7. Minimum selection on ranges

The standard is not a limitation on ideas

When I joined A9 I wrote a little generic function and it was not using STL iterators. It used some other iterator like thing, but, totally different. We wrote it, tested it, sent it in and one of the wise people who works for us looked at it and sent back the reply, “you guys have to learn about STL iterators”, which I really appreciate. In Silicon Valley you have to be careful about saying you’re an expert. You might be talking to the guy who invented it.

That guy’s attitude is that if there is something in the standard, that’s the only thing you should use. This is false. Use STL only when it fits. If you come up with some other idea (many ideas in the world) try it.

Minimum element in a range

Now we want to apply min to a range. Suppose we want to find an item with smallest price. We need an interface which allows us to take as many items as you like and find the minimum. In STL it’s called std::min_element

The algorithm is very simple, but there are details we need to learn about ranges. We will show an implementation and then talk about it:

template <typename I, typename Compare>
// I is a ForwardIterator
// Compare is StrictWeakOrdering on ValueType(I)
I min_element(I first, I last, Compare cmp) {
  if (first == last) {
    return last;
  }

  I min_el = first;
  ++first;

  while (first != last) {
    if (cmp(*first, *min_el)) {
      min_el = first;
    }
    ++first;
  }

  return min_el;
}

Iterator conventions

We need to talk about iterator conventions. Notice that min_element doesn’t return the value itself, but the iterator pointing to the minimum element. Why? Because we probably want to update the value.

Suppose I’m a manager and I want to look for the worst-performing guy and then fire him (joke). I don’t want his value, I want a handle on him. I want an iterator. Do you see my point? You very often want to do things with what you find.

There is another reason. The range might be empty¹. In which case we return the last iterator.

first and last are maybe bad names, but that’s what they are. They are hard to change, because I called them that everywhere. last actually doesn’t mean last. last means one after the last element². In order to define a sequence you need to point past the last. Because you want to able to work with empty ranges.

        a b c d
first   ^
last    ^

Suppose last is actually last and you point to the same place. That indicates a range of one element. There is no way to do zero.

Later, we will look at algorithms for partitioning. We want to partition good people from bad people. After the partition we will return a pointer which separates good from bad. There will be first, last, and middle. The middle partitions between the first bad and first good. But there may be no good, or no bad elements. We need to able to return an empty range.

In C++ this is a standard convention. Some people in the world of Java and Python are slowly realizing that maybe it has something to do with mathematics and not with C++. But it will take decades before people fully realize that you have to always go past the end. Mathematically you need semi-open intervals. They are denoted like so:

[i, j)

Or in our terms:

[first, last)

Forward iterator

What kind of thing is I. We indicated it was a ForwardIterator. There is also InputIterator. Both are for iterating forward, as opposed to random jumps. What is an InputIterator? Input iterators describe algorithms which go through the river once³. It’s like they’re reading stuff from the wire⁴. Imagine the various kinds of streams⁵.

But, in our algorithm we store a previous position in the variable min_el. Therefore the things which go through the wire will not work. So it must be a ForwardIterator.

Finding min and max together

How many comparisons do we need to find the min of five elements? Four. In general why do we need n - 1 comparisons. Why no more? We don’t need to compare an element with itself. Why no fewer? Maybe we could do it in n - 2 with a clever algorithm? The simple argument to remember is that n - 1 guys have to lose. We’re finding a winner in our competition. If a person didn’t play in a competition, if he didn’t lose, we cannot eliminate him. We need to eliminate all but one.

How many comparisons if we need to find minimum and maximum together? Obviously we could do 2n - 2, but what about fewer⁶? The idea is very simple. Assume that we worked up to the middle of of our range and we have a running min and a running max. The temptation is take the next element compare him with min and with max. That’s very sad because very often we will do two comparisons and then discount the other. So for one element we will be doing two comparisons. The trick is to pick two new elements, the next element, and the one after next. We know nothing about them. We compare them with one comparison. Then we know that one of them is a potential min and the other is a potential max. Then we compare potential min with the current min and potential max with current max. So we need to do three comparisons for every pair, so roughly speaking we need 3n/2.

Let’s do it. This can be a little bit tricky. Do not start writing things from the beginning because you don’t know what goes in the beginning. Try to write code from the middle. Try to write the body of the loop. That’s where I described the algorithm to start with. Then go back and make the loop invariants on which we depend true.

template <typename I, typename Compare>
// I is a ForwardIterator
// Compare is StrictWeakOrdering on ValueType(I)
std::pair<I, I> minmax_element(I first, I last, Compare cmp) {
  if (first == last) {
    return std::make_pair(last, last);
  }

  I min_el = first;
  ++first;
  if (first == last) {
    return std::make_pair(min_el, min_el);
  }

  I max_el = first;
  I next = first;
  ++next;

  if (cmp(*max_el, *min_el)) {
    std::swap(min_el, max_el);
  }

  while (first != last && next != last) {
    // min_el contains the current min
    // max_el contains the current max
    // next and first and the next pair of elements to examine
    I potential_min = first;
    I potential_max = next;

    if (cmp(*potential_max, *potential_min)) {
      std::swap(potential_max, potential_min);
    }

    if (cmp(*potential_min, *min_el)) {
      min_el = potential_min;
    }

    if (!cmp(*potential_max, *max_el)) {
      max_el = potential_max;
    }

    ++next;
    first = next;
    ++next;
  }

  if (first != last) {
    // odd elements, one left over

    if (cmp(*first, *min_el)) {
      min_el = first;
      // first < min_el <= max_el, so we don't need to check the next case
    } else if (!cmp(*first, *max_el)) {
      max_el = first;
    }
  }

  return std::make_pair(min_el, max_el);
}

Example:

int a[] = { 1, 2, 3, 4, 5, 4, 3, 2, 1, 8, 7, 6, 2 };
size_t n = sizeof(a) / sizeof(int);
auto pair = minmax_element(a, a + n, std::less<int>());

This algorithm was invented by Ira Pohl of UC Santa Cruz, in the mid-seventies⁷. He also proved it was optimal. It’s also practically good. It was added to the standard in C++11.

Code

• minmax_range.h
• test_minmax.cpp