Re:Non-Universality in Computation: The Myth of the Universa...
by frisco on Thursday April 16, 2009 @12:02PM
(#565338)
It sounds like he's stating that a UTM can't simulate an arbitrary lambda function then, since that returns a variable function (a second-order UTM: a UTM can't simulate a UTM, which makes it not a UTM?).
In order to evaluate a lambda you need to be able to compute an arbitrary function of arbitrary variables; with infinite resources, you should be able to process infinite variables? Surely there's work done on UTMs and lambdas.
Re:Non-Universality in Computation: The Myth of the Universa...
by davidmathers on Thursday April 16, 2009 @01:07PM
(#565438)
It sounds like he's stating that a UTM can't simulate an arbitrary lambda function then
The concept doesn't make sense. Lambda functions don't do anything, they're just notations. The process of beta reduction is what causes things/computation to happen. Beta reduction rewrites all the lambda functions, hence they're no longer the same functions. What you're saying is something like: a computer can't compute "computation".
Re:Non-Universality in Computation: The Myth of the Universa...
by frisco on Thursday April 16, 2009 @01:14PM
(#565450)
Example: a function accepts as input the source of another program (assume it's computable and halts). Thus, the original function is variable; can it be simulated by a UTM for any arbitrary valid input? I'm probably using the wrong terminology here, but the overall logic is what I'm getting at.
Re:Non-Universality in Computation: The Myth of the Universa...
by davidmathers on Thursday April 16, 2009 @01:20PM
(#565459)
The input to a function can't change (it's a fixed set). If the input changes then you've defined a new function. That's the mathematical definition of a function. You're talking about a set/class of functions, so you would need 1 UTM per function.
Sure. But that's not what Turing was talking about. Give a UTM a computable function, and the UTM will compute it. Specify an infinite class of functions F(t), and of course a UTM will have trouble.
There's no contradiction here, just a really weird way of defining "function".