Puzzle 59: Totally Rooked - The Zotmeister — LiveJournal
solving the puzzle of life one entry at a time
Apr. 6th, 2012
02:03 pm - Puzzle 59: Totally Rooked
Be careful, Michael.
- KITT, Knight Rider
See Puzzle 41 for instructions. This puzzle, made for Mike "projectyl" Sylvia, just serendipitously happens to include something permitting me to make a horrible pun of a reference to an Eighties TV show - which is irresistible, naturally, so there you go. The nightrider is a so-called "fairy chess" piece, something that typically only exists in chess puzzles. ...Oh, hey, look, this is a puzzle based on chess! Well, okay then! It moves as a knight can, but can leap more than once, as long as all leaps are in the same direction (think vectors) and all cells it leaps into on the way are empty (think backgammon - each step the piece takes has to "touch down" onto a free spot). For example, pretend the cell to the immediate right of the '0' on the top row had a nightrider. It would be able to "rook" exactly seven cells - its own, two off to the left, and four downward. Obviously, all upward directions leave the grid; the two remaining rightward directions immediately hit walls after just one knight move; heading downward (and angling left) has no problem hopping OVER walls a couple times, but is stopped before it would LAND on one on the leftmost column; the last direction supports two leaps before the grid would be exited.
As it turns out, the nightrider is a frighteningly suitable piece for Totally Rooked. The way it interacts with rooks, grid layouts, and the wall-sharing rule is alarmingly robust. It proved to be no problem whatsoever making the nightriders of an unknown quantity as well as the rooks. What lies below is a stupendously difficult puzzle that acts upon a lot of those revelations. In fact, for a variant of what is often thought of as Nikoli's easiest puzzle type - bar none - this may well be the most difficult puzzle on my journal to date, and perhaps for the forseeable future, as although this is humanly solvable entirely with deductive logic, it is truly right on the border of infeasibility. Good luck - you'll need it. - ZM
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