could indeed only fit the first 39 digits of PI. And even then you haven't really stored PI, because you haven't stored where the comma is supposed to go.