graph equivalence 1: the overall algorithm

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I want to talk about some interesting graph processing I've done recently. As a part of a bigger problem, I needed to collate a few millions of (not that large) graphs, replacing every set of equivalent graphs with a single graph and a count. I haven't found much on the internets about the graph equivalency, all I've found is people asking about it and other people whining that it's a hard problem. Well, yeah, if you brute-force it, it's obviously a factorial-scale problem. But I believe I've found a polynomial solution for it, with not such a high power, and I want to share it. It's kind of simple in the hindsight but it took me four versions of the algorithm and a couple of months of background thinking to get there, so I'm kind of proud of it (NOTE: it's NOT a general solution of the graph isomorphism, see P.P.P.S. below).

But first, the kind of obvious thing that didn't take me any time to think about at all: if you're going to collate a large number of values that have a large degree of repeatability among them, you don't want to compare each one to each one. You want to do at least a logarithmic search, or even better a hash table. Which means that you need to not just compare for equality or inequality but to impose an order, and/or generate a value that can be used as a hash. If you do the factorial-complexity comparison of two graphs, you get neither, all you know is whether they are equal but you don't have any order between the unequal graphs, nor anything you can use as a hash. Instead, you need a signature: something that would serve as a "standard representation" of an equivalency class of graphs and that can be compared in a linear time (you could even represent the signatures as strings, and compare them as strings if you want). And then you can easily impose order and compute hashes on these signatures.

The real problem then is not to compare the graphs but to find such signatures of the graphs. What should these signatures be? If we can order the nodes of a graph based purely on their place in the topology of the graph, that would work as a signature. A graph might have multiple nodes with the same topology (i.e. belonging to the same equivalence class), then any reordering of these nodes would provide the same signature. In a graph with N nodes (vertices) and L links (edges), this signature could be represented as a string of N+L elements, and thus compared in a linear time.

The complication is that if there are multiple equivalence classes, there might be interdependencies between there classes. For example, consider an untagged graph in the shape of a cross (A...I here are not the proper tags on the nodes but some temporary random names that we assign for the sake of discussion):

.
      I
      |
      H
      |
E--D--A--B--C
      |
      F
      |
      G

The nodes (B, D, F, H) are equivalent among themselves, and the nodes (C, E, G, I) are also equivalent among themselves. So we could interchange their positions within each equivalence class. But once we pick the positions for nodes of one class, the positions of the other class become fixed. Once we assign a certain position in the signature to the node B, we have to assign a certain related position to the node C. We could swap the positions of nodes B and D, but then we have to also swap the positions of nodes C and E accordingly.


First let me tell the final algorithm I came up with, and then I'll explain why it works. The algorithm uses hashes, so let's also initially pretend that there are no such things as hash collisions, and then I'll show how such collisions can be resolved.

The algorithm works on the graphs that may have tags (or labels, whatever term you prefer) on the nodes and links. The links may be directional, and this can also be expresses as a tag on them. The nodes might have some tags on the "ports" where the links attach, such tags can also be "moved" to the links. The algorithm uses the "node-centric" view of the tags on the links, i.e. the specific tag used by it depends on which node is used as a viewpoint. For example, suppose we have nodes A and B and a directional link with tag X that goes from port 1 on A to port 2 on B:

.
        X 
A[1] -------->[2]B

When looking from the node A, such a link can be seen as having the tag "O, 1, X, 2". "O" stands for "outgoing", then the local port, tag of the link itself, the remote port. When looking from the node B, the same link will have the tag "I, 2, X, 1".

The nodes get ordered into the signature by assigning them the sequential IDs starting from 0. The algorithm starts with no IDs assigned, and assigns them as it works.

The algorithm computes the hashes of the graph's topology around each node, up to the distance (you can also think of it as radius) D. The distance 0 includes the tag on the node itself and the tags on the links directly connected to it. The distance 1 also includes the immediately neighboring nodes and links connected to them, and so on. For each node, we keep growing the distance by 1 and computing the new hashes. It's easy to see that when we get to the distance D=N, we're guaranteed to include the topology of the graph to all the nodes. But a general graph is not a tree, it also contains the "redundant" links, and to include the nodes on the other side of all the redundant links, one more step would be necessary, to D=N+1. The inclusion of these nodes for the second time is needed to include the full topology of these redundant links.

However for most graphs the maximal radius will be smaller.  If each node is connected to each node, the whole graph will be included at D=1. So we don't have to iterate all the way to N, instead we can keep the set of the included nodes for each hash, and we can stop growing when this set includes all the nodes in the graph.

Let's call these hashes H, and the node sets S.

The hash H(0) and set of nodes S(0) for the distance 0 around a node is computed as follows:

1. Include this node itself into the set of nodes S(0).
2. If the node has an ID assigned, include this ID into the hash and stop. Otherwise continue with the next steps.
3. Include the tag of the node itself into the hash H(0).
4. Order the links connected to this node in some fixed way (for example, by comparing the components of their tags in the order shown in the example above: the direction, local port, tag of the link itself, remote port).
5. Include the tags of the links in this order into the hash H(0).

The hash H(D+1) and set of nodes S(D+1) for the distance D+1 around a node is computed by induction as follows:

1. Combine S(D) of this node and of all the directly connected nodes to produce S(D+1).
2. If the node has an ID assigned, assume H(D+1) = H(0) and stop. Otherwise continue with the next steps.
3. Include the hash H(0) at distance 0 of the node itself into H(D+1).
4. Order the links connected to this node in the same way by the tags, but with an addition of the information about the nodes at the other end of the link: if some links have equal tags, order them according to the previous hash H(D) of the nodes at the other ends of these links.
5. Include the tags of the links and H(D) of the nodes at the other ends of the links into H(D+1).

The rest of the algorithm then is:

1. While there are any nodes with unassigned ID:
  1.1. Repeat for D=0...N+1
    1.1.1. For each node, compute H(D), S(D).
    1.1.2. Order nodes in the ascending order by (ID, H(D)), the unassigned IDs go after all the assigned IDs.
    1.1.3 If all S(D)==S(D-1), break out of the loop (it means that the whole topology is included).
    1.1.4. See if any of the nodes with unassigned ID have unique H(D). If any of them do, assign the IDs for them, continuing from the last assigned ID, and going in the order of their H(D). Re-compute their H(0...D) based on the newly assgined ID.
  1.2.  If there are still any nodes with unassgined ID, take one arbitrary node with the lowest H(D) and assign the next ID in sequence to it.

Let's estimate the run time:

At least one ID gets assigned on each iteration of the outer loop, so there would be at most N iterations. The inner loop would also have at most N+1 iterations. The most expensive operations in it are the computation of the next H(D) and the sorting. The computation of H(D) would go through N nodes and would do the sorting of the links connected to this node that would take O(EperN*log(EperN)) steps. The sorting of the nodes would take O(N*log(N)) steps. So to cover both these cases, let's take O(N*N*log(N)) as the worst case that would cover them both. So the total comes to O(N^4 * log(N)), which is much better than the factorial.

The resolution of the hash collisions would add to this but on the other hand, some optimizations are possible.

When we compute H(D+1), the links can be sorted by their tags only once, with the further sorting only among the links with the same tags. This sorting is needed to make sure that these links are treated as a set, that no matter in which order the remote nodes are originally seen, the set of remote nodes with the matching H(D) would produce the same H(D+1). But there is another, more hacky but faster way to achieve that: combine the hashes of nodes at equivalent links in a commutative way, such as by addition, before including them into H(D+1). This would remove one power of N, and the total would be O(N^3 * log(N)).

Also, as soon as a node gets its ID, its hash becomes fixed, so there is no need to compute it any more. And we can skip computing the set S for them too. This means that the other nodes would not necessarily include the whole graph into their set S, since the propagation would stop at the nodes with the assigned IDs. But that's fine, because the inner loop would stop anyway when all the sets S stop changing. So the inner loop has to work only on the nodes that have no ID assigned yet.

More on the hash collision resolution, and on why this works at all, in the next installment(s).

P.S. The description of how to compare signatures is also in the next installment.

P.P.S. An actual implementation example can be found in https://github.com/tensorflow/tensorflow/blob/ab8e195d2e0978c21234a5632d4fabf47535eda1/tensorflow/core/grappler/graph_analyzer/sig_node.h (and the nearby files).

P.P.P.S. The algorithm has turned out to be incorrect in the general case, see the part 7 for the explanation. But it's still works for the graphs that are typical for description of the flow control computations: labelled, directional, and without the particularly tricky loops.

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SQL "With" statement

I've recently learned about the WITH statement in SQL that had apparently been there since the 1999 standard. It pretty much allows you to define the "temporary variables" (or "temporary views") in the SQL query for common sub-expressions, like this:

WITH
   name1 AS (SELECT ...),
   name2 AS (SELECT ...)
SELECT... FROM name1, name2, whatever ...

This solves a major problem with the SQL statements, especially for self-joins on such pre-computed data, and lets you do fork-and-join in a single statement. I'm surprised that it wasn't used all over the place for the CEP systems.

Comments and names

I've read an opinion that if the names of the methods are self-explanatory, the comments describing these methods are not needed.

Well, sometimes it's so, but in general I strongly disagree with this idea. The names and comments have differnt purposes. The point of the name is to give the method a designation that is difficult to confuse with the other methods. The point of the comment is to explain the meaning of a method. Mixing them doesn't do well.

The attempts to make the names more self-explanatory back-fire in two ways: the code becomes longer, and the names become less distinct. And the worst part, it still doesn't explain enough, you can't understand such an uncommented method until you read its implementation. A good comment must explain the various considerations that went into the development of the method, its pre-conditions and post-conditions, and the exact meaning of parameters. There is simply no way to fit it into a name.

In a way, a method name is similar to an acronym in the texts. Or, well, not necessarily an acronym but some  short meaningful name. The first time you use this acronym or word, you explain what it means. When you define a method, you define what it means in the comment, and give it a short name. Then, as you use the word in the rest of the text, you use the method name in the program. And if the reader forgets what it means, he can always look up the explanation. If you used the full multi-word definition throughout your text, it would grow completely unreadable.

waiting for a condition: another way

I've recently encountered another way of doing synchronization without condition variables in the Abseil library: their mutex has the method Await(). It provides an alternative to the construction (copied from their description):

mu.Lock();
... // arbitrary code A
while (!f(arg)) {
  mu.cv.Wait(&mu);
}
... // arbitrary code B
mu.Unlock();

as

mu.Lock();
... // arbitrary code A
mu.Await(Condition(f, arg));
... // arbitrary code B
mu.Unlock();

Here the condition is given in the argument of Await() as a functor, an object that keeps the related context and executes a function in this context, in the new-fangled C++14 code that's usually given as a lambda.

This works by evaluating all the currently requested conditions on unlock, and if one of them becomes true, waking up the appropriate thread. Looks kind of wasteful on the first sight but there are good sides to it too.

One of the typical problems with the classic condition variables is "how fine-grained should they be made"?  Should there be a separate condition variable for every possible condition or can the conditions be mixed sometimes on the same variable? The Await() solves this problem: it can easily allocate a separate variable per condition, and do it right on the stack in Await().  The obvious problem with multiple conditions sharing the same condition variable is that then on a signal multiple threads would wake up, some of them will find their condition and continue, some will not find it and go back to sleep, creating the churn. Await() clearly solves this problem.

But it can do one better too. Another problem is when the right thread wakes up but has to go back to sleep because before it woke up some other thread snuck in, got the mutex and ruined the condition. Await() can guarantee that it won't happen by simply keeping the mutex locked but passing its ownership to the waiting thread. Then when the waiting thread wakes up, it doesn't even need to check the condition again, because it's been already checked for it. This is one of those times where having your own low-level thread library pays off, you couldn't do such a trick with a Posix mutex.

So when they say that it about evens out on performance with the classic condition variables, I think there are good reasons to believe that.

removing the duplication in Bayesian & AdaBoost

I've been thinking some more on how to remove the duplication between the Bayesian events/AdaBoost partial hypotheses. Here are some notes. They're not some final results, just some thinking aloud.

I think the problem of removing the commonality can be boiled down to the following: Suppose we have an event A that gives the prediction with some effect a on the training cases, and an event B that gives the prediction b on the training cases. These predictions have some part ab (that's a 2-letter identifier, not multiplication of a and b) in common. Then the effects can be written as:

A: a = a' + ab
B: b = b' + ab

Where a' and b' are the differences between the original a and ab, and b and ab. Here the operator "+" means not the addition but application of the events in the deduction one after another. So "a + b" means "apply a, then apply b".  The reason why I chose the sign "+" is that this combined application should act like addition. I.e. have the commutative and distributive properties like

x + y = y + x
(x + y) + z = x + (y + z)

No, I don't have a proof why they do. But the problem with the AdaBoost hypotheses that I'm trying to solve is that the result there depends on the order of partial hypotheses (i.e. Bayesian events). I want to be able to decompose these events in such a way as to make the result independent of the order, hence the commutative requirement. The exact operation that satisfies these rules will need to be found but for now we can follow through the results of these rules.

So, going by these rules, if we apply both events, that will be:

a + b = a' + ab + b' + ab = a' + b' + 2*ab

The part ab gets applied twice. To remove this duplication we've got rid of one copy of ab. We can do this by splitting the original two events into three events A', B' and AB (again, that's a 2-letter identifier, not a multiplication):

A': a'
B': b'
AB:ab
a = a' + ab; b = b' +ab

The new event AB gets the common part, while A and B lose it and become A' and B'. Then with all events applied we get the result without duplication:

a' + b' +ab

If we want to add the fourth event C, we've got to get its overlap with all 3 previous events. This requires a whole bunch of multi-letter identifiers:

a'c would be the overlap between a' andc
b'c would be the overlap between b' and c
abc would be the overlap between ab and c

And the double-primes would be the "remainders":

a' = a'' + a'c
b' = b'' + b'c
ab = ab' +abc
c = c' + a'c + b'c +abc

Then the result of applying all these events without duplication will be:

a'' + a'c + b'' + b'c + ab' + abc + c'

This gives the outline of the solution but it still has two problems: the number of events after splitting grows exponentially (a power of 2) with the number of the original events, and the concrete meaning of the operation "+" needs to be defined. And actually there is one more operation that needs to be defined: what does the "overlap" mean and how do we compute the value of ab from the values of a and b?

The answers to both problems might be connected. One way to define the overlap would be to associate each training case with exactly one event that predicts it well. Then when we split a and b into a', b', and ab, we would be splitting the whole set of training cases into 3 parts: one part gets predicted well by both A and B, one only by A, and one only by B.

Then the problem with the exponential growth can be answered in a two-fold way. First, the number of the events after splitting can't grow higher than the number of the training cases. Second, we can put an artificial cap on the number of events that we're willing to use (Emax), and select up to this many events, ones that predict the most training cases each. We can then either sweep the remaining training cases under the carpet, saying that they are one-offs that would cause overfitting, or split them somehow with partial strength among the events that get included.

The last sentence also suggests the third way of tackling the growth: if we define some way to do the splitting, instead of multiplying the events we could just split the power they have on a particular training case. But I think this would create issues in case of the partial confidence during the execution of the model that can be handled better with the combined events.

Returning to the definition of "+", I couldn't figure out yet how to make such an operation directly in AdaBoost. Maybe it can be done through logarithms, that will need some more thinking. It requires the ability to say that some training case doesn't get affected by a partial hypothesis/event. But there seems to be an easy way to do it in the Bayesian way: mark the confidence of that event for that training case as 0.5. And then these transformed events/partial hypotheses can be fed into AdaBoost too.

This fits very well with the underlying goal of boosting: to find the approximately equal rate of the predominantly correct votes for each training case. After the transformation there will be exactly one right vote for each training case, and the rest of votes will be "undecided" (or at least sort of, subject to the limitations introduced by the mixing of training cases).

Another consequence is that such splitting would make the produced events fully independent of each other: each event would have an exactly 50% correlation with each other event, meaning that they can be applied in any order. And this is exactly what we've been aiming for.

So it all looks good and simple but I'm not sure yet if I've missed something important that will take the whole construction apart. There might well be.

P.S. After a little thinking I've realized that the idea of number of events being limited by the number of training cases is pretty much equivalent to the computation directly on the training cases by weight.

AdaBoost conjunctions & XOR

I've been thinking recently about the better ways to extract the commonality from two Bayesian events (hope to find time to write about that soon), and that brought me back to the XOR problem: how to handle the situation when two events indicate a certain hypothesis when they are the same and another hypothesis when they are different.

BTW, I took a class on machine learning recently, and it turns out that even with the neural network this problem doesn't have a great solution. The neural networks are subject to the random initial state  before the training, and sometimes they do find the good combinations and sometimes they don't. This tends to get solved more reliably by the human intervention: a data scientist stares at the data, maybe does a few training attempts to see what happens, and then manually adds a new event that represents the XOR of the original events.

So, thinking about it, I've remembered again about the concept of conjunctions that I've read in the AdaBoost book. I picked up the book and tried to find the conjunctions but couldn't: they're not anywhere in the index. Fortunately, an Internet search turned up the online copy of the book, and the conjunctions are in the chapter 9.

The idea there is that the quality of AdaBoost partial hypotheses can be improved by forming them as conjunctions of two or more raw partial hypotheses. The idea is to take the two best choices and try a conjunction on them. If it gets better, try adding the third one.

For a simple example, consider the data with two parameters, with values 1 collected in the upper-right quarter, and 0 in the remaining three quarters:

.
              ^y
          0   |   1
              |
       -------+------>
              |      x
          0   |   0

The two partial hypotheses (x > 0) and (y > 0) taken together but separately will identify the upper right quarter quite decently. Each of them will have a 75% success rate. But the values in the upper-left and lower-right corner will get one vote "for" and one vote "against" from these hypotheses and will be left undecided. I actually think that the classic AdaBoost would not be able to decide well for them. It would probably crowd the field with the extra hypotheses in the upper-right corner like (x > 1), (y > 1), (x > 2), (y > 2) and so on so that the upper-left and lower-right quarters would get better but the parts of the upper-right close to the boundary would get worse. If we take the classic Bayesian approach and consider that the prior probability of encountering a 0 is 75% (assuming that all the quarters are filled at the same density), the "undecided" result would leave the probability of 0 at the same 75%, and it would be a vote for 0. So this is probably a good example of how AdaBoost could benefit from using the prior probabilities.

The conjunctions are good at handling such situations: if these partial hypotheses are combined into one conjunction (x > 0) & (y > 0), we get a hypothesis with 100% success.

They can be used to handle the XOR situation as well. Let's look at the situation with 1s in upper-right and lower-left corners:

.
              ^y
          0   |   1
              |
       -------+------>
              |      x
          1   |   0

The two partial hypotheses formed by conjunctions, ((x > 0) & (y > 0)) and ((x < 0) & (y < 0)), would each give the 75% success rate. The quadrants with 0s would get correctly voted by both hypotheses but 1s would get one vote for and one vote against.  But the similar hypotheses can be added for the other two quarter: !((x > 0) & (y < 0)), !((x < 0) & (y > 0)). These would add two correct votes for the quadrants with 1s, and two opposite voted canceling each other for the 0s, and taken together these 4 hypotheses can identify everything correctly.

An interesting thing to note here is that the raw hypotheses like (x > 0) aren't the best ones at selectivity in this example.  They're actually the worst ones, having the exact 50% correctness. But then again, pretty much all the hypotheses with vertical and horizontal boundaries here will have about 50% selectivity. Hunting for slightly better than 50% by selecting boundaries that fit the random density fluctuations would actually make things worse, not hitting the right center. I actually see no way to select the right boundaries by choosing the raw hypotheses independently. Only taken together in a conjunction they become optimizable.

The other way to do it would be to use XORs instead of conjunctions. The case with two diagonal quadrants of 1s is trivial with XORs, since it can be described perfectly by one XOR, just as the case with one quadrant of 1s is trivial with conjunctions. Let's look at the one quadrant with XOR:

.
              ^y
          0   |   1
              |
       -------+------>
              |      x
          0   |   0

If we use one XORed hypotheses, !((x > 0) XOR (y > 0)) and two simple ones (x > 0) and (y > 0), each of the three will have the 75% success rate. The upper-right corner will get all three votes for 1, and the rest of the quadrants will get two votes for the right value 0 and one against it, so the right vote would win in them too.

The way I described it, it looks like the XORed hypothesis in this example isn't clearly better than the simple ones. But that's not how it will work with AdaBoost. After the two simple hypotheses are picked, the relative weights of the upper-left and lower-right quadrants would grow, and the XOR clearly is better at fitting them.

Overall, it looks like XORing is better at producing the useful combined hypotheses than conjunctions: the conjunctions required 4 complex hypotheses to fit the XOR case but XOR required only 3 hypotheses to fit the conjunction case, 2 of them simple hypotheses.

Boosting book reference

I've accidentally found the Freund and Schapire's book on AdaBoost online:

https://mitpress.mit.edu/sites/default/files/titles/content/boosting_foundations_algorithms/toc.html

Great for searching.

Reversible computing

I've read an article in the IEEE magazine about he reversible computing: https://spectrum.ieee.org/computing/hardware/the-future-of-computing-depends-on-making-it-reversible

I've never thought about it before, that the normal switching in the circuits dissipates the energy continuously, and the way to fix it would be to make sure that the energy is always shuffled around and never dissipated (although of course there will always be the "friction losses"). I haven't found that much other information on the Internet. So I've started thinking about how such a reversible computing can be implemented, based on the example from the article. And I think I've come up with a decently simple implementation on the digital level, not sure why the article says that it's complicated. Of course, it might end up extremely complicated in the implementation, I know nothing about that. So, here is what I've come up with:

The first obvious problem was how do the values ("billiard balls" from the mechanical analogy) get returned back? That has a kind of obvious fundamental solution: in all the digital automata the information fundamentally goes in circles like this:

.
          +-----+ 
          |     |---> output   
          |     |   
          |     |    +------+
          |     |    |      |
input --->|logic|--->|memory|--+
          |     |    |      |  |
      +-->|     |    +------+  |
      |   +-----+              |              
      +------------------------+

Well, mostly goes in circles, there are inputs and outputs. So the next problem is how to make the inputs and outputs work. Note that the inputs don't come from nowhere, and outputs don't go into nowhere, instead they connect to the other circuits. So the inputs and outputs can be also made in the form of cycles, with the interaction between the output of one circuit and input of another circuit:

.
output --->-+ | +-<-- recycle
            | > |
recycle <---+ | +---> input

The output/input interaction can be represented as a function with the inputs:

x - the output value
1 - the constant 1

and outputs:

rx - recycling of the output value
i - the input sent to the circuit
ri - recycling of the input value

The truth table for this function will be:

x  1  |  rx i  ri
==================
0  1  |  0  0  1
1  1  |  1  1  0

As you can see, the number of 1s in the output always matches the number of 1s in the inputs. If x=1, the constant 1 gets shipped to the input, otherwise it goes to the recycling and back into the constant.

The next problem is the memory: it has to store a value that might be 0 or 1, which is obviously not symmetric. But it looks like the basic solution for that is easy: for each memory cell keep 2 sub-cells, one of them storing the value, and another one its complement. So there will always be one packet of energy stored, and the value would be determined by the sub-cell where it's stored. Then the contents of one sub-cell would be sent out for the next step of computations, and the other sub-cell will stay until the new value gets stored. I guess this could also be expressed equivalently as a constant 1 that gets connected to one or another output of the memory unit.

How would the recycling be done? For that we've got to connect the 1s in the results back to where they came from. And ultimately they all come from the constant 1s, either introduced in the computations or use din the logic of the inputs. For all I can tell, by the time the bits get recycled, they get really mixed up, and there is no easy way to put it back. The only universal way I see to put them back would be to order all the original "constant 1s" and all the recycled values (the specific order doesn't matter, just there has to be some order) and then do the logic like this:

If the 1st recycled value is 1, put it into the 1st "constant 1".
If the 2nd recycled value is 1 {
  If the 1st "constant 1" is still 0, put the 2nd recycled value into it,
  else put the 2nd recycled value into the 2nd "constant 1".
}
If the 3rd recycled value is 1 {
  If the 1st "constant 1" is still 0, put the 3rd recycled value into it,
  else if the 2nd "constant 1" is still 0, put the 3rd recycled value into it,
  else put the 3rd recycled value into the 3rd "constant 1".
}

and so on. As you can see, the logic can be parallelized to some extent but still will require as many steps as there are input values. So for the large circuits it becomes quite slow. Maybe this can be optimized in some way for some specific applications but that would be quite difficult. Maybe that's what they mean when they say that the design of the reversible circuits is difficult.

But I think that there is a simple solution: make each individual circuit small. Maybe limit it to 2 inputs and possibly one additional constant 1. That would limit the recycling to 3 steps, which is reasonably fast. Then use the input/output method described above to connect these small circuits into the large circuits. Then as the inputs propagate through the layers of these simple circuits, the early layers would compute their recycling in parallel with that propagation, so overall there will be very little penalty.

Note that there would be no memory between these layers of small circuits, it would be mostly layers of the small circuits, with one memory layer at the end.

The classic digital electronics has the implementation of functions like AND and OR with a large number of inputs, not just 2, that get computed in the same time as a function with 2 inputs, which comes quite handy. This won't be that easy for the reversible circuits. But there probably is no point in trying to do this for the reversible circuits in the first place: even if a function of N arguments would get computed in one step, then it would still have to do the recycling in N steps. On the other hand, such a function of N arguments can be built as a tree hierarchy of 2-argument functions, of the depth log2(N). This tree would be computed in log2(N) steps and the recycling of the last element will take 5 steps (the recycling of the previous elements could be done in parallel). The number 5 comes from a logical operation on 2 inputs producing 3 outputs to stay reversible, as given in an example in the article (and I suspect that it's an universal property of such operations), plus 2 inputs having 1 recycling output each. So the total time would be (log2(N)+5) which is better than N. Even if we include the I/O overhead, that would still be (2*log2(N)+5) which is still better than N.

Thus I think that the basic design of the reversible circuits on the digital level with the "billiard ball analogs" looks not that difficult. It might be that I've missed something and got everything wrong. And of course, the implementation on the level of the underlying electronic elements like transistors might be very difficult. I'm still trying to figure out how it could be done, and since my knowledge about the transistors is fairly limited, it's difficult for me. I have some ideas but I'd need to think some more about them before writing them down (and they might be completely wrong anyway).

A bit on a side subject, about the logic. In the article they use the the example of function [(NOT A) AND B] which looks kind of unconventional (but maybe they have some good reasons, one of them being the convenience of producing both the result and its complement). Well, there is more than one way to create  a set of operations that can represent all the boolean functions. One is the set of (AND, OR, NOT), then (AND-NOT) and (OR-NOT) do everything in one operation(and perhaps the [(NOT A) AND B] does that too, I forgot how to check that), and yet another set is (XOR, constant 1). The XOR gate looks more straightforward to me, it was the first one I've thought of. And it turns out that XOR is quite popular in the reversible computing, it's known as the Toffoli gate. But in the end it probably doesn't matter much. Or maybe the best way is to use the set of (AND, OR, NOT). The reason for that being that I think OR of 2 arguments can get by with only 2 outputs, and so does NOT, only AND requiring 3 outputs. Which means one fewer recycling step after the OR, which means a slightly higher performance and about 10% (half of the saved 20%) simpler recycling circuitry compared to the situation when all the circuitry is built of one universal function like (AND-NOT) or (OR-NOT) or (XOR).

AdaBoost 10: a better metric?

Here is one more consequence I've noticed from the application of AdaBoost approach to the Bayesian one: it can be applied the other way around too.

I.e. AdaBoost provides the list of partial hypotheses and weights for them. But at some point it tends to start repeating the same hypotheses over and over again, because a set of hypotheses taken together gives some gain, and applying it multiple times seemingly increases this gain. This gain is subject to the same problems as the Bayesian computations with the duplicate events: it increases the confidence only seemingly. So the same solution can be applied to it:

After the set of partial AdaBoost hypotheses has been computed, they can be re-ordered in such a way as to remove the duplication, by selecting on every step the partial hypotheses with the smallest effect. This "run order" can be kept and updated on every step, instead of the AdaBoost's "selection order". So when the sequence will start repeating, a duplicate hypothesis would add nothing to the cumulative effect of the "run order", and at this point the selection algorithm should try a different hypothesis (or simply stop).

I guess, in case if the selection algorithm doesn't stop but tries to do the next best pick (and then maybe another one if the second choice is no good either), it becomes fundamentally more complicated: on each iteration of AdaBoost it should select not simply the hypotheses that maximizes the effect of the whole sequence when appended to the end of it but one that maximizes the effect of the whole sequence when inserted at the position in the sequence where it adds the least effect. Each such insert could also cause a re-ordering of the part of the sequence that follows it.

Which sounds even more heavy-weight than the original AdaBoost selection of the hypothesis. On the other hand, since the AdaBoost selection is already so heavy-weight, maybe this would make little extra overhead. I don't know how to analyze it yet.

Bayes 27 & AdaBoost: another problem with partial confidence

The solution from the previous installment can handle the partial confidence but it has its own issue. To demonstrate it, let's look at a run with the table from the part 26 but where both of the fully-correlated events E1 and E2 get a confidence of 0.7:

$ perl ex26_01run.pl -n -p 1 tab24_01.txt in27_01_02.txt 
Original training cases:
!      ,       ,E1          ,E2          ,E3          ,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E1             ,E2             ,E3             ,
hyp1   ,3.00000,0.66667^0.66667,0.66667^0.66667,0.40000^0.33333,
+      ,       ,               ,0.50000^0.50000,0.40000^0.33333,
hyp2   ,3.00000,0.33333^0.33333,0.33333^0.33333,0.60000^0.66667,
+      ,       ,               ,0.50000^0.50000,0.60000^0.66667,
--- Initial weights
hyp1    w=3.00000 p=0.50000
hyp2    w=3.00000 p=0.50000
--- Applying event E1, c=0.700000
hyp1    w=1.70000 p=0.56667
hyp2    w=1.30000 p=0.43333
--- Applying event E2, c=0.700000
hyp1    w=0.91800 p=0.60554
hyp2    w=0.59800 p=0.39446
--- Applying event E3, c=1.000000$ perl ex27_01run.pl -n -p 1 tab24_01.txt in27_01_02.txt
hyp1    w=0.36720 p=0.50579
hyp2    w=0.35880 p=0.49421

The probabilities move after applying E1, after E2 they move again. The interesting question is, should have they moved after E2? It really depends on how E1 and E2 have been measured: are their confidences derived independently or tracking back to the same measurement?

The code assumes that they are independent, so if both are independently measured at the confidence of 0.7, together they provide a higher confidence. But if they both come from the same test, this assumption is wrong.

If they came from the same test then E2 carries no additional information after E1 has been processed. In this case we should have treated the value C(E2)=0.7 (i.e. C(E2)=C(E1)) as having the meaning "I don't know", and then the differences from it would carry the new information.

But this approach has its own problems. First of all, it's fine if the confidence values C(E) represent the measured probabilities. But if they represent some expert estimation then an expert saying "I don't know" and giving the confidence of 0.5 would shift the probabilities instead of leaving them unchanged. Well, it it could be re-scaled automatically but then how do we know that the expert means "I don't know" and not "I've carefully considered everything and I think that it's a 50-50 chance"? So it sounds like getting it right would require using two separate numbers: the absolute confidence that tells how sure we are of the estimation, and separately the estimation. Which would create all kinds of messes.

The second problem, if we want to do such a re-scaling, to which value do we re-scale? It's reasonably easy if two events are copies of each other but what if the correlation is only partial? There actually is an a rather easy answer to the problem: the needed value would be P(E), the probability that the event is true at that point (after the previous events have been considered). I've actually went and modified the code to compute the P(E) based on the weights only to find out that I've forgotten an important property of the AdaBoost logic. Let me show its output:

$ perl ex27_01run.pl -n -p 1 tab24_01.txt in27_01_02.txt
Original training cases:
!      ,       ,E1          ,E2          ,E3          ,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E1                     ,E2                     ,E3                     ,
hyp1   ,3.00000,0.66667^0.66667@0.66667,0.66667^0.66667@0.66667,0.40000^0.33333@0.50000,
+      ,       ,                       ,0.50000^0.50000@0.50000,0.40000^0.33333@0.50000,
hyp2   ,3.00000,0.33333^0.33333@0.33333,0.33333^0.33333@0.33333,0.60000^0.66667@0.75000,
+      ,       ,                       ,0.50000^0.50000@0.50000,0.60000^0.66667@0.75000,
--- Initial weights
hyp1    w=3.00000 p=0.50000
hyp2    w=3.00000 p=0.50000
--- Applying event E1, c=0.700000
Expected P(E)=0.500000
hyp1    w=1.70000 p=0.56667
hyp2    w=1.30000 p=0.43333
--- Applying event E2, c=0.700000
Expected P(E)=0.513333
hyp1    w=0.91800 p=0.60554
hyp2    w=0.59800 p=0.39446
--- Applying event E3, c=1.000000
Expected P(E)=0.598615
hyp1    w=0.36720 p=0.50579
hyp2    w=0.35880 p=0.49421

The computation table now has one more element printed after "@", the estimation of event's probablity for the particular branch based on the previous events. During the computation these values get mixed together based on the weight of the branches. But notice that when applying E2, the message says:

Expected P(E)=0.513333

It's 0.513333, not 0.7! The reason is two-fold. First, AdaBoost works by adjusting the weights of the training cases such as to bring the probability of the last seen event to 0.5. It just does the correction opposite to what I was trying to achieve. Second, these are the wrong probabilities in the table. These probabilities are computed per-branch based on the absolute confidence. They don't care if the actual confidence was 0.7 or 0.3, these values are the same from the standpoint of the absolute confidence. But this difference matters a lot when trying to predict P(E)!

Instead we need to honestly compute the P(E) based on the previous events, not per the AdaBoostian table but per the original training cases. But if we do that, we end up again with a table of exponential size. The trick with ignoring all the events but the last few might still work though. The idea would be similar: since we try to order the closely-correlated events close to each other, the previous few events would be the ones with the strongest effects. And if the farther events aren't closely correlated, they won't have a big effect anyway, so they can be ignored. Maybe I'll try this next. Or maybe it's simply not worth the trouble.

Bayes 26 & AdaBoost: more on the partial confidence

I think I've found at least a partial solution to the problem of partial confidence that avoids going exponential.

It grows right out of the problem definition: the problem really exists only for the closely related events. If two events are unrelated, then the confidence of one wouldn't affect the other in any noticeable way. But for the related events, the first one takes up the bulk of the effect, and then the effect of the following events becomes greatly diminished. If two events are actually the same (like E1 and E2 in tab24_01.txt in the part 24), the second event gets stripped of any effect whatsoever. But if at the execution time we find that we have no information about the first event and good information about the second event, it's too late to make use of the second event because its probabilities are already hardcoded in the table and condemn it to having no effect. But if we could change the probability at run time then we could use the second event instead. Better yet, the probabilities of the following events would be unaffected because they don't care which one of the duplicate events provided the information. For a sequence of not equal but closely correlated events, if the first one is excluded from the table preparation, the others still seem to converge in their effects to something close to the original sequence, and the following events don't get affected that much.

Since the logic in the part 24 assumes that the closely-related events would be placed in the sequence next to each other, the solution might be to compute an event's table entry in multiple versions, depending on whether the values of a limited number of the previous events are known or not. It would still be exponential for this number of previous events. But since we expect a fast convergence and hope that not too many event values will be unknown, this number of previous events can be small, and the exponent of it would be reasonably small too.

By the way, I've come up with a name for the K(E) described in the previous part: the "absolute confidence". The confidence C(E) has the range from 0 showing that we're completely sure  that the event is false, though 0.5 showing that we have no idea, to 1 showing that the event is true. The absolute confidence

K(E) = abs(2*C(E)  - 1)

has the range from 0 showing that we have no idea to 1 showing that we're completely sure that the event is true or false.

I had to draw a picture to figure it out how it would work. Suppose that we want to take 3 previous events into the consideration. The datapoints with the probability values for these events can then be seen as nodes of a tree, with the nodes for each event aligned vertically under the event name:

E1         E2       E3       E4        E5
                      +------o
                    0/
              +-----o
            0/      1\
            /         +------o
      +----o
     /      \         +------o
   0/       1\      0/
   /          +-----o
  /                 1\
 /                    +------o 
o                              
 \                            0+-------o
  \                   +------o<
   \                0/        1+-------o
   1\          +----o
     \        /     1\        0+-------o
      \     0/        +------o<
       \    /                 1+-------o
        +--o                  
            \                 0+-------o
            1\        +------o<
              \     0/        1+-------o
               +----o
                    1\        0+-------o
                      +------o<
                              1+-------o

E1 still has only one probability value because it has no preceding events. But E2 has two possible probabilities, depending on whether E1 was known (K(E1)=0) or unknown (K(E1)=1). E3 has four probabilities, depending on the possible combinations of E1 and E2. And E4 has eight probabilities.

When we come to E5, it can't just continue branching exponentially.It should depend on only three previous events, so we drop the dependency on E1. This is done by abandoning the subtree with K(E1)=0 and continuing branching the subtree with K(E1)=1. This is a reasonable thing to do because we assume that the combination {0, 0, 0} "should never happen" (i.e no more than 3 events in a row would be unknown, otherwise use a higher value of depth), and the other mirroring pairs of branches of {0, b, c} and {1, b, c} would converge to about the same probability value. Thus the events starting from E5 would also have 8 probability values each.

Then when we do the computation on the built table, we can have any K(E) in range [0, 1], so we do an interpolation between the branches. Suppose we're computing E4, and have 8 branches {E1, E2, E3} to interpolate. We do it by adding them up with the following weights:

W{0, 0, 0} = (1-K(E1))*(1-K(E2))*(1-K(E3))
W{0, 0, 1} = (1-K(E1))*(1-K(E2))*K(E3)
W{0, 1, 0} = (1-K(E1))*K(E2)*(1-K(E3))
W{0, 1, 1} = (1-K(E1))*K(E2)*K(E3)
W{1, 0, 0} = K(E1)*(1-K(E2))*(1-K(E3))
W{1, 0, 1} = K(E1)*(1-K(E2))*K(E3)
W{1, 1, 0} = K(E1)*K(E2)*(1-K(E3))
W{1, 1, 1} = K(E1)*K(E2)*K(E3)

Or the other way to look at it is that we start with the weight of 1 for every branch, then for E1 multiply the weight of every branch where K(E1)=0 by the actual 1-K(E1), and where K(E1)=1 by the actual K(E1). Then similarly for E2, multiply the weight of every branch where K(E2)=0 by the actual 1-K(E2), and where K(E2)=1 by the actual K(E2). And so on. The sum of all the weights will always be 1.

The physical meaning of this interpolation is that we split the computation into multiple branches and finally add them up together. Since the computation is additive, we can put them back together right away rather than waiting for the end.

And since I was implementing the partial confidences in the final computation, I've also implemented the support for the partial confidences in training: it's done by logically splitting the training case into two weighted sub-cases with the event values of 0 and 1.

All this is implemented in the script ex26_01run.pl, as a further modification of ex24_01run.pl. Without the extra parameters it works just as ex24_01 does. The new parameter -p sets the number of previous events whose absolute confidence affect the current event. Here is an example of a run with the same training table as before and the same input:

$ perl ex26_01run.pl -n tab24_01.txt in24_01_01.txt 
Original training cases:
!      ,       ,E1          ,E2          ,E3          ,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E1             ,E2             ,E3             ,
hyp1   ,3.00000,0.66667^0.66667,0.50000^0.50000,0.40000^0.33333,
hyp2   ,3.00000,0.33333^0.33333,0.50000^0.50000,0.60000^0.66667,
--- Initial weights
hyp1    w=3.00000 p=0.50000in a row 
hyp2    w=3.00000 p=0.50000
--- Applying event E1, c=1.000000
hyp1    w=2.00000 p=0.66667
hyp2    w=1.00000 p=0.33333
--- Applying event E2, c=1.000000
hyp1    w=1.00000 p=0.66667
hyp2    w=0.50000 p=0.33333
--- Applying event E3, c=1.000000
hyp1    w=0.40000 p=0.57143
hyp2    w=0.30000 p=0.42857

The option -n disables the normalization of the weight of the hypotheses, just as before. The output is the same as ex24_01 produced. In the training data E1 and E2 are duplicates, so the probabilities computed for E2 in the adaboostian table (P(H|E) and P(~H|~E), they are separated by "^") are at 0.5, meaning that it has no effect.

And here is the same training data but with support for one previous unknown event, and the input data having E1 unknown (i.e. C(E1) = 0.5):

$ perl ex26_01run.pl -n -p 1 tab24_01.txt in26_01_01.txt 

Original training cases:
!      ,       ,E1          ,E2          ,E3          ,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E1             ,E2             ,E3             ,
hyp1   ,3.00000,0.66667^0.66667,0.66667^0.66667,0.40000^0.33333,
+      ,       ,               ,0.50000^0.50000,0.40000^0.33333,
hyp2   ,3.00000,0.33333^0.33333,0.33333^0.33333,0.60000^0.66667,
+      ,       ,               ,0.50000^0.50000,0.60000^0.66667,
--- Initial weights
hyp1    w=3.00000 p=0.50000
hyp2    w=3.00000 p=0.50000
--- Applying event E1, c=0.500000
hyp1    w=1.50000 p=0.50000
hyp2    w=1.50000 p=0.50000
--- Applying event E2, c=1.000000
hyp1    w=1.00000 p=0.66667
hyp2    w=0.50000 p=0.33333
--- Applying event E3, c=1.000000
hyp1    w=0.40000 p=0.57143
hyp2    w=0.30000 p=0.42857

Here E2 and E3 have not one pair of probabilities in the adaboostian table but two pairs, as shown in the branching picture above. The extra branches are shown on the continuation lines that start with a "+". The branches for E2 show that if E1 is known (the second branch), E2 will have no effect as in the previous example, but if E1 is unknown (the first branch), E2's probabilities would be the same as for E1, so E2 would replace it and converge to the same values.

As you can see in the run log, after E1 with confidence 0.5 (i.e. absolute confidence 0) is applied, both hypotheses still have the equal weight. But then after E2 gets applied, the weights and probabilities of hypotheses become the same as in the previous example, thus the knowledge of E2 compensating for the absent knowledge of E1.

An interesting thing to notice is that the probabilities for E3  are the same in either case. This is because this table can handle only one unknown event in a row, which means that at least one of E1 or E2 should be known, and E2 is able to fully compensate for the unknown E1, so both ways converge to the same probabilities of E3.

Now some information about how it all is implemented:

Some of the arrays have gained an extra dimensions, to be able to keep track of the multiple branches.The weight of the training case {wt} has become an array instead of a scalar, and the probabilities {ppos} and {pneg} of the hypotheses became 2-dimensional arrays.

The function compute_abtable() that computes the adaboostean table had gained an extra nested loop that goes through all the branches. As the final step of processing every event it prunes the branches coming from the previous event and then splits each of these branches in two. The support of the partial confidence in the training data is done by replacing the statement

if ($c->{tc}[$i] >= 0.5) {
    $c->{wt} /= $abhyp{$hyp}->{ppos}[$i];
} else {
    $c->{wt} /= (1. - $abhyp{$hyp}->{pneg}[$i]);
} 

with

push @{$c->{wt}}, 
    $c->{tc}[$i] * $oldwt->[$br] / $abhyp{$hyp}->{ppos}[$i][$br]
    + (1. - $c->{tc}[$i]) * $oldwt->[$br] / (1. - $abhyp{$hyp}->{pneg}[$i][$br]);

The function apply_abevent() that does the computation based on the table has been updated to manage the branches during this step. It had gained the extra argument of an array that keeps the history of the absolute confidence of the few previous events. This history is used to compute the weights of the branches:

for (my $br = 0; $br <= $maxbr; $br++) {
    my $w = 1.; 
    for (my $bit = 0; $bit <= $#$evhist; $bit++) {
        if ($br & (1 << $bit)) {
            $w *= $evhist->[$bit];
        } else {
            $w *= 1. - $evhist->[$bit];
        }   
    }   
    push @brwt, $w; 
}   

Then an update based on the current event is computed for each branch, and the weight of the hypothesis is updated per these weighted bracnhes:

my $p = 0.; 

# each branch is weighed per the observed history
for (my $br = 0; $br <= $maxbr; $br++) {
    $p += $brwt[$br] * ( 
        $conf * $hyp->{ppos}[$evi][$br]
        + (1. - $conf) * (1. - $hyp->{pneg}[$evi][$br])
    );  
}   

$hyp->{cwt} *= $p; 

And of course the functions that print the tables and normalize the data have been updated too.

All this stuff works on the small examples. I'm not sure yet if it would work well in the real world, I'd need a bigger example for that, and I don't have one yet. Also, I've noticed one potential issue even on the small examples but more about it in the next installment.

Neuron in the Bayesian terms, part 3

I want to elaborate on why the second trick is so important. Basically, because in general the necessary condition of the chances changing on an event in the opposite way (by dividing or multiplying by the same number) is not true. It takes a specifically defined event to make it true.

In general if an event is true, the chance of a hypothesis after applying this event is:

Chance(H|E) = P(H/E) / P(~H|E)

The probabilities change on an event as follows:

P(H|E) = P(H) * P(E|H) / P(E)
P(~H|E) = P(~H) * P(E|~H) / P(E)

So then the chance changes (after substituting and reducing the formula):

Chance(H|E) = Chance(H) *   P(E|H) / P(E|~H)

If an event is false, the chance changes in a similar but different way:

Chance(H|~E) = P(H/~E) / P(~H|~E)
= Chance(H) *   P(~E|H) / P(~E|~H) 

Being able to do the "opposites" requires that

Chance(H|E) / Chance(H) = 1/ ( Chance(H|~E) / Chance(H) )

Chance(H|E) / Chance(H) = Chance(H) / Chance(H|~E)


Which then means after doing the substitution:

P(E|H) / P(E|~H) = P(~E|~H) /  P(~E|H)

Which is not always true. But if we take a page from the AdaBoost's book and define an event such that 

P(E|H) = P(~E|~H)
and
P(E|~H) =  P(~E|H)

Then it all magically works. Again, this magic is not needed for the Bayesian computation in general, it's needed only to fit it to the formula used in the neural networks and in AdaBoost.

There are versions of AdaBoost that use the different multipliers for when x[i] < 0 and x[i] > 0. Which means that

l[i]^x[i]

gets replaced with

if (x[i] > 0) {
  lpos[i]^x[i]
} else {
  lneg[i]^x[i]
}

AdaBoost in this situation still uses the same definition of E=(x == y) for the reasons of its internal logic,  but if we don't require this logic then we can define E=(x > 0) and easily implement the chance computation for it, since the requirement of both multipliers being the opposites would disappear.

This opens two possibilities for variations of the neurons:

(1) One with still the definition E=(x==y) but different multipliers for x[i] of different sign.
(2) One with the definition E=(x > 0)

A network with such neurons would obviously take twice the memory space for its trained multipliers (but not for the data being computed). I'm not even sure if it would be trainable at all, especially the version (2), since AdaBoost has good reasons to keep its definition of E for the convergence. But maybe it would, and considering that AdaBoost in this modification converges faster, maybe the training of the neural network would converge faster too. Might be worth a try.

emergents

I've attended a talk about the commercial use of the pre-trained neural networks. The idea is that you take a pre-trained network, throw away the top couple of layers, then put the blank layers on top instead and train only these new layers for your purpose.

The presenter gave an example of taking the Google's general image classifier network and teaching it to recognize your specific images instead: "huggable" vs "non-huggable". It was smart enough to recognize that a knitted cactus is huggable while a real one is not.

Well, it's kind of not surprising: the top layers of the classifier network are trained to recognize the various kinds of images but to do that the layer below it must produce the characteristics that allow to recognize these various kinds of images. If the original classifier has the wide knowledge (and it does), the previous layer would know what is important to recognize a very wide rage of images. They call the output of that previous layer "the emergents". And then when you use them as inputs, the new top layer would have an easy job classifying them for the new goal. You could probably even do a Bayesian classifier on top of them without much difference.

Neuron in the Bayesian terms, part 2

I've been trying to describe in proper formulas my insight that a neuron in neural networks is a kind of a Bayesian machine(OK, this is apparently known to the people who do the neural networks professionally but it was an insight for me), and I think that I've finally succeeded. Let's start from the beginning:

A neuron implements the function

y = nonlinear(sum(w[i] * x[i]))

Here y is the output of the neuron, x[i] is the array of its inputs, w[i] is the array of some weights determined by training. Let's define the range of y and x[i] as [-1, 1] (if the values are in the other ranges, they can be trivially scaled to the range [-1, 1]  by adjusting the weights w[i]).

The nonlinear function is generally some kind of a sigmoid function that matches what we do at the end of the Bayesian computations: if the probability of a hypothesis is above some level, we accept it as true, i.e. in other words pull it up to 1, if it's below some equal or lower level we reject it, i.e. pull it down to 0, and if these levels are not the same and the probability is in the middle then we leave it as some value in the middle, probably modified in some way from the original value. Except of course that this nonlinear function deals not with the probabilities in range [0, 1] but with the values in range [-1, 1], -1 meaning "false" and 1 meaning "true".

The Bayesian formula uses the multiplication, not addition, so the first trick here is that one can be converted into another using a logarithm. To do that let's introduce a new value l[i], of which w[i] will be the logarithm:


l[i] = exp(w[i])
w[i] = ln(l[i])


This can be substituted into the expression for the sum:

sum (w[i] * x[i]) = sum( ln(l[i]) * x[i] )

Using the rule ln(a)*b = ln(a^b), the sum can be converted further:

sum( ln(l[i]) * x[i] ) = sum( ln(l[i]^x[i]) )

Then using the rule ln(a) +ln(b) = ln(a * b) the sum can be converted into a product:

sum( ln(l[i]^x[i]) ) = ln( product( l[i]^x[i] ) )

Now let's see how this product can be connected to the Bayesian computation. The classic Bayesian formula for probabilities is:

 P(H|E) = P(H) * P(E|H) / P(E)

So as each Bayesian event E[i] gets applied, we can say that the final probability will be:

P(H) = P0(H) * product( P(E[i]|H) / P(E[i]) )

Well, technically I maybe should have used another letter instead of i for the index, maybe j, but since the i in both formulas will be soon connected to each other, I'll just use it in both places from the start.

Instead of computing with probabilities we will be computing with chances. In case if you haven't used the chances before nor seen my mentioning of them in the previous posts, a chance is a relation of the probability of the hypothesis being true to the probability of it being false. For a simple example, if we have 5 doors, with prizes behind 2 of them and no prizes behind the remaining 3, then the chance of getting a prize by opening a random door is 2 to 3, or 2/3 (but the probability is 2/5). In a formula this gets expressed as:

Chance(H) = P(H) / P(~H) = P(H) / (1 - P(H))

So if we have two mutually exclusive hypotheses H and ~H, the probabilities for them will be:

P(H) = P0(H) * product( P(E[i]|H) / P(E[i]) )
P(~H) = P0(~H) * product( P(E[i]|~H) / P(E[i]) )


And the chances will be:


Chance(H) = P(H) / P(~H) 
= ( P0(H) * product( P(E[i]|H) / P(E[i]) ) ) / ( P0(~H) * product( P(E[i]|~H) / P(E[i]) ) )
= (P0(H)/P0(~H)) * product( P(E[i]|H) / P(E[i]|~H) )


If the initial probabilities of both hypotheses are equal, P0(H) = P0(~H) = 0.5, then their relation will be 1 and can be thrown out of the formula:
 
Chance(H) = product( P(E[i]|H) / P(E[i]|~H) )

Well, almost. The consideration above glanced over the question of what do we do if the event is false? The answer is that in this case the factor in the product should use the probabilities of ~E instead of E: 

Chance(H) = product( 
  if (E is true) {
    P(E[i]|H) / P(E[i]|~H)
  } else {
    P(~E[i]|H) / P(~E[i]|~H)
  }
)

Now comes the turn of the second major trick: what kind of events to use for the Bayesian formula. We'll use the same kind of events as for AdaBoost, the event being "x[i] accurately predicts y", so if H is "y > 0" and thus ~H is "y < 0", the conditional probability will be:

P(E[i]|H) = P(y == x[i])
P(~E[i]|~H) = P(y == x[i])

An interesting property of this definition is that the conditional probabilities of these events get computed across the whole range of the training data, without differentiation between x[i] < 0 and x[i] > 0. This means that

P(E|H) = P(~E|~H)
P(E|~H) = P(~E|H)

We then use the general property that

P(E|H) = 1 - P(~E|H)
P(E|H~) = 1 - P(~E|~H)

and get

P(E|~H) = P(~E|H) = 1 - P(E|H)

So after substituting all these equivalencies the computation of the chances becomes nicely symmetrical:

Chance(H) = product( 
  if (E is true) {
    P(E[i]|H) / (1 - P(E[i]|H))
  } else {
    (1 - P(E[i]|H)) / P(E[i]|H)
  }
)

When x[i] is in range [-1, 1], and especially if x[i] is in the set {-1, 1}, the if/else can be replaced by the power of x[i]: when x[i]==1, it will leave the expression as-is, when x==-1, the expression will be "turned over":

Chance(H) = product(  ( P(E[i]|H) / (1 - P(E[i]|H)) )^x[i] )

We can now interpret the original formula in terms of the chances. That formula was:

y = nonlinear(sum(w[i] * x[i]))
= nonlinear( ln( product( l[i]^x[i] ) ) )

Here we can substitute the product:

y = nonlinear( ln( Chance(y > 0) ) )

A logarithm is a very convenient thing to apply on the chance to obtain a symmetric range of values (-infinity, +infinity) centered at 0 instead of [0, +infinity) "centered" at 1. And if the weights are properly selected, the actual range of y will be not (-infinity, +infinity) but [-1, 1].

This substitution of the product will mean:

l[i] = P(E[i]|H) / (1 - P(E[i]|H))
w[i] =  ln( P(E[i]|H) / (1 - P(E[i]|H)) )

So the original weights in the neuron have the "physical meaning" of the logarithms of the chance multipliers.

OS support for thread stopping

I've recently had done some thinking about how can an OS make the shutdown of a multithreaded application or its part easier, to stop the threads quickly but relatively nicely on request. I think a very helpful support can be pretty easy to implement:

• Have system call to request the shutdown of an OS-level thread.
• This call would make any system calls that could block non-blocking (except for the thread synchronization calls, like mutex or condition variable operations), would interrupt any current such blocking calls, and return an error. This includes the timed sleeps, and even the thread synchronization calls with timeouts would return an error immediately instead of sleeping. Optionally, it might also just fail any file operations altogether, possibly except for close(). Note that the file descriptors as such wouldn't be revoked altogether, they would just kind of stop working in one thread, but the other threads would still work.
• A problem with this might be in such things as logging: when your thread is stopping, it might still write the useful log files. Revoking these log files would be  Well, one solution for this would be to buffer the logs in memory and write from a separate thread. Another might be to keep some kind of exception list.
• The threads might also beget threads, and keeping track of all of them is complicated. So there is a need for grouping them, like the process groups in the OS, or Triceps Triead fragments, and then stopping the whole group.

Yet another separate approach might be to say "but if you want the ability to stop something, why don't you run it as a separate process and kill if when needed?". This would make a lot of sense for such things as PAM authentication modules.

There are problems with the separate processes too. But after some thinking, they seem to be fairly easy to solve.

When you think about processes, usually a separate binary comes to mind but it doesn't have to. You can just do a fork() and run from the same binary, and you even get the whole address space copy-on-write out of the box. Though I guess it might be interesting to NOT copy-on-write the whole address space. Instead just allocate some communication area for passing the arguments and results, and get rd of the rest, maybe even re-running the run-time loader from scratch in the new context (that would be kind of like the Java class loader spaces).

Communication with a separate process is kind of painful. But if you map an anonymous memory segment, that segment can be inherited by the child process. Indeed, should be inherited to act as the communication region (truly shared, not COW) even if the rest of address space is dropped. And then you can happily create the mutexes and condition variables in it for synchronization. The arguments and results would have to be serialized to pass through this region but it shouldn't be such a big deal. It would be also possible to keep some large data dictionaries in this communication space.

Of course the callbacks wouldn't work across the process boundary. But there is a solution for this too: create a local thread pool (or simply a thread) for callbacks that would get the requests from the remote process though a portion of the same communications region, execute the callback, and return the results back there.

I think this looks feasible even without any new technology needed, just needs a helper library to implement the interface. This isn't even a new concept, it's a classic microkernel architecture without the common memory space. Yeah, copying the data in the serialized form adds an overhead but it's not that much overhead compared to say JVM, and people use Java and .NET all over the place with no regrets.

Bayes 25 & AdaBoost: the partial confidence problem

The code in the previous example works only with the full-confidence events, those that have the confidence of 0 or 1. Unfortunately I haven't found a good way to handle the partial confidence yet. I've found a way but it has problems. Let me show you.

Previously I was treating the partial confidence as a superposition of 0 and 1. This approach cannot be properly used here because it doesn't cover the connections between the events. To give an example, in the previous post the code had correctly found that in tab24_01.txt the events E1and E2 are the same, so after E1 is applied, E2 becomes ignored (it's AdaBoost probabilities become 0.5). So now if we then do a computation and find that E1 is unknown (i.e. its confidence is 0.5) but E2 is known, E2 will still be ignored because its table entry says to always ignore it.

The correct way should handle this situation properly: if E1 is unknown, this should bring up the table values for E2 to compensate for it and treat E2 as E1 was formerly.

The obvious way to do it is to treat the partial confidence as a superposition of "the event is unknown" and "the event is known". So if the event is known, we process it as before. If the event is unknown, we would follow the unknown branch and then apply all the following events as if this events wasn't there. And if the event is partially known, we would split the weights of all the outcomes proportionally and handle one part with the known branch and another one with the unknown branch.

So the confidence of 0.7 would mean a superposition of 0.4 of unknown and 0.6 of the known value "1". The confidence 0.1 would mean a superposition of 0.8 of unknown and 0.2 of the known value "0". So if we denote C(E) as the event confidence, K(E) as the known weight in the superposition, and U(E) as the unknown weight in the superposition, we can write them as:

K(E) = abs(C(E) - 0.5) * 2
U(E) = 1 - K(E)

The downside of this obvious approach is that the table and computation becomes more complex. As we go through more events, two branches become four, then eight, then sixteen, and so on. The formerly quadratic complexity O(Noutcomes * Nevents) becomes exponential O(Noutcomes*Nevents*2^Nevents). I don't know yet if there is a better way.

Maybe it would be possible to build a table of "gradients" between the events: i.e. if the confidence of the event A changes, how are the probabilities of another event B affected. The problem though is that this gradient depends on all the events in the sequence between A and B, so tis idea doesn't seem to really fly.

Maybe another idea is workable, that would reduce the exponential cost to a cubic one: after splitting the superposition, instead of following two branches, just take the probabilities for each following event from both branches and mix them by the weight of the branches:

P(H|Ei) = K(Ej) * P(H|KjEi) + U(Ej) * P(H|UjEi)

It's better than exponential but still not great, with the run time O(Noutcomes*Nevents*Nevents). And the table of these pre-computed values would still seem to be exponential, each event depending on all the possible combinations of the previous events.

Bayes 24: AdaBoost as a solution to the correlated events

When writing the whitepaper, I've realized that AdaBoost is very good at finding the cross-correlation of its partial hypotheses (that map to the Bayesian events), and it always picks the next step with the minimal cross-correlation. The Bayesian computations do have an issue with cross-correlation, as described in the whitepaper and near the end of Part 8 of my notes: if there are multiple events that duplicate or mirror each other, all of them are treated as independent and drive the probabilities off. This looked like AdaBoost could give a hint of how to solve this problem, so I've started thinking about it, and I think I've found a solution.

In retrospect the solution is pretty obvious: instead of computing the Bayesian model directly, treat it as an AdaBoost model whose set of the partial hypotheses equals to the set of the Bayesian events, with an extra limitation that each one of them must be used exactly once. If there are multiple equivalent events and they happen to go one after another, AdaBoost will treat the first one as real, and the following ones will get their error rate of 0.5 and will be effectively no-ops.

Well, if there are other events interspersed in between these equivalent copies, they won't be no-ops. These other events will shift the weights of the training cases, so that the second an further copies will find themselves drifted away from 0.5. The good news here is that AdaBoost is pretty good at removing the partial correlation too but the bad news is that the order matters, to fully remove the duplication the equivalent events should go one after another. And in general, it looks like the best removal of the cross-correlation happens when the closely related events follow closely one after another. This kind of order can be implemented if we reverse the AdaBoost preference of the partial hypotheses and make it always pick the one that has its training error rate closest to 0.5. Since (unlike the classic AdaBoost) the set of partial hypotheses is limited and each element from it must be used only once, the better ones will eventually be used as well.

To experiment with these ideas, I wrote the code example ex24_01run.pl. It doesn't reorder the events, just computes the AdaBoost values for whatever order it's given, allowing to experiment with the various orders. But it has a couple of advanced features from the later chapters of the book on AdaBoost: it does the multi-class (but single-label) version that also partitions the error rate computation for  the sign of the value of the partial hypothesis.

The multi-class means that it can work not only with two outcomes (a yes/no decision) but with multiple possible outcomes (i.e. Bayesian hypotheses), just like the normal Bayesian table. And just like the normal Bayesian table, these outcomes must be mutually exclusive, so each case may have only one outcome (single-label). The multi-class logic works in the way that is similar to the one described in Part 10: the resulting table is just a combination of the individual per-outcome tables. From the standpoint of each outcome, the single-label property means that the rest of the rows of the table taken together represent the opposite of this outcome, so there is no need to compute that opposite explicitly.

The partitioning of the error rate computation not only makes the result better but also comes very handy to prove that the opposite of one outcome is equivalent to the rest of the rows of the table. The AdaBoost training error e is the ratio of the total weights of the training cases where the value of the event E (that is, the Bayesian event, in AdaBoost terms it's a partial hypothesis) is not equal to the value of the outcome (Bayesian hypothesis) H. And we'll compute it separately for E (i.e. E=1) and ~E (i.e. E=0). It ends up expressable as probabilities that can be computed by summing and dividing weights W of the training cases:

e(H,E) = P(~H|E) = (sum for cases i where H(i)=0 and E=1 (Wi))/(sum for cases j where E=1 (Wj))
e(H,~E) = P(H|~E) = (sum for cases i where H(i)=1 and E=0 (Wi))/(sum for cases j where E=0 (Wj))

So if we pick a particular outcome and denote it H1, we can show that its opposite ~H1 is equivalent to the sum of the rest of hypotheses H2...HN by showing that

P(~H1|E) = sum for i in [2, N] (P(Hi|E))

which after removing the common denominator means

(sum for cases j where H1(E)=0 and E=1 (Wj)) = (sum for i in [2, N] (sum for cases k where Hi(E)=1 and E=1 (Wk))

The single-label property means that each case is marked with exactly one outcome, and all the cases where the outcome H1 is absent must have some other outcome Hi present, so these sums are indeed equal.

So it is all self-consistent. The key part of the computation is:

for my $hyp (values %abhyp) {
        if ($conf) {
            $hyp->{wt} *= $hyp->{ppos}[$evi];
        } else {
            $hyp->{wt} *= (1. - $hyp->{pneg}[$evi]);
        }
    }

And the matching part of the table building is:

foreach my $c (@case) {
        my $hyp = $c->{hyp}[0];
        # this code really handles only the TC values of 0 and 1,
        # for the arbitrary values it would have to do an interpolation
        if ($c->{tc}[$i] >= 0.5) {
            $c->{wt} /= $abhyp{$hyp}->{ppos}[$i];
        } else {
            $c->{wt} /= (1. - $abhyp{$hyp}->{pneg}[$i]);
        }   
    }   

The training undoes the weight change that will be done in the computation.

Now let's try some examples. Here is the example that I used in the whitepaper:

$ perl ex24_01run.pl -n tab24_01.txt in24_01_01.txt
Original training cases:
!      ,       ,E1          ,E2          ,E3          ,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E1             ,E2             ,E3             ,
hyp1   ,3.00000,0.66667^0.66667,0.50000^0.50000,0.40000^0.33333,
hyp2   ,3.00000,0.33333^0.33333,0.50000^0.50000,0.60000^0.66667,
--- Applying event E1, c=1.000000
hyp1    w=2.00000 p=0.66667
hyp2    w=1.00000 p=0.33333
--- Applying event E2, c=1.000000
hyp1    w=1.00000 p=0.66667
hyp2    w=0.50000 p=0.33333
--- Applying event E3, c=1.000000
hyp1    w=0.40000 p=0.57143
hyp2    w=0.30000 p=0.42857

Here the option "-n" means "don't normalize the weights" (or when normalized to the sum of 1 they will be the same as probabilities). The two values printed in the AdaBoostian table separated by a "^" are the P(H|E) and P(~H|~E).

As you can see, the result is not correct either (because of the XOR problem) but it's better than with the plain Bayes version that gave hyp1 the probability 0.667. And since E1 and E2 are the same, E2 ended up with the values of 0.5 in the table, and applying it didn't change the weight.

If we change the order of events, E2 gains some use back:

$ perl ex24_01run.pl -n tab24_02.txt in24_01_01.txt 
Original training cases:
!      ,       ,E1          ,E3          ,E2          ,
hyp1   ,1.00000,1.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E1             ,E3             ,E2             ,
hyp1   ,3.00000,0.66667^0.66667,0.40000^0.33333,0.47368^0.48276,
hyp2   ,3.00000,0.33333^0.33333,0.60000^0.66667,0.52632^0.51724,
--- Applying event E1, c=1.000000
hyp1    w=2.00000 p=0.66667
hyp2    w=1.00000 p=0.33333
--- Applying event E3, c=1.000000
hyp1    w=0.80000 p=0.57143
hyp2    w=0.60000 p=0.42857
--- Applying event E2, c=1.000000
hyp1    w=0.37895 p=0.54545
hyp2    w=0.31579 p=0.45455

Rather surprisingly, the result ended up more correct. Or maybe it's not that surprising: after all, AdaBoost is designed to gain more information from available data.

Yet another interesting result get produced from repeating the same pair of events multiple times:

$ perl ex24_01run.pl -n tab24_03.txt in24_03_01.txt 
Original training cases:
!      ,       ,E01         ,E02         ,E11         ,E12         ,E21         ,E22         ,E31         ,E32         ,E41         ,E42         ,E51         ,E52         ,E61         ,E62         ,E71         ,E72         ,E81         ,E82         ,E91         ,E92         ,
hyp1   ,1.00000,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,
hyp1   ,1.00000,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,0.00000/1.00,1.00000/1.00,
hyp2   ,1.00000,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,0.00000/1.00,
AdaBoostian table:
!      ,       ,E01            ,E02            ,E11            ,E12            ,E21            ,E22            ,E31            ,E32            ,E41            ,E42            ,E51            ,E52            ,E61            ,E62            ,E71            ,E72            ,E81            ,E82            ,E91            ,E92            ,
hyp1   ,3.00000,0.66667^0.66667,0.40000^0.33333,0.47368^0.48276,0.49694^0.49526,0.49916^0.49946,0.49990^0.49985,0.49997^0.49998,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,
hyp2   ,3.00000,0.33333^0.33333,0.60000^0.66667,0.52632^0.51724,0.50306^0.50474,0.50084^0.50054,0.50010^0.50015,0.50003^0.50002,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,0.50000^0.50000,
--- Applying event E01, c=1.000000
hyp1    w=2.00000 p=0.66667
hyp2    w=1.00000 p=0.33333
--- Applying event E02, c=1.000000
hyp1    w=0.80000 p=0.57143
hyp2    w=0.60000 p=0.42857
--- Applying event E11, c=1.000000
hyp1    w=0.37895 p=0.54545
hyp2    w=0.31579 p=0.45455
--- Applying event E12, c=1.000000
hyp1    w=0.18831 p=0.54242
hyp2    w=0.15886 p=0.45758
--- Applying event E21, c=1.000000
hyp1    w=0.09400 p=0.54159
hyp2    w=0.07956 p=0.45841
--- Applying event E22, c=1.000000
hyp1    w=0.04699 p=0.54149
hyp2    w=0.03979 p=0.45851
--- Applying event E31, c=1.000000
hyp1    w=0.02349 p=0.54147
hyp2    w=0.01990 p=0.45853
--- Applying event E32, c=1.000000
hyp1    w=0.01175 p=0.54146
hyp2    w=0.00995 p=0.45854
...

--- Applying event E92, c=1.000000
hyp1    w=0.00000 p=0.54146
hyp2    w=0.00000 p=0.45854

AdaBoost converges pretty quickly to the weights that make both events get the probability 0.5 and stop the further changes.

This seems to work, although it needs more testing. One more limitation of this implementation is that it works only with the events that have the value of either 0 or 1, not anything in between. I'll talk more about this in the next post.