DiEbLog

Bernoulli vs. Clark Kent

2010-10-10T23:33:00.000-07:00

When discussing the consequences of their Horizontal No Free Lunch theorem, the authors W. Dembski and R. Marks state:

Because the active entropy is strictly negative, any uninformed assisted search will on average perform worse than the baseline search.

Here, the baseline search is the blind search. But what is an assisted search? As an example, the authors have introduced the perfect search

an assisted search guaranteed to succeed

Let's look at an example: a shell game - three shells, one pea. The operator is a fair man, who will hide the pea with probability 1/3 under any of the three shells.

The players are Bernoulli and Clark Kent. Knowing the paper of W. Dembski and R. Marks on the HNFLT, Bernoulli makes a guess at random, using a uniform distribution on the shells.

Clark Kent just picks the shell which hides the pea (remember - Clark Kent is Superman...), thereby performing a perfect search.

Who fares better? According to W. Dembski and R. Marks, both do equally well: observing the picking pattern of Clark Kent, they conclude that Clark Kent uses a uniform measure on the search space - as Bernoulli does: unfortunately, the mathematics of their paper has no way to deal with a strategy which is influenced by the position of the target (as any assisted search would be.)

Draft of a letter to JACIII

2010-09-24T08:00:00.000-07:00

In their article The Search for a Search¹ the authors William Dembski and Robert Marks seem to assume that any search on a space induces a probability measure on the search space, as they write in section 3.1 (Active Information):

"Let U denote a uniform distribution on Ω characteristic of an unassisted search and φ the (nonuniform) measure of Ω for an assisted search. Let U(T) and φ(T) denote the probability over the target set T ∈_[sic]Ω."

Such an induced measure gives the probability to find a subset T of Ω when performing the corresponding search. But for the assisted searches mentioned in the text (a "perfect search" - "indicating an assisted search guaranteed to succeed" - and another search "indicating an assisted search guaranteed to fail"), such probability measures obviously do not exist! Thus these cases are not covered by their Theorem 1 (Horizontal No Free Lunch), as the Kullback-Leibler-distance is defined only for probability measures.

But perhaps the authors intended their theorem only to be valid for those searches which induce a measure on the search space - indeed, they use the term uninformed assisted searches when describing their results. Unfortunately, this term is not defined - neither in the current text, nor in their previous publications Conservation of Information in Search - Measuring the Cost of Success² or Efficient Per Query Information Extraction from a Hamming Oracle³.

Unfortunately, even in this case, the theorem does not work: assuming the easiest nontrivial case of guessing an element of a set with two elements (tossing a coin), Ω = {head, tail}, the two strategies naming an element at random (uniform distribution on

\Omega\,

) and sticking with one element (dirac distribution) give the same success rate for a fair coin. The same result occurs when averaging over the probability of showing head (assuming that this probability is uniformly distributed on [0,1]), contradicting the statement of William Dembski and Robert Marks:

"Because the active entropy is strictly negative, any uninformed assisted search (φ) will on average perform worse than the baseline search". (Highlighting mine)

In my opinion, the fundamental problem stems from identifying the search space Ω_(n) and the space of (deterministic) searches. This is doubtless elegant, but does not seem to work.

References:

¹W. A. Dembski and R. J. Marks II, "The Search for a Search: Measuring the Information Cost of Higher Level Search," Journal of Advanced Computational Intelligence and Intelligent Informatics, Vol. 14, No. 5, pp. 475-486, September 2010

²W. A. Dembski and R. J. Marks II, "Conservation of Information in Search: Measuring the Cost of Success," IEEE Trans. on Systems, Man and Cybernetics A, Systems and Humans, Vol. 39, No. 5, pp. 1051-1061, September 2009

³W. Ewert, G. Montanez, W. A. Dembski, and R. J. Marks II, "Efficient Per Query Information Extraction from a Hamming Oracle," Proc. of the 42nd Meeting of the Southeastern Symposium on System Theory, IEEE, University of Texas at Tyler, pp. 290-297, March 7-9, 2010

What is wrong with public debate?

2010-09-01T12:16:00.000-07:00

R. Marks just informed me of his policy not to engage in correspondence with anyone publicly critical of him or his work and that independent of the validity or invalidity of the details of the exchange, these things are best discussed thoroughly before any public pronouncements.

Indeed, I wrote to R. Marks about my concerns regarding the Horizontal No Free Lunch theorem. And I sent the links to my blog entries. And these blog entries are critical of his work. But what is the problem?

my blogging could be totally unsubstantiated. In this case he really should ignore it - for any valid theory, you'll find thousands of idiots blogging about imagined problems.
my entries could be well-thought through, but ridiculously wrong. In this case, I made a fool out of myself in public- and he could criticize me for doing so.
there maybe something true in my blog entries: if these grains of truth aren't buried in filthy language, insults, etc., why should they be ignored?

I just don't get it.

Horizontal No Free Lunch Theorem

2010-08-31T08:55:00.001-07:00

In my previous post I criticized the Horizontal No Free Lunch theorem for searches consisting of at least two queries. But even in the case of a single assisted query, I can't get my head around it.

Dembski and Marks define an assisted query as any choice from Ω₁ that provides more information about the search environment or candidate solutions than a blind search. As an example, they give for instance, we might imagine an Easter egg hunt where one is told “warmer” if one is moving away from an egg and “colder” if one is moving toward it.

The ultimate assisted search is no search at all, the unveiled Easter egg. So, imagine a blind (A), a drunken (B), and a seeing, sober (C) chess-player, and ask them to locate the White Dame (♕) on a chess board. (A) will be find it with a probability of 1/64, (C) with a probability of 1, and lets assume for (B) a probability of 1/2.

Then (A) represents the blind search, (B) an assisted search and (C) what the authors call a perfect search on our search space Ω = {a1,a2,....,h8} (|Ω|=64).

Imagine that ♕ is positioned on a1, so T₁ = {a1}. Then we have three probability measures, α_a1, β_a1, and γ_a1 for each one of our players, which allow us to calculate the probability for each one to find T₁:

α_a1(T₁)= 1/64

β_a1(T₁) = 1/2

γ_a1(T₁) = 1

And we can calculate - according to (3.1) (10) the active information of these measures for various sets and combinations, e.g.
I₊(β_a1|α_a1(T₁)){a1} = log₂1/2 - log₂1/64 = 5.

So, the drunken player has more active information on the whereabouts of ♕ than the blind one.

I₊(γ_a1|α_a1(T₁)){b1} = log₂0 - log₂1/64 = -∞

A more surprising result: if ♕ is on b1, but (C) sees it on a1, he is much less informed than (A).

But of course, γ_a1 isn't the right measure if ♕ is on b1, we would expect γ_b1: the idea of an assisted search seems to be that it assists to the right direction - for our Easter egg hunt: if your helper directs you always to the same place, whether there is an egg or not, he is no helper indeed, he's just a nuisance:

Trivially, a search helped in this way can't be differed from an unassisted search.

But I'm afraid that exactly what W. Dembski and R. Marks do in their Horizontal No Free Lunch Theorem: let's take the partition T~ = {{a1},{a2},...,{h8}} of our Ω₁ above. What does

Σ α_a1{T_i} × I₊(γ_a1|α_a1){T_i}

mean? Nothing else than that our seeing, sober man errs 63 times out of 64 when looking for ♕. That just doesn't make sense.

The meaningful expression would have been:

Σ α_{T_i}{T_i} × I₊(γ_{T_i}|α_{T_i}){T_i}

And while the first sum adds up to -∞ the latter one is not negative at all: it's just 6.

nota bene: I think this is the idea behind Rob's (R0b?) comment to my previous post. So, thanks Rob!

Search for Search Paper Finally Out

2010-08-29T22:44:00.000-07:00

William A. Dembski informs us at Uncommon Descent that the paper A Search for a Search:Measuring the Information Cost of Higher Level Search (a collaboration with Robert Marks II) is finally published. To things are remarkable about his post:

The comments are turned off
It's so restrained - wasn't this paper once announced as a nail to the coffin of Darwinism?

One obvious flaw I found in an earlier draft of the paper is addressed: instead of talking about any searches, they are now talking about searches without repetition.

But I've still a(t least one) problem with this paper, which may be resolved by more careful reading over the next few days. As I wrote to Robert Marks:

I was a little bit irritated that the proof of the HNFLT still uses the
Kullback-Leibler distance, as I can't see how a non-trivial search-space
(i.e., a search spaces for a search existing from at least two queries)
can be exhaustively partitioned in a meaningful way.

Here's is an example I used earlier: Imagine a shell game with three shells where you are allowed to have two guesses. To put it more formally:

The search space Ω₁ is {1,2,3}

In section 2.1. Blind and Assisted Queries and Searches, Dembski and Marks describe how such searches can be construed as a single query when the search space is appropriately defined. They introduce an augmented search space Ω_Q, existing from the sequences without repetition of length Q, i.e., the number of queries. In our case:

The augmented search space Ω₂ is {(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)}

Of course, the target has to be changed accordingly to T_Q, where T_Q ∈ [sic] Ω_Q consists of all the elements containing the original target.

So, if in the original game the shell with the pea was No. 1, in our new space, T₂ = {(1,2),(1,3),(2,1),(3,1)}. All of this seems to be sensible

Let's have a look at a search strategy, for instance:

First, look under a random shell
Second, look under another random shell

By such a strategy, a probability measure is introduced on Ω₂, and the probability of a successful search for T₂ can be seen immediately: it's |T₂|/|Ω₂| = 4/6 = 2/3. No surprise here.

In fact, any search strategy can be seen as a measure on Ω_Q - and that's what Dembski and Marks are doing in the following. My problem: not any subset of Ω_Q can be seen as a reasonable target for a search: The complement of T₂ doesn't represent any target in the original space Ω₁! And though this set can be measured by the measure introduced above (2/6), this measure doesn't make sense in the context of a search.

And that's what I don't understand about the Horizontal No Free Lunch Theorem (3.2). The first sentence is:

Let φ and ψ be arbitrary probability measures and let T~={T_i}_i=1^N be an exaustive partition of Ω all of whose partition elements have positive probability with respect to ψ

How can Ω₂ partitioned in a sensible way? And why should I try to compare two search strategies on sets which they will never look for?

A silent death of the Horizontal no Free Lunch theorem

2009-10-23T13:13:00.000-07:00

The only reply I got for my concerns regarding W. Dembski's and R. Marks's article The Search for a Search - and its Horizontal No Free Lunch theorem - was an email from R. Marks, stating that he'd look into it.
Now, two weeks later, R. Marks took the article from his list of publications. The Evolutionary Informatics lab has still an announcement, basically the abstract of the article:

Many searches are needle-in-the-haystack problems, looking for small targets in large spaces. In such cases, blind search can stand no hope of success. Success, instead, requires an assisted search. But whence the assistance required for a search to be successful? To pose the question this way suggests that successful searches do not emerge spontaneously but need themselves to be discovered via a search. The question then naturally arises whether such a higher-level “search for a search” is any easier than the original search. We prove two results: (1) The Horizontal No Free Lunch Theorem, which shows that average relative performance of searches never exceeds unassisted or blind searches. (2) The Vertical No Free Lunch Theorem, which shows that the difficulty of searching for a successful search increases exponentially compared to the difficulty of the original search.

The given link doesn't lead to the article any longer, but to their earlier opus Conservation of Information in Search - neither the term vertical nor the term horizontal appear in this article.

In his first podcast with Casey Luskin, W. Dembski promised that this article was something like a nail to the coffin of evolution. I assume that W. Dembski will explain how he was wrong with this assessment in his next interview...

One if by land, and two if by sea...

2009-10-23T00:58:00.000-07:00

Or, in the Intelligent Design version: Two lanterns contain the same information as one lantern, ergo, they come by land...

Horizontal No Free Lunch Theorem

2009-10-22T03:07:00.000-07:00

You keep using that word. I do not think it means what you think it means

Inigo Montoya

In their upcoming paper The Search for a Search: Measuring the Information Cost of Higher Level Search W. Dembski and R.Marks introduce a Horizontal No Free Lunch Theorem.

At Uncommon Descent, I tried to reach W. Dembski with the following post ~~which is - of course - awaiting moderation at the moment~~.

The Horizontal No free Lunch Theorem doesn't work for a search of length m > 1 for a target T ⊆ Ω:

Let Ω be a finite search-space, T ⊆ Ω the target - a non-empty subset of Ω. A search is a (finite) sequence of Ω-valued random variables (φ₁, φ₂, ..., , φ_m). A search is successful, if φ_n ∈ T for one n, 1 ≤ n ≤ m.

I suppose we do agree here. Now, we look at a search Φ as a Ω^m-valued random variable, i.e., Φ := (φ₁, φ₂, ..., , φ_m).

When is it successful? If we are still looking for a T ⊆ Ω we can say that we found T during our search if

Φ ∈ Ω^m \ (Ω \ T)^m

Let's define Θ as the subspace of Ω^m which exists from the representations of targets in Ω, i.e.,

Θ := {Ω^m \ (Ω \ T)^m|T non-empty subset of Ω}

Obviously, Θ is much smaller than Ω^m.

But this Θ is the space of feasible targets. And if you take an exhaustive partition of Θ instead of Ω^m in Theorem III.1 Horizontal No Free Lunch, you'll find that you can indeed have positive values for the active entropy as defined in the same theorem.

But that's not much of a surprise, as random sampling without repetition works better than random sampling with repetition.

But if you allow T to be any subset of Ω^m, your results get somewhat trivial, as you are now looking at m independent searches of length 1 for different targets.

The searches which you state as examples in this paper and the previous one all work with a fixed target, i.e., elements of Θ. You never mention the possibility that the target changes between the steps of the search (one possible interpretation of taking arbitrary subsets of Ω^m into account).

So, I'm faced with two possibilities:

You didn't realize the switch from stationary targets to moving ones when you introduced searching for an arbitrary subset of Ω^m
You realized this switch to a very different concept, but chose not to stress the point.

W. Dembski's and R. Marks's The Search for a Search and my major problem with it

2009-10-20T21:48:00.000-07:00

William Dembski and Robert Marks announced a new paper The Search for a Search: Measuring the Information Cost of Higher Level Search (draft) which should appear in a peer-reviewed journal at any moment.

My problem: When I read the term search for a target without any qualifier, I imagine that

the target is fix
the search is over when the target is found the first time

A model for this: a game of Battleship against a random generator with this additional rules:

all ships are of size one
you don't know how many ships there are
the game is over with the first hit

So, if your search space is Ω then there are 2^|&Omega| - 1 different targets.

But with this kind of target, the math of W. Dembski's and R. Marks's paper doesn't work: They claim that you can't find a strategy for that works better on average than random search. In the Battleship example: there is no better strategy than shooting at random each turn until your first hit. But this is obviously wrong: A better strategy is not to hit a square twice.

But W. Dembski and R. Marks proved their claim - how so? Well, they redefined the concept of target. The altered rules for the Battleship:

the position and the number of the ships changes each round
to succeed, you have to hit a ship each round

For this version - and not for the one above - they proved all their results. The problem: That's not search as we now it...

The technical details: In section III, B, they formalize a search

Suppose now Ω denotes an arbitrary search space. In general, assume Ω is a compact metric space with metric D, though for most theorems and calculations assume Ω is finite (in which case D is a discrete metric). Let X₁, X₂,...,X_n denote Ω-valued random variables that are independent and identically distributed with uniform probability distribution U. X₁, X₂,...,X_m therefore represents a blind search of Ω. Next, let Y₁, Y₂,...,Y_m denote Ω-valued random variables that are arbitrarily distributed. X₁, X₂,...,X_n therefore characterizes an assisted search of Ω.

That's our first version of Battleship - I've no problem with this section. But the next paragraph states

If we now let Ω^m denote the m-fold Cartesian product of Ω, then X₁, X₂,...,X_m and Y₁, Y₂,...,Y_m are m-valued random variables. Because the uniform probability on Ω^m is just the m-fold product measure of U, X₁, X₂,...,X_m) is uniformly distributed on
= Ω^m. Thus, by attending solely to the probabilistic
structure of searches, we represent the blind search := (X₁, X₂,...,X_m) as having a m-dimensional uniform distribution with respect to a metric D~ that suitably extends to the metric D on Ω. Moreover, we represent the assisted search := (Y₁, Y₂,...,Y_m ), in searching for an arbitrary subset of , as having distribution on .
But all these tildes are superﬂuous. Dropping them, we see that the probabilistic connection between a blind search and an assisted search can be entirely characterized in terms of a search space , a metric D that induces a compact topology on Ω, a target T included
in Ω, a uniform probability measure U on induced by D, and an arbitrary probability measure φ on Ω. U represents the probabilistic structure of a blind search, and φ represents the probabilistic structure of an assisted search.

Now, is a subset of Ω^m - and there are 2^{|Ω| ^m} -1 non-empty subsets! A successful search means that ∈ , i.e., Y_n ∈ T_n for all 1 ≤ n ≤ m. So, a stationary target would be not represented by

Tⁿ

as one may naively expect, but by

T×Ω×...×Ω
∪ Ω×T×Ω×...×Ω
∪ Ω×Ω×T×Ω×...×Ω
∪...
∪ &Omega×Ω×...×&Omega×T

but most of the targets don't fit this pattern.

Of course you can claim that knowing that you target doesn't move is additional information. But I haven't encountered a definition of search which expects you to be right on target each time...

P.S.: I tried to reach R. Marks via email - and I left an entry at W. Dembski's blog...

P.P.S.: some details.

I wrote a letter to W. Dembski and R. Marks

2009-10-17T04:43:00.000-07:00

After reading their article Conservation of Information in Search:
Measuring the Cost of Success in some detail, I decided to write to W. Dembski and R. Marks, just to hint them to some minor miscalculations: So, they have the chance to get at least the math straight in their peer-reviewed paper.

At uncommondescent.com, I tried to draw attendance of Willima Dembski to two minor errors in your IEEE paper ''Conservation of Information in Search – Measuring the Cost of Success'':

Could you correct the sign errors in equation (27) and on page 1057, left column, for Q?

More problematic is the next line: you speak about the active information per query, but you calculate the active information per generation, as you drop the factor -2 from Q ~ -2log(L)/log(1-2µ) and so reach

I+ ~ L/log(L) * log(1-2µ) instead of
I+ ~ L/(2*log(L)) * log(1-2µ).

So, I+ should be µ*L/ln(L) and not 2*µ*L/ln(L).

(I stumbled upon this when I calculated my approximation I+ ~ µ*L/H_((1-b)L) for the average number of queries Q ~ H_((1-b)L)/µ )

And there is a little problem with Fig. 2: all the paths seem to start with no correct bits at all - while the math in ch. III-2 assumes that half of the bits are correct.

Let L=100, µ=0.00005:

1. if you use b(eta) = .5 (as in your calculation) Q ~ 92,100 and I+ ~ 0.00109. Then, you should take other paths for your pic

2. if you use b(eta) = 0 (as in you figure) Q ~ 106,000 and I+ ~ 0.000943

The number in your text is I+ ~0.0022. That's wrong in any case.

Yours
Di***** Eb**

Don't get me wrong: I don't want to be just nit-picking. One sign error is no big deal, an even number of sign errors cancels itself out :-)
And the factor two? It's at least the same order of magnitude. (And no one know what magnitude is to be expected for active information, so even a factor 100 would not have been easily spotted). And I didn't mention all the inaccuracies in their references' section....

Code I'm not proud of...

2009-10-12T14:06:00.000-07:00

I was asked for the code I used to make my little pics. Well, it's not pretty, it's just a few functions in R. For example:


weasel.matrix.solo <- function(target.length=5, alphabet.length=27,mut.prob=.04,mut.effect=FALSE){
if(mut.effect) mut.prob <- alphabet.length*mut.prob/(alphabet.length-1)
M <- matrix(integer((target.length+1)*(target.length+1)),nrow=target.length+1)
for(k in 1:(target.length)){
 verschlechtert <- integer(2*target.length+2)
 verbessert <- verschlechtert
 verschlechtert[1:k + target.length +1] <- dbinom(0:(k-1),k-1,mut.prob*(alphabet.length-1)/alphabet.length)
 verbessert[1:(target.length+2-k) + target.length + 1] <- dbinom(0:(target.length+1-k),(target.length+1-k),mut.prob/alphabet.length)
 for(kk in 1:(target.length+1)){
  M[k,kk] <-sum( verbessert[(0:target.length) + target.length+3-k]*verschlechtert[0:target.length +target.length+3-kk])
 }
}
M[target.length+1,target.length+1] <- 1
return(M)
}

This creates a transition matrix for Dawkins's algorithm with a single child. target.length is the length of the target phrase, alphabet.length is self-explaining, mut.prob is the probability with which a letter is mutated: if mut.effect==FALSE, such a mutation has no effect with probability 1/N, i.e., it's a change to the same letter :-)


weasel.matrix.cum <- function(M, pop.size = 10){
 target.length <- dim(M)[1]-1
 M1 <- M
 for(k in 1:(target.length)){
  M1[k,] <- cumsum(M[k,])
 }
 M1 <- M1^pop.size
 for(k in 1:(target.length))
 M1[k,] <- M1[k,] - c(0,M1[k,-target.length-1]) 
 return(M1)
}

creates a transition matrix for a whole population of size pop.size, given a transition matrix M for a single-child process.

The last thing which is needed is this little snippet:

erw.exakt <- function(M,start.vektor){
 
 M.size <- dim(M)[1]
 M <- M[-M.size, -M.size]
 M <- solve(diag(M.size-1)-M)
 M <- apply(M,1,sum)
 return(sum(start.vektor[1:(M.size-1)] * M))
}

Here, the expected value of generations for a transition matrix M is calculated, using a initial distribution for start.vektor (mostly Bin(target.length,mut.prob) distributed - or just the first unit vector).

If I ever prettify this code, I'll take Tom English on his generous offer to put it somewhere on the net.

Another approach

2009-10-12T11:37:00.000-07:00

I completed this little project here.

I thought about different ways to present my ideas on W. Dembski's and R. Marks's article Conservation of Information in Search: Measuring the Cost of Success, and tried a few out in the last posts. The section which I'm interested in is the description of three evolutionary algorithms they give on pp. 1055-1057. Here's another approach for III E - G:

Original Text	Annotiations
E. Partitioned Search
Partitioned search [12] is a “divide and conquer” procedure best introduced by example.	The name partitioned search seems to be an invention of R. Marks and W. Dembski. The reference is made to Dawkins's book The Blind Watchmaker in which the phrase can't be found. (See Tom English's blog)
Consider the L = 28 character phrase `METHINKS∗IT∗IS∗LIKE∗A∗WEASEL`. (19)	This is indeed a phrase which is used in Dawkins's book - in an algorithm with which Dawkins explained the idea of cumulative selection.
Suppose that the result of our ﬁrst query of L =28 characters is `SCITAMROFN∗IYRANOITULOVE∗SAM`. (20)	An example of the egregious humorous skills of Dembski and Marks: backwards, we get: `masevolutionaryinformatics` That's no problem, as the first phrase of the algorithm can be arbitrarily chosen
Two of the letters {E, S} are in the correct position. They are shown in a bold font. In partitioned search, our search for these letters is ﬁnished.	At least, that's the way Dembski's and Marks's search works
For the incorrect letters, we select 26 new letters and obtain `OOT∗DENGISEDESEHT∗ERA∗NETSIL`. (21)	`listenarethesedesignedtoo` hilarious. And an sign that we don't see the output of an actual program, but something imagined to be a run of their algorithm. BTW, the fitness function would have to encode the position* of the correct letters, and the possible outcomes of the function wouldn't be a totally ordered set, but only partially ordered (what's better: `METHINKSITISXXXXXXXXXXXXXX` or `XXXXXXXXXXXXXXLIKEAWEASEL`). That's at least unusual, and perhaps a reason that no one else uses partitioned search*.
Five new letters are found, bringing the cumulative tally of discovered characters to {T, S,E, ∗,E, S,L}. All seven characters are ratcheted into place. The 19 new letters are chosen, and the process is repeated until the entire target phrase is found.	This ratcheting into place is special for the algorithm: the algorithm described in Dawkins's book doesn't show it.
Assuming uniformity, the probability of successfully identifying a specified letter with sample replacement at least once in Q queries is 1 − (1 − 1/N)^Q, and the probability of identifying all L characters in Q queries is q = (1 − (1 − 1/N)^Q)^L (22)	Yep, that's true. But why not doing a little math and giving us the expected number of queries? That would be a little bit less banal. And it's helpful if you want compare different algorithms
For the alternate search using purely random queries of the entire phrase, a sequence of L letters is chosen. The result is either a success and matches the target phrase, or does not. If there is no match, a completely new sequence of letters is chosen. To compare partitioned search to purely random queries, we can rewrite (5) as p =1 − (1 − (1/N)^L)^Q (23)	yup.
For L =28 and N =27 and moderate values of Q,we have p << q corresponding to a large contribution of active information. The active information is due to knowledge of partial solutions of the target phrase. Without this knowledge, the entire phrase is tagged as “wrong” even if it differs from the target by one character.	So, how big is this active information? For p, it was calculated in section III, A as I₊(p) = log(Q), and using the same approximation, we get I₊(q) = L log(Q) (that's only true-ish for small values of Q and large alphabets...)
The enormous amount of active information provided by partitioned search is transparently evident when the alphabet is binary. Then, independent of L, convergence can always be performed in two steps. From the ﬁrst query, the correct and incorrect bits are identified. The incorrect bits are then complemented to arrive at the correct solution. Generalizing to an alphabet of N characters, a phrase of arbitrary length L can always be identified in, at most, N − 1 queries. The ﬁrst character is offered, and the matching characters in the phrase are identiﬁed and frozen in place. The second character is offered, and the process is repeated. After repeating the procedure N − 1 times, any phrase characters not yet identiﬁed must be the last untested element in the alphabet.	Wow, the hangman's game. In detail.
Partitioned search can be applied at different granularities. We can, for example, apply partitioned search to entire words rather than individual letters. Let there be W words each with L/W characters each. Then, partitioned search probability of success after Q queries is p_W=(1 − (1 − N^−L/W)^Q)^W. (24)	What's that all about? Imagine an alphabet of 32 letters - including {A,B,...,Z,*} and our weasel-phrase. Then the phrase could also be encoded by 28 5-bit words. One 5-bit word is only correct, if all 5 bits are correct. Therefore, we get the same expressions for N=32, L=28, W=28 and N=2, L=140 and W=28.
Equations (22) and (23) are special cases for W = L and W =1.If N^−L/W << 1, we can make the approximation p_W ≈ Q^WN^−L from which it follows that the active information is I₊ ≈ W log₂ Q. (25) With reference to (6), the active information is that of W individual searches: one for each word.	So, for W=L, we get I₊(q) = L log(Q). My interpretation: virtually, this algorithm provides you with L fitness functions, one for each letter, indicating whether it is correct or not. BTW: I think it is surprising how the fitness functions get ignored during the whole article...

Evolutionary Algorithms in Conservation of Information in Search

2009-10-11T05:33:00.000-07:00

work in progress :-)

In their paper Conservation of Information in Search: Measuring the Cost of Success, William A. Dembski, Senior Member, IEEE, and Robert J. Marks II, Fellow, IEEE present three algorithms which could loosely be described as evolutionary. Idiosyncratically, they are labeled

Partitioned Search
Optimization by Mutation, and
Optimization by Mutation With Elitism

These procedures are discussed section II, part E and F (pp. 1055-1057), and some additional math is done in the appendix.

The algorithms and the examples given

The three algorithms are introduced or illustrated with an example, for which the parameters can be found in this table:

Algorithm	class	alphabet	length of target	Mutation	fitness function
Partitioned Search	none	{*,A,B,...Z} (27 el.)	28	every incorrect letter is mutated	position of correct letters
Optimization by Mutation	(1,2) ES	{0,1} (2 el.)	100	each letter is flipped with probability µ=0.00005	number of correct letters
Optimization by Mutation With Elitism	(1+1) ES	{0,1} (2 el.)	131^*	exactly one letter is flipped	number of correct letters

^*Why 131b for the last algorithm? I could understand 133 ~ 28* log₂27, but 131?

References

Four references are given in the section under discussion:

[3] M. Aigner, Discrete Mathematics. Providence, RI: Amer. Math. Soc., 2007.
[12] R. Dawkins, The Blind Watchmaker: Why the Evidence of Evolution Reveals a Universe Without Design. New York: Norton, 1996.
[32] D. J. C.MacKay, Information Theory, Inference and Learning Algorithms. Cambridge, U.K.: Cambridge Univ. Press, 2002.
[34] A. Papoulis, Probability, Random Variables, and Stochastic Processes, 3rd ed. New York: McGraw-Hill, 1991, pp. 537–542.

The first one is unproblematic - though superfluous: it is given as a reference for the Lth harmonic number. The three other references are either problematic or at least indicating sloppy work:

1. Partitioned search [12] is a “divide and conquer” procedure best introduced by example. Richard Dawkins doesn't use the expression "partitioned search" in his book.

2. When μ << 1, the search is a simple Markov birth process [34] that converges to the target. The page numbers of the reference don't fit: it should be pp. 647-650, not pp. 537–542. But at least, page numbers are given!

3. To illustrate, consider a single-parent version of a simple mutation algorithm analyzed by McKay [32].David MacKay analyzes explicitly multi-parented algorithms, there is not much of a resemblance with Dembski's and Marks's procedure. And annoyingly, again, page numbers are omitted. MacKay's book has over 600 pages, only a few in Chapter III, 19 : Why have Sex? Information Acquisition and Evolution are dedicated to evolutionary algorithms.

But one has to be grateful for small things: At least these references weren't garbled like

[7] S. Christensen and F. Oppacher, “What can we learn from no free lunch? A ﬁrst attempt to characterize the concept of a searchable,” in Proc. Genetic Evol. Comput., 2001, pp. 1219–1226.

Of course, it's about the concept of a searchable function. Not the first time, R. Marks and W. Dembski got that wrong.
Ann.: Tom English explained that errors in references are common in the engineering literature. So, I'll stop classifying the references of Dembski's and Marks's article as either superfluous or erroneous for now.

I'm annoyed.

2009-10-10T08:36:00.001-07:00

Perhaps I'm easily annoyed. But every time I read the paper Conservation of Information in Search - Measuring the Cost of Success, I'm getting, well, pretty much distressed. I wanted to write a concise rebuke of this article, but you can't be concise: there is so much disturbing in it, often little things. Take for instance this innocuous little sentence

To illustrate, consider a single-parent version of a simple mutation algorithm analyzed by McKay [32].

The reader may be interested in McKay's analysis of a simple mutation algorithm, and follows the reference:

D. J. C.MacKay, Information Theory, Inference and Learning Algorithms. Cambridge, U.K.: Cambridge Univ. Press, 2002.

Thankfully, David J.C. MacKay (he seems to prefer this spelling) put his book online. And yes, it is a book, a textbook with over 600 pages. It's covering a wide range, so at least, Dembski and Marks should have given the pages which they thought to be relevant. Which they thought to be relevant is a key phrase, because a skimming of the book doesn't render anything resembling an analysis of a simple mutation algorithm like the one in the article: yes, there is Chapter III, 19 :Why have Sex? Information Acquisition and Evolution - but this is explicitly about algorithms with multiple parents, it just doesn't fit a single parent version - MacKay assumes for instance that the fitness of the parents is normally distributed, something you just can't have for a single parent version. So why do Marks and Dembski give MacKay as a reference? I have no idea.

And it's going on

Just checking the next reference: [34] A. Papoulis, Probability, Random Variables, and Stochastic Processes, 3rd ed. New York: McGraw-Hill, 1991, pp. 537–542. for this little piece of marginal information

When μ << 1, the search is a simple Markov birth process [34] that converges to the target.

You may say: At least, they are giving a page number. But it's the wrong one! Take out your 3rd edition of Papoulis's very good introduction to the theory of probability, and you'll find Markov birth processes on pp. 647 - 650: why can't they get at least the details right?

Which parameter should Dembski have used?

2009-10-03T02:05:00.000-07:00

In the previous post I talked about Dawkins's weasel algorithm as understood by virtually everyone. Even William Dembski and Robert Marks describe something similar in their paper Conservation of Information in Search: Measuring the Cost of Success. But it's not the algorithm Partioned Search introduced on p. 1055 (and claimed by William Dembski to represent Dawkins's weasel), it's an algorithm called Random Search of p. 1056. The differences to the weasel of Dawkins:

the alphabet consists of two letters, not 27.
the target length is 100, not 28
the size of a generation is quite low: it's two
the mutation probability per letter is very small: µ = 0.00005

The value S=2 for the population size and µ = 0.00005 for the effective mutation rate are chosen to be able to do a little math - but they result in quite a great number of generations (~55,000) and queries (~110,000) on average.
Which values would a real programmer have used? Fooling around with his program, tweaking the parameter S and µ a little bit, he would soon have found out, that he should alter the original values considerably:

The greater the size of the population, the greater the optimal size of the mutation rate.

The smallest number of queries (1576) on average is achieved with S = 9 and µ = 0.009. Any practitioner would have come up with values near to these, as they make the program considerably faster. Sadly, these values doesn't allow for an application of Dembski's and Marks's elegant estimation of the number of correct letters as a function of generations.

What's interesting: While Dawkins's weasel had a range of population sizes which preferred virtually the same mutation probability, this doesn't happen in this realization. We'll see that this effect depends mainly on the size of the alphabet...

Which parameters did Dawkins use for his weasel?

2009-10-02T13:10:00.000-07:00

If R. Dawkins used a weasel algorithm as understood by virtually everyone who read his book (and tried to program it), he has two parameters to play with:

the size of the population S and
the probability µ to change a letter

So, which values did he use? That, I can't answer. But I can say which values he should have used:

Obviously, the greater the size of the population for a fixed µ, the smaller the number of generations it takes to find the target. On the other hand, for a fixed population size, a similar rule for µ can't be given - one probably wants to keep the parent in the next generation, so µ has to be quite small for small populations, and can be bigger for great ones.

Here, the coloured lines represent the expected values of the number of generations for different sizes of population as a function of µ. At each curve, I added the minimal expected number - and I joined these minima by a black line. So, a population size of 200 results at best in 49 generation on average - and this minimum is reached for µ ~ 0.044.

But if you want to have a fast algorithm, you'd look at the number of queries, not the number of generations, i.e., the total number of children, or how often the fitness function has to be calculated:

(Yep, I like this pic :-) Here, you can see that for getting the best results, the size of the population is surprisingly small: 20 individuals. But though it takes over 325 generations on average to reach the target, the fitness value for only ~6514 individuals has to be calculated - that is, if you chose µ ~ 0.041.
Another interesting effect: for population sizes from 20 to 200, the optimal µ doesn't change very much, it's just a little bit over 0.04.

Of course, no one does these calculations before running the program: you write your weasel and play with it. But the natural behaviour of this algorithm will lead you to population sizes from 20 to 200, and a µ of 0.04 - 0.05.

The Original Weasels?

2009-09-27T01:07:00.000-07:00

W. Dembski introduces two programs in his post The original weasels at his blog Uncommon Descent. These programs were sent to him by someone called Oxfordensis, and W. Dembski states

These are by far the best candidates [for being Dawkins's original program] we have received to date.

Let's have a short look:

Weasel 1

Here's how this program works for the target "METHINKS IT IS LIKE A WEASEL":

Chose random string
copy string 100 times, each time changing exactly one letter at random
select string with most correct letters in place
is the number of correct letters 28? Then STOP. Otherwise:
goto 2

The mutation procedure is interesting: While most programs I've seen mutate the parent string by changing each letter with a certain probability, here, exactly one letter is changed - but I've seen implementations of this version, too. The letter-based mutation seems to be more natural, while the other one has mathematical advantages: the neighbourhood of a string s, i.e., the strings s can mutate into, becomes small: it isn't the whole set of size 27²⁸ ≈ 1.20 * 10⁴⁰ anymore, but only of size 27 * 28 = 756...
Another advantage: you only have to call your random generator twice for each child, while a straight-forward implementation of the letter-mutating process take (1+μ)*28 of these costly calls per child - though this number can be reduced to (1+28*μ) calls per child....
So, one parameter less to tweak: though the number of children is set to 100 in this program, this can be changed easily.

But which value to chose for the remaining parameter - the size of the population? The bigger the size of the population, the less generations (black line) it takes to complete the search. So, make it big? No, not necessarily: commonly, you want your program to be fast. Therefore, you should try to reduce the number of queries (blue line), i.e, the number of mutated strings - which is the number of evaluations of the fitness function, too. Mutating at random, and calculating the fitness function, that's what's taking time.
I suppose if you fool around with the program for a while, you'll quickly find out that the optimal size of generations in the sense above is about 50 - and an exact calculation leads to an average of 3931 queries for a population size of 49 as an optimum (σ = 648.04).

Weasel 2

Well, this beast is boring. BORING. Look how it works for the target "METHINKS IT IS LIKE A WEASEL":

Chose random string
make new string by copying old string one time, changing exactly one letter at random
chose string from old and new string with most correct letters in place
is the number of correct letters 28? Then STOP. Otherwise:
goto 2

It's just a randomized version of the hangman game, taking 2969 steps on average (σ = 959.2). And it's what W. Dembski and R. Marks call Optimization by Mutation With Elitism in their current paper:

3) Optimization by Mutation With Elitism: Optimization by
mutation with elitism is the same as optimization by mutation
in Section III-F2 with the following change. One mutated
child is generated. If the child is better than the parent, it
replaces the parent. If not, the child dies, and the parent
tries again. Typically, this process gives birth to numerous
offspring, but we will focus attention on the case of a single
child. We will also assume that there is a single bit ﬂip per
generation.

Conclusions

Could these be the programs used by R. Dawkins in his book The Blind Watchmaker? The first one would fit his algorithm, and with a population size, it would take 47.6 generations on average (σ= 11.7). And what's about the BBC video? If the second version was used for this, I'd regard it at best as a visual effect, as it doesn't match Dawkins's algorithm. And yes, it is (implicitly) latching :-)

Random Mutation

2009-09-23T05:36:00.000-07:00

In their paper Conservation of Information in Search: Measuring the Cost of Success William Dembski and Robert Marks present an algorithm which resembles Dawkins's weasel (the algorithm I talked about before - at length :-)
It isn't the algorithm Partioned Search introduced on p. 1055, it's the algorithm Random Mutation on p. 1056. The differences

the alphabet consists of two letters, not 27.
the target length is 100, not 28
the size of a generation is quite low: it's two
the mutation probability per letter is very small: µ = 0.00005

Many programmers have implemented weasel-like programs, and the runs I've seen usually use a µ between 0.04 and 0.05, while the generation size is greater than 20.

So, why do W. Dembski and R. Marks use their values? Well, in the appendix of their paper, they calculated an approximation for the number of correct letters depending on the number of generations. This approximation discards terms quadratic in µ, so to use it, not only has µ << 1, but the size of a generation has to be small, too.

But should they use these values? I don't think so: As usual, I modeled their algorithm as a Markov chain and could calculate some exact values. The expected number of generations to get to the target for their choice of parameters is 55,500. (see pic 1)

Some fooling around with their program should have shown that this number can be reduced dramatically by using an more appropriate value for µ. The optimum lies near to µ ~ 0.0005, which takes only 10,600 generations.

And though their approximation doesn't work as fantastically as for the parameter ten times smaller, the numbers are not more than 2.5% off. (see pic 2)

A New Weasel

2009-09-19T15:25:00.000-07:00

At Uncommon Descent, W. Dembski presents us with a new weasel: Someone with the name Oxfordensis programmed years ago two Pascal versions.
The interesting aspect: In contrast to the usual weasels - as discussed in the earlier posts - not every letter is changed with a certain probability in this program, but exactly one letter is changed in each child. Here, the word change is used loosely, as with a probability of 1/27, no letter is changed at all.
Mathematically, this is easier to handle: the transition matrix of the corresponding Markov process gets sparser...
And you have only one parameter which can be changed: the size of the generation.
If I had any reader he may ask himself: Does this algorithm latch? Well, the pic provides the answer: it depends :-)
And here are some values for the probability that the program will show a change to the worse during a run of the program, depending on the size of the population:

99.999 %
98.485 %
83.879 %
55.944 %
30.008 %
14.071 %
6.168 %
2.649 %
1.137 %
0.491 %
0.214 %
0.094 %

So, even with a generation size of one hundred children, one in two hundred runs will show that this algorithm doesn't latch - it's not what W. Dembski and R. Marks describe as a partitioned search.

On Oracles and Weasels

2009-08-26T23:47:00.000-07:00

It's once again about Richard Dawkins's weasel algorithm. In his book The Blind Watchmaker, Dawkins gives the following description.

Dawkins describes his algorithm in the following way:

It again begins by choosing a random sequence of 28 letters, just as before:

WDLTMNLT DTJBKWIRZREZLMQCO P

It now 'breeds from' this random phrase. It duplicates it repeatedly, but with a certain chance of random error - 'mutation' - in the copying. The computer examines the mutant nonsense phrases, the 'progeny' of the original phrase, and chooses the one which, however slightly, most resembles the target phrase, METHINKS IT IS LIKE A WEASEL. [...] the procedure is repeated.

In my previous posts, I explained how this can be transformed into a program:

chose random string
copy string n times with mutation. NOTE: at this step you don't know which letters are correct in place, so no letter is safe from being mutated!
chose best fitting string. NOTE: best fitting seems to be generally understood to be the string with the most correct letters, the fitting is expressed by a number between 0 and 28
Stop, if the number of correct letters is 28, otherwise
goto 2

Some people (including W. Dembski, I think) are baffled by step 3: seemingly, the program has to know the best string and can extract information about it. Therefore, I restate this algorithm slightly:

chose random string
copy string n times with mutation.
ask oracle which string it likes best
Stop, if the number of correct letters is 28, otherwise
goto 2

Evaluating a fitness function for an algorithm is like questioning an oracle. Even Robert Marks's and Richard Dembski's web site says so - though they don't think it through.

So, how does the oracle work in Dawkins's case? You present it with a list of strings and it just tells you which string is best ("The computer examines the mutant nonsense phrases, the 'progeny' of the original phrase, and chooses the one which, however slightly, most resembles the target phrase, METHINKS IT IS LIKE A WEASEL."). You don't have to know how the oracle deems one string better than the other: it could be that it prefers the string with the most letters correctly in place, or the one which shares the longest substring with its (hidden) target, or perhaps it looks how many letters are right from the begin of the phrase, preferring METHINKS IT IS FROGLIKE XXX to BETHINKS IT IS LIKE A WEASEL, or preferring correct vowels to correct consonants, it could be a combination of these on Fridays, and Saturdays something else....

Could Dembski's algorithm with such an oracle? No. Here is a his algorithm:

chose random string
ask oracle which letters are in place
if all letters are in place: STOP Otherwise:
mutate letters which are not in place
goto 1

What Dembski exploits is the knowledge about the inner processes of the oracle: but knowing how it works allows for a deterministic approach - the Hangman game - which terminates in a few steps...

White Screen & Denise O'Leary

2009-08-26T09:46:00.000-07:00

My edits at Uncommon Descent obviously aren't welcomed anymore. What a pity: Now there are at two active threads over there discussing Dawkins's and Dembski's weasel algorithms (as described earlier in A Story of Two Weasels, The Latching Weasel, and Dawkins's weasel: it isn't that bad)

Denise O'Leary asked:

Make the code for the program public. Perhaps Richard Dawkins himself or his friends at RichardDawkins.net can finally provide this code (apparently a program written in BASIC).

I tried to enlighten her. After all, the program is now over 20 years old, and may be rotting in a heap of 5 1/4" floppies...

Dawkins describes his algorithm in the following way:

It again begins by choosing a random sequence of 28 letters, just as before:

WDLTMNLT DTJBKWIRZREZLMQCO P

It now 'breeds from' this random phrase. It duplicates it repeatedly, but with a certain chance of random error - 'mutation' - in the copying. The computer examines the mutant nonsense phrases, the 'progeny' of the original phrase, and chooses the one which, however slightly, most resembles the target phrase, METHINKS IT IS LIKE A WEASEL. [...] the procedure is repeated.

This is enough to replicate his program:
1. chose random string
2. copy string n times with mutation. NOTE: at this step you don't know which letters are correct in place, so no letter is safe from being mutated!
3. chose best fitting string. NOTE: best fitting seems to be generally understood to be the string with the most correct letters, the fitting is expressed by a number between 0 and 28
4. Stop, if the number of correct letters is 28, otherwise
5. goto 2

The parameters which you can chose is the number of copies n, and the probability that a letter in a string is mutated, p. You may even chose another procedure of mutation - but keep in mind the NOTE of step 2.

It's really basic to realize steps 1 - 5 in the programming language of your choice...

While the edit above made it through Uncommon Descent's gates, I'm not allowed to post the follow-up:

Just to elaborate: The interesting part is what Atom describes as oracle, i.e., the application of the fitness function. In Dawkins description we read

The computer examines the mutant nonsense phrases, the 'progeny' of the original phrase, and chooses the one which, however slightly, most resembles the target phrase

An oracle - a black box - which accomplishes this just hints to one string when presented with a population of strings. No further information has to be exchanged. It is not necessary to know how the oracle defines the best string: It could just hint you to the one with the greatest number of correct letters - or perhaps the one which shares the longest substring with the target phrase.

And that's rather a pity: W. Dembski is a guy who is interested in information, and the amount of information which is transferred from a environment into the algorithm. He should be interested in the flow of this information: How is this information exchanged?
The answer: Via the conversation with the oracle. Could his partition search algorithm work with an oracle a described above? No, as it is not enough to name the best string, the position of the correct letters in the string has to be named, too.
Does Dawkins's algorithm work with such an oracle? Obviously...

"Your comment is awaiting moderation...."

2009-08-26T04:33:00.000-07:00

Perhaps ~~I should be grateful that I'm not blocked at Uncommon Descent~~. Update: Perhaps I should have been grateful that I was not blocked at Uncommon Descent.
But it's rather annoying to wait for a couple of hours to get your comment through moderation while a discussion is ongoing: my edit contributed at 2:52 am today (Aug 26, 2009) has still not appeared over six hours later, though there were fifteen other comments in the meantime. What is so hard in moderating the following?

I took the string
SCITAMROFN*IYRANOITULOVE*SAM
and calculated a next generation using Dawkins's algorithms with populations of 10,50 and 100 - and mutation rates of .04, .05 and .1. The tenth string in the list is the second generation given in the paper of Mark and Dembski. The differences with the first generation are in bold face:

1. SCITAMROFN*IYRANOIEULOVE*SAM
2. SCITAMROFN*IYRANOITULOGE*SAM
3. ECITAMRI*N*IYZANOITULOVE*SAM
4. SCITAMROFN*IYRANOITUL*VE*SAM
5. SCITAMROFN*IYRANOITULOVE*SEM
6. SCITAMOOLNOIYRAMOITULOVE*SEM
7. SCITANROFN*IYYANOITULOVE*SAM
8. SCITIMROFN*JYRANOITULOVE*SAM
9. SCITAMROFN*ICRHNOITSLOWE*SAV
10. OOT*DENGISEDESEHT*ERA*NETSIL

Can anyone spot a difference in the design of the strings? Anyone? KF? Anyone?

I'll stay tuned :-)

Update: Now I've two comments waiting in the queue:

232 - DiEb - 08/26/2009 - 9:39 am Your comment is awaiting moderation.
I try to get involved in the discussion, but my last edit (#213) is now in moderation for nearly seven hours…

Update: After eight hours, the comment got through. But I seem to be blocked from now on....

The Latching Weasel

2009-08-25T05:27:00.000-07:00

I described Dawkins's ''weasel algorithm'' in a previous posts. Some claim that in the process the correct letters are ''explicitly or implicitly'' latched. That's not true, the algorithm isn't necessarily monotonously non-decreasing: a change to the worse from one generation to the next at least once while running the program is possible.
I calculated the probabilities for two case: a population of ten and a population of fifty - both times with a mutation rate of 4%.

10: 95.7 %
50: 0.0000026 %

With a small population, the effect will most probably be observed. But even with the bigger population, a programmer may spot the event - if he is lucky enough.

Why did I chose this population? A population of ten seems to be used by R. Dawkins in his video of 1987, while a population size of fifty was probably used for his book in 1986.
In the first case, it takes 1306 generations on average (σ=924), in the second case, 139 generations (σ=48). So it is quite probable that observe such a reversal, as seen on the video, two. For the bigger population such an occurence is at least more likely than winning a lottery....

Here is a complete view for various rates of mutation ( 1% - 10%) and populations from 1 to 100:

Dawkin's weasel - it isn't that bad...

2009-08-24T06:02:00.000-07:00

Richard Dawkins writes about his weasel alogrithm:

Although the monkey/Shakespeare model is useful for explaining the distinction between single-step selection and cumulative selection, it is misleading in important ways.

But it is useful is just another way: It helps to illustrate the importance of a sensible choice of the rate of mutation!
I calculated the expected number of generations for ten different rates of mutation per letter: 1%, 2%,..., 10%. As the graph - and the table - shows, the algorithm works best with a moderate rate of 4% or 5%.

	25	50	100
	size of the population
1 %	534	274	145
2 %	335	175	96
3 %	280	148	83
4 %	263	139	79
5 %	266	140	79
6 %	296	146	82
7 %	434	159	88
8 %	1394	179	96
9 %	15313	229	107
10 %	398,256	429	123

That's not much of a surprise, but contrasts with the algorithm W. Dembski declared to be the weasel: his version works best with a maximal rate of mutation...

A story of two weasels

2009-08-24T05:17:00.000-07:00

In his boock ''The Blind Watchmaker'', Richard Dawkins presents a small program which became known as ''Dawkins Weasel''. He wanted to show the mechanisms of mutation and selection. William Dembski has criticized Dawkins program, and claimed to have done so in his Conservation of Information in Search: Measuring the Cost of Success: Though he doesn't name Dawkins in this essay, in his blog he states:

In it we critique, for instance, Richard Dawkins METHINKS*IT*IS*LIKE*A*WEASEL (p. 1055). Question: When Dawkins introduced this example, was he arguing pro-Darwinism? Yes he was. In critiquing his example and arguing that information is not created by unguided evolutionary processes, we are indeed making an argument that supports ID.''

But does he even use the algorithm which Richard Dawkins had proposed? This is very doubtful...

Description of the algorithms

Here is the lengthy description which Dawkins gave in his book ''The Blind Watchmaker'':

So much for single-step selection of random variation. What about cumulative selection; how much more effective should this be? Very very much more effective, perhaps more so than we at first realize, although it is almost obvious when we reflect further. We again use our computer monkey, but with a crucial difference in its program. It again begins by choosing a random sequence of 28 letters, just as before:

WDLTMNLT DTJBKWIRZREZLMQCO P

It now 'breeds from' this random phrase. It duplicates it repeatedly, but with a certain chance of random error - 'mutation' - in the copying. The computer examines the mutant nonsense phrases, the 'progeny' of the original phrase, and chooses the one which, however slightly, most resembles the target phrase, METHINKS IT IS LIKE A WEASEL. In this instance the winning phrase of the next 'generation' happened to be:

WDLTMNLT DTJBSWIRZREZLMQCO P

Not an obvious improvement! But the procedure is repeated, again mutant 'progeny' are 'bred from' the phrase, and a new 'winner' is chosen. This goes on, generation after generation. After 10 generations, the phrase chosen for 'breeding' was:

MDLDMNLS ITpSWHRZREZ MECS P

After 20 generations it was:

MELDINLS IT ISWPRKE Z WECSEL

By now, the eye of faith fancies that it can see a resemblance to the target phrase. By 30 generations there can be no doubt:

METHINGS IT ISWLIKE B WECSEL

Generation 40 takes us to within one letter of the target:

METHINKS IT IS LIKE I WEASEL

And the target was finally reached in generation 43. A second run of the computer began with the phrase:

Y YVMQKZPFfXWVHGLAWFVCHQXYOPY,

passed through (again reporting only every tenth generation):

Y YVMQKSPFTXWSHLIKEFV HQYSPY

YETHINKSPITXISHLIKEFA WQYSEY

METHINKS IT ISSLIKE A WEFSEY

METHINKS IT ISBLIKE A WEASES

METHINKS IT ISJLIKE A WEASEO

METHINKS IT IS LIKE A WEASEP

and reached the target phrase in generation 64. m a third run the computer started with:

GEWRGZRPBCTPGQMCKHFDBGW ZCCF

and reached METHINKS IT IS LIKE A WEASEL in 41 generations of selective 'breeding'.

And this is Dembski's weasel-algorithm:

Partitioned search s a "divide and conquer" procedure best introduced by example. Consider the L =28 character phrase

METHINKS ∗ IT ∗ IS ∗ LIKE ∗ A ∗ WEASEL. (19)

Suppose that the result of our ﬁrst query of L =28 characters is

SCITAMROFN ∗ IYRANOITULOVE ∗ SAM. (20)

Two of the letters {E, S} are in the correct position. They are shown in a bold font. In partitioned search, our search for these letters is ﬁnished. For the incorrect letters, we select 26 new letters and obtain

OOT ∗ DENGISEDESEHT ∗ ERA∗NETSIL. (21)

Five new letters are found, bringing the cumulative tally of discovered characters to {T, S,E, ∗,E, S,L}. All seven characters are ratcheted into place. The 19 new letters are chosen, and the process is repeated until the entire target phrase is found.

Well, at least it is short.

Obvious differences

While Dawkins mentioned a couple of strings (it duplicates it repeatedly), Dembski only mentions one string.
The mutation rate of Dawkins example seems to be much lower than the Dembski's, at least judging from the first two generations:
Dawkins:
1) WDLTMNLT*DTJBKWIRZREZLMQCO*P
2) WDLTMNLT*DTJBKSIRZREZLMQCO*P

Dembski:
1) SCITAMROFN*IYRANOITULOVE*SAM
2) OOT*DENGISEDESEHT*ERA*NETSIL

(changes in bold font)
Dawkins gives the number of generations for three experiments: 43, 64, and 41. The expected number of generations G, i.e, E(G), for Dembski's algorithm is 104.55 (=Σ_k(1-(1-(1/N)^k)^L), where N is the size of the alphabet and L the number of letters of the phrase, i.e. N = 27, L = 28 .

Claimed differences

Dembski always stressed that in the algorithm a letter once found is ''latched'' or ''ratcheted''. And the strings given as examples in Dawkins's book doesn't exclude that ''latching'' is happening. But Dawkins stated repeatedly that in his algorithm latching doesn't take place.

Conclusion

Even this superficial comparison of the two algorithms shows that both are very different: so to say two kinds of weasels.