Comment by nodja

The way I think about it is that it's unreasonable for a compute graph with a static number of operations to be able to answer both y=a*10 and something like y=((((x+x)*(x+1))/((2*x)+2))+((x*(x+3))/(x+3))-((x*x)/(x+1))+((x*x)/(x+1))-((x*(x+3))/(x+3))) in a single forward pass. Tokens are essentially a unit of work and can also be used for intermediate steps, not just final results.