18Sets Appeal

Earlier [Sets as Collective Data] we introduced sets. Recall that the elements of a set have no specific order, and ignore duplicates.If these ideas are not familiar, please read Sets as Collective Data, since they will be important when discussing the representation of sets. At that time we relied on Pyret’s built-in representation of sets. Now we will discuss how to build sets for ourselves. In what follows, we will focus only on sets of numbers.

We will start by discussing how to represent sets using lists. Intuitively, using lists to represent sets of data seems problematic, because lists respect both order and duplication. For instance,
check:
[list: 1, 2, 3] is [list: 3, 2, 1, 1]
end
fails.

In principle, we want sets to obey the following interface:Note that a type called Set is already built into Pyret, so we won’t use that name below.
<set-operations> ::=
 mt-set :: Set is-in :: (T, Set -> Bool) insert :: (T, Set -> Set) union :: (Set, Set -> Set) size :: (Set -> Number) to-list :: (Set -> List)
We may also find it also useful to have functions such as
insert-many :: (List<T>, Set<T> -> Set<T>)
which, combined with mt-set, easily gives us a to-set function.

Sets can contain many kinds of values, but not necessarily any kind: we need to be able to check for two values being equal (which is a requirement for a set, but not for a list!), which can’t be done with all values [Comparing Functions]; and sometimes we might even want the elements to obey an ordering [Converting Values to Ordered Values]. Numbers satisfy both characteristics.

18.1Representing Sets by Lists

In what follows we will see multiple different representations of sets, so we will want names to tell them apart. We’ll use LSet to stand for “sets represented as lists”.

As a starting point, let’s consider the implementation of sets using lists as the underlying representation. After all, a set appears to merely be a list wherein we ignore the order of elements.

18.1.1Representation Choices

The empty list can stand in for the empty set—
type LSet = List
mt-set = empty
and we can presumably define size as
fun size<T>(s :: LSet<T>) -> Number:
s.length()
end
However, this reduction (of sets to lists) can be dangerous:
1. There is a subtle difference between lists and sets. The list
[list: 1, 1]
is not the same as
[list: 1]
because the first list has length two whereas the second has length one. Treated as a set, however, the two are the same: they both have size one. Thus, our implementation of size above is incorrect if we don’t take into account duplicates (either during insertion or while computing the size).

2. We might falsely make assumptions about the order in which elements are retrieved from the set due to the ordering guaranteed provided by the underlying list representation. This might hide bugs that we don’t discover until we change the representation.

3. We might have chosen a set representation because we didn’t need to care about order, and expected lots of duplicate items. A list representation might store all the duplicates, resulting in significantly more memory use (and slower programs) than we expected.

To avoid these perils, we have to be precise about how we’re going to use lists to represent sets. One key question (but not the only one, as we’ll soon see [Choosing Between Representations]) is what to do about duplicates. One possibility is for insert to check whether an element is already in the set and, if so, leave the representation unchanged; this incurs a cost during insertion but avoids unnecessary duplication and lets us use length to implement size. The other option is to define insert as linkliterally,
insert = link
and have some other procedure perform the filtering of duplicates.

18.1.2Time Complexity

What is the complexity of this representation of sets? Let’s consider just insert, check, and size. Suppose the size of the set is $$k$$ (where, to avoid ambiguity, we let $$k$$ represent the number of distinct elements). The complexity of these operations depends on whether or not we store duplicates:
• If we don’t store duplicates, then size is simply length, which takes time linear in $$k$$. Similarly, check only needs to traverse the list once to determine whether or not an element is present, which also takes time linear in $$k$$. But insert needs to check whether an element is already present, which takes time linear in $$k$$, followed by at most a constant-time operation (link).

• If we do store duplicates, then insert is constant time: it simply links on the new element without regard to whether it already is in the set representation. check traverses the list once, but the number of elements it needs to visit could be significantly greater than $$k$$, depending on how many duplicates have been added. Finally, size needs to check whether or not each element is duplicated before counting it.

Do Now!

What is the time complexity of size if the list has duplicates?

One implementation of size is
fun size<T>(s :: LSet<T>) -> Number:
cases (List) s:
| empty => 0
if r.member(f):
size(r)
else:
1 + size(r)
end
end
end

Let’s now compute the complexity of the body of the function, assuming the number of distinct elements in s is $$k$$ but the actual number of elements in s is $$d$$, where $$d \geq k$$. To compute the time to run size on $$d$$ elements, $$T(d)$$, we should determine the number of operations in each question and answer. The first question has a constant number of operations, and the first answer also a constant. The second question also has a constant number of operations. Its answer is a conditional, whose first question (r.member(f) needs to traverse the entire list, and hence has $$O([k \rightarrow d])$$ operations. If it succeeds, we recur on something of size $$T(d-1)$$; else we do the same but perform a constant more operations. Thus $$T(0)$$ is a constant, while the recurrence (in big-Oh terms) is

\begin{equation*}T(d) = d + T(d-1)\end{equation*}

Thus $$T \in O([d \rightarrow d^2])$$. Note that this is quadratic in the number of elements in the list, which may be much bigger than the size of the set.

18.1.3Choosing Between Representations

Now that we have two representations with different complexities, it’s worth thinking about how to choose between them. To do so, let’s build up the following table. The table distinguishes between the interface (the set) and the implementation (the list), because—owing to duplicates in the representation—these two may not be the same. In the table we’ll consider just two of the most common operations, insertion and membership checking:
 With Duplicates Without Duplicates insert is-in insert is-in Size of Set constant linear linear linear Size of List constant linear linear linear
A naive reading of this would suggest that the representation with duplicates is better because it’s sometimes constant and sometimes linear, whereas the version without duplicates is always linear. However, this masks a very important distinction: what the linear means. When there are no duplicates, the size of the list is the same as the size of the set. However, with duplicates, the size of the list can be arbitrarily larger than that of the set!

Based on this, we can draw several lessons:
1. Which representation we choose is a matter of how much duplication we expect. If there won’t be many duplicates, then the version that stores duplicates pays a small extra price in return for some faster operations.

2. Which representation we choose is also a matter of how often we expect each operation to be performed. The representation without duplication is “in the middle”: everything is roughly equally expensive (in the worst case). With duplicates is “at the extremes”: very cheap insertion, potentially very expensive membership. But if we will mostly only insert without checking membership, and especially if we know membership checking will only occur in situations where we’re willing to wait, then permitting duplicates may in fact be the smart choice. (When might we ever be in such a situation? Suppose your set represents a backup data structure; then we add lots of data but very rarely—indeed, only in case of some catastrophe—ever need to look for things in it.)

3. Another way to cast these insights is that our form of analysis is too weak. In situations where the complexity depends so heavily on a particular sequence of operations, big-Oh is too loose and we should instead study the complexity of specific sequences of operations. We will address precisely this question later [Halloween Analysis].

Moreover, there is no reason a program should use only one representation. It could well begin with one representation, then switch to another as it better understands its workload. The only thing it would need to do to switch is to convert all existing data between the representations.

How might this play out above? Observe that data conversion is very cheap in one direction: since every list without duplicates is automatically also a list with (potential) duplicates, converting in that direction is trivial (the representation stays unchanged, only its interpretation changes). The other direction is harder: we have to filter duplicates (which takes time quadratic in the number of elements in the list). Thus, a program can make an initial guess about its workload and pick a representation accordingly, but maintain statistics as it runs and, when it finds its assumption is wrong, switch representations—and can do so as many times as needed.

18.1.4Other Operations

Exercise

Implement the remaining operations catalogued above (<set-operations>) under each list representation.

Exercise

Implement the operation
remove :: (Set<T>, T -> Set<T>)
under each list representation (renaming Set appropriately. What difference do you see?

Do Now!

Suppose you’re asked to extend sets with these operations, as the set analog of first and rest:
one :: (Set<T> -> T)
others :: (Set<T> -> T)
You should refuse to do so! Do you see why?

With lists the “first” element is well-defined, whereas sets are defined to have no ordering. Indeed, just to make sure users of your sets don’t accidentally assume anything about your implementation (e.g., if you implement one using first, they may notice that one always returns the element most recently added to the list), you really ought to return a random element of the set on each invocation.

Unfortunately, returning a random element means the above interface is unusable. Suppose s is bound to a set containing 1, 2, and 3. Say the first time one(s) is invoked it returns 2, and the second time 1. (This already means one is not a function.) The third time it may again return 2. Thus others has to remember which element was returned the last time one was called, and return the set sans that element. Suppose we now invoke one on the result of calling others. That means we might have a situation where one(s) produces the same result as one(others(s)).

Exercise

Why is it unreasonable for one(s) to produce the same result as one(others(s))?

Exercise

Suppose you wanted to extend sets with a subset operation that partitioned the set according to some condition. What would its type be? See [REF join lists] for a similar operation.

Exercise

The types we have written above are not as crisp as they could be. Define a has-no-duplicates predicate, refine the relevant types with it, and check that the functions really do satisfy this criterion.

18.2Making Sets Grow on Trees

Let’s start by noting that it seems better, if at all possible, to avoid storing duplicates. Duplicates are only problematic during insertion due to the need for a membership test. But if we can make membership testing cheap, then we would be better off using it to check for duplicates and storing only one instance of each value (which also saves us space). Thus, let’s try to improve the time complexity of membership testing (and, hopefully, of other operations too).

It seems clear that with a (duplicate-free) list representation of a set, we cannot really beat linear time for membership checking. This is because at each step, we can eliminate only one element from contention which in the worst case requires a linear amount of work to examine the whole set. Instead, we need to eliminate many more elements with each comparison—more than just a constant.

In our handy set of recurrences [Solving Recurrences], one stands out: $$T(k) = T(k/2) + c$$. It says that if, with a constant amount of work we can eliminate half the input, we can perform membership checking in logarithmic time. This will be our goal.

Before we proceed, it’s worth putting logarithmic growth in perspective. Asymptotically, logarithmic is obviously not as nice as constant. However, logarithmic growth is very pleasant because it grows so slowly. For instance, if an input doubles from size $$k$$ to $$2k$$, its logarithm—and hence resource usage—grows only by $$\log 2k - \log k = \log 2$$, which is a constant. Indeed, for just about all problems, practically speaking the logarithm of the input size is bounded by a constant (that isn’t even very large). Therefore, in practice, for many programs, if we can shrink our resource consumption to logarithmic growth, it’s probably time to move on and focus on improving some other part of the system.

18.2.1Converting Values to Ordered Values

We have actually just made an extremely subtle assumption. When we check one element for membership and eliminate it, we have eliminated only one element. To eliminate more than one element, we need one element to “speak for” several. That is, eliminating that one value needs to have safely eliminated several others as well without their having to be consulted. In particular, then, we can no longer compare for mere equality, which compares one set element against another element; we need a comparison that compares against an element against a set of elements.

To do this, we have to convert an arbitrary datum into a datatype that permits such comparison. This is known as hashing. A hash function consumes an arbitrary value and produces a comparable representation of it (its hash)—most commonly (but not strictly necessarily), a number. A hash function must naturally be deterministic: a fixed value should always yield the same hash (otherwise, we might conclude that an element in the set is not actually in it, etc.). Particular uses may need additional properties: e.g., below we assume its output is partially ordered.

Let us now consider how one can compute hashes. If the input datatype is a number, it can serve as its own hash. Comparison simply uses numeric comparison (e.g., <). Then, transitivity of < ensures that if an element $$A$$ is less than another element $$B$$, then $$A$$ is also less than all the other elements bigger than $$B$$. The same principle applies if the datatype is a string, using string inequality comparison. But what if we are handed more complex datatypes?

Before we answer that, consider that in practice numbers are more efficient to compare than strings (since comparing two numbers is very nearly constant time). Thus, although we could use strings directly, it may be convenient to find a numeric representation of strings. In both cases, we will convert each character of the string into a number—e.g., by considering its ASCII encoding. Based on that, here are two hash functions:
1. Consider a list of primes as long as the string. Raise each prime by the corresponding number, and multiply the result. For instance, if the string is represented by the character codes [6, 4, 5] (the first character has code 6, the second one 4, and the third 5), we get the hash
num-expt(2, 6) * num-expt(3, 4) * num-expt(5, 5)
or 16200000.

2. Simply add together all the character codes. For the above example, this would correspond to the has
6 + 4 + 5
or 15.

The first representation is invertible, using the Fundamental Theorem of Arithmetic: given the resulting number, we can reconstruct the input unambiguously (i.e., 16200000 can only map to the input above, and none other). The second encoding is, of course, not invertible (e.g., simply permute the characters and, by commutativity, the sum will be the same).

Now let us consider more general datatypes. The principle of hashing will be similar. If we have a datatype with several variants, we can use a numeric tag to represent the variants: e.g., the primes will give us invertible tags. For each field of a record, we need an ordering of the fields (e.g., lexicographic, or “alphabetical” order), and must hash their contents recursively; having done so, we get in effect a string of numbers, which we have shown how to handle.

Now that we have understood how one can deterministically convert any arbitrary datum into a number, in what follows, we will assume that the trees representing sets are trees of numbers. However, it is worth considering what we really need out of a hash. In Set Membership by Hashing Redux, we will not need partial ordering. Invertibility is more tricky. In what follows below, we have assumed that finding a hash is tantamount to finding the set element itself, which is not true if multiple values can have the same hash. In that case, the easiest thing to do is to store alongside the hash all the values that hashed to it, and we must search through all of these values to find our desired element. Unfortunately, this does mean that in an especially perverse situation, the desired logarithmic complexity will actually be linear complexity after all!

In real systems, hashes of values are typically computed by the programming language implementation. This has the virtue that they can often be made unique. How does the system achieve this? Easy: it essentially uses the memory address of a value as its hash. (Well, not so fast! Sometimes the memory system can and does move values around [REF garbage collection]. In these cases computing a hash value is more complicated.)

18.2.2Using Binary Trees

Because logs come from trees.

Clearly, a list representation does not let us eliminate half the elements with a constant amount of work; instead, we need a tree. Thus we define a binary tree of (for simplicity) numbers:
data BT:
| leaf
| node(v :: Number, l :: BT, r :: BT)
end

Given this definition, let’s define the membership checker:
fun is-in-bt(e :: Number, s :: BT) -> Boolean:
cases (BT) s:
| leaf => false
| node(v, l, r) =>
if e == v:
true
else:
is-in-bt(e, l) or is-in-bt(e, r)
end
end
end
Oh, wait. If the element we’re looking for isn’t the root, what do we do? It could be in the left child or it could be in the right; we won’t know for sure until we’ve examined both. Thus, we can’t throw away half the elements; the only one we can dispose of is the value at the root. Furthermore, this property holds at every level of the tree. Thus, membership checking needs to examine the entire tree, and we still have complexity linear in the size of the set.

How can we improve on this? The comparison needs to help us eliminate not only the root but also one whole sub-tree. We can only do this if the comparison “speaks for” an entire sub-tree. It can do so if all elements in one sub-tree are less than or equal to the root value, and all elements in the other sub-tree are greater than or equal to it. Of course, we have to be consistent about which side contains which subset; it is conventional to put the smaller elements to the left and the bigger ones to the right. This refines our binary tree definition to give us a binary search tree (BST).

Do Now!

Here is a candiate predicate for recognizing when a binary tree is in fact a binary search tree:
fun is-a-bst-buggy(b :: BT) -> Boolean:
cases (BT) b:
| leaf => true
| node(v, l, r) =>
(is-leaf(l) or (l.v <= v)) and
(is-leaf(r) or (v <= r.v)) and
is-a-bst-buggy(l) and
is-a-bst-buggy(r)
end
end
Is this definition correct?

It’s not. To actually throw away half the tree, we need to be sure that everything in the left sub-tree is less than the value in the root and similarly, everything in the right sub-tree is greater than the root.We have used <= instead of < above because even though we don’t want to permit duplicates when representing sets, in other cases we might not want to be so stringent; this way we can reuse the above implementation for other purposes. But the definition above performs only a “shallow” comparison. Thus we could have a root a with a right child, b, such that b > a; and the b node could have a left child c such that c < b; but this does not guarantee that c > a. In fact, it is easy to construct a counter-example that passes this check:
check:
node(5, node(3, leaf, node(6, leaf, leaf)), leaf)
satisfies is-a-bst-buggy # FALSE!
end

Exercise

Fix the BST checker.

With a corrected definition, we can now define a refined version of binary trees that are search trees:
type BST = BT%(is-a-bst)
We can also remind ourselves that the purpose of this exercise was to define sets, and define TSets to be tree sets:
type TSet = BST
mt-set = leaf

Now let’s implement our operations on the BST representation. First we’ll write a template:
fun is-in(e :: Number, s :: BST) -> Bool:
cases (BST) s:
| leaf => ...
| node(v, l :: BST, r :: BST) => ...
... is-in(l) ...
... is-in(r) ...
end
end
Observe that the data definition of a BST gives us rich information about the two children: they are each a BST, so we know their elements obey the ordering property. We can use this to define the actual operations:
fun is-in(e :: Number, s :: BST) -> Boolean:
cases (BST) s:
| leaf => false
| node(v, l, r) =>
if e == v:
true
else if e < v:
is-in(e, l)
else if e > v:
is-in(e, r)
end
end
end

fun insert(e :: Number, s :: BST) -> BST:
cases (BST) s:
| leaf => node(e, leaf, leaf)
| node(v, l, r) =>
if e == v:
s
else if e < v:
node(v, insert(e, l), r)
else if e > v:
node(v, l, insert(e, r))
end
end
end
In both functions we are strictly assuming the invariant of the BST, and in the latter case also ensuring it. Make sure you identify where, why, and how.

You should now be able to define the remaining operations. Of these, size clearly requires linear time (since it has to count all the elements), but because is-in and insert both throw away one of two children each time they recur, they take logarithmic time.

Exercise

Suppose we frequently needed to compute the size of a set. We ought to be able to reduce the time complexity of size by having each tree cache its size, so that size could complete in constant time (note that the size of the tree clearly fits the criterion of a cache, since it can always be reconstructed). Update the data definition and all affected functions to keep track of this information correctly.

But wait a minute. Are we actually done? Our recurrence takes the form $$T(k) = T(k/2) + c$$, but what in our data definition guaranteed that the size of the child traversed by is-in will be half the size?

Do Now!

Construct an example—consisting of a sequence of inserts to the empty tree—such that the resulting tree is not balanced. Show that searching for certain elements in this tree will take linear, not logarithmic, time in its size.

Imagine starting with the empty tree and inserting the values 1, 2, 3, and 4, in order. The resulting tree would be
check:
insert(4, insert(3, insert(2, insert(1, mt-set)))) is
node(1, leaf,
node(2, leaf,
node(3, leaf,
node(4, leaf, leaf))))
end
Searching for 4 in this tree would have to examine all the set elements in the tree. In other words, this binary search tree is degenerateit is effectively a list, and we are back to having the same complexity we had earlier.

Therefore, using a binary tree, and even a BST, does not guarantee the complexity we want: it does only if our inputs have arrived in just the right order. However, we cannot assume any input ordering; instead, we would like an implementation that works in all cases. Thus, we must find a way to ensure that the tree is always balanced, so each recursive call in is-in really does throw away half the elements.

18.2.3A Fine Balance: Tree Surgery

Let’s define a balanced binary search tree (BBST). It must obviously be a search tree, so let’s focus on the “balanced” part. We have to be careful about precisely what this means: we can’t simply expect both sides to be of equal size because this demands that the tree (and hence the set) have an even number of elements and, even more stringently, to have a size that is a power of two.

Exercise

Define a predicate for a BBST that consumes a BT and returns a Boolean indicating whether or not it a balanced search tree.

Therefore, we relax the notion of balance to one that is both accommodating and sufficient. We use the term balance factor for a node to refer to the height of its left child minus the height of its right child (where the height is the depth, in edges, of the deepest node). We allow every node of a BBST to have a balance factor of $$-1$$, $$0$$, or $$1$$ (but nothing else): that is, either both have the same height, or the left or the right can be one taller. Note that this is a recursive property, but it applies at all levels, so the imbalance cannot accumulate making the whole tree arbitrarily imbalanced.

Exercise

Given this definition of a BBST, show that the number of nodes is exponential in the height. Thus, always recurring on one branch will terminate after a logarithmic (in the number of nodes) number of steps.

Here is an obvious but useful observation: every BBST is also a BST (this was true by the very definition of a BBST). Why does this matter? It means that a function that operates on a BST can just as well be applied to a BBST without any loss of correctness.

So far, so easy. All that leaves is a means of creating a BBST, because it’s responsible for ensuring balance. It’s easy to see that the constant empty-set is a BBST value. So that leaves only insert.

Here is our situation with insert. Assuming we start with a BBST, we can determine in logarithmic time whether the element is already in the tree and, if so, ignore it.To implement a bag we count how many of each element are in it, which does not affect the tree’s height. When inserting an element, given balanced trees, the insert for a BST takes only a logarithmic amount of time to perform the insertion. Thus, if performing the insertion does not affect the tree’s balance, we’re done. Therefore, we only need to consider cases where performing the insertion throws off the balance.

Observe that because $$<$$ and $$>$$ are symmetric (likewise with $$<=$$ and $$>=$$), we can consider insertions into one half of the tree and a symmetric argument handles insertions into the other half. Thus, suppose we have a tree that is currently balanced into which we are inserting the element $$e$$. Let’s say $$e$$ is going into the left sub-tree and, by virtue of being inserted, will cause the entire tree to become imbalanced.Some trees, like family trees [REF], represent real-world data. It makes no sense to “balance” a family tree: it must accurately model whatever reality it represents. These set-representing trees, in contrast, are chosen by us, not dictated by some external reality, so we are free to rearrange them.

There are two ways to proceed. One is to consider all the places where we might insert $$e$$ in a way that causes an imbalance and determine what to do in each case.

Exercise

Enumerate all the cases where insertion might be problematic, and dictate what to do in each case.

The number of cases is actually quite overwhelming (if you didn’t think so, you missed a few...). Therefore, we instead attack the problem after it has occurred: allow the existing BST insert to insert the element, assume that we have an imbalanced tree, and show how to restore its balance.The insight that a tree can be made “self-balancing” is quite remarkable, and there are now many solutions to this problem. This particular one, one of the oldest, is due to G.M. Adelson-Velskii and E.M. Landis. In honor of their initials it is called an AVL Tree, though the tree itself is quite evident; their genius is in defining re-balancing.

Thus, in what follows, we begin with a tree that is balanced; insert causes it to become imbalanced; we have assumed that the insertion happened in the left sub-tree. In particular, suppose a (sub-)tree has a balance factor of $$2$$ (positive because we’re assuming the left is imbalanced by insertion). The procedure for restoring balance depends critically on the following property:

Exercise

Show that if a tree is currently balanced, i.e., the balance factor at every node is $$-1$$, $$0$$, or $$1$$, then insert can at worst make the balance factor $$\pm 2$$.

The algorithm that follows is applied as insert returns from its recursion, i.e., on the path from the inserted value back to the root. Since this path is of logarithmic length in the set’s size (due to the balancing property), and (as we shall see) performs only a constant amount of work at each step, it ensures that insertion also takes only logarithmic time, thus completing our challenge.

To visualize the algorithm, let’s use this tree schematic:
 p / \ q   C / \ A   B
Here, $$p$$ is the value of the element at the root (though we will also abuse terminology and use the value at a root to refer to that whole tree), $$q$$ is the value at the root of the left sub-tree (so $$q < p$$), and $$A$$, $$B$$, and $$C$$ name the respective sub-trees. We have assumed that $$e$$ is being inserted into the left sub-tree, which means $$e < p$$.

Let’s say that $$C$$ is of height $$k$$. Before insertion, the tree rooted at $$q$$ must have had height $$k+1$$ (or else one insertion cannot create imbalance). In turn, this means $$A$$ must have had height $$k$$ or $$k-1$$, and likewise for $$B$$.

Suppose that after insertion, the tree rooted at $$q$$ has height $$k+2$$. Thus, either $$A$$ or $$B$$ has height $$k+1$$ and the other must have height less than that (either $$k$$ or $$k-1$$).

Exercise

Why can they both not have height $$k+1$$ after insertion?

This gives us two cases to consider.

18.2.3.1Left-Left Case

Let’s say the imbalance is in $$A$$, i.e., it has height $$k+1$$. Let’s expand that tree:
 p / \ q   C / \ r   B / \ A1  A2
We know the following about the data in the sub-trees. We’ll use the notation $$T < a$$ where $$T$$ is a tree and $$a$$ is a single value to mean every value in $$T$$ is less than $$a$$.
• $$A_1 < r$$.

• $$r < A_2 < q$$.

• $$q < B < p$$.

• $$p < C$$.

Let’s also remind ourselves of the sizes:
• The height of $$A_1$$ or of $$A_2$$ is $$k$$ (the cause of imbalance).

• The height of the other $$A_i$$ is $$k-1$$ (see the exercise above).

• The height of $$C$$ is $$k$$ (initial assumption; $$k$$ is arbitrary).

• The height of $$B$$ must be $$k-1$$ or $$k$$ (argued above).

Imagine this tree is a mobile, which has gotten a little skewed to the left. You would naturally think to suspend the mobile a little further to the left to bring it back into balance. That is effectively what we will do:
 q / \ r     p / \   / \ A1  A2 B  C
Observe that this preserves each of the ordering properties above. In addition, the $$A$$ subtree has been brought one level closer to the root than earlier relative to $$B$$ and $$C$$. This restores the balance (as you can see if you work out the heights of each of $$A_i$$, $$B$$, and $$C$$). Thus, we have also restored balance.

18.2.3.2Left-Right Case

The imbalance might instead be in $$B$$. Expanding:
 p / \ q   C / \ A   r / \ B1  B2
Again, let’s record what we know about data order:
• $$A < q$$.

• $$q < B_1 < r$$.

• $$r < B_2 < p$$.

• $$p < C$$.

and sizes:
• Suppose the height of $$C$$ is $$k$$.

• The height of $$A$$ must be $$k-1$$ or $$k$$.

• The height of $$B_1$$ or $$B_2$$ must be $$k$$, but not both (see the exercise above). The other must be $$k-1$$.

We therefore have to somehow bring $$B_1$$ and $$B_2$$ one level closer to the root of the tree. By using the above data ordering knowledge, we can construct this tree:
 p / \ r   C / \ q   B2 / \ A   B1
Of course, if $$B_1$$ is the problematic sub-tree, this still does not address the problem. However, we are now back to the previous (left-left) case; rotating gets us to:
 r /    \ q      p / \    / \ A   B1 B2  C
Now observe that we have precisely maintained the data ordering constraints. Furthermore, from the root, $$A$$’s lowest node is at height $$k+1$$ or $$k+2$$; so is $$B_1$$’s; so is $$B_2$$’s; and $$C$$’s is at $$k+2$$.

18.2.3.3Any Other Cases?

Were we a little too glib before? In the left-right case we said that only one of $$B_1$$ or $$B_2$$ could be of height $$k$$ (after insertion); the other had to be of height $$k-1$$. Actually, all we can say for sure is that the other has to be at most height $$k-2$$.

Exercise

• Can the height of the other tree actually be $$k-2$$ instead of $$k-1$$?

• If so, does the solution above hold? Is there not still an imbalance of two in the resulting tree?

• Is there actually a bug in the above algorithm?