def qs4(ls):
'''
And we can even take away the accum. Notice, however, the final comma, making
the statement a tuple.
'''
pivot = middle(ls)
iff(len, (qs4, {_ < pivot}) + [pivot] + (qs4, {_ > pivot})),
def fib4(x):
'''
And last but not least, we are enclosing this in a singleton tuple with no
reference to x (note the , at the end). We can do this because the compiler
infers that _ can only refer to x, since there are no other variables in the
scope of the function.
'''
iff(_ > 1, (fib4, _ - 1) + (fib4, _ - 2)),
def qs3(ls):
'''
And we can also take away the call to p.
'''
pivot = middle(ls)
(ls, iff(len, (qs3, {_ < pivot}) + [pivot] + (qs3, {_ > pivot})))
def qs2(ls):
'''
Now, we take away the return statement for space.
'''
pivot = middle(ls)
p(ls, iff(len, (qs2, {_ < pivot}) + [pivot] + (qs2, {_ > pivot})))
def fib1(x):
'''
Here, however, we use the iff macro. So for a conditional fArg cond, and a
statement, iff(cond,statement) evaluates as:
{cond:statement,
'else':_}
So, we only apply the sum to x if it's above 1.
'''
return p(x, iff(_ > 1, (fib1, _ - 1) + (fib1, _ - 2)))
def qs1(ls):
'''
Now, we are going to apply the iff macro. So for a conditional fArg cond, and a
statement, iff(cond,statement) evaluates as:
{cond:statement,
'else':_}
.
So, if len evaluates as True (nonzero), then we return the statement, otherwise
we return the empty list.
'''
pivot = middle(ls)
return p(ls, iff(len, (qs1, {_ < pivot}) + [pivot] + (qs1, {_ > pivot})))
def fib3(x):
'''
And now, we are even stripping out the pype call, which the compiler can handle.
'''
(x, iff(_ > 1, (fib3, _ - 1) + (fib3, _ - 2)))
def fib2(x):
'''
Now, we are stripping out the return statement, since the compiler includes it.
'''
p(x, iff(_ > 1, (fib2, _ - 1) + (fib2, _ - 2)))