# [Retros] Retros Digest, Vol 146, Issue 3

Dominique Forlot dominique.forlot at wanadoo.fr
Fri Sep 2 06:31:42 EDT 2016

```To specify my thought on the problem of the matt by "knight alone" :

because of mat(∅,∅,c) : « Knights alones» is unsound we can Fit out the problem in a mat(a,b,c) where:
the sum ( a +b*sqr(2) ) would be minimized and c would be as great as possible

we can now searching for a matt with a knignt alone (white or black) witch minimise the adverse's moves execute by another piece than a knignt.
as this Mat(3,0,7)

the spécification became :

Compose a sound checkmate PG wicth "maximize" the geometric score = c*sqr(5) - (α + β*sqr(2))

Dominique.

> Message du 02/09/16 07:42
> De : retros-request at janko.at
> A : retros at janko.at
> Copie à :
> Objet : Retros Digest, Vol 146, Issue 3
>
> I wrote:
>
> < What about *minimum* geometric length? Here it may
> < be possible to prove that one has the absolute minimum (if it's small
> < enough that anything beyond the search bounds must be longer).
>
> Francois Labelle  replied:
>
> > You're right, it's possible.
>
> > The minimum geometric length for a sound checkmate PG is
> > 9.656854 = (4,4,0), so the proof of optimality only required searching
> > up to ply 9.
>
> > Only one sound checkmate PG achieves it: [...]
>
> Great!
>
> andrew buchanan  writes:
>
> > > The second-shortest is 10.242641 = (6,3,0)
>
> > > So I guess someone could ask:
> > > "Compose a sound checkmate PG with geometric length less than 10".
>
> > This unnecessarily removes the case 10.
>
> > I would prefer the more whimsical stipulation:"Compose a sound
> > checkmate PG with geometric length less than 10.24264."
>
> Normally "geometric length at most 10", though "length less than
> 6 + sqrt(18)" hints at the runner-up length too.
>
> NDE
>
>
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