Newsgroup: it.hobby.enigmi
Titolo: Polipercubi [update]

Aggiornata la pagina con le cose dette anche qua:
http://www.geocities.com/liviozuc/polyhypercubes.html
Gli eptaipercubi 4D in 4D sono 1316.
Forse mi fermo qua. Son già contento.
Certo che gli ottoipercubi mi attraggono, ma sembrano fuori
dalla mia capacità di calcolo. Il loro numero sarà compreso tra
8699 (nr. 8-cubi 4D in spazio multi-D) e 57444 che sono gli
ottoipercubi ottenuti accrescendo di un ipercubo i 1316 epta.
Saranno sì e no una dozzina di migliaia.
Ora bisognerebbe confrontare ciascuno di quei 57444 con
ciascun altro per eliminare i doppioni. Per ciascun confronto
bisogna ruotarne uno 192 volte, fanno 316 776 546 432 di
confronti e ogni coso da confrontare è una matrice booleana
di 8x8x8x8.
Mi è venuto in mente di misurarli per poter confrontare tra
loro solo quelli che hanno le stesse dimensioni:
1x1x1x8 2
1x1x2x4 5
1x1x2x5 71
1x1x2x6 63
1x1x2x7 53
1x1x3x3 33
1x1x3x4 386
1x1x3x5 447
1x1x3x6 97
1x1x4x4 344
1x1x4x5 197
1x2x2x2 3
1x2x2x3 374
1x2x2x4 990
1x2x2x5 841
1x2x2x6 202
1x2x3x3 2980
1x2x3x4 5616
1x2x3x5 1468
1x2x4x4 1025
1x3x3x3 2462
1x3x3x4 1789
2x2x2x2 396
2x2x2x3 6220
2x2x2x4 3801
2x2x2x5 617
2x2x3x3 15144
2x2x3x4 6330
2x3x3x3 5488
Ma come vedete le cose non migliorano un gran che.
Se qualcuno ha altre idee, me lo faccia sapere.
Ciao!!!  :o)
Livio

1x1x1x8 2
ce ne sono due;-)  e me lo spieghi perche' due?
-- 
Ciao, rez	||  -- GNU/Linux 2.4.25 su Slackware 9.1  ||

 ce ne sono due;-)  e me lo spieghi perche' due?
Dal processo di accrescimento ne nascono due, uno
aggiungendo un cubetto da una parte e uno dall'altra.
Così come di L ne nascono 12 dalla I + 1 dalla L(7).
Sta al prog di sfoltimento scoprire che sono uguali
ed eliminarli. Questo è il metodo classico di conteggio
dei polimini. Ha anche il vantaggio di poter collaudare
il sw sui livelli più bassi, scrivendolo in modo parametrico.
Quando ho contato i decamini (2D) ho usato un altro
metodo: ho chiesto a questo ng di trovare la figura più
piccola che potesse contenerli tutti e 4655 e poi ho
mosso all'interno 12 quadrati eliminando gli insiemi
disgiunti e poi quei pochi sovrapponibili. Ma nel caso
4D ho visto che questo metodo non era affatto efficiente.
Può essere che ce ne sia qualcun altro.
Ciao!!!  :o)
Livio

 "simple" side-forms?
Yes.  The polynomial for cubes is (x^6 + 3x^4 + 12x^3 + 8x^2) / 24.
An Attempt at a Simple Explanation of Basic Polya Enumeration.
We wish to count the number of different ways to color some shape. If
we use x colors on a shape with y sides, then at first glance we might
think there are x^y ways to color it.  (For instance, using x=2 colors
to color a domino (y=2 ends) there are 2^2=4 colorings: RR RB BR BB.)
But wait! Some of these colorings are the same. (In the domino example
RB and BR are the same: we can turn the domino around)  How do we keep
from counting these multiple times?  The answer is to count all of the
colorings multiple times, and then divide by the number of times we
counted them.  For a domino colored with x colors, the number x^2 counts
almost all colorings twice, except for those that use only one color.
Those are counted only once.  So: count them again.  There are x of
them.  The total number of different ways to color a domino with two
colors is (x^2 + x)/2.  How can we generalize this?  The basic idea
is to consider the "symmetries" of the object we are trying to color.
Consider a more complicated example, say, a square (a one-sided square -
that is, one that can _not_ be flipped over.)  If we color the (y=4) corners
of a square with x=5 colors the term x^y counts most colorings four times,
like the ones here (each coloring in each column is the same):
AB    AB    CD
DC    CD    EE
BC    BD    DE
AD    AC    CE
CD    DC    EE
BA    BA    DC
DA    CA    EC
CB    DB    ED
But some colorings have not been counted four times, like these:
AA    AB
AA    BA
      BA
      AB
The first has only been counted once, the second has been counted twice.
There are 5 (x) colorings like the first one, and 20 like the second
one. [5 choices for A, (5-1) choices for B].  To count every coloring four
times, we would need the first type three more times and the second type
two more times.  The equation is
(5^4 + 3*5 + 2*5(5-1))/4 = (5^4 + 5 + 2*5^2)/4
The key observation is found by examining the symmetries of our square.
It can be rotated by 0 degrees, 90 degrees, 180 degrees, or 270 degrees.
By careful thought or mathematical proof, one can
convince oneself that we fail to count a coloring in the term x^y only
when some symmetry of the object leaves that coloring totally unchanged.
(For instance, the square AB
                          BA is left unchanged if we rotate it 180 degrees).
This means we can turn our counting around (Burnside's Lemma): For each
symmetry of our object, we count the number of colorings that are left
unchanged by that symmetry.  We add these all up and divide by the
number of symmetries, the result is the number of different colorings.
The number of colorings left unchanged by a symmetry is x^c where c
is the number of cycles in the permutation representation of the
symmetry.  For instance, in the case of a two-sided square (one we
_can_ flip over) the symmetry "reflect about the line from upper left
to lower right) has three cycles (if we are coloring the corners.)
The upper left corner maps to itself.  The lower right corner maps
to itself.  The upper right corner maps to the lower left corner, which
maps back to the upper right.  That's three.  On the other hand, that
same symmetry has only two cycles if we are talking about coloring the
edges of the square: The top goes to the left which goes back to the
top, and the right goes to the bottom which goes back to the right.
To work out a couple of examples in full, consider coloring the edges
of one-sided triangles.  We have have three symmetries: Rotate by 0
degrees (R0), R120 and R240.  There are x^3 colorings left unchanged
by R0.  (three cycles: each edge goes to itself).  There are x colorings
left unchanged by R120 (one cycle: each edge maps to the next).  There are
x left unchanged by R240 (again, one cycle.)  The total number of colorings
is (x^3 + 2x)/3.  If we instead do two-sided triangles, we must add three
reflections, each has two cycles (because a reflection of a triangle swaps
two edges [the first cycle] and leaves the third unchanged [the second cycle].
Thus, for coloring the edges of two-sided triangles, we have
(x^3 + 3x^2 + 2x)/6 colorings.
What you call "complex" edges work out automatically: If you look
at the Xominoes which have 5 edge "colors" MFSZ-.  How many colorings
are left unchanged (for example) by flipping a square around it's vertical
axis?  Well, we can put anything on the left edge that we like [and put
it's reflection on the right].  For the top and the bottom, we can use
M, F, or -, but not S or Z (since M, F, and - stay the same if we reflect
them through the middle, but S and Z aren't).  Thus, there are
5*3*3 colorings that aren't changed by that reflection.  If you work
it out, you'll find each term in the polynomials that I sent you before
comes from one of the symmetries of the object.  One more example: The
cube polynomial listed above:
(x^6 + 3x^4 + 12x^3 + 8x^2) / 24
The cube has 24 symmetries (6 choices for the top face, 4 for the front).
If you look for them all, you find that they are all rotations.  R0,
the identity rotation accounts for the term x^6.  There are three
rotations of 180 degrees around axis through the center of the faces.
Each of these has 4 cycles.  [For instance, rotation 180 degrees about
the axis through the center of the top and bottom faces has the following
cycles: (top) (bottom) (left,right) (front, back).] That's the term
3x^4. There are six rotations of 90 or 270 degrees about those same
axis, each of these has three cycles. [Example, 90 degrees about
axis through center of top to center of bottom.  (top) (bottom)
(front,left,back,right)] That's 6x^3.  There are 8 rotations around
axis through diagonally opposite vertices of the cube [4 axis, rotation
by 120 or 240].  Each of these has two cycles, which is the term 8x^2.
There are 6 axis through midpoints of opposite edges, rotation by
180 degrees about such an axis gives three cycles, which is another
6x^3.  Putting them all together, we get the above polynomial.
Polya enumeration goes much further, actually, but I'm not going to
explain it in detail.  By putting in a little more information about
the cycles each symmetry has, you can come up with a polynomial that
[when expanded] will let you look up (for example) the number of
5-colorings of the faces of a cube that use red twice, blue once,
and green three times.
Hope this was helpful.
 -Chris
--
Prof. Chris Hartman   
http://tigger.cs.uaf.edu/hartman
 
har...@arsc.edu
University of Alaska, Fairbanks  and  Arctic Region Supercomputing

aggiungendo un cubetto da una parte e uno dall'altra.
Certo.. domanda sciocca chissa` che pensavo, perche'
e` proprio quello che fa e deve fare il programma.
Ps. Interessantissimo anche quel che dici nell'altro
reply, poi guardo anche quel link (ma ho una connessione
lenta), TNX :-)


 Cos'e` OEIS eccetera?
http://www.research.att.com/~njas/sequences/
ciao,
g.

 
http://www.research.att.com/~njas/sequences/
Oh, finalmente ti vedo! Ci mancavi in quel party!
Ciao!!!  :o)
Livio

 il terzo asse invertito.
Intendo le coordinate nell'ordine XYZT. Prendiamo solo
le prime 3, trasliamo di -1 e osserviamo i 2 oggetti in 3D:
(0,0,0)
(0,1,0)
(1,0,0)
(1,0,1)
(0,0,0)
(0,1,0)
(1,0,0)
(1,0,-1)
Abbiamo detto che se cambiamo il segno ad un numero
dispari di coordinate otteniamo una *riflessione* mentre
se lo facciamo ad un numero pari otteniamo una *rotazione*
di pigreco. E' evidente che in 3D possiamo ottenere l'uno
dall'altro soltanto cambiando il segno ad un numero dispari
di coordinate, quindi trattasi di *riflessione*.
Questo ragionamento è lo stesso in 2D e in 4D e in nD.
Andiamo in 4D:
(X,Y,Z,T)
(0,0,0,0)
(0,1,0,0)
(1,0,0,0)
(1,0,1,0)
(0,0,0,0)
(0,1,0,0)
(1,0,0,0)
(1,0,-1,0)
Ora cambiamo il segno a Z  e a T. (!)
Abbiamo cambiato il segno ad un numero pari di
coordinate, quindi abbiamo fatto una *rotazione*
di pigreco in 4D. Messa giù così è quasi banale, vero?
Ciao!!!  :o)
Livio

di pigreco in 4D. Messa giù così è quasi banale, vero?
Si`.
Dunque lo fai, o meglio lo fai fare al programma, solo
lavorando sulle matrici.
Con questo metodo e` allora immediato vedere la famosa
storia Magenta vs Mattone:
MAGENTA
0  0  0  0
1  0  0  0
0  1  0  0
0  0  1  0
0  1  0 +1
con scambio col 1-3 [x <-> z] e segno col 4
[inversione asse t], passa a
MATTONE
0  0  0  0
0  0  1  0
0  1  0  0
1  0  0  0
0  1  0 -1
indistinguibile.
All'anima! e` schiacciante come superiorita`! :-)
Be', sempre piu` interessante:-)  quando ho un po' di
tempo provo a vivisezionare le matrici.. ma non credo
se ne potra` cavare altro, voglio dire: mi sa che per
una formola campa cavallo.
