On Tue, Nov 6, 2012 at 12:45 AM, Oscar Benjamin
On 5 November 2012 21:00, Daniel Fischer wrote
and the former is not even correct, because there is exactly one permutation of an empty list.
I'm trying to convince myself that this is logically necessary but it still seems consistent in my mind that the set of permutations of an empty list is empty. I guess maybe if you posit that a sequence should always be contained by the set of its permutations then you reach this conclusion.
A permutation of a set is a bijection from the set to itself. There is one map from the empty set to itself, the empty map, that is a bijection. The number of permutations of a set of n elements is n!, and 0! = 1 should make you expect that there is one permutation of an empty set.
Trace the execution of very simple cases (empty lists, singleton lists,
lists
with two elements) by hand with pencil and paper. That's the most instructive and fruitful way.
Check the results of simple cases against what you know the result ought to be.
I see what you mean now.
Single-step through the evaluation of simple cases in the ghci debugger if necessary.
I need to work out how to use this. I've just found why I was unable to use it before: it only works if the file is interpreted. The quick fix was to delete all files created by the compiler. Is there a way that I can tell ghci to just ignore those files and load my program as interpreted for debugging?
You can
ghci> :load *ModuleName to get it interpreted.