Here is a ‘difficult but not unsolvable’ Sudoku from David Eppstein’s paper Nonrepetitive paths and cycles in graphs with application to Sudoku.

As always I try to solve Sudokus without having to use backtracking (that is, making a guess and working from there on to a solution or a contradiction in which case one uses the other option). Clearly, this is not well defined. When one starts solving Sudokus one often resorts to backtracking but after a while one discovers rules which seem to avoid backtracking (but in a sense are still). For example, if two cells in a same block (or row or column) can only be filled with two numbers one can use this fact by forbidding other numbers to occupy those cells. However, this is a mini-backtracking strategy. Still, I allow all such rules. More precisely, any formal rule is non-backtracking in my dictionary. In Eppstein’s paper there is a good summary of the rules most people apply when starting a Sudoku. He calls them the ‘local rules’. Here they are

• If a digit x has only one remaining cell that it can be placed in, within some row, column, or square, then we place it in that cell. Any potential positions of x incompatible with that cell (because they lie in the same row, column, or square) are removed from future consideration.
• If a cell has only one digit x that can be placed in it, we place x in that cell. Incompatible positions for x are removed from future consideration.
• If some three cells, formed by intersecting a row or column with a square, have three digits whose only remaining positions within that row, column, or square are among those three cells, we prevent all other digits from being placed there. We also remove positions for those three forced digits outside the triple but within the row, column, or square containing it.
• If the cells of a square that can contain a digit x all lie in a single row or column, we eliminate positions for x that are outside the square but inside that row or column. Similarly, if the cells that can contain x within a row or column all lie in a single square, we eliminate positions that are inside that square but outside the row or column.
• If two digits x and y each share the same two cells as the only locations they may be placed within some row, column, or square, then all other digits must avoid those two cells.
  
• If the placement of digit x in cell y can not be extended to a placement of nine copies of x covering each row and column of the grid exactly once, we eliminate cell y from consideration as a placement for x.
• If the placement of a digit x in cell y within a single row, column, or square can not be extended to a complete solution of that row, column, or square, then we eliminate that placement from consideration.

But even if one manages to use all these rules (and frankly I only use a subset) one might get stuck. I don’t know how many cells you can fill in the above problem with these local rules, I’m afraid I only managed … At such moments, the bivalue Sudoku-graph may come in handy. Eppstein defines this as follows

In this graph, we create a vertex for each cell of the Sudoku grid that has not yet been filled in but for which we have restricted the set of digits that can fill it to exactly two digits. We connect two such vertices by an edge when the corresponding two cells both lie in a single row, column, or square, and can both be filled by the same digit; the label of the edge is the digit they can both be filled by.

Eppstein then goes on to define new rules (each of which is a mini-backtracking strategy) which often help to crack the puzzle. Here are Eppstein’s ‘global rules’

• If an edge in the bivalue graph belongs to a nonrepetitive cycle, the digit labeling it must be placed at one of its two endpoints, and can be ruled out as a potential value for any other cell in the row, column, or square containing the edge.
• If the bivalue graph has a cycle in which a single pair of consecutive edges has a repeated label, that label can not be placed at the cell shared by the two edges, so that cell must be filled by the other of its two possible values.
• If the bivalue graph contains two paths, both starting with the same label from the same cell, both ending at cells in the same row, column, or square, and such that in the two ending squares the values not occurring on the incident edge labels are equal, then the cell at the start of the paths can not be filled by the start label of the paths, and must be filled by the other of its two possible values.

For example, in the above problem it is not hard to verify that the indicated places X,Y and Z form a nonrepetitive cycle in the bivalue graph so applying the first global rule we have two choices of filling these places (one leading to a solution, the other to a contradiction)

In fact, it turns out that making this choice is enough to solve the puzzle by simple local rules. So, if I change the original puzzle by filling in the cell X

you will have no problem solving the puzzle.

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Artin, arxiv, puzzle, sudoku

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Posted in games
Written on Tue, 27 December 2005 at 12:11 pm
Tags: Artin, arxiv, puzzle, sudoku
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