*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 Ω*As an example, they give

_{1}that provides more information about the search environment or candidate solutions than a blind search.*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 T

_{1}= {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 T

_{1}:

- α
_{a1}(T_{1})= 1/64 - β
_{a1}(T_{1}) = 1/2 - γ
_{a1}(T_{1}) = 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}(T

_{1})){a1} = log

_{2}1/2 - log

_{2}1/64 = 5.

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

I

_{+}(γ

_{a1}|α

_{a1}(T

_{1})){b1} = log

_{2}0 - log

_{2}1/64 = -∞

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 Ω

_{1}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

_{Ti}{T

_{i}} × I

_{+}(γ

_{Ti}|α

_{Ti}){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!

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