# Complexity

• The whole point of the big-O/Ω/Θ stuff was to be able to say something useful about algorithms.
• So, let's return to some algorithms and see if we learned anything.
• Consider this simple procedure that sums a list (of numbers, we assume):
• First: is the algorithm correct? Does it solve the problem specified?
• Second: is it fast?
• To evaluate the running time of an algorithm, we will simply ask how many “steps” it takes.
• In this case, we can count the number of times it runs the += line.
• For a list with $$n$$ elements, it takes $$n$$ steps.
• Or is counting the += line the right thing to do?
• When implementing the for loop, each iteration requires an add (for the loop index) and a comparison (to check the exit condition). We should count those.
• Also, the variable initialization and return steps.
• So, $$3n+2$$ steps.
• But, not all of those steps are the same.
• How long does an x86 ADD instruction take compared to a CMP or RET instruction?
• Will the compiler keep both i and total in registers, or will one/both be in RAM? (A factor of ~10 difference.)
• How do those instructions interact in the pipeline? Which can be sent through parallel pipelines in the processor and executed concurrently?
• The answer to those is simple: I don't know and you don't either.
• That's part of the reason we're asking about algorithms, not programs.
• But both $$n$$ and $$3n+2$$ are perfectly reasonable proposals for the answer.
• Deciding between them requires more knowledge about the actual implementation details than we have.
• Good thing we have the function growth notation.
• Remember: this is easy for $$n=5$$ elements. A good or bad algorithm will both be fast then.
• We want to know how the algorithm behaves for large $$n$$.
• Finally our answer: the sum procedure has running time $$\Theta(n)$$.
• We'll say that this algorithm has time complexity $$\Theta(n)$$, or “runs in linear time”.
• Both $$n$$ and $$3n+2$$ are $$\Theta(n)$$, and so is any other “exact” formula we could come up with.
• The easy answer (count the += line) was just as correct as the very careful one.
• The big-Θ notation hides all of the details we can't figure out anyway.
• Another example: print out the sum of each two numbers in a list.
• That is, given the list [1,2,3,4,5], we want to find 1+2, 1+3, 1+4, 1+5, 2+3, 2+4,….
• Pseudocode:
• For a list with $$n$$ elements, the for j loop iterates $$n-1$$ times when it is called with i==0, then $$n-2$$ times, then $$n-3$$ times,…
• So, the total number of times the print step runs is \begin{align*} (n-1)+(n-2)+\cdots+2+1 &= \sum_{k=1}^{n-1} k\\ &= \frac{n(n-1)}{2}\\ &= \frac{n^2}{2}-\frac{n}{2}\,. \end{align*}
• If we had counted the initialization of the for loops, counter incrementing, etc, we might have come up with something more like $$\frac{3}{2}n^2 + \frac{1}{2}n + 1$$.
• Either way, the answer we give is that it takes $$\Theta(n^2)$$ steps.
• Or, the algorithm “has time complexity $$\Theta(n^2)$$” or “has $$\Theta(n^2)$$ running time” or “has quadratic running time”.
• The lesson: when counting running time, you can be a bit sloppy.
• We only need to worry about the inner-most loop(s), not the number of steps in there, or work in the outer levels.
• … because they are going to disappear anyway as constant factors and lower-order terms when they go into a big-O/Ω/Θ anyway.

# Average and Worst Case

• Consider a linear search: we want to find an element in a list and return its (first) position, or -1 if it's not there.
• How many steps there?
• The answer is: it depends.
• If the thing we're looking for is in the first position, it takes $$\Theta(1)$$ steps.
• If it's at the end, or not there, it takes $$\Theta(n)$$ steps.
• The easiest thing to count is usually the worst case: what is the maximum steps required for any input of size $$n$$?
• The worst case is that we go all the way to the end of the list, but don't find it and return -1.
• The only line that makes sense to count here is the if line. It's in the inner-most loop, and is executed for every iteration.
• We could also count the number of comparisons made: the == and the implicit comparison in the for loop.
• That is either $$n$$ or $$2n+1$$ steps, so $$\Theta(n)$$ complexity.
• The other useful option is the average case: what is the average steps required over all inputs of size $$n$$?
• Much harder to calculate, since you need to consider every possible input to the algorithm.
• Even if we assume the element is found, the possible number of comparisons are:
• On average, the number of comparisons is: $\frac{2+4+\cdots+2n}{n} = n+1\,.$
• Again, we have $$\Theta(n)$$ complexity.
• … but it's a good thing we checked. Some algorithms are different.

• We have said that these running times are important when it comes to running times of algorithm.
• But we are throwing away a lot of information when we look only at big-O/Ω/Θ.
• The lower-order terms must mean something.
• The leading constants definitely do.
• Assuming one million operations per second, this is the approximate running time of an algorithm given running time, with an input of size $$n$$:
• Maybe that gives a little idea why we'll only worry about complexity
• … at least at first.
• A summary:
• If you can get $$O(\log n)$$ life is good: hand it in and go home.
• $$O(n\log n)$$ is pretty good: hard to complain about it.
• $$O(n^k)$$ could be bad, depending on $$k$$: you won't be solving huge problems. These are polynomial complexity algorithms for $$k\ge 1$$.
• $$\Omega(k^n)$$ is a disaster: almost as bad as no algorithm at all if you have double-digit input sizes. These are exponential complexity algorithms for $$k\gt 1$$.
• A problem that has a polynomial-time algorithm is called tractable.
• No polynomial time algorithm: intractable.
• There is a big category of problems that nobody has a polynomial-time algorithm for, but also can't prove that none exists: the NP-complete problems.
• Some examples: Boolean satisfiability, travelling salesman, Hamiltonian path, many scheduling problems, Sudoku (size $$n$$).
• If you have an algorithm with a higher complexity than necessary, no amount of clever programming will make up for it.
• No combination of these will make a $$O(n^2)$$ algorithm faster than an $$O(n\log n)$$: faster language, better optimizer, hand-optimization of code, faster processor.
• Important point: the complexity notations only say things about large $$n$$.
• If you always have small inputs, you might not care.
• Algorithms with higher complexity class might be faster in practice, if you always have small inputs.
• e.g. Insertion sort has running time $$\Theta(n^2)$$ but is generally faster than $$\Theta(n\log n)$$ sorting algorithms for lists of around 10 or fewer elements.
• The most important info that the complexity notations throw away is the leading constant.
• There is a difference between $$n^2$$ instructions and $$100n^2$$ instructions to solve a problem.
• Once you have the right big-O, then it's time to worry about the constants.
• That's what clever programming can do.
• When we're talking about algorithms (and not programming), the constants don't usually matter much.
• It's rare to have an algorithm with a big leading constant.
• So it's not really possible to decide between the algorithms.
• Usually it's a choice between $$4n\log n$$ or $$5n\log n$$: you probably have to implement, compile, and profile to decide for sure.
• Example: sorting algorithms. There are several algorithms to sort a list/array.
• Insertion/Selection/Bubble Sorts: $$\Theta(n^2)$$.
• Merge/Heap Sorts: $$\Theta(n\log n)$$.
• Quicksort: $$\Theta(n\log n)$$ average case but (very rarely) $$\Theta(n^2)$$ worst case.
• But quicksort is usually faster in practice.
• … except when it isn't.
• Several recent languages/libraries have implemented a heavily-optimized mergesort (e.g. Python, Perl, Java ≥JDK1.3, Haskell, some STL implementations) instead of Quicksort (C, Ruby, some other STL implementations).

# Space Complexity

• We have only been talking about running time/speed so far.
• It also makes good sense to talk about the complexity of other things.
• Most notably, memory use by an algorithm.
• An algorithm that uses $$\Theta(n^{3})$$ space is bad. Maybe as bad as $$\Theta(n^{3})$$ time.
• An algorithm that uses $$O(1)$$ extra space (in addition to space needed to store the input) is called in-place.
• e.g. selection sort is in-place, but mergesort ($$\Theta(n)$$ extra space) and Quicksort ($$\Theta(\log n)$$ extra space, average case) aren't.