Into how many pieces can you cut this string?

Python, 88 75 73 bytes

lambda x:max(sum((a+.5-m)*(a+.5-n)<0for m,n in zip(x,x[1:]))for a in x)+1

Just a straightforward lambda

---

Just to show another approach:

Pyth, 28 27 bytes

heSmsmq@S+k(d)1dC,QtQm+b.5Q

This one's roughly equivalent to

lambda x:max(sum(a+.5==sorted(n+(a+.5,))[1]for n in zip(x,x[1:]))for a in x)+1

applied to the input list from STDIN. Try it on the online interpreter.

Pyth: 31 30 29 28 24 23 character (Python 68 chars)

heSmsm&<hSkdgeSkdC,tQQQ

Try it here: Pyth Compiler/Executor

It expects a list of integers as input [-1, 3, 1, -2, 5, 2, 3, 4]

It's a straightforward translation of my Python program:

lambda s:1+max(sum(min(a)<i<=max(a)for a in zip(s,s[1:]))for i in s)

Old solution: Pyth 28 char

Just for archiving reasons.

heSmsm<*-dhk-dek0C,tQQm+b.5Q

A corresponding Python code would be:

f=lambda x:1+max(sum((i-a)*(i-b)<0for a,b in zip(x,x[1:]))for i in [j+.5 for j in x])

APL, 16 14 bytes

1+⌈/+/2≠/∘.≤⍨⎕

Thanks to @ngn for saving 2 bytes.

The ⎕ is actually a box character, not a missing-font error. You can try the program at tryapl.org, but since ⎕ is not fully supported there, you have to replace it by the input value:

    1+⌈/+/2≠/∘.≤⍨ ¯1 3 1 ¯2 5 2 3 4
6

Explanation

The program is best explained with the example input s = ¯1 3 1 ¯2 5 2 3 4, which is taken from STDIN by ⎕. First, we compute the ≤-outer product of s with itself using ∘.≤⍨. This results in a Boolean matrix whose ith row tells which elements of s are less than or equal to s[i]:

1 1 1 0 1 1 1 1
0 1 0 0 1 0 1 1
0 1 1 0 1 1 1 1
1 1 1 1 1 1 1 1
0 0 0 0 1 0 0 0
0 1 0 0 1 1 1 1
0 1 0 0 1 0 1 1
0 0 0 0 1 0 0 1

The occurrences of 0 1 and 1 0 on row i mark places where the string passes over the point s[i] + 0.5. We match on these on every row using 2≠/, "reduce 2-sublists by ≠":

0 0 1 1 0 0 0
1 1 0 1 1 1 0
1 0 1 1 0 0 0
0 0 0 0 0 0 0
0 0 0 1 1 0 0
1 1 0 1 0 0 0
1 1 0 1 1 1 0
0 0 0 1 1 0 1

What remains is to take the sums of the rows with +/

2 5 3 0 2 3 5 3

and one plus the maximum of these with 1+⌈/:

6

The result is automatically printed to STDOUT in most APL implementations.