Generalized knights

Define an “m-n-knight” as a piece which moves m squares vertically, then n squares horizontally; or n squares vertically, then m squares horizontally; and can jump over all intervening pieces in doing so. m and n are each whole numbers in the range 0 through 7.

Observations:

  • An m-n knight is the same thing as an n-m knight (commutativity).
  • A 2-1-knight is just a conventional chess knight.
  • Any m-n knight has at most 8 possible moves at any given time.
  • An m-n knight may have as few as 0 possible moves, even on an empty board, but how did it get there?

Note 1: A 5-0 knight can move directly from b1 to g1, but unlike a rook, can jump over its own or enemy pieces in doing so.

Note 2: A 0-0 knight moves by simply having the player thump it, but it does not actually move to a different square. This can be useful in positions where the player wishes to lose a tempo (e.g. to gain the opposition or something similar).

Rule Variation 1: When promoting to an m-n knight, the player must state what m and n are. OR:

Rule Variation 2: When promoting to an m-n knight, the player must state what m and n are only when it makes its first move after promotion.

The second variation is interesting, because any threat to promote is also a mate threat. Just promote to an (undesignated) m-n knight, then on the next move, declare m and n to be the number of squares, vertically and horizontally respectively, that the m-n knight must move in order to land on top of the enemy king.

Hmm. Let’s go with Rule Variation 1.


Now define a root-RR knight as a piece that functions as an m-n knight for some m and n where m-squared plus n-squared equals RR. RR can be any whole number from 0 through 98 for which appropriate m and n can be found. (For example, a 7-7-knight is a root-98 knight.)

Note the Pythagorean formula here. A 2-1 knight is a root-5 knight because 2 squared plus 1 squared equals 5. To look at it another way, the distance from the center of the g1 square to the center of the f3 square is the square root of 5 (where 1 represents the distance between the centers of two horizontally or vertically adjacent squares).

Can a root-RR knight be an m-n knight for more than one pair m-n? As it turns out, there are only two possibilities. A root-25 knight can be either a 5-0 knight or a 4-3 knight. (Remember the famous 3-4-5 right triangle from 10th grade geometry?) And a root-50 knight can be either a 7-1 knight or a 5-5 knight. No other such “duals” exist.


Several years ago, one of the players in Helen Warren’s U.S. Masters tournament proposed a puzzle involving root-50 knights. (These could move either 7 and 1, or 5 and 5.) It was a mate in three or something like that. This conversation came up in the skittles room before the tournament started. I had showed up to soak up some master atmosphere. (Obviously, I was not qualified to enter the tournament.)

The following year, I again visited the skittles room of the U.S. Masters, and while milling around, I recognized a familiar voice. “Hey, I remember you, you’re the square root of 50 guy!” “Yes, I am the square root of 50 guy.” And he proceeded to show the same problem again, to anybody would listen.

I never did catch the name of that player. He had an Eastern European accent, and was probably an IM or something like that.

Would anybody out there have an inkling as to who this player might have been?

Bill Smythe

You overlooked the number of squares each such knight could move to.
For instance:
If M and N are both at least four then such knights have only two squares to move to.
If they are both equal then the knights are limited to squares of one color (also true if they are both even or both odd).
If one is zero and the other is at least four then they have only four squares they can move to.

An easy way out of your variation 2 problem is to say that a promoted knight cannot be declared as the type that would attack the square the king was on because that would mean the last move by the opponent was illegal and you cannot gain an advantage by failing to point out illegal moves. It does, however mean that any promotion threat would have a very good chance of being able to take the most powerful enemy piece the following move (not absolute because the promotion might be met by a double check that would require a king move in response instead of a knight move - it does still leave the option of capturing the most powerful enemy piece after the next enemy move that is not a double-check). Since the easy-way-out precludes naming any certain m-n value that could capture a king, how quickly can the the opposing king make the requisite number of moves to force the not-yet-moved knight to be deemed an m=0/n=0 knight? There are 36 possible unique combinations for m and n (since m-n is deemed to be the same as n-m).

hmmm

Type of knight / # of squares it can reach
1-0 / 64
2-0 / 16
3-0 / 9
4-0 / 4
5-0 / 4 (2 if promoted on d/e files)
6-0 / 4 (2 if promoted on c/d/e/f files)
7-0 / 4 (2 if promoted on b/c/d/e/f/g files)

1-1 / 32
2-2 / 8
3-3 / 5 if promoted on a/b/g/h, 4 if promoted on d/e, 3 if promoted on c/f
4-4 / 2
5-5 / 2 (1 if promoted on d/e)
6-6 / 2 (1 if promoted in c/d/e/f)
7-7 / 2 (1 if promoted on b/c/d/e/f/g)

1-3 / probably 32 (three 1-3 moves can equal one 1-1 move, such as a1 → b4 → e3 → b2)
2-4 / probably 16 (a1->c5 ->g3 → c1 is a 2-0 move)
3-5 / possibly 30 (a1 ->d6 → g1 → b4 is a 3-1 move but the center squares are unreachable and I expect a lot of other difficulties)
4-6 / 3 if promoted on a/b/g/h (a1 - e7 and a1 - g5), 2 if promoted on c/d/e/f
5-7 / 3 if promoted on a/h, 2 if promoted on b/c/f/g, 1 if promoted on d/e

1-2 / 64
2-3 / possibly 64 (a1 → c4 → e1 → b3 is a 1-2 move)
3-4 / possibly 64 but I’m guessing less (a1 → d5 → g1 → c4 is a 2-3 move)
4-5 / 3 if promoted on a/h (a1-e6 and a1-f5), 5 if promoted on b/g (b1-f6-a2-e7 and b1-g5), 7 if promoted on c/f (c1-g6-b2-f7-a3-e8 and c1-h5), 6 if promoted on d/e (d1-h6-c2-g7-b3-e8)
5-6 / 3 if promoted on a/h (a1-f7 and a1-g6), 5 if promoted on b/g (b1-g7-a2-f8 and b1-h6), 4 if promoted on c/f (c1-h7-b2-g8) 1 if promoted on d/e (d1-i7 is not available)
6-7 / 3 if promoted on a/h, 2 if promoted on b/g, 1 if promoted on c/d/e/f

That leaves 1-4, 2-5, 3-6, 4-7, 1-5, 2-6, 3-7, 1-6, 2-7, 1-7

fixed 5-6 and 6-7

Until you complete your research, consider the following question:

Which of the various m-n knights is/are the most powerful?

Bill Smythe

The 0-1, 1-2, 1-4, 2-3 (probably) and 3-4 (possibly) knights can reach all 64 squares.
The 0-1 knight can go to either 2 squares (corners), 3 squares (other 24 edge squares) or 4 squares for a total of 220 start-end combinations.
1-2 can go to 2 squares (4 corners), 3 squares (8 squares like b1, a2, etc), 4 squares (other 16 edge squares plus b2, g2, b7, g7), 6 squares (other 16 squares one in from the edge), or 8 squares (16 center squares) for a total of 336 start-end combinations.
2-3 can go 2 squares (4 sets of the 2x2 corner squares = 16), 3 squares (c/f1&2&7&8, a/b/g/h3&6 = 16), 4 squares (d/e1&2&7&8, a/b/g/h4&5, c/f3&6 = 20), 6 squares (c/f4&5, d/e3&6 = 8 ), or 8 squares (4 center squares) for a total of 240 start-end combinations.
3-4 can go 2 squares (4 sets of the 3x3 corner squares = 36), 4 squares (the four center squares - e4 can reach a1, a7, b8, h8), or 3 squares (e1-e3 cannot reach a lower-numbered rank) for a total of 152 start-end combinations.
1-4 has at most four squares at a time (e4 can reach f8, d8, a5, a3) and a total of 204 start-end combinations.
2-5 has at most four squares at a time (c3 can reach a8, c8, h5, h1), is immobile on the four center squares (so they are unreachable) and a total of 144 start-end combinations.

I have serious doubts about a 3-4 knight being able to reach all 64 squares and some mild doubts about a 2-3 knight being able to do so. Even if we grant that they can, the 1-2 knight can reach all of the squares and has the largest number of possible moves on the board. Next would be the 2-3 knight followed by the 1-4 knight, then the 0-1 knight and finally - maybe the 3-4 knight and then the 2-5 knight.