Polymorphic Recursion in OCaml

In Chapter 10 of Purely Functional Data Structures, Okasaki describes recursive types that are non-uniform. In this post we’ll learn more about these types, how to implement them in OCaml and see an example by studying the implementation of Random Access Binary Lists using such a construct.

Uniform recursive type

As an example of uniform recursive data structure we have a simple list

Cons(1, Cons(2, Cons(3, Nil)))

Which has a recursive type, for example

type 'a myList = Nil | Cons of 'a * 'a myList

Each element of the list is either Nil (terminal) or it has a value of a polymorphic type 'a, followed a recursive list also of type 'a.

Non-uniform recursive type

Now, say that the type of the recursive list is not the same as the current list? Then we have a non-uniform polymorphic recursive type, for example:

type 'a seq = Nil | Cons of 'a * ('a * 'a) seq

We’ll name this a sequence. A int seq would have the value in the first node would have type int, but the element from the second node would have type (int, int), the third type ((int, int), (int, int)) and so on. This structure is equivalent to a complete binary tree, where the i-th element of seq represents the i-th level of the tree.

An example of value with this type is:

Cons (1, Cons ((2, 3), Cons (((4, 5), (6, 7)), Nil)))

We need a special syntax to type recursive functions that take recursive non-uniform types, because the type of the function called recursively might be a different polymorphic type than the caller. OCaml by default tries to infer the generic types of the function and bind them to specific instances [2]. For example, in

let f: 'a list -> 'a list = fun x -> 13 :: x

OCaml will assume 'a is int and will compile fine. We can see this by pasting that code in the command line, utop.

utop # let f: 'a list -> 'a list = fun x -> 13 :: x;;
val f : int list -> int list =

The function will then not be polymorphic anymore. To prevent OCaml from auto-binding specific type instances, we can use a special syntax introduced in OCaml 3.12 [3]

utop # let f3: 'a. 'a list -> 'a list = fun x -> 13 :: x;;

This time we’ll get a compilation error:

Error: This definition has type int list -> int list which is less general than ‘a. ‘a list -> ‘a list

The important thing is that this allow us binding the recursive calls with different types. According to the Jane Street Tech Blog [3]:

Note that a polymorphic type annotation holds inside the body of a recursive definition as well as outside, allowing what is known as polymorphic recursion, where a recursive call is made at some non-trivial instantiation of the polymorphic type.

So for example, we can write this function to calculate the size of a sequence:

The problem with this structure is that it can only represent lists of size in the form of 2^k - 1. To work around that, we allow some items to not hold any elements at all, so that each item corresponds to a digit in the binary representation of the size of the list.

type 'a seq = Nil | Zero of ('a * 'a) seq | One of 'a * ('a * 'a) seq

For example, we can now represent a list of 10 elements, as

Zero(One((1, 2), Zero(One(((3, 4), (5, 6)), ((7, 8),(9, 10))), Nil))))

mandelbrot

Sequence binary random access list

We can use a sequence to implement a random access binary access list in a concise way.

Insertion

Inserting an element at the beginning is analogous to incrementing the binary number, that is, starting from the least significant digit, if it’s a zero, we make it one, if it’s one we make it a 0, and carry over a 1, by adding it to the next digit.

The carry over process is simple in this structure. Because the type of an item following an item of type 'a is ('a, 'a), to merge the element to be added with the existing element, we simply make a tuple and pass it to the recursive call.

Head and Tail

Removing or retrieving the first element is analogous to decrementing a binary number. If the digit is one, we make it zero and return the element and the resulting list. If it’s a zero, we make a recursive call to get the next available element. However since the returned element is of type ('a, 'a), and our return type is 'a, we only use the first value of the pair.

Implementing head and tail using popAux is now trivial

Lookup

Finding an element can be done by transforming the problem into smaller instances.

It helps to look at some simple examples. Let’s consider 3 cases.

Case 1. If we had a single element, we either return it if the index is 0, or throw if it’s larger than that.

0: (0)

Case 2. If we have 2 elements,

0: ()
1: (0, 1)

Notice that when we go from the first level to the second, all items “doubled” in size, so we can “transform” this to the single element case by treating pairs as a single element, but since the index has individual elements granularity, we need to transform it by halving it. We reduced it to Case 1.

If our initial index was either 0 or 1, it’s now 0, and we found our element in level 1.


1: (0)

The problem is that we need to return a single element at level 0, not a pair, so we need to undo the transformation. We can use the parity of the original index will to decide which side of the pair to return. If it’s even, we return the first element, otherwise the second one.

Case 3. If we have 3 elements,
0: (0)
1: (1, 2)

and our index is larger than 0, we move to the next level but we need to account for the level we’re skipping, so the transformation of index would look like:
0: ()
1: (0)

which is reduced to Case 2.

These 3 cases can be used to find elements larger than 3. For example, say we have 10 elements and want to find the element at position 6:

0: ()
1: (0, 1)
2: ()
3: (((2, 3), (4, 5)), ((6, 7), (8, 9)))

Step 1. This is Case 2. We need to transform this by treating pairs as elements and halving the index:

1': (0)
2': ()
3': ((1, 2), (3, 4))

Note how this reduced the problem of finding the element at position 3 of a list with size 5. Step 2. We now are in case 3, where we need to skip the current element:

1': ()
2': ()
3': (((0), (1)), ((2), (3)))

Our index is now 2. Step 3. we go one level down

2: ()
3: (0, 1)

With an index of 1. Step 4. Finally, we halve it once again and we finally found the right level that contains our index.

3: (0)

We now need to recurse back to find exactly which element on that level to pick. On Step 4, we can see our index 1 was on the right side of the pair in level 3, so we pick the right side, that is, ((6, 7), (8, 9)).

On Step 3, our index 2 was on the left size of the innermost pair, that is (6, 7). On Step 2, we skipped the current element but didn’t change levels, so there’s no need to choose an element from the pair. Finally, on Step 1, the index 6 was on the left side of the innermost pair, which should return the element with a corresponding index 6.

In general, we can tell which side of the innermost pair to pick by observing that the indexes are ordered from left to right in a given level. And because every level has an even number of elements, we can assume that the first index in the level – or the first element in the first pair – is even. Hence the parity of the index is sufficient to determine which side of the pair to pick.

With this algorithm in mind, the lookup function is quite compact:

The update follows a similar idea as the lookup, but the problem is that we need to return the updated level when returning from the recursion. That is, we need to update the level before returning.

To accomplish that, we can pass a callback, the updater, that encodes which pair we would pick at each level. We start with a function that simply returns the element to be updated

(fun _ -> element)

Then, at each level we create a new updater, which applies the previous updater on the left or right side of the pair, depending on the parity of the index:

When we finally find the level that has our index, we can apply the function, which has the effect of “narrowing” down the elements from a given level to a single element, replacing the value at the target index and then returning the updated elements when returning from the recursion.

After applying the updater, we return the updated level recursively.

Structural Decomposition

Okasaki introduces this implementation in the context of Structural Decomposition, which is a technique for creating data structures from incomplete ones. In this example, the raw implementation of the sequence can only represent lists of size 2^k, but modeling each node in the sequence to be zero or one, zero not having any elements, we can work around the size restriction.

Conclusion

The implementation of random access binary access list using sequences is very neat, but very hard to understand.

One of the major selling points of shorter code is that it tends to contain less bugs and also less corner cases. On the other hand, if the code is also harder to read and understand, it might be harder to spot bugs.

This post helped me understand a bit more about OCaml’s type system. Digging a little also led me to interesting topics such as Parametric Polymorphism [4] and Existential vs. Universally quantified types [5].

References

[1] Purely Function Data Structures – Chris Okasaki
[2] Jane Street – Ensuring that a function is polymorphic
[3] Ensuring that a function is polymorphic in Ocaml 3.12
[4] Wikipedia – Parametric polymorphism
[5] Existential vs. Universally quantified types in Haskell

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Numerical Representations as inspiration for Data Structures

Book

In this chapter Okasaki describes a technique for developing efficient data structures through analogies with numerical representations, in particular the binary and its variations.

We’ve seen this pattern arise with Binomial Heaps in the past. Here the author presents the technique in its general form and applies it to another data structure, binary random access lists.

Binary Random Access Lists

These lists allows efficient insertion at/removal from the beginning, and also access and update at a particular index.

The simple version of this structure consists in distributing the elements in complete binary leaf trees. A complete binary leaf tree (CBLF) is one that only stores elements only at the leaves, so a tree with height i, has 2^(i+1)-1 nodes, but only 2^i elements.

Consider an array of size n, and let Bn be the binary representation of n. If the i-th digit of Bn is 1, then we have a tree containing 2^i leaves. We then distribute the elements into these trees, starting with the least significant digit (i.e. the smallest tree) and traversing the tree in
pre-order.

For example, an array of elements (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) has 11 elements, which is 1011 in binary. So we have one tree with a single leave (1), a tree with 2 leaves (2, 3) and another containing 8 leaves (4, 5, 6, 7, 8, 9, 10, 11).

We use a dense representation of the binary number, by having explicit elements for the 0 digit. Here’s a possible signature for this implementation:

Inserting a new element consists in adding a new tree with a single leave. If a tree already exists for a given size, they have to be merged into a tree of the next index. Merging two CBLFs of the same size is straightforward. We just need to make them children of a new root. Since elements are stored in pre-order, the tree being inserted or coming from carry over should be the left child.

Looping back to our example, if we want to insert the element 100, we first insert a tree with a single leaf (100). Since the least significant digit already has an element, we need to merge them into a new tree containing (100, 1) and try to insert at the next position. A conflict will arise with (2, 3), so we again merge them into (100, 1, 2, 3) and try the next position. We finally succeed in inserting at position 2, for a new list containing trees like (100, 1, 2, 3) and (4, 5, 6, 7, 8, 9, 10, 11).

The complexity of inserting an element is O(log n) in the worst case which requires merging tree for all digits (e.g. if Bn = 111...111). Merging two trees is O(1).

Removing the first element is analogous to decrementing the number, borrowing from the next digit if the current digit is 0.

Searching for an index consists in first finding the tree containing the index and then searching within the tree. More specifically, because the elements are sorted beginning from the smallest tree to the largest, we can find the right tree just by inspecting the number of elements in each tree until we find the one whose range includes the desired index. Within a tree, elements are stored in pre-order, so we can find the right index in O(height) of the tree.

After finding the right index, returning the element at that index is trivial. Updating the element at a given index requires rebuilding the tree when returning from the recursive calls.

Okasaki then proposes a few different numbering systems that allow to perform insertion/removal in O(1) time. Here we’ll only discuss the less obvious but more elegant one, using skew binary numbers.

characters

Skew Binary Random Access Lists

A skew binary number representation supports the digits 0, 1 and 2.

The weight of the i-th digit is 2^(i+1) - 1. In its canonical form, it only allows the least significant non-zero digit to be 2.

Examples:

table-v2

Decimal and Skewed Binary

It’s possible to show this number system offers a unique representation for decimal numbers. See the Appendix for a sketch of the proof and an algorithm for converting decimals to skewed binary numbers.

Incrementing a number follows these rules:

  • If there’s a digit 2 in the number, turn it into 0 and increment the next digit. By definition that is either 0 or 1, so we can safely increment it without having to continue carrying over.
  • Otherwise the least significant digit is either 0 or 1, and it can be incremented without carry overs.

The advantage of this number system is that increments (similarly, decrements) never carry over more than once so the complexity O(1), as opposed to the worst-case O(log n) for regular binary numbers.

A skew binary random access list can be implemented using this idea. We use a sparse representation (that is, not including 0s). Each digit one with position i corresponds to a tree with (2^(i+1) - 1) elements, in this case a complete binary tree with height i+1. A digit 2 is represented by two consecutive trees
with same weight.

Adding a new element to the beginning of the list is analogous to incrementing the number, which we saw can be done in O(1). Converting a digit 0 to 1 or 1 to 2, is a matter of prepending a tree to a list. To convert a 2 to 0 and increment the next position, we need to merge two trees representing it with the element to be inserted. Because each tree is traversed in pre-order, we make the element the root of the tree.

Elements are inserted in pre-order in each tree, so when searching for an
index, we can first find the right tree by looking at the tree sizes and within a tree we can do a “binary search” in O(height) of the tree.

Binomial Heaps

In this chapter, this technique is also applied to improve the worst case runtime of insertion of binomial heaps. The implementation, named Skewed Binomial Heap, is on github.

Conclusion

This chapter demonstrated that binary representations are a useful analogy to come up with data structures and algorithms, because they’re simple. This simplicity can lead to inefficient running times, though. Representations such as skewed binary numbers can improve the worst case of some operations with the trade-off of less intuitive and more complex implementations.

Appendix A – Proof

Sketch of the proof. First, it’s obvious that two different decimals cannot map to the same binary representation. Otherwise the same equation with the same weights would result in different values. We need to show that two binary representations do not map to the same decimal.

Suppose it does, and let them be B1 and B2. Let k be the largest position where these number have a different digit. Without loss of generality, suppose that B1[k] > B2[k].

First case. suppose that B1[k] = 1, and B2[k] = 0 and B2 doesn’t have any digit 2. B1 is then at least M + 2^{k+1} - 1, while B2 is at most M + \sum_{i = 1}^{k} (2^{i} - 1) which is M + 2^{k + 1} - k (M corresponds to the total weight of digits in positions > k). This implies that B2 < B1, a contradiction.

Second case. suppose that B1[k] = 1, but now B2 does have a digit 2 at position j. It has to be that j < k. Since only zeros follow it, we can write B2‘s upper bound as

M + \sum_{i = j + 1}^{k} (2^{i} - 1) + 2^{j + 1} - 1

Since 2(2^{j + 1} - 1) < 2^{j + 2} - 1, we have

\sum_{i = j + 1}^{k} (2^{i} - 1) + 2^{j + 1} - 1 < \sum_{i = j + 2}^{k} (2^{i} - 1) + 2^{j + 2} - 1

We can continue this argument until we get that B2 is less than M + 2(2^{k} - 1) which is less than M + 2^{k + 1} - 1, B1.

Third case. Finally, suppose we have B1' such that B1'[k] = 2, and B2'[k] = 1. We can subtract 2^{k+1} - 1 from both and reduce to the previous case. ▢

Appendix B – Conversion algorithm

Converting from a decimal representation to a binary one is straightforward, but it’s more involved to do so for skewed binary numbers.

Suppose we allow trailing zeros and have all the numbers with k-digits. For example, if k=2, we have 00, 01, 02, 10, 11, 12 and 20. We can construct the numbers with k+1-digits by either prefixing 0 or 1, and the additional 2 followed by k zeros. For k=3, we have 000, 001, 002, 010, 011, 012, 020, 100, 101, 102, 110, 111, 112, 120 and finally 200.

More generally, we can see there are 2^(k+1) - 1 numbers with k digits. We can construct the k+1 digits by appending 0 or 1 and then adding an extra number which is starts 2 and is followed by k zeros, for a total of 2^(k+1) - 1 + 2^(k+1) - 1 + 1 = 2^(k + 2) - 1, so we can see this invariant holds by induction on k, and we verify that is true for the base, since for k = 1 we enumerated 3 numbers.

This gives us a method to construct the skewed number representation if we know the number of its digits say, k. If the number is the first 2^(k) - 1 numbers, that is, between 0 and 2^k - 2, we know it starts with a 0. If it’s the next 2^(k) - 1, that is, between 2^k - 1 and 2^(k+1) - 3, we know it starts with a 1. If it’s the next one, exactly 2^(k+1) - 2, we know it starts with a 2.

We can continue recursively by subtracting the corresponding weight for this digit until k = 0. We can find out how many digits a number n has (if we’re to exclude leading zeros) by find the smallest k such that 2^(k+1)-1 is greater than n. For example, for 8, the smallest k is 3, since 2^(k+1)-1 = 15, and 2^(k)-1 = 7.

The Python code below uses these ideas to find the skewed binary number representation in O(log n):

One might ask: why not OCaml? My excuse is that I already have a number theory repository in Python, so it seemed like a nice addition. Converting this to functional code, in particular OCaml is easy.

This algorithm requires an additional O(log n) memory, while the conversion to a binary number can be done with constant extra memory. My intuition is that this is possible because the weights for the binary numbers are powers of the same number, 2^k, unlike the skewed numbers’ weights. Is it possible to work around this?

Lazy Rebuilding

Book

In this chapter Okasaki starts with a common technique data structures use to achieve efficient complexity times. A classic example are balanced binary search trees, which, on one hand, need to be balanced to avoid degenerated cases (think of a binary tree which can degenerate to a path), but on the other hand, balancing is usually too expensive to perform at every operation.

The tradeoff is to allow a certain degree of “imbalance”, such that the big-O complexity of operations does not degenerate, and to let us perform re-balances only every so often, in a way that leads to an efficient amortized complexity. Structures such as AVL trees and Red-black trees make use of this technique.

The general technique is named batched rebuilding. The main issue with it though is that we saw that amortized analysis does not play well persistent data structures.

To address that problem, a technique was developed by Mark Overmars, called global rebuilding.

Global Rebuilding

The basic idea is to keep two copies of a structure in parallel, one of which we perform incremental updates (write) and the other is used for read operations. After a few operations, we copy the write version into the read-only version. The idea is that the incremental updates are cheap, but might put the structure in a state that is not suitable for reading. This allows for distributing the costs across operations such that each operation has worst case efficient bounds.

The major downside of this technique is the complexity and corner cases, because we also need to account for changes to the structure that renders the write copy of the structure invalid.

We’ll now describe yet another implementation of queues that use this technique, developed by Robert Hood and Robert Melville. Similar to other queue implementations, this structure maintains a front and rear lists, the latter in reverse order. It also has the invariant that the rear list cannot be larger than the front.

When that happens we must perform a rotation, which consists in reversing the rear queue and appending it to the end of the front queue. We cannot perform this operation piecemeal, because we only have efficient access to the first element of a list. The operation we can do partially is concatenating the reverse of a list to the beginning of another list, that is, given xs and ys, we can do ~xs + ys, where ~ represents the reverse of a list. Note we can also reverse a list, that is xs to ~xs piecemeal by having ys = [].

Now, to achieve our goal which is xs + ~ys, we reverse both xs and ys, then concatenate the reverse of ~xs to ~ys:

1) Start with xs and ys
2) Reverse both: ~xs and ~ys
3) Then (~(~xs)) + ~ys which is xs + ~ys

We can perform these operations one step at a time by using a “state machine”, we first start with a state Reversing(xs, ys) which will reverse xs and ys at the same time into ~xs and ~ys, at which point we switch to the state Appending(~xs, ~yx) which will concatenate the xs into ~ys, so then we get to Done(xs + ~ys).

The problem of keeping a separate state is that it might not be accurate by the time it’s done. In particular, if we remove an element from the front, the Appending step can be shortcut. We can keep a count of how many of the elements in ~xs of Appending are still present. If, by the time we’re appending the number of present elements goes to 0, the last elements of ~xs (that is, the first elements of xs) have been removed, so they do not need to be transferred to ~ys.

A possible implementation of the states is:

The Idle is just a placeholder to make sure the state machine can keep going to the next state even when we’re not currently performing a rotation. The state transition definition is given by:

One important detail here is that Appending ({validCount = 0; front; rear}) must appear before the other matching rule for Appending, because the other one includes this.

When we remove an element from the front of the queue we need to update the number of valid elements in the front of the rotation state:

Similar to the Real-time queues, we can guarantee constant time worst case for all the operations if we execute the state machine twice for each operation. The check() function verifies if we need a rotation and also executes the nextStep() function twice.

The other operations are then straightforward. The only thing is that pop() must call the invalidate() function because it’s decreasing the size of front:

The full, up-to-date implementation in on github.

Lazy Rebuilding

As we can see, the Hood-Melville implementation is quite involved. Okasaki argues that making use of lazy evaluation and memoization simplifies the implementation. We can indeed see that the implementation of Real-time queues is much cleaner.

One simplification is that we don’t have to sync the write copy with the read copy. We just need to make sure to execute items from the schedule by the time they are needed. Memoization will take care of the rest.

Another advantage in this case is that reversing and concatenating lazily evaluated lists does not require an intermediate reversal of the front, which means that we can remove the front element without the need to invalidate any items in the schedule.

The author provided more examples of lazy rebuilding for deques (double-ended queues) by first presenting an amortized version using the Banker’s method and then a lazy rebuilding version. The Banker’s deque and Real-time deque are also on my github.

Conclusion

In this chapter the technique behind Real-time queues was formalized. I really enjoy the organization of the book in which the author presents a data structure and from there he extracts a generic pattern of technique that can be applicable to other cases.

When I was studying data structures I don’t recall learning about general techniques underlying existing data structures, such as batched rebuilding for AVL trees and Red-black trees. That would have been interesting.

Eliminating Amortization

Book

In this chapter Okasaki presents techniques to convert persistent data structures with amortized bounds to worst-case bounds. A few reasons to want worst-case guarantees include:

  • Real-time systems, where an expensive operation can cause the system to miss a hard deadline, even if it’s efficient when amortized over time.
  • Parallel systems, in which we want avoid one processor to take more time than the other if it happens to execute the expensive operation.

The intuition is to make the theoretical amortized analysis part of the actual implementation. In the amortized analysis, we saw either through the Banker’s method or the Physicist method the idea of paying debt ahead of time so by the time an expensive operation takes place, we could show it’s cost had already been distributed throughout previous operations.

To eliminate the amortization, we can literally pay off these costs when running the algorithm. We do it through a structure called schedule, which contains the unevaluated operations on the data structure. Whenever we perform an operation, we evaluate one or more items of the schedule. Due to memoization, by the time we actually need the evaluated structure, it will be evaluated.

Often the amortized analysis can dictate how the execution of the schedule is performed. For example, if the analysis tells us to pay off 2 units of debt on every action, that can be translated to executing 2 items from the schedule on every action.

The main challenge in this conversion is to modify the data structure in such a way it can be partially executed.

We’ll discuss an example using queues and then binomial heaps.

Real-time Queues

For the queue structure, the costly operation is the rotation that takes place when we need to combine the rear with the front. It’s a monolithic operation, so we need to change that.

Let’s start by defining the structure. It’s similar to the stream queue, except that we don’t need the rear to be a stream and we have a schedule field, which is also a stream. The idea is that whenever a rotation happens, it will be “scheduled” to be executed little by little.

The rotation() function is the most complicated part of the structure. The invariant is that we only call this function when |rear| = |front| + 1.

We construct a stream with the elements of rear reversed, newSchedule and on the return step of the recursion we append the elements of front to that stream.

Note that this is performed lazily, so a call to rotate only executes one step.

Now we have the execution of the schedule. It serves two purposes. One is to execute an item from the schedule and the other is to trigger a rotation when the is schedule empty.

The schedule being empty is a proxy to when |rear| = |front| + 1. Why? Because when the schedule is populated, it has the same size of front (they’re both assigned the same stream), and rear is empty. Whenever we insert an element, increasing the size of rear by one, or remove an element, reducing the size of front by one, we decrease the difference between |front| - |rear| by 1, and so the size of the schedule is decreased by 1.

Implementation-wise, maybe it would be more clear if we checked for the empty schedule outside and only called exec() is it was not empty.

With these in place, the usual operations for queue are straightforward. The ones that mutate the queue, push() and pop(), need to make sure to call exec().

Scheduled Binomial Heaps

We discussed Binomial Heaps in a previous post. The idea is that a binomial heap is a list of binomial trees, an a binomial tree is defined recursively based on it’s rank.

The heap is represented by a binary number (as a list of digits). If the k-th digit of the number is 1, it indicates the existence of a binomial tree of rank k (containing 2^(k-1)). A heap with n elements, has a unique representation, which is the binary representation of n.

For example, if n = 10, then the binary representation of the heap is 1010, meaning it contains a binomial tree of rank 2 (2 elements), and one with rank 4 (8 elements), holding 10 elements in total.

Inserting an element into the heap is analogous to incrementing a binary number by 1. We first create a binomial tree with rank 0 containing that element, and try to insert it into the beginning of the digits list. If the position we want to insert is occupied, we need to merge the trees, to generate a tree with a higher rank, which is analogous to the carry over operation when adding two binary numbers.

The schedule is a list of all the numbers generated when any operation is performed.

The structure for the heap is then the following:

As we discussed above, the insertion is analogous to incrementing the binary number. But because the digits are a stream, we need to deal with lazy evaluation:

Executing the schedule consists in evaluating one digit from the first number on the schedule. The key is to avoid evaluating the part of the number that already has been evaluated. It’s possible to show this happens when we find the first digit one. The intuition is that the trailing zeros from the current number were 1’s before the increment, so there was a mutation (linking) while the remaining digits were not modified by that operation.

For example, if we have the number 101011, an increment causes it to become 101100. The two trailing zeros in 101100 correspond to a linking of the binomial tree.

Finally, inserting a new element consists in creating a binomial tree of ranking 0, insert it, and then execute the schedule.

The full code is available on github.

Lessons learned

Because we now have several different implementations of queues, I wanted to implement tests that were applicable to any queue implementing an “interface”.

One way to do that is to put the interface in a separate module, IQueue.ml and have each implementation require it as their module type:

To make the test work with any module implementing the IQueue interface, we can use a functor (module transformation) and

Other things I’ve learned were matching lazy patterns. Matching a lazily evaluated variable with the keyword lazy forces the evaluation, so we can use a cleaner syntax, for example:

Another syntactic construct that helps with legibility is the record. The examples in Okazaki’s book use tuples for most of composed structs, but I prefer the more verbose and explicit records.

Finally, another lesson learned is that adding an insertion operation to the Stream API is likely wrong. The idea of the stream is that is avoids evaluation of its tail, so the whole block has to lazily evaluated.

For example, in the queue implementation, in the first block, we will evaluate all the rotation and then make the result lazy, which is not what we want.

Conclusion

Eliminating evaluation is a very interesting technique, but it has limited application is practice. It complicates the implementation and, as I learned, it can be tricky to spot bugs (for example, the one in which we’re evaluating the rotate() function) that would only show up if we profile the data structure.

Largest sets of subsequences in OCaml

I’ve noticed that there is this set of words in English that look very similar: tough, though, through, thought, thorough, through and trough. Except thought, they have one property in common: they’re all subsequence of thorough. It made me wonder if there are interesting sets of words that are subsequences of other words.

Word cloud made with Jason Davie’s tool (https://www.jasondavies.com/wordcloud/)

This post is an attempt to answer a more general question: given a list of words, what is the largest set of these words such that they’re subsequences of a given word?

A word A is a subsequence of a word B if A can be obtained by removing zero or more characters from B. For example, “ac” is a subsequence of “abc”, so is “bc” and even “abc”, but not “ba” nor “aa”.

A simple algorithm to determine if a word A is a subsequence of another is to start with 2 pointers at the beginning of each word, pA and pB. We move pB forward until pA and pB point to the same character. In that case we move pA forward. A is a subsequence of B if and only if we reach the end of A before B. We could then iterate over each word W and find all the words that are subsequences of W. If the size of the dictionary is n, and the size of the largest word is w, this algorithm would run in O(n^2 w).

For English words, we can use entries from /usr/share/dict/words. In this case, n is around 235k (wc -l /usr/share/dict/words), so a quadratic algorithm will take a while to run (around 5e10 operations).

Another approach is to generate all subsequences of words for a given word W and search the dictionary for the generated word. There are O(2^w) subsequences of a word of length w. If we use a hash table, we can then do it in O(n w 2^w). In /usr/share/dict/words, the length of the largest word, w, is 24.

Running a calculation with the numbers (R script), the number of high-level operations is 4e10, about the same order of magnitude as the quadratic algorithm.

Distribution using ggplot2

A third approach is to use a trie. This data structure allows us to store the words from the dictionary in a space-efficient way and we can search for all subsequences using this structure. The trie will have at most 2e6 characters (sum of characters of all words), less because of shared prefixes. Since any valid subsequence has to be a node in the trie, the cost of search for a given word cannot be more than the size of the trie t, so the complexity per word is O(\min(2^w, t)). A back of envelope calculation gives us 2e9. But we’re hoping that the size of the trie will be much less than 2e6.

Before implementing the algorithm, let’s define out trie data structure.

The Trie data structure

A trie is a tree where each node has up to |E| children, where |E| is the size of the alphabet in consideration. For this problem, we’ll use lower case ascii only, so it’s 26. The node has also a flag telling whether there’s a word ending at this node.

Notice that in this implementation of trie, the character is in the edge of the trie, not in the node. The Map structure from the stlib uses a tree underneath so get and set operations are O(log |E|).

The insertion is the core method of the structure. At a given node we have a string we want to insert. We look at the first character of the word. If a corresponding edge exists, we keep following down that path. If not, we first create a new node.

To decide whether a trie has a given string, we just need to traverse the trie until we either can’t find an edge to follow or after reaching the end node it doesn’t have the hasEntry flag set to true:

This and other trie methods are available on github.

The search algorithm

Given a word W, we can search for all its subsequences in a trie with the following recursive algorithm: given a trie and a string we perform two searches: 1) for all the subsequences that contain the first character of current string, in which case we “consume” the first character and follow the corresponding node and 2) for all the subsequences that do not contain the first character of the current string, in which case we “consume” the character but stay at the current node. In pseudo-code:

Search(t: TrieNode, w: string):
    Let c be the first character of w.
    Let wrest be w with the first character removed

    If t contains a word, it's a subsequence of the
    original word. Save it.

    // Pick character c
    Search(t->child[c], wrest)

    // Do not pick character c
    Search(t, wrest)

The implementation in OCaml is given below:

Experiments

Our experiment consists in loading the words from /usr/share/dict/words into a trie, and then, for each word in the dictionary, look for its subsequences. The full code is available on github.

The code takes 90 seconds to run on my laptop. Not too bad but I’m still exploring ways to improve the performance. One optimization I tried is to, instead of returning an explicit list of strings as mentioned in the search implementation, return them encoded in a trie, since we can save some operations due to shared prefixes. I have that version on github, but unfortunately that takes 240 seconds to run and requires more memory.

Another way is to parallelize the code. The search for subsequences is independent for each word, so it’s an embarrassingly parallel case. I haven’t tried this path yet.

The constructed trie has 8e5 nodes or ~40% of the size of sum of characters.

Subsequences of “thorough”

The question that inspired this analysis was finding all the subsequences of thorough. It turns out it has 44 subsequences, but most of them are boring, that is, single letter or small words that look completely unrelated to the original word. The most interesting ones are those that start with t and have at least three letters. I selected some of them here:

  • tho
  • thoo
  • thoro
  • thorough
  • thou
  • though
  • thro
  • throu
  • through
  • thug
  • tog
  • tou
  • toug
  • tough
  • trough
  • trug
  • tug

The word with most subsequences is pseudolamellibranchiate, 1088! The word cloud at the beginning of the post contains the 100 words with the largest number of subsequences. I tried to find interesting words among these, but they’re basically the largest words – large words have exponentially more combination of subsequences, and hence the chance of them existing in the dictionary is greater. I tried to come up with penalization for the score:

1) Divide the number of subsequences by the word’s length. This is not enough, the largest words still show on top.
2) Apply log2 to the number of subsequences and divide by the word’s length. In theory this should account for the exponential number of subsequences of a word. This turns out to be too much of a penalization and the smallest word fare too well in this scenario.

I plotted the distribution of number of subsequences by word lengths. We can see a polynomial curve but with increased variance:

Generated with this ggplot2

In the chart above, we’d see all points with the same x-value in a single vertical line. One neat visualization trick is to add noise (jitter) so we also get a sense of density.

If we use a box plot instead, we can see a quadratic pattern more clearly by looking at the median for each length.

Generated with this ggplot2

Given this result, I tried a final scoring penalization, by dividing the number of subsequences by the square of the length of the word, but it’s still not enough to surface too many interesting words. Among the top 25, streamlined is the most common word, and it has 208 subsequences.

One interesting fact is that the lowest scoring words are those with repeated patterns, for example: kivikivi, Mississippi, kinnikinnick, curucucu and deedeed. This is basically because we only count unique subsequences.

Conclusion

This was a fun problem to think about and even though it didn’t have very interesting findings, I learned more about OCaml and R. After having to deal with bugs, compilation and execution errors, I like OCaml more than before and I like R less than before.

R has too many ways of doing the same thing and the API is too lenient. That works well for the 80% of the cases which it supports, but finding what went wrong in the other 20% is a pain. OCaml on the other hand is very strict. It doesn’t even let you add an int and a float without an explicit conversion.

I learned an interesting syntax that allows to re-use the qualifier/namespace between several operations when chaining them, for example:

I also used the library Batteries for the first time. It has a nice extension for the rather sparse String module. It allows us to simply do open Batteries but that overrides a lot of the standard modules and that can be very confusing. I was scratching my head for a long time to figure out why the compiler couldn’t find the union() function in the Map module, even though I seemed to have the right version, until I realized it was being overridden by Batteries. From now on, I’ll only use the specific modules, such as BatString, so it’s easy to tell which method is coming from which module.

References

OCaml

  • [1] OCaml Tutorials > Map
  • [2] Strings – Batteries included
  • [3] Using batteries when compiling

R

  • [1] R Tutorial – Histogram
  • [2] Creating plots in R using ggplot2 – part 10: boxplots
  • [3] My R Cheat sheet

Amortization and Persistence via Lazy Evaluation

In this chapter Okasaki works around the problem of doing amortized analysis with persistent data structures because the amortized analysis assumes in place modification while for persistent data structures (partial) copies are made. The intuition is that lazy evaluation, which comes with memoization and avoids recomputation, solves this problem.

He adapts the Banker’s and Physicists’s methods to work with lazy evaluated operations and applies them to a few structures including Binomial Heaps, Queues and Lazy Pairing Heaps. In this post we’ll only cover the examples of the Queues using both methods.

We’ll first introduce some concepts and terminology, then we’ll present a queue implementation using lazy evaluation that allows us analyzing it under persistence. Following that we’ll explain the Banker’s and Physicist’s methods and prove that the implementation for push/pop has an efficient amortized cost.

Persistence as a DAG

An execution trace is a DAG where nodes represent the operations (e.g. updates to a data structure) and an edge from nodes v to v' indicates that the operation corresponding to v' uses the output of the one corresponding to v.

For example, if we have this set of operations:

let a = push 0 newEmpty
let b = push 1 a
let c = pop b
let d = push 2 b
let e = append c d
let f = pop c
let g = push 3 d 

The corresponding execution graph is:

Execution Graph

The logical history of an operation v is the set of all operations it depends on (directly or indirectly, and including itself). Equivalently, in terms of the DAG, it’s the set of nodes that have a directed path to v.

A logical future of an operation v is any directed path from v to a terminal node.

A framework for analyzing lazy evaluated data structures

We need to introduce a few more concepts and terminology. Note: we’ll use suspension and lazy operation interchangeably, which can be evaluated or forced.

The unshared cost of an operation is the time it would take to execute it if it had already been performed and memoized before, so if the operation involves any expression that is lazy, that expression would be O(1).

The shared cost of an operation is the time it would take it to execute (force) all the suspensions created (but not evaluated) by the operation.

The complete cost is the sum of the shared and unshared costs. Alternatively, the complete cost of an operation is the time it would take to execute the operation if lazy evaluation was replaced by strict. To see why, first we note that the unshared costs have to be paid regardless of laziness. Since we’re assuming no laziness, the operation has to pay the cost associated with the suspension it creates, which corresponds to the shared costs. Note that under this assumption we wouldn’t need to account for the cost of forcing suspensions created by previous operations because in theory they have already been evaluated.

When talking about a sequence of operations, we can break down the shared costs into two types: realized and unrealized costs. The realized costs are the shared costs from suspensions were actually forced by some operation in the sequence. Example: say that operations A and B are in the sequence and A creates a suspension, and then B forces it. The cost for B to force it is included in the realized cost. The unrealized costs are the shared costs for suspensions that were created but never evaluated within the sequence. The total actual cost of a sequence of operations is the sum of the realized costs and the unshared costs.

Throughout a set of operations, we keep track of the accumulated debt, which starts at 0 at the beginning of the sequence. Whenever an operation is performed, we add its shared cost to it. For each operation, we can decide how much of this debt we want to pay. When the debt of a suspension is paid off, we can force it. The amortized cost of an operation is its unshared cost plus the amount of debt it paid (note that it does not include the realized cost). Note that as long as we always pay the cost of a suspension before it’s forced, the amortized cost will be an upper bound on the actual cost.

This framework simplifies the analysis for the case when a suspension is used more than once by assuming that its debt was paid off within the logical history of when it was forced, so we can always analyze a sequence of operations and don’t worry about branching. This might cause the debt being paid multiple times, but it simplifies the analysis.

The author uses the term discharge debit as synonym of pay off debt. I find the latter term easier to grasp, so I’ll stick with it throughout this post.

Let’s introduce an example first and then proceed with the explanation of the Physicist’s method and the corresponding analysis of the example.

The Stream Queue

To allow efficient operations on a queue in the presence of persistence, we can make some of the operations lazy. Recall in a previous post we defined a queue using two lists. To avoid immediate computation, a natural replacement for lists is using its lazy version, the stream data structure, which we also talked about in a previous post.

For the list-based queue, the invariant was that if the front list is empty, then the rear list must be empty as well. For the stream queue, we have a tighter constraint: ‘front’ must be always greater or equal than ‘rear’. This constraint is necessary for the analysis.

The definition of the stream queue is the following:

We store the lengths of the streams explicitly for efficiency.

We’ll be using the Stream developed in the previous chapter, so we’ll refer to the module Stream2 to avoid ambiguity with the standard Stream module.

Inserting an element at the end of the queue is straightforward, since the rear stream represents the end of the queue and is reversed:

The problem is that inserting at rear can cause the invariant of the queue to be violated. check() changes the structure so to conform to the invariant by potentially reversing rear and concatenating with front:

Removing an element from the queue requires us to evaluate the first element of the front stream. Again, the invariant can be violated in this case so we need to invoke check() again:

The complete code for the stream queue is on Github.

Analysis using the Banker’s Method

The idea of the Banker’s Method is basically define an invariant for the accumulated debt and a strategy for paying it off (that is, decide how much debt each operation pays off). Then we show that whenever we need to force a suspension, the invariant guarantees that the accumulated debt has been paid off. One property of the Banker’s method is that it allows associating the debt to specific locations of the data structure. This is particularly interesting for streams, because it contains multiple (nested) suspensions, so we might force parts of this structure before we paid the debt associated with the entire structure.

By inspection, we can see that the unshared cost of both push and pop are O(1). It’s obvious in the case of push, and in the case of pop, in theory check could take O(m) where m is the size of the queue, but since Stream2.concat() and Stream2.reverse() are both lazy, and hence memoized, they are not included in the unshared costs.

To show that the amortized cost of both operations is O(1), we can show that paying off O(1) debt at each operation is enough to pay for the suspension before it is forced. For the queue, we also need to associate the debt with parts of the data structure, so that we could force the suspension of only some parts of it (for example, on the stream we can evaluate only the head, not necessarily the entire structure).

We now define an invariant that should be respected by all operations. Let d(i) be the debt at the i-th node on the front stream, and D(i) = \sum_{j=0}^{i} d(j) the accumulated debt up to node i. The invariant is:

D(i) \le \min(2i, |f| - |r|)

This constraint allows us evaluating the head at any time, because D(0) = 0, which means its debt has been paid off. The second term in min(), guarantees that if |f| = |r| the entire stream can be evaluated, because D(i) = 0 for all i.

The author then proves that by paying off one debt in push() and two debt units in pop() is enough to keep the debt under the constraint.

Queue with suspended lists

Because the Physicist’s method cannot assign costs to specific parts of the data structure, it doesn’t matter if the structure can be partially forced (like streams) or if it’s monolithic. With that in mind, we can come up with a simpler implementation of the queue by working with suspended lists instead of streams. Only the front list has to be suspended because the cost we want to avoid, the reversal of the back list and concatenation to the front list, happens on the front list.

On the other hand, we don’t want to evaluate the front list when we perform a peek or pop, so we keep a evaluated version of the front list too.

The signature of the structure is as follows:

As mentioned in the code above, the invariants we want to enforce is that the front list is never smaller than the rear list

and that the evaluated version of the front list is never empty if the lazy version still has some elements.

The push and pop operations are similar to the other versions of queue, but since we mutate the structure, we might need to adjust it to conform to the invariants:

Finally, because of our second invariant, peeking at the queue is straightforward:

The complete code for the suspended queue is on Github.

Analysis using the Physicist’s Method

We’ve seen the Physicist’s Method in a previous post when we’re ignore the persistence of the data structures. We adapt the method to work with debits instead of credits. To avoid confusion, we’ll use \Psi to represent the potential function. \Psi(i), represents the accumulated debt of the structure at step i. At each operation we may decide to pay off some debit, which will be then included in the amortized cost. We have that \Psi(i) - \Psi(i - 1) is the increase in debt after operation i. Remember that the shared cost of an operation corresponds to the increase in debt if we don’t pay any of the debt. Thus, we can find out how much debt was paid off then by s_i - \Psi(i) - \Psi(i - 1), where s(i) is the shared costs of operation i. Let u(i) and c(i) be the unshared and complete costs of the operation i. Given that, by definition, c(i) = u(i) + s(i), we can then express the amortized cost as:

a_i = cc_i - (\Psi(i) - \Psi(i - 1))

To analyze the suspended queue we need to assign values to the potentials such that by the time we need to evaluate a suspension the potential on the structure is 0 (that is, the debt has been paid off). For the suspended queues we’ll use the following potential function:

\Psi(q) = \min(2|w|, |f| - |r|)

Where w is the forcedFront, f is lazyFront and r is rear.

We now claim that the amortized cost of push is at most 2. If we push and element that doesn’t cause a rotation (i.e. doesn’t violate |f| \ge |r|), then |r| increases by 1, and the potential decreases by 1. No shared is incurred and the unshared cost, inserting an element at the beginning of rear is 1, hence the amortized cost for this case is 1 – (-1) = 2. If it does cause a rotation, then it must be that after the insertion |f| = m and |r| = m + 1. After the rotation we have |f| = 2*m + 1 and |r| = 0, but w hasn’t changed and cannot be larger than the original f, so the potential function is at most 2*m. The reversal of r costs m + 1 and concatenating to a list of size x costs x (discussed previously), plus the cost of initially appending an element to read, so the unshared cost is 2*m + 2. No suspensions were created, so the amortized cost is given by 2*m + 2 - (2*m) = 2.

Our next claim is that the amortized cost of pop is at most 4. Again, if pop doesn’t cause a rotation, |w| decreases by 1, so the potential is reduced by 2. The unshared cost is 1, removing an element from w, and the shared cost, 1 comes from the suspension that lazily removes the head of lazyFront. The amortized cost is 2 – (-2) = 4. Note that if w = 0, we’ll evaluate f, but the ideas is that it has been paid off already. If the pop operation causes a rotation, then the analysis is similar to the push case, except that the complete cost is must account for the shared cost of lazily reming the head of lazyFront, so it’s 2*m + 3, for an amortized cost of 3.

Note that when the suspensions are evaluated, the potential is 0, either when |f| = |r| or |w| = 0.

Conclusion

In this post we covered a simple data structure, the queue, and modified it to be lazy evaluated and can show, with theory, that it allows for efficient amortized costs. The math for proving the costs is not complicated. The hardest part for me to grasp is to get the intuition of how the analysis works. The analogy with debt is very useful.

Meta: Since wordpress code plugin doesn’t support syntax highlighting for OCaml, I’m experimenting with the Gist plugin. Other advantages is that Gists allow comments and has version control!

Amortization Analysis

In Chapter 5 of Purely Functional Data Structures, Okasaki introduces the amortization analysis as preparation for the complexity analysis of functional data structures.

The idea of amortized analysis is to prove the complexity of an operation when the average cost per operation is less than the worst case cost of each individual operation.

In this post we’ll perform analysis of 2 data structures and show that their amortized running time is much better than their worst case. There are 2 methods presented in the book, the Banker’s Method and the Physicist Method. I found the Physicist easier to work with, so I’ll only use that. Also, I find it easy to understand concepts with an examples, so we’ll first introduce the data structure and then introduce the analysis.

Queue with lists

An easy way to implement the queue data structure is by using two lists, one representing the beginning (left) and the other the end (right). Inserting/removing an element from the beginning of a list in OCaml is O(1), while doing it at the end is O(n). For this reason, we store the end of the queue reversed in the right list. (In imperative languages the opposite is true, insertions/removals at the end are O(1) and at the beginning O(n), so the left array that is reversed).

x

Queue using lists representing [1, 2, 3, 4, 5, 6]

Insertions are performed by appending an element to the beginning of the right array. Removing from the front of the queue requires removing the first element of the left array. When the left array is empty, we need to reverse the contents of the right array and move it to the left.

We work with an invariant, in which the left array is only empty if the right array is empty as well. That requires a slight change to the removal, where we perform the reversal and move of the right array when the left array has only one element and is about to become empty.

I’ve implemented this idea in Ocaml.

Analysis – The Physicist Method

If we only look at the worst case, we will get that removing from a queue is an O(n) operation. We can improve this by looking at the amortized cost. One way to perform this analysis is via the physicist method, in which we define a potential (or cost) function \Phi(d) for the data structure d. In the queue case, we can define it to be the size of the right list.

The amortized cost of step i, a_i is defined by

a_i = t_i + \Phi(d_i) - \Phi(d_{i-1})

Where t_i is the actual cost (in number of operations)

The intuition is that some costly operations such as reversing the list, which takes O(n) operations, can only happen after O(n) insertions happened, so the average cost per operation is O(1). Translating it to the potential analogy, each insersion operation increases the potential of the data structure by 1, so by the time a revert happens, the right list is emptied and that causes a loss of potential of n, which had been accumulated and that offsets the cost of the operation t_i.

We can get the accumulated amortized cost of the operations up to a specific step by summing them, and we’ll get

A_i = T_i + \Phi(d_i) - \Phi(d_0)

If we choose \Phi(d) such that \Phi(d_i) \ge \Phi(d_0), for all i, then A_i is an upper bound of T_i. Since we assume d_0 is the empty queue, \Phi(d_0) = 0 and we have \Phi(d_i) \ge 0, this property is satisfied.

One important remark is that this analysis is only applicable to non-persistent data structures, which is not the case in OCaml. We present the OCaml code as exercise, but the correct analysis will done in later chapters.

Splay Trees

Splay Tree is a binary search tree. The insertion of an element e consists in spliting the current tree in 2 subtrees, one of all elements less or equal than e, which we’ll call small and another will all elements larger than e, which we’ll call big. We then make e the root of tree and small the left subtree and big the right subtree.

Partition

Partition

The partition operation is O(height of tree). The intuition is, when looking for the small subtree, imagine we’re analysing a node x. If x >= e, then we know all elements in the left subtree are also >= e, so are all the elements in the right subtree, so we can discard it and only look at the left subtree. The same idea applies to looking for the big subtree.

In an ideal scenario, the tree will be well balanced in such a way that the height of the tree is O(log n), and so the insertion operation. But that’s not always the case, so we introduce a balancing strategy. Whenever we follow 2 lefts in a row when searching for the big subtree, we perform a right rotation. Similarly, when following 2 rights in a row when searching for the small subtree, we perform a left rotation.

These re-balancing are enough to make partition and also insertion an O(log n) amortized time, as we shall see in the next section.

Rotation

Rotation

The operation for finding the minimum element is trivial. To delete the minimum element, we also perform rotations. Since we only traverse the left paths, we only have to do a left rotation.

I’ve implemented these ideas in OCaml.

Amortized Analysis of Splay Trees

We first define a potential function. Given a tree t and a node x \in t, let t(x) be the subtree of t rooted in x. Then \phi(t(x)) = \log(|t(x)| + 1) be a potential function where |t(x)| the number of nodes of the subtree of which x is root, and let the potential of a tree t, \Phi(t), be the sum of the sub-potentials of all nodes of t, that is,

(1) \Phi(t) = \sum_{x \in t} \phi(t(x)).

The amortized equation of calling partition() is the actual cost of the partition, plus the potential of the resulting trees (small and big) minus the potential of the original tree. Let’s denote small(t) as s, and big(t) as b.

(2) \mathcal{A}(t) = \mathcal{T}(t) + \Phi(s) + \Phi(b) - \Phi(t)

We can prove that \mathcal{A}(t) \le 1 + 2 \log(|t|) (see appendix), which means the amortized cost of the partition operation is O(\log n). It implies that insertion also has an amortized cost of O(\log n).

The same idea can be used to prove the amortized costs of the deletion of the minimum element. Okasaki claims that splay trees are one of the fastest implementations for heaps when persistence is not needed.

Conclusion

I found amortized analysis very hard to grasp. My feeling is that the amortized analysis is too abstracted from the real problem, making it not very intuitive. It’s also not clear how one picks the potential function and whether there’s a choice of potential that might result in a better than the “actual” amortized cost.

Appendix

As we can see in code, there are four cases to consider: left-left, left-right, right-left and right-right. The left-* and right-* are symmetric, so let’s focus on the left-left and left-right cases. For the left-left case, suppose we perform the following partition:

Left-left case of partition()

Left-left case of partition()

We start with (2)

(3) \mathcal{A}(t) = \mathcal{T}(t) + \Phi(s) + \Phi(b) - \Phi(t)

Since = \mathcal{T} corresponds to the number of recursive calls to partition, \mathcal{T}(t) = 1 + \mathcal{T}(u), so

(4) = 1 + \mathcal{T}(u) + \Phi(s) + \Phi(b) - \Phi(t)

The recursive call to u yields (s’, b’), so we have the equivalence, \mathcal{A}(u) = \mathcal{T}(u) + \Phi(s') + \Phi(b') - \Phi(u),

(5) = 1 + \mathcal{A}(u) - \Phi(s') - \Phi(b') + \Phi(u) + \Phi(s) + \Phi(b) - \Phi(t)

Since, s = s’,

(6) = 1 + \mathcal{A}(u) - \Phi(b') + \Phi(u) + \Phi(b) - \Phi(t)

and \Phi(b') - \Phi(b) = \phi(b(x)) + \phi(b(y)) + \Phi(c) + \Phi(d), and \Phi(t) - \Phi(u) = \phi(t(x)) + \phi(t(y)) + \Phi(c) + \Phi(d), we can cancel out some terms and simplify the equation above to

(7) = 1 + \mathcal{A}(u) - \phi(b(x)) + \phi(b(y)) - \phi(t(x)) - \phi(t(y))

by induction, \mathcal{A}(u) \le 1 + 2 \phi(u),

(8) \le 2 + 2\phi(u) + \phi(b(x)) + \phi(b(y)) - \phi(t(x)) - \phi(t(y))

Since u is a subtree of t(y), \phi(u) < \phi(t(y)) and since the nodes in b is a subset of t, \phi(b(y)) \le \phi(t(x))

(9) < 2 + \phi(u) + \phi(b(x))

It’s possible to show that, given x, y, z such that y + z \le x, then 1 + \log y + \log z < 2 \log x. We have that |t| = |u| + |c| + |d| + 2, while |b(x)| = |c| + |d| + 1, thus, |u| + |b(x)| < |t| and by the relation above, 1 + \log (|u|) + \log (|b(x)|) + 1 < 2 \log (|t|) + 1, \phi(u) + \phi(b(x)) + 1 < 2 \phi(t), which yields

(10) \mathcal{A}(t) < 1 + 2 \phi(t).

It’s very subtle, but in this last step we used the assumption of the rotation, in particular in |b(x)| = |c| + |d| + 1, if we hadn’t performed the rotation, |b(x)| = |b| + |c| + |d| + 2, and |u| + |b(x)| < |t| wouldn’t hold!

For the left-right case, we can do a similar analysis, except that now we’ll apply the recursion to c, and the resulting partition will look different, as depicted below:

Left-right case of partition()

Left-right case of partition()

In this case, (8) will become

(11) \le 2 + 2\phi(c) + \phi(b(x)) + \phi(b(y)) - \phi(t(x)) - \phi(t(y))

We then use that \phi(c) < \phi(t(y)) and \phi(t(y)) < \phi(t(x)) this will allow us to cancel out some terms to

(12) < 2 + \phi(b(x)) + \phi(b(y))

Similar to what we've done before, we can show that |b(x)| + |b(y)| = |s| + |b| = |t| since the partitions of |t| are mutually exclusive. We can still use the result from the other case and get (10).

References

[1] Wikipedia – Amortized Analysis