Sudoku box indices start position

问题

I'm implementing a sudoku solver using backtracking. It reads a sudoku board in the form:

027800061000030008910005420500016030000970200070000096700000080006027000030480007

I know how I can figure the elements on a column by doing index % 9 (and then doing a simple arithmetic progression of ratio 9) as well the elements on a line by using index/9 (and then by adding one until I get every one of them), where index is a number in the range [0,80].

What I cannot figure out is how to get the starting index of a box if I have an element's index in that box.

So I googled and I got: http://jakevdp.github.io/blog/2013/04/15/code-golf-in-python-sudoku/

This guy is getting the starting index in a box like this:

start = 27 * int(i / 27) + 3 * int((i % 9) / 3) Where i is the index in my list of elements.

I cannot understand how he figured out that formula and how can I deduce it myself, so please explain that to me.

I understand the list comprehension that comes after this formula, it all makes sense just not this formula.

PS: I write this to learn Haskell, but it really doesn't matter, since now I want to get the gist of that formula.

回答1:

index means index into your list. blockRow, blockCol and blockIndex refer to row/column/index of the block start. All divisions are integer divisions (rounding down to next integer).

index = row*9 + col

row = index / 9
col = index % 9

blockRow = (row / 3) * 3
blockCol = (col / 3) * 3

blockRow = (index / 9 / 3) * 3 = (index / 27) * 3
blockCol = (index % 9 / 3) * 3

blockIndex = (blockRow*9) + blockCol = ((index / 27) * 3 * 9) + (index % 9 / 3) * 3  = 
(index / 27) * 27 + 3 * (index % 9 / 3)

回答2:

Sorry to necro-bump this, but here's a solution someone may find interesting (provided you can read lisps; Clojure in this case). It returns the indices of each "sector" of a sudoku board, and is general enough to be used for many different board sizes. standard-9-sector-indices assumes the board is split into 9 sectors, so the board width/height should be a multiple of 3. get-sector-indices can be used for any board size however:

(defn get-sector-indices [board-width sector-top-left-pos sector-dimensions]
  (let [[sw sh] sector-dimensions
        [tx ty] sector-top-left-pos]
    (for [y (range ty (+ ty sh))
          x (range tx (+ tx sw))]
      (+ x (* y board-width)))))

(defn standard-9-sector-indices [board-dimensions]
  (let [[bw bh] board-dimensions
        [sw sh :as sd] (map #(/ % 3) board-dimensions)]
       (for [y (range 0 bh sh)
             x (range 0 bw sw)]
         (get-sector-indices bw [x y] sd))))

The dimension arguments should be a vector/list representing a [x y] pair.

(standard-9-sector-indices [9 9])

Returns:

((0 1 2 9 10 11 18 19 20)
 (3 4 5 12 13 14 21 22 23)
 (6 7 8 15 16 17 24 25 26)
 (27 28 29 36 37 38 45 46 47)
 (30 31 32 39 40 41 48 49 50)
 (33 34 35 42 43 44 51 52 53)
 (54 55 56 63 64 65 72 73 74)
 (57 58 59 66 67 68 75 76 77)
 (60 61 62 69 70 71 78 79 80))

which are the indices of each sector of a standard Sudoku board (tested).

来源：https://stackoverflow.com/questions/16872773/sudoku-box-indices-start-position

标签

algorithm

language-agnostic

sudoku