Hacker News Reader

Since x + y = z can't be reversed without preserving input, this would have to use additional memory. What if you then have z + a = b? Can you now throw out x/y, or does it have to be reversible all the way back to the beginning?

View on HN

You start with {x, y, z=0, a, b=0} and then reversibly compute z+=x; z+=y, b+=z; b+=a to end up with {x, y, z=x+y, a, b=x+y+a} from which you could reversibly compute the starting state again with b-=a, etc.

Thanks for this

Right, but when can you throw out your initial values? Because if this is going to be usable you have to free up memory sometime.

It goes something like this:

1) Perform some reversible operation

2) Copy answer to another location (At some cost in heat)

3) Perform reverse operation, which rolls back memory to starting point

When you're done you have zeroed out memory, and the only costly operation was the copying of the answer.

You can't. If you want to zero the memory holding the initial values, you have to exchange it with some other memory. This is how current ('non'-reversable) computers work: exchanging junk data with the large number of known-value bits constituting a low-temperature and/or fixed-voltage environment via diffusion (eg, attaching a bit storage location to GND to clear it by diffusing its change into the GND bus + waste heat).

Reply on news.ycombinator.com