Ãchecs et Maths
Nous avons posté ceci hier soir sur le site de France-Ãchecs. Espérons que cette idée puisse inspirer quelques compositeurs de problèmes.
__________
> Bonsoir à tous,
> En retombant sur le célèbre problème de Smullyan (photo) où le Roi blanc est invisible (...), je me suis demandé si ce concept pouvait faire l’objet d’une PCPJ intéressante (rappelons aux néophytes que PCPJ signifie « Plus Courte Partie Justificative » : on demande donc de trouver, à partir de la position de départ d’une partie d’échecs orthodoxe, l’enchaînement de coups blancs et noirs le plus rapide menant au diagramme – enchaînement unique de coups, idéalement, lequel, quand il est subtil, évoquera une fine mécanique suisse et horlogère).
Voici la légende du diagramme suivant :
PCPJ avec une seule pièce invisible sur l’échiquier :
La réponse est facile à trouver :
1.e4–d5
2.exd—Dixd (le i derrière D dit que la Dame noire est invisible – elle est en d5).
La position ci-après, légendée pareil, résulte-t-elle d’un enchaînement unique de coups ? Si oui, lequel ? (C’est à peine plus difficile que l’exemple précédent.)
PCPJ avec une seule pièce invisible sur l’échiquier :
rnbqk1nr/pppp1ppp/8/4p3/8/P7/2PPPPPP/1RBQKBNR
Attention, une pièce invisible est une pièce tout à fait normale — elle est juste invisible.
Solution ci-dessous (passer la souris entre les étoiles).
*1.Ca3—e5 2.Tb1—FxC 3.bxF (et a2 invisible)*
Si ce concept est original, il promet, je crois (certaines PCPJ laissant deux pièces invisibles sur l’échiquier pourraient être amusantes à concevoir et difficiles à résoudre).
_______________________________________________________________________
_______________________________________________________________________
And now for the mathematical part.
A few ideas for a few sequences (the first two are not of gigantic interest).
A] Underline the smallest digit of S; the succession of the underlined digits is the succession of the digits of S itself. S is the lexicographically earliest sequence of distinct terms > o with this property. By construction, we refuse the presence of the digit zero in any term of S.
S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 22, 14, 33, 23, 24, 15, 44, 34, 35, ...
Lets underline (in yellow here) according to the above rule:
S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 22, 14, 33, 23, 24, 15, 44, 34, 35, ...
Ok, so far so good. But what will happen when we will need to extend S with a term T having the digit "9" as its smallest one – the integers 9 to 99 being already present in S? We will extend S at that point with 999, of course. But the three 9s of 999 must be "described" too. We will need three successive terms T for S whose smallest digit, again, must be 9; those terms will be 9999, 99999, 999999... In the long run, S will be suffocated by those 9s.
B] Underline the largest digit of S; the succession of the underlined digits is the succession of the digits of S itself. S is the lexicographically earliest sequence of distinct terms > o with this property. By construction, we refuse the presence of the digit zero in any term of S.
S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 111, 1111, 11111, 111111,...
S is already suffocated by 1s.
C] Now the fun part.
Say B is the largest digit of a(n) + a(n+1). The successive Bs reproduce the successive digits of S. By construction, we refuse the presence of the digit zero in any term of S.
If I am not wrong we get:
This will be https://oeis.org/draft/A363420 if accepted.
D] Lexicographically earliest sequence of distinct terms > 0 such that n is a substring of a(n) + a(n+1):
If I am not wrong again we get:
This will be https://oeis.org/draft/A363872 if accepted.
____________
Next day update :
Amiram Eldar was quick to correct the above start; here is what Amiram's program found:
S = 1, 9, 3, 10, 4, 11, 5, 2, 6, 13, 87, 23, 89, 24, 90, 25, 91, 26, 92, 27, 93, 28, 94, 29, 95, 30, 96, 31, 97, 32, 98, 33, 99, 34, 100, 35, 101, 36, 102, 37, 103, 38, 104, 39, 105, 40, 106, 41, 7, 42, 8, 43, 109, 44, 110, 45, 111, 46, 12, 47, 113, 48, 14, 49, 15,...
Many thanks Amiram (and Marcus); this will be soon in the OEIS, here.
(I guess this is https://oeis.org/A063733)
F] Hello Math-Fun
we want a(n)+a(n+1) to be a substring of the concatenation [a(n);a(n+1)].
This works with this S:
S = 1, 10, 99, 899, 8099, 1819, 6377, 1336,…
But is S the lexicographically earliest sequence of distinct terms > 0 with this property?
Best,
Ã.
1+10=11
10+99=109
99+899=998
899+8099=8998
8099+1819=9918
1819+6377=8196
6377+1336=7713
…
____________________
Hans H. was quick to lexico-beat me:
EA: "S = 1, 10, 99, 899, 8099, 1819, 6377, 1336, ..."
--> S = 1, 10, 99, 889, 8009, 1101, 9089, 80718, ...
____________________
... and Maximilian H. confirmed:
> I get the following :
0, 1, 10, 99, 889, 8009, 1101, 9089, 80718, 100284, 183899, 206021, ...
with PARI/gp:
ok(a,k)=my(c=a*10^logint(k*10,10)+k); k=10^logint(10*a+=k,10);
until(a>c\=10, c%k==a&& return(1))
U=[a=0]; for (n=1,20, print1(a", "); for(k=1,oo, !set search(U,k) &&
ok(a,k) && (a=k) && break); U=set union(U,[a]))
PS: seems to confirm Hans' result, received with delay...
____________________
Then, a few minutes ago (Monday, July 3rd, 4pm Brussels time, Belgium), I received this extraordinary link by Hans:
S = 1, 10, 99, 889, 8009, 1101, 9089, 80718, 100284, 183899, 206021, 396118, 215703,
354632, 108578, 469891, 229021, 61195, 34146, 7321, 13817, 3536, 1825, 749, 167,
508, 324, 2096, 4337, 2958, 2870, 4171, 12941, 16470, 30560, 25465, 21056, 35296,
17665, 35927, 23345, 10106, 548, 279, 516, 1094, 3228, 5302, 16331, 14817, 2425,
1826, 859, 5091, 4049, 945, 3590, 2312, 816, 871, 6291, 2837, 958, 4890, 4014, 10086,
55459, 3934, 5411, 6209, 3101, 7069, 2145, 2378, 1403, 1774, 2473, 2259, 334, 3009,
6656, 49842, 34392, 4856, 826, 1791, 6125, 6439, 2845, 1671, 5044, 38839, 45106,
5959, 3378, 406, 215, 316, 315, 205, 327, 414, 3729, 3564, 2078, 5779, 2013, 1096,
5448, 2835, 738, 3092, 6169, 812, 431, 2695, 2561, 3052, 2169, 4778, 3005, 2216,
4204, 16212, 5038, 3291, 5867, 925, 1590, 4314, 27113, 44021, 167655, 369272, 323451,
21670, 45375, 8383, 24018, 16163, 152, 371, 342, 86, 571, 140, 262, 361, 251, 261,
352, 169, 526, 2094, 2117, 5425, 19772, 52755, 2494, 1673, 1474, 2806, 3548, 1264,
1377, 2395, 1556, 4008, 4435, 30873, 42552, 12660, 13941, 25471, 21650, 28636, 8182,
10028, 18153, 13163, 3168, 3668, 3015, 2205, 3105, 2105, 3216, 3093, 6269, 701,
6315, 9649, 39748, 9168, 7709, 1434, 1985, 3349, 142, 280, 525, 1994, 2228, 57,
14,...
Many thanks to both of you, Hans and Maximilian!
If accepted, this will enter the OEIS in a couple of weeks: https://oeis.org/draft/A359482
Ce commentaire a été supprimé par l'auteur.
RépondreSupprimerCe commentaire a été supprimé par l'auteur.
RépondreSupprimer