IIh',l~--"El,t=+n,,~ut+l[,,.-~+am=:l'i"l<=4n+,~nnu,,~+.-1,+,<.--~ Michael Kleber and Ravi Vakil, Editors !

uch has been said of the affinity understood using the terminology and M between mathematics and : techniques of combinatorics. We also The two domains of human thought where relate a few of these ideas with practi- very limited sets of rules yield inex- cal endgame technique (see Diagrams haustible depths, challenges, frustra- lff., 10, 11). Mathematical tions and beauty. Both fields support a The latter half of the article shows venerable and burgeoning technical lit- some remarkable chess problems fea- erature and attract much more than turing the knight or knights. Most Knioht their share of child prodigies. For all "practical" chess players have little pa- that, the intersection of the two do- tience for the art of chess problems, Noam D. Elkies mains is not large. While chess and which has evolved a long way from its Richard P. Stanley mathematics may favor similar mind- origins in instructive exercises. But the sets, there are few places where a same formal concerns that may deter chess player or analyst can benefit the over-the-board player give some from a specific mathematical idea, problems a particular appeal to math- such as the symmetl7 of the board and ematicians. For instance, we will of most pieces' moves (see for instance exhibit a position, constructed by This column is a place for those bits of [24]) or the combinatorial game theory P. O'Shea and published in 1989, where of Berlekamp, Conway, and Guy (as in White, with only and knight, has contagious mathematics that travel [4]). Still, when mathematics does find just one way to force mate in 48 (the from person to person in the applications in chess, striking and in- current record). We also show the community, because they are so structive results often arise. longest known legal game of chess that elegant, suprising, or appealing that is determined completely by its last Introduction move (discovered by R0sler in 1994)- one has an urge to pass them on. This article shows several mathemati- which happens to be by Contributions are most welcome. cal applications that feature the knight to a knight. and its characteristic (2,1) leap. It is based on portions of a book tentatively Algebraic notation titled Chess and Mathematics, cur- We assume that the reader is familiar rently in preparation by the two au- with the , but we assume thors of this article, which will cover very little knowledge of . all aspects of the interactions between (The reader who knows, or is willing chess and mathematics. Mathemati- to accept as intuitively obvious, that cally, the choice of (2,1) and of the 8 • king and win against king, or 8 board may seem to be a special case even against king and knight if there is of no particular interest, and indeed we no immediate , will have no diffi- shall on occasion indicate variations culty following the analysis.) The and generalizations involving other reader will, however, have to follow leap parameters and board sizes. But the notation for chess moves, either by long experience points to the standard visualizing the moves on the diagram knight's move and size as or by setting up the position on the felicitous choices not only for the game board. Several notation systems have of chess but also for puzzles and prob- been used; the most common one lems involving the board and pieces, in- nowadays, and the one we use here, is cluding several of our examples. "algebraic notation," so called because We will begin by concentrating on of the coordinate system used to name Please send all submissions to the puzzles such as the knight's tour. Many the squares of the board. In the re- Mathematical Entertainments Editor, of these are clearly mathematical prob- maining paragraphs of this introduc- Ravi Vakil, Stanford University, lems in a very thin disguise (for in- tory section we outline this notation Department of Mathematics, Bidg, 380, stance, a closed knight's tour is a system. Readers already fluent in alge- Stanford, CA 94305-2125, USA Hamiltonian circuit on a certain graph braic notation may safely skip to the e-mail: [email protected],edu ~), and can be solved or at least better next section, A .

22 TIlE MATHEMATICAL INTELLIGENCER 2003 SPRINGER VERLAG NEW YORK Each square on the 8 x 8 board is analysis begins with a Black move, we natural try is 1.KXc2, eliminating one uniquely determined by its row and use " . . . " to represent the previous and imprisoning two of Black's column, called "rank" and "file" re- White move; thus "1 . . . Nd3!" is the remaining three men in the corner. But

spectively. The ranks are numbered same first Black move. 1 . . . Nd3! breaks the blockade (Dia- from 1 to 8, the files named by letters A few further refinements are gram 2a). Black threatens nothing but a through h. In the initial array, ranks needed to subsume promotion and controls the el. The rules 1 and 2 are occupied by White's pieces , and to ensure that every move of chess do not allow White to pass the and pawns, ranks 8 and 7 by Black's; is uniquely specified by its notation. move; unable to go to el, the king must both queens are on the d-file, and both For instance, if Black were to move move elsewhere and release Black's kings on the e-file. Thus, viewed from first in Diagram 1 and promoted his c2- men. After 2.K• d3 (or any other move) white's side of the board (as are all the pawn to a queen (giving ), we Kbl followed by 3... alQ, Black wins diagrams in this article), the ranks are would write this as 1... clQ+, or more easily. numbered front bottom to top, the files likely 1 . . . clQ+?, because we shall from left to right. We name a square by see that after 2.Kxcl White can draw. Diagram 2a its column followed by the row; for in- Short and long castling are notated 0-0 stance, the White king in Diagram 1 be- and 0-0-0 respectively. If the piece and low is at d2. Each of the six kinds of destination square do not specify the chessmen is referred to by a single let- move uniquely, we also give the de- ter, usually its initial: K, Q, R, B, P are parture square's file, rank, or both. An king, queen, , , and pawn extreme example: Starting from Dia- (often lowercase p is seen for pawn). gram 9, "Nbl" uniquely specifies a We cannot use the initial letter for the move of the c3 knight. But to move it knight because K is already the king, to d5 we would write "Ncdh" (because so we use its phonetic initial, N for other knights on the b- and f-files could kNight. For instance, Diagram 1 can be also reach dh); to a4, "N3a4" (not described as: White Kd2, Black Kal, "Nca4" because of the knight on c5); Nf2, Pa2, Pc2. and to e4, "Nc3e4" (why?). To notate a chess move we name the piece and its destination square, inter- A Chess Endgame polating "x" if the move is a capture. We begin by analyzing a relatively sim- Diagram 2b For pawn moves the P is usually sup- ple chess position (Diagram 1). This pressed; for pawn captures, it is re- may look like an endgame from actual placed by the pawn's file. Thus in Dia- play, but is a composed position--an grain 11, Black's pawn moves are ""--created (by NDE) notated a2 and a • b2 rather than Pa2 to bring the key point into sharper and P • b2. We follow a move by "+" focus. if it gives check, and by 'T' or "?" if we regard it as particularly strong or weak. Diagram 1 In some cases 'T' is used to indicate a thematic move, i.e., a move that is es- sential to the "theme" or main point of the problem. As an aid to following the analysis, moves are numbered consecutively, from the start of the game or from the diagram. For instance, we shall begin Returning to Diagram 1, let us try in- the discussion of Diagram 1 by con- stead 1.Kcl! This still locks in the Black sidering the possibility 1.KXc2 Nd3!. Kal and Pa2, and prepares to capture Here "1" indicates that these are the Pc2 next move, for instance 1 . .. White's and Black's first moves from Nd3+ 2.K• arriving at Diagram 2a the diagram; "K• means that the with Black to move. White has in effect White king captures the unit on c2; and succeeded in passing the move to "Nd3!" means that the Black knight White to move Black by taking a detour from d2 to c2. moves to the unoccupied square d3, Now it is Black who cannot pass, and and that this is regarded as a strong White, reduced to , can do any move restores the White king's ac- move (the point here being that Black no better than draw, and even that with cess to el. For instance, play may con- prevents 2.Kcl even at the cost of let- difficulty: Black will surely win if either tinue 2... Nb4+ 3.Kcl, reaching Dia- ting White capture the knight), when pawn safely promotes to a queen. A gram 2b. Black is still bottled up. If it

VOLUME 25, NUMBER 1, 2003 23 were White to move in Diagram 2b, to reach Diagram 2b with White to the same color as the one occupied by White would have to release Black with move; and again Black could aim for the knight. Kdl or Kd2 and lose; but again Black another square that controls c2. But REMARK. Our analysis would reach the must move and allow White back to c2, each of these squares is one or three same conclusions if the Black pawn on for instance 3... Nd3+ 4.Kc2 and we knight moves away from d3, so again c2 were removed from Diagrams 1 and are back at Diagram 2a. would yield a closed path of odd length 3b; we included this superfluous pawn So white does draw--at least if through d3. only as bait to make the wrong choice Black obligingly shuttles the knight be- Can Black thus pass the move back of c2 more tempting. tween d3 and b4 to match the white to White? For that matter, what should king's oscillations between cl and c2. White do in Diagram 3b? Does either Puzzle 1. For which rectangular boards But what if Black tries to improve on Kcl or K• draw, or is White lost re- (if any) does part (i) or (ii) of the this? While the king is limited to those gardless of this choice? Lemma fail? That is, which ~5~,,,~ are not two squares, the knight can roam over The outcome of Diagram 2a thus connected, or not bipartite? (All puz- almost the entire board. For instance, hinges on the answer to the following zles and all diagrams not explicated in from Diagram 2a Black might bring the problem in graph theory: the text have solutions at the end of knight to the far corner in m moves, this article.) Let ~5 = ~s,s be the graph whose ver- reaching a position such as Diagram tices are the 64 squares of the 8 • 8 3a, and then back to d3 in n moves. If Knight's Tours and the chessboard and whose edges are the m + n is odd, then Black will win since Thirty-Two Knights pairs of squares joined by a knight's it will be White's turn to move. Instead The graph ~5 arises often in problems move. Does ~ have a cycle of odd of d3, Black can aim for b3 or e2, which and puzzles involving knights. For in- length through d3? also control cl; but each of these is two stance, the perennial knight's tour puz- knight moves away from d3, so we get Likewise White's initial move in Dia- zle asks in effect for a Hamiltonian an equivalent parity condition. Alter- gram 3b and the outcome of this path on ~5; a "re-entrant" or "closed" natively, Black might try to reach b4 endgame comes down to the related knight's tour is just a Hamiltonian cir- from d3 in an even number of moves, question concerning the same graph ~5: cuit. The existence of such tours is classical--even Euler spent some time Diagram 3a What are the possible parities of lengths constructing them, finding among oth- of paths on ~ from h8 to cl or c2? ers the elegant centrally symmetric The answers result from the following tour illustrated in Diagram 4 (from [9, ~///////z basic properties of ~5: p. 191]): LEMMA. (i) The graph ~ is connected. Diagram 4 (ii) The graph is bipartite, the two parts comprising the 32 light squares and 32 dark squares of the chess- board. Proof'. Part (i) is just the familiar fact that a knight can get from any square

on the chessboard to any other square. I Y/ Part (ii) amounts to the observation that every knight move connects a light Diagram 3b and a dark square. COROLLARIES. (1) There are no knight cycles of odd length on the chessboard. (2) Two squares of the same color are connected by knight-move paths oj A closed knight's tour constructed by Euler even length but not of odd length; two squares of opposite color are con- The extensive literature on knight's nected by knight-move paths of odd tours includes many examples, which, % length but not by paths of even length. when numbered along the path from 1 We thus answer our chess ques- to 64, yield semi-magic squares (all row tions: White draws both Diagram 1 and and column sums equal 260), some- Diagram 3b by starting with Kcl. More times with further "magic" properties, generally, for any initial position of the but it is not yet known whether a fully Black knight, White chooses between magic knight's tour (one with major di- White to move cl and c2 by moving to the square of agonals as well as rows and columns

24. THE MATHEMATICALINTELLIGENCER summing to 260), either open or closed, set of more than at least three squares Puzzle 2. What happens if m,n are can exist. must include two of the san~e color. Co- both odd, or if m --< 2 or n --< 2? More generally, we may ask for cliques are more interesting: how many Are Diagram 6 and its complement the Hamiltonian circuits on %~,,, for other pairwise nonattacking knights can the only maximal cocliques? Yes, but this m,n; that is, for closed knight's tours chessboard accommodate? ~ We follow is harder to show. One elegant proof, on other rectangular . A Golomb ([21], via M. Gardner [9, p. given by Greenberg in [21], invokes the necessary condition is that ~,,,,, be a 193]). Again the fact that c~ is bipartite existence of a closed knight's tour, connected graph with an even number suggests the answer (Diagram 6): such as Euler's Diagram 4. In general, of vertices. Hence we must have 2Jmn on a circuit of length 2M the only sets and both m,n at least 3 (cf. Puzzle 1). Diagram 6 of M pairwise nonadjacent vertices are But not all %,,n satisfying this condi- the set of even-numbered vertices and tion admit Hamiltonian circuits. For in- the set of odd-numbered ones on the stance, one easily checks that ~J3,4 is circuit. Here M = 32, and the knight's not Hamiltonian. Nor are ~,6 and ~5~,s, tour in effect embeds that circuit into but ~g~,J0 has a Hamiltonian circuit, as (6, so a fortiori there can be at most does G3,,~ for each even n > 10. For in- two cocliques of size M on ~5--and we stance, Diagram 5 shows a closed have already found them both! knight's tour on the 3 • 10 board: Of course this proof applies equally to any board with a closed knight's tour: Diagram 5 on any such board the light- and dark- squared subsets are the only maximal cocliques. Conversely, a board for which there are further maximal cocliques can- 32 mutually nonattacking knights not support a closed knight's tour. For A closed knight's tour on the 3 x 10 board example, any 4 x n board has a nfixed- Diagram 7 color maxnual coclique, as illustrated for n = 11 in Diagram 8. There are sixteen such tours (ignoring the board symmetries). More gener- Diagram 8 ally, enumerating the closed knight's tours on a 3 X (8 + 2n) board yields a sequence 16, 176, 1536, 15424, . . . sat- isfying a constant linear recursion of N degree 21 that was obtained indepen- dently by Donald Knuth and NDE in April 1994. See [23, Sequence A070030]. A third maximal knight coclique on the 4 x In 1997, Brendan McKay first com- 11 board puted that there are 13267364410532 (more than 1.3 x 10 t3) closed knight's This yields possibly the cleanest proof tours on the 8 • 8 board ([19]; see also that there is no closed knight's tour on [23, Sequence A001230], [26]). A one-factor in a 4 • n board for any n. (According We return now from enumeration to to Jelliss [14], this fact was known to existence. After c~,},,~ the next case is It is not hard to see that we cannot do Euler and first proved by C. Flye ~54,,~. This is tricMer: the reader might better: the 64 squares may be parti- Sainte-Marie in 1877; Jelliss attributes try to construct a closed knight's tour tioned into 32 pairs, each related by a the above clean proof to Louis Posa.) on a 4 • 11 board, or to prove that knight move, and then at most one none exists. We answer this question square from each pair can be used (Di- WARNING: the existence of a closed later. agram 7). This is Patenaude's solution knight's tour is a sufficient but not nec- What of maximal cliques and co- in [21]. Such a pairing of c9 is called a essary condition for the existence of cliques on q3? A clique is just a collection "one-factor" in graph theory. Similar only two maximal knight cocliques. It of pairwise defending (or attacking) one-factors exist on all ~3m,,~when 21ran is known that an m x n board supports knights. Clearly there can be no more and m,n both exceed 2; they can be a closed tour if and only if its area inn than two knights, again because ~5 is bi- used to show that in general a knight is an even integer > 24 and neither m partite: two squares of the same color coclique on an m • n board has size at nor n is 1, 2, or 4. In particular, as noted cannot be a knight's move apart, and any most ran~2 for such m,n. above there are no closed knight's

~Burt Hochberg jokes (in [11, p. 5], concerning the analogous problem for queens) that the answer is 64, all White pieces or all Black: pieces of the same color can- not attack each other! Of course this joke, and similar jokes such as crowding several pieces on a single square, are extraneous to our analysis.

VOLUME 25, NUMBER 1 2003 25 tours on the 3 • 6 and 3 x 8 boards, We already noted that ~5, being bi- pawn and giving White time for 4.N x a2 though as it happens on each of these partite, has no cycles of odd length. and a draw. On other Black moves boards the only maximal knight co- (We also encountered the nonexis- from Diagram 10a White resumes con- cliques are the two obvious mono- tence of 3-cycles as "~5 has no cliques trol of a2 with 3.Ncl or 3.Nb4; for in- chromatic ones. of size 3".) Thus the girth (minimal cy- stance 2... Kc2 3.Nb4+ or 2... Kc3 cle length) of ~5 is at least 4. In fact the 3.Ncl Kb2 (else Na2+) 4.Nd3+! etc. More about ~: girth is exactly 4, as shown for instance Note that the White king was not Number, Girth, and the in Diagram 10. needed. Knight Metric Another classic puzzle asks: How many Diagram 10 NOTE TO ADVANCED CHESS PLAYERS: it knights does it take to either occupy or might seem that the knight does need defend every square on the board? In a bit of help after 1.Nb4 ICol!?, when graph theory parlance this asks for the either 2.Na2? or 2.Nd3? loses (in the "domination number" of <6.2 For the latter case to 2... a2), but Black has standard 8 • 8 board, the symmetrical no threat so White can simply make a solution with 12 knights (Diagram 9) random ("waiting") king move. But this has long been known: is not necessary, as White could also draw by thinking (and playing) out of Diagram 9 the a2-b4-d3-cl-a2 box: 1.Nb4 Kbl 2.Nd5! If now 2... a2 then 3.Nc3+ is a new drawing , and otherwise White plays 3.Nb4 and resumes the square dance.

White to move draws Puzzle 5. Construct a position where this Nd5 resource is White's only way Diagram 10a to draw.

WARNING: This puzzle is hard and re- quires considerably more chess back- ground than anything else in this arti- cle. The construction requires some delicacy: it is not enough to simply All unoccupied squares controlled the White king, because then % White can play 2.Na2 with impunity; on Puzzle 3. Prove that this solution is the other hand if the White king is put unique up to reflection. in (so that it has some legal moves, but all of them lose), then the The knight domination number for direct 1... a2 2.Nxa2 Kxa2 wins for chessboards of arbitrary size is not Black. known, not even asymptotically. See Even more important for the prac- [9, Ch. 14] for results known at the time After 2 Nd3! tical chess player is the distance func- for square boards of order up to 15, tion on .~, which encodes the number most dating back to 1918 [1, Vol. 2, p. This square cycle is important to of moves a knight needs to get from 359]. If we ask instead that every endgame theory: a White knight trav- any square to any other. The diame- square, occupied or not, be defended, eling on the cycle can prevent the pro- ter (maximal distance) on ~ is 6, then the 8 • 8 chessboard requires 14 motion of the Black pawn on a3 sup- which is attained only by diagonally knights. On an m • n board, at least ported by its king. To draw this opposite corners. This is to be ex- mn/8 knights are needed, since a position White must either the pected, but shorter distances bring knight defends at most 8 squares. pawn or capture it, even at the cost of some surprises. The table accompa- Puzzle 4. Prove that mn/8 + O(m + the knight. The point is seen after nying Diagram 11 shows the distance n) knights suffice. HINT: Treat the light l.Nb4 Kb3 2.Nd3! (reaching Diagram from each vertex of ~5 to a corner and dark squares separately. 10a) a2 3.Ncl+!, "forking" king and square:

2This terminology is not entirely foreign to the chess literature: A piece is said to be "dominated" when it can move to many squares but will be lost on any of them. (The meaning of "many" in this definition is not precise because domination is an artistic concept, not a mathematical one.) The introduction of this term into the chess lexicon is attributed to Henri Rinck ([12, p. 93], [16, p. 151]). The task of constructing economical domination positions, where a few chessmen cover many squares, has a pronounced combinatorial flavor; the great composer of endgame studies G.M. Kasparyan devoted an entire book to the subject, Domination in 2545 Endgame Studies, Progress Publishers, Moscow, 1980.

26 THE MATHEMATICALINTELLIGENCER Puzzle 6. Determine the knight dis- Diagram 13 5 4 5 4 5 4 5 6 tance from (0,0) to (re,n) on an infinite board as a function of the integers m,n. 4 3 4 3 4 5 4 5 Further Puzzles 3 4 3 4 3 4 5 4 We continue with several more puzzles 2 3 2 3 4 3 4 5 that exploit or extend the above dis- cussion. 3 2 323434 Puzzle 7. How does White play in Di- agran~ 12 to force checkmate as quickly 2 1 432345 as possible against any Black defense? /

3 4* 1 23434 Yes, it's White who wins, despite hav- ing only king and pawn against 15 Black ~3232345 men. Black's men are almost paralyzed, N with only the queen able to move in its The 4! shortest knight paths from dl to d7 Diagram 11 comer prison. White must keep it that way: if he ever moves his king, Black ii. Now suppose we reqmre each will his e2-pawn by promoting knight to be defended exactly once. it, bring the Black army to life and soon What is the largest number of overwhehn White. So White must move knights on the 8 x 8 board satisfy- only the pawn, and the piece that it will ing this constraint, and what are all % the maximal configurations? promote to. That's good enough for a draw, but how to actually win? Puzzle 10. A "" is a (3,1) leaper, that is, an unorthodox that Puzzle 8. (See Diagram 13.) There are moves from (x,y) to one of the squares exactly 24 = 4! paths that a knight on (x_+3, y_+l) or (x-+l, y+-3). (A dl can take to reach d7 in four moves; knight is a (2,1) leaper.) Because there plotting these paths on the chessboard are eight such squares, it takes at least yields a beautiful projection of (the 1- mn/8 camels to defend every square, skeleton of) the 4-dimensional hyper- White loses occupied or not, on an m x n board. cube! Explain. Are mn/8 + O(m + n) sufficient, as in The starred entry is due to the board Puzzle 9. We saw that there is an es- Puzzle 4? edges: a knight can travel from any sentially unique maximal configuration square to any diagonally adjacent square of 32 mutually non-defending knights Synthetic Games in two moves except when one of them on the 8 x 8 board. The remainder of this article is devoted is a comer square. But the other irregu- to composed chess problems featuring i. Suppose we allow each knight to be larities of the table at short distances do knights. not depend on edge effects. Anywhere defended at most once. How many A synthetic game [13] is a chess on the board, it takes the otherwise ag- more knights can the board then ac- game composed (rather than played) ile knight three moves to reach an or- commodate? to achieve some objective, usually in a thogonally adjacent square, and four minimal number of moves. Ideally the moves to travel two squares diagonally. Diagram 12 solution should be unique, but this is This peculiarity must be absorbed by very rare. Failing this, we can hope for any chess player who would learn to an "almost unique" solution, e.g., one play with or against knights. One con- where the final position is unique, but sequence, known to endgame theory, is not the move order. For instance, the shown in Diagram 11, which exploits shortest game ending in checkmate by both the generic irregularity and the spe- a knight is 3.0 moves: 1.e3 Nc6 2.Ne2 cial comer case. Even with White to Nd4 3.g3 Nf3 mate. White can vary the move, this position is a win for Black, order of his moves and can play e4 who will play.., a2 and . . . alQ. One and/or g4 instead of e3 and g3. The might expect that the knight is close Black knight has two paths to f3. The enough to stop this, but in fact it would biggest flaw, however, is that White take it three moves to reach a2 and four could play c3/c4 instead of g3/g4, and to reach al, in each case one too many. | Black could mate at d3. At least all 72 In fact this knight helps Black by block- solutions share the central feature that ing the White king's approach to al! White to play and mate as quickly as possible White incarcerates his king at its home

VOLUME 25, NUMBER 1, 2003 27 square. A better synthetic game in- Diagram 14 earliest of all proof games, while Dia- volving a knight is given in Puzzle 11. gram 16 is considerably more chal- lenging. Diagram 17 features a differ- Puzzle ll. Construct a game of chess ent kind of impostor. Note that it has in which Black White on A two solutions; it is remarkable how Black's fifth move by promoting a each solution has a different impostor. pawn to a knight. The complex and difficult Diagram 18 illustrates the Frolki~l theme: the mul- Proof Games tiple capture of promoted pieces. Dia- A very successful variation of syn- gram 19 shows, in the words of Wilts thetic games that allows unique solu- and Frolkin [28, p. 53], that "the seem- tions are proof games, for which the ingly indisputable fact that a knight length n of the game and the final po- cannot lose a is not quite un- sition P are specified. For the condi- ambiguous." tion (P,n) to be considered a sound problem, there should be a unique Position after Black's 4th move. How did the Puzzle 15. Construct a proof game game in n moves ending in P. (Some- game go? ending with mate by a pawn promoting times there will be more than one so- to a knight without a capture on the lution, but solutions should be related Five other proof game problems in- mating move (6.0). in some thematic way. Here we will volving knights are presented in Puzzles only consider conditions (P,n) that 12 through 16. The minimum known Puzzle 16. Construct a proof game are uniquely realizable, with the ex- number of moves for achieving the game ending with mate by a pawn promoting ception of Diagram 17.) is given in parentheses. (We repeat, the to a knight with no captures by the mat- The earliest proof games were com- game must be uniquely realizable from ing side throughout the game (7.0). posed by the famous "Puzzle King" Sam the number of moves and final position.) Loyd in the 1890s but did not have Diagram 15 unique solutions; the earliest proof Puzzle 12. Construct a proof game . game meeting today's standards seems without any captures that ends with to have been composed by T. R. Daw- mate by a knight (4.5). son in 1913. Although some interesting proof games were composed in subse- Puzzle 13. Construct a proof game quent years, the vast potential of the ending with mate by a knight malting a subject was not suspected until the fan- capture (5.5). tastic pioneering efforts of Michel Cail- laud in the early 1980s. A close to com- Puzzle 14. Construct a proof game plete collection of all proof games ending with mate by a pawn promoting published up to 1991 (around 160 prob- to a knight (5.5). lems) appears in [28]. There is a remarkable variant of Puz- Let us consider some proof games zle 14. Rather than having the game de- related to knights. We mentioned ear- termined by its final position and num- lier that the shortest game ending in After Black's 4th. How did the game go? ber of moves, it is instead completely mate by knight has length 3.0 moves. determined by its last move (including None of the 72 solutions yield proof the move number)! This is the longest Diagram 16 games with unique solutions; i.e., every known game with this property. terminal position has more than one x i way of reaching it in 3.0 moves. It is Puzzle 14'. Construct a game of chess therefore natural to ask for the least with last move 6.gxf8N mate. number n (either an integer or half- integer) for which there exists a The above proof games focus on uniquely realizable game of chess in n achieving some objective in the mini- moves ending with checkmate by mum number of moves. Many other knight, i.e., given the final position, proof games in which knights play a there is a unique game that reaches it key role have been composed, of which in n moves. Such a game was found in- we give a sample of five problems. Di- dependently by the two authors of this agrams 15, 16, and 17 feature "impos- article in 1996 for n = 4.0, which is tors"--some piece(s) are not what they surely the minimum. The final position seem. The first of these (Diagram 15) is shown in Diagram 14. is a classic problem that is one of the After Black's 12th. How did the game go?

28 THE MATHEMATICALINTELLIGENCER Diagram 17 Diagram 20 should be dual-free, which means that Black has at least one method of de- fending which forces each White move ! uniquely if White is to achieve his ob- jective. The objective of checkmate can be combined with other condi- tions, such as White having only one unit besides his king. The ingenious Di- agram 21 shows the current record for a "knight minimal," i.e., White's only unit besides his king is a knight. For other length records, as well as many i N NAN other tasks and records, see [20]. Diagram 21 After White's 13th. How did the game go? Mate in one Two solutions! history of the game. It is only assumed Diagram 18 that the prior play is legal; no assump- tion is made that the play is "sensible." Proof games are a special class of retro problems. We will give only one illus- tration here of a retro problem that is not a proof game. It is based on con- siderations of parity, a common theme ~y///>.- whenever knights are involved. Dia- gram 20 is a mate in one. A chess prob- lem with this stipulation almost invari- ably involves an element of retrograde analysis, such as determining who has Mate in 48 the move. In a problem with the stipu- lation "Mate in n," it is assumed that Paradox white moves first unless it can be The term "paradox" has several mean- After White's 27th move. How did the game proved that Black has the move in or- ings in both mathematics and ordinary go? der for the position to be legal. discourse. We will regard a feature of Diagram 19 a (or chess game) as Length Records paradoxical if it is seemingly opposed X In length records, one tries to con- to common sense. For instance, com- J struct a position that maximizes the mon sense tells us that a material ad- A number of moves that must elapse be- vantage is beneficial in winning a chess fore a certain objective is satisfied. The game or mating quickly. Thus sacrifice most obvious and most-studied objec- in an orthodox chess problem (i.e., a tive is checkmate. In other words, how direct mate or study) is paradoxical. Of large can n be in a problem with the course it is just this paradoxical ele- objective "mate in n" (i.e., White to play ment that explains the appeal of a sac- and checkmate Black in n moves)? rifice. Another common paradoxical Chess problem standards demand that theme is underpromotion. Why not the solution should be unique if at all promote to the strongest possible possible. It is too much to expect, es- piece, namely, the queen? This theme pecially for long-range problems, that is related to that of sacrifice, because After Black's 10th move. How did the game White has a unique response to every in each case the player is forgoing ma- go? Black move for White to achieve his terial. To be sure, underpromotion to objective. In other words, it is possible knight in order to win, draw, or check- Retrograde Analysis for Black to defend poorly and allow mate quickly is not so surprising (and In retrograde analysis problems (called white to achieve his objective in more has even occurred a fair number of retry problems for short), it is neces- than one way, or even to achieve it ear- times in games), since a knight can sary to deduce information from the lier than specified. The correct unique- make moves forbidden to a queen. Tim current position concerning the prior ness condition is that the problem Krabbd thus remarks in [15] that

VOLUME 25, NUMBER 1, 2003 29 knighting hardly counts as a true "un- other knights. Similarly the time-wast- solution to #341] and called the method derpromotion. "3 Nevertheless, knight ing 5.hXg8N 6.Nh6 7.Nxf7 of Diagram of "buttons and strings," is to form a promotions can be used for surprising 19 seems paradoxical--why not save a graph whose vertices are the squares purposes that heighten the paradoxical move by 5.hxgSB and 6.bxf7+? of the board, with an edge between two effect. vertices if the problem piece (here a Helpmate knight) can move from one vertex to Diagram 22 shows four knight sacri- In a helpmate in n moves, Black moves the other. For Diagram 25, the graph is rices, all promoted pawns, with a total first and cooperates with White so that just an eight-cycle (with an irrelevant of five promotions to knight. Diagram White mates Black on White's nth isolated vertex corresponding to the 23 shows a celebrated problem com- move. If the number of solutions of a center square of the board). Diagranl posed by Sam Loyd where a pawn pro- helpmate is not specified, then there 26 is a representation of the problem motes to a knight that threatens no should be a unique solution. For a long that makes it quite easy to see that the pieces or checks and is hopelessly out time it was thought impossible to con- minimum number of moves is sixteen of play. For some interesting com- struct a sound helpmate with the theme (eight by each color), achieved for in- ments by Loyd on this problem, see [27, of Diagram 24, featuring knight promo- stance by cyclically moving each p. 403]. tions. Note that the first obstacle to knight four steps clockwise around the overcome is the avoidance of check- eight-cycle. If a White knight is added Diagram 22 mating White or stalemating Black. The at bl and a Black knight at b3, then composer of this brilliant problem, Ga- somewhat paradoxically the minimum bor Cseh, was tragically killed in an ac- number of moves is reduced to eight! cident in 2001 at the age of 26. A variation of the stipulation of Dia- gram 25 is the problem presented as Diagram 24 Puzzle 17, whose solution is a bit tricky N N|174 and essentially unique. Diagram 25

White to play and win the knights in a minimum number Diagram 23 t of moves Diagram 26 a2 :~ Helpmate in 10

Piece Shuffle In piece shuffles or permutation tasks, a rearrangement of pieces is to be achieved in a minimum number of moves, sometimes subject to special b3 ~L ~-~ '~2 a3 conditions. They may be regarded as special cases of "moving counter prob- lems" such as given in [2, pp. 769-777] The graph corresponding to Diagram 25 or [3, pp. 58-68]. A classic example in- Mate in 3 volving knights, going back to Guarini Puzzle 17. In Diagram 25 exchange the in 1512, is shown in Diagram 25. The knights in a minimum number of move Note that the impostors of Diagrams knights are to exchange places in the sequences, where a "move sequence" is 15-17 may also be regarded as para- mininmm number of moves. (Each an unlimited number of consecutive doxical, because we're trying to reach White knight ends up where a Black moves by the same knight. the position as quickly as possible, and knight begins, and vice versa.) The sys- For some more sophisticated prob- it seems a waste of time to move tematic method for doing such prob- lems similar to Diagram 25, see [10, pp. knights into the original square(s) of lems, first enunciated by Dudeney [3, 114-124]. The most interesting piece

3More paradoxical are underpromotions to rooks and bishops, but we will not be concerned with them here.

30 THE MATHEMATICAL INTELLIGENCER shuffle problems connected with the parity is a knight-move away from ceeds the above lower bound by 2. game of chess (though not focusing on exactly one of the squares with co- 7. (adapted from Gorgiev) To win, knights) are due to G. Foster [5, 6, 7, ordinates (2x,2y) with x ~- y rood White must promote the pawn to a 8], created with the help of his com- 4. Intersecting this lattice with an knight, capture the pawns on b5 puter program WOMBAT (Work Out m x n chessboard yields ran~16 + and c4, and then mate with Nxb3 Matrix By Algorithmic Techniques). O(m + n) knights that cover all when the Black queen is on al. odd squares at distance at least 3 Thus Nxb3 must be an odd-num- Puzzle Answers, Hints, from the nearest edge. Thus an ex- bered move. Therefore 1.h4, 2.h5, and Solutions tra O(m + n) knights defend all 3.h6, 4.h7, 5.hSN does not work be- 1. The graph ~m,~ is connected for the odd squares on the board. The cause all knight paths from h8 to m = n = 1 (only one vertex) and same construction for the even b3 have odd length. Since the not connected for m = n = 3 (the squares yields a total of mn/8 + knight cannot "lose the move," the central square is an isolated ver- O(m + n). pawn must do so on its initial move: tex). With those two exceptions, 1.h3!, followed by 6.hSN!, 7.Nf7, ~m,n is connected if and only if Diagram 27 8.Nd6, 9.Nxb4, 10.Nd6, ll.NXc4. m > 2 and n > 2. Every ~Sm,, is bi- 12.Na5. At this point the Black partite, except ~1,1 (empty parts queen is on a2, having made 11 not allowed); each non-connected moves from the initial position; graph ~,,,,, is bipartite in several whence the conclusion: 12... Qal ways except for q~l,2 = q~2,1. 13.Nxb3 mate. (We omitted from 2. If m= lorn= l then ~Jm,. is dis- Gorgiev's original problem the connected, so the maximal co- initial move 1.Ki2XNel Qa2-al, clique is the set of all mn vertices. @ which only served to give Black his The graph ~J2,. (or ~J,~,2) decom- entire army in the initial position poses into two paths of length and thus maximize the material Ln/2J and two of length F:n/2]. It disparity; and we moved a Black thus has a one-factor if and only if pawn from c5 to b5 to make the so- 4in , and otherwise has cocliques of lution unique, at some cost in size >n; the maximal coclique size strategic interest.) isn + 6where 6 E {0, 1, 2} andn --- White to move draws 8. Recall that a knight's move joins + ~ rood 4. If m and n are odd in- squares differing by one of the tegers greater than 1 then the max- 5. One such position is shown in Di- eight vectors (_+1, _+2) or (_+2, imal coclique size of ~Sm,,, is (ran + agram 27. Once the a-pawn is gone, _+ 1), and check that to get some 1)/2, attained by placing a knight the position is a theoretical draw four of those to add to (0,6) we on each square of the same parity whether Black plays fx g2 + (Black must use the four vectors with a as a comer square of an m • n can do no better than stalemate positive ordinate in some order. board. One can prove that this is against Kxg2, Khl, Kg2 etc.) or f2 Thus, to reach d7 from dl (or, maximal by deleting one of these (ditto after Ke2, Kfl, etc.), or lets more generally, to travel six squares and constructing a one- White play gxf3 and Kg2 and then squares north with no obstruction factor on the remaining mn- 1 jettison the f-pawn to reach the from the edges of the board) in vertices of ~.,,,,. same draw that follows fxg2+. four moves, the knight must move 3. Each of the four 2 x 2 comer sub- But as long as Black's a-pawn is on once in each of its four north-go- boards requires at least three the board, White can move only ing directions. Therefore each path knights, and no single knight may the knight since gxt'3 would liber- corresponds to a permutation of occupy or defend squares in two ate Black's bishop which could the four vectors (_+ 1,2) and (_+ 2,1). different subboards. Hence at least then force White's knight away The number of paths is thus 4! = 4 3 = 12 knights are needed. For (for instance 1.Nb4 Kbl 2.gxf3? 24, and drawing them all yields the three knights to cover the {al, bl, g2+! 3.Kxg2 Bd6 4.Nd5 Kb2) and image of the 4-cube under a pro- a2, b2} subboard, one of them must safely promote the a-pawn. Black's jection taking the unit vectors to be on c3; likewise if3, f6, c6 must pawn on f3 could also be on h3 (-+1,2) and (-+2,1). Instead of dl be occupied if 12 knights are to with the same effect. and d7 we could also draw the 24 suffice. It is now easy to verify that 6. The distance is an integer, congru- paths from a4 to g4 in four moves Diagram 9 and its reflection are the ent to m + n mod 2, that equals or to get the same picture. Not b2 and only ways to place the remaining 8 exceeds each of !mi/2, !nl/2, and f6, though: besides the 24 paths of knights so as to cover the entire (jm~ + inl)/3. It is the smallest such Diagram 13 there are other four- chessboard. integer except in the cases already move journeys, for instance b2-d3- 4. ([3, #319, p. 127]) On an infinite noted of (m,n) = (0,-+1), (-+1,0), f4-h5-f6. chessboard, each square of odd or (-2,_+2), when the distance ex- 9. (i) The maximum is still 32 (though

VOLUME25, NUMBER112003 31 there are many more configura- Diagram 28 Diagram Solutions tions that attain this maximum). Diagram 14. (N. Elkies, R. Stanley, To show this, it is enough to prove 1996) 1.c4 Na6 2.c5 NXc5 3.e3 a6 that at most 8 knights can fit on a 4.Ne2 Nd3 mate. 4 • 4 board if each is to be de- Diagram 15. (G. Schweig, Tukon, 1938) fended at most once. This in turn 1.Nc3 d6 2.Nd5 Nd7 3.Nxe7 Ndf6 can be seen by decomposing ~4.4 4.Nxg8 Nxg8. The impostor is the as a union of four 4-cycles (Dia- knight at g8, which actually started gram 28), and noting that only two out at b8. knights can fit on each 4-cycle. Diagram 29 (ii) Once again, the maximum is Diagram 16. (U. Heinonen, The Prob- 32, this time with a new configu- lemist 1991) 1.c4 Nf6 2.Qa4 Ne4 ration (Diagram 29) unique up to 3.Qc6 Nxd2 4.e4 Nb3 5.Bh6 Na6! reflection! (But note that this con- 6.Nd2 Nb4 7.Rcl Nd5 8.Rc3 Nf6 9.Rf3 figuration has a cyclic group of 4 Ng8 10.Rf6 Nc5 11.f4 Na6 12.Ngf3 symmetries, unlike the elementary NbS. Here both Black knights are im- abelian 2-group of symmetries of postors, as they have exchanged the maximal coclique (Diagram places! For a detailed analysis of this 6).) That this is maximal follows N problem, see [16, pp. 207-209]. from the first part of this puzzle. For uniqueness, our proof is too Diagram 17. (D. Pronkin, Die Schwalbe, long to reproduce here in full; it 1985, 1st prize) 1.b4 Nf6 2.Bb2 Ne4 proceeds as follows. In any 32- 3.Bf6 exf6 4.b5 Qe7 5.b6 Qa3 6.bXa7 knight configuration, each of the Bc5 7.axb8B Ra6 8.Ba7 Rd6 9.Bb6 Kd8 10.Ba5 b6 11.Bc3 Bb7 12.Bb2 four 4 x 4 corner subboards must Solution of Puzzle 8(ii) contain eight knights, two on each Kc8 13.Bcl. of its four 4-cycles. We analyze 1.Nc3 Nf6 2.Nd5 Ne4 3.Nf6+ exf6 cases to show that it is impossible 11. 1.d3 e5 2.Kd2 e4 3.Kc3 eXd3 4.b3 4.b4 Qe7 5.b5 Qa3 6.b6 Bc5 7.bxa7 for two knights in different sub- dXe2 5.Kb2 eXdlN mate. White b6 8.aXbSN Bb7 9.Na6 0-0-0 1O.Nb4 boards to defend each other. We can play d4 instead of d3 (so Black Rde8 ll.Nd5 Re6 12.Nc3 Rd6 13.Nbl. then show that Diagram 29 and its plays e x d4) and can vary his move This problem illustrates the reflection are the only ways to fit order, but the final position is be- theme: a piece leaves its original four eight-knight configurations lieved to be unique. This game first square to be sacrificed somewhere into an 8 x 8 board under this con- appeared in [17]. else, then a pawn promotes to ex- straint. 12. (G. Forshind, Retros Mailing List, actly the same piece which returns 10. Yes, mn/8 + O(m + n)camels suf- June 1996) l.e3 f5 2.Qf3 Kff to the original square to replace the fice. The camel always stays on 3.Bc4+ Kf6 4.Qc6+ Ke5 5.Nf3 sacrificed piece. In the first solution squares of the same color. The mate. the bishop at cl is phoenix, while in squares of one color may be re- 13. (G. Wicklund, Retros Mailing List. the second it is the knight at bl! As garded on a chessboard in its own October 1996) 1.Nf3 e6 2.Ne5 Ne7 if this weren't spectacular enough, right, tilted 45 ~ and magnified by a 3.Nxd7 e5 4.Nxf8 Bd7 5.Ne6 Rf8 Black castles in the second solution factor of ~/2--in other words, 6.Nxg7 mate. but not the first. multiplied by the complex number 14. (P. R6ssler, Problemkiste, August Diagram 18. (M. Caillaud, Th~mes-64, 1 + i. On this board, the camel's 1994 (version)) 1.h4 d5 2.h5 Nd7 1982, 1st prize) 1.a4 c5 2.a5 c4 3.a6 move amounts to the ordinary 3.h6 Ndf6 4.hxg7 Kd7 5.Rh6 Ne8 c3 4.aXb7 a5 5.Ra4 Na6 6.Rc4 a4 7.b4 knight's move since 3 + i = (2 - 6.gxf8N mate. a3 8.Bb2 a2 9.Na3 alN! 10.Nb5 Nb3 i)(1 + i). We can thus adapt our 14'. See solution to Puzzle 14. 11.cXb3 c2 12.Be5 clN! 13.Bb8 solution to Puzzle 4. Explicitly, on 15. (G. Donati, Retros Mailing List, Nd3+ 14.eXd3 e5 15.Qg4 e4 16.Ke2 an infinite chessboard each square June 1996) 1.h4 g6 2.Rh3 g5 3.Re3 e3 17.Kt3 e2 18.Ke4 eXflN! 19.Nf3 with both coordinates odd is a g• 4.1"3 h3 5.Kf2 h2 6.Qel hlN Ng3+ 20.hxg3 h5 21.Rel h4 22.Re2 camel's move away from exactly mate. h3 23.Nel h2 24.Qh3 hlN! 25.g4 one square of the form (4x,8y). 16. (O. Heimo, Retros Mailing List, Ng3+ 26.fxg3 Ne7 27.Nd6 mate. An Thus camels at (4x + a, 8x + b) June 1996) 1.d4 e5 2.d• d5 amazing four promotions by Black to {0,1}) cover the entire board (a,b E 3.Qd4 Be6 4.Qb6 d4 5.Kd2 d3 6.Kc3 knight, all captured! without duplication, and the inter- d2 7.a3 dlN mate. section of this configuration with 17. al-c2, cl-b3-al, c3-a2-cl-b3, a3-bl- Diagram 19. (A. Frolkin, Shortest Proof an m • n board covers all but c3-a2-cl, c2-a3-bl-c3, al-c2-a3, b3- Games, 1991) 1.g4 e5 2.g5 Be7 3.g6 O(m + n) of its squares. cl. Seven move sequences. Bg5 4.gxh7 Qf6 5.hxg8N! Rh4 6.Nh6

3 ~' THE MATHEMATICAL INTELLIGENCER Re4 7.Nxf7 Kxf7 8.Nc3 Kg6 9.Na4 15.aSN and wins, as White can play cealed, and the most difficult key- Kh5 1O.c3 g6. If 5.hXg8B? Rh4 19 Nxf3 mate just after 18... blQ. move that could be selected" [18, p. 6.BXf7+ Kxf7 7.Nc3 Re4 8.Na4 Kg6 For the history of this problem, see 156]. For further problems by Loyd 9.c3 Kh5, then White must disturb his [25, pp. xxi-xxii]. featuring distant knight promotion, position before 10... g6. A knight is see [27, pp. 402-403]. Diagram 23. (S. Loyd, Holyoke Tran- able to "lose a tempo" by taking two script, 1876) 1.bXa8N! Kxg2 2.Nb6, Diagram 24. (G. Cseh, StrateGems, moves to get from g8 to fT, while a followed by 3.a8Q (or B) mate. Note 2000, 1st prize) 1.hlN! Nd6 2.h2 Nf5 bishop must take one or at least that a knight is needed to prevent 2 3.Ng3+ NXg3 4.hlN! Ne2 5.fxe2+ three moves. 9 . . B• A queen or bishop pro- Kg2! (not 5... Kxe2?, since Black's Diagram 20. (V. A. Korolikov, Sehach, motion at move one would be stale- tenth move would then check White) 1957) White's knights are on squares mate, and a rook promotion leads 6.flN! Rc7 7.Bg3 Rxc3 8.Bxe5 RXc2 of the same color and hence have nowhere. Normally a key move of 9.Bg7 Rxcl 10.bxclN! BXg7 mate. made an odd number of moves in all. capturing a piece is considered a se- Four promotions to knight by Black. The White rooks and the White king rious flaw because it reduces Black's have made an even number of moves, strength. Here, however, the capture REFERENCES and White has made one pawn move. seems to accomplish nothing so it is [1] W. Ahrens, Mathematische Unterhaltun- No other White units (i.e., the queen acceptable. Loyd himself says "[ill gen und Spiele, Teubner, Leipzig, 1910 and bishops) have moved. Hence the capture seems a hopeless move (Vol. 1) and 1918 (Vol. 2). White has made an even number of 9 . .then it is obviously well con- [2] E. R. Berlekamp, J. H. Conway, and R. K. moves in all. Similarly, Black has made an odd number of moves. Since White moved first, it is currently Black's move, so Black mates in one with 1... N Xc2 mate.

Diagram 21. (P. O'Shea, The Prom lemist, 1989, 1st prize) 1.Ne5 blN+ (the only defense to 2.Nf3 mate) 2.Ka2 Nd2 3.Kal Nb3+ 4.Kbl Nd2§ 5.Ka2. If Black moves either knight then checkmate is immediate, so 5 . . . a3 is forced. Now White and Black repeat the maneuver Kal, Nb3+, Kbl, Nd2+, Ka2 (any pawn moves by Black would just hasten the end): 8.Ka2 a4 ll.Ka2 a5 14.Ka2 a6. Then 15.Kal Nb3§ 16.Kbl a2+ 17.K• Nd2. This maneuver gets re- peated until all of Black's a-pawns are captured: 44.Kxa2 Nd2 45.Kal Nb3§ 46.Kbl Nd2+ 47.Ka2. Finally Black must allow 48.Nf3 mate or 48.Ng4 mate!

Diagram 22. (H. M. Lommer, Szachy, 1965) White cannot allow Black's rook at e8 to stay on the board, but how does White prevent Black from being stalemated without releasing the sleeping units in the hl comer? 1.fxe8N d3 2.Nf6 (not 2.Nd6? eXd6, and stalemate cannot be prevented without releasing the hl corner) g• f6 (capturing with the other pawn merely hastens the end) 3.g5 f xg5 4.g7 g4 5.g8N g3 6.Nf6 exf6 7.Kg6 f5 8.e7 f4 9.e8N f3 10.Nd6 c• 11.c7 d5 12.c8N d4 13.Nb6 afb6 14.a7 b5

VOLUME 25, NUMBER 1, 2003 33 Guy, Winning Ways, vol. 2, Academic [12] D. Hooper and K. Whyld, The Oxford placed on a chessboard in such a way that Press, London/New York, 1982. Companion to Chess, Oxford, Oxford Uni- no knight attacks any other?"), with solu- [3] H. E. Dudeney, Amusements in Mathe- versity Press, 1984. tions by R. Patenaude and R. Greenberg, matics, Dover, New York, 1958, 1970 [13] G. P. Jelliss, Synthetic Games, Septem- Amer. Math. Monthly 71 no. 2 (Feb. 1964), (reprint of Nelson, 1917). ber 1998, 22 pp. 210-211. [4] N. D. Elkies, On numbers and endgames: [14] G. P. Jelliss, Knight's Tour Notes: Knight's [22] R. J. Nowakowski, ed., Games of No Combinatorial game theory in chess end- Tours of Four-Rank Boards (Note 4a, 30 Chance, MSRI Publ. #29 (proceedings of games, in [22], pp. 135-150. November 2001), http://home.freeuk.net/ the 7/94 MSRI conference on combinato- [5] G. Foster, Sliding-block problems, Part 1, ktn/4a.htm rial games), Cambridge, Cambridge Uni- The Problemist Supplement 49 (Novem- [15] T. Krabb& Chess Curiosities, George versity Press, 1996. ber, 2000), 405-407. Allen & Unwin Ltd., London, 1985. [23] N. J. A. Sloane, The On-Line Encyclope- [6] G. Foster, Sliding-block problems, Part 2, [16] J. Levitt and D. Friedgood, Secrets of Spec- dia of Integer Sequences, on the Web at The Problemist Supplement 51 (March, tacular Chess, Batsford, London, 1995. http://www.research.att.com/-njas/se- 2001 ), 430-432. [17] C. D. Locock, Manchester Weekly Times, quences. [7] G. Foster, Sliding-block problems, Part 3, December 28, 1912. [24] L. Stiller, Multilinear Algebra and Chess The Problemist Supplement 54 (Septem- [18] S. Loyd, Strategy, 1881. Endgames, in [22], pp. 151-192. ber, 2001), 454. [19] B. McKay, Comments on: Martin Loeb- [25] M.A. Sutherland and H. M. Lommer, 1234 [8] G. Foster. Sliding-block problems, Part 4, bing and Ingo Wegener, The Number of Modem End-Game Studies. Dover, New The Problemist Supplement 55 (Novem- Knight's Tours Equals 33,439,123,484, York, 1968. ber, 2001), 463-464. 294--Counting with Binary Decision Dia- [26] G. T6rnberg, "Knight's Tour", http://wl. [9] M. Gardner. Mathematical Magic Show, grams, Electronic J. Combinatorics, http:// 859.telia.com/- u85905224/knight/eknig Vintage Books, New York, 1978. www.combinatorics.org/Volume 3/Com- ht.htm. [10] J. Gik, Schach und Mathematik, MIR, ments/v3ilr5.html. [27] A. C. White, Sam Loyd and His Chess Moscow, and Urania-Verlag, Leipzig/Jena/ [20] J. Morse, Chess Problems: Tasks and Problems, Whitehead and Miller, 1913; Berlin, 1986; translated from the Russian Records, London, Faber and Faber, 1995; reprinted (with corrections) by Dover, New original published in 1983. second ed., 2001. York, 1962. [11] B. Hochberg, Chess Braintwisters, Ster- [21] I. Newman, problem E 1585 ("What is the [28] G. Wilts and A. Frolkin, Shortest Proof ling, New York, 1999. maximum number of knights which can be Games, Gerd Wilts, Karlsruhe, 1991.

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