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光和尘

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🔖 算法精确覆盖DLX 算法

前言

很早就想要补一下舞蹈链和精确覆盖算法,却一直各种拖延,趁着最近有空,又重新翻开了刘汝佳的书。大学的时候看过几次,甚至照着书里的思路手敲了一遍并通过了例题,但对于算法的原理一直有些不求甚解。很早以前,一位同学告诉我说“现在看不懂的东西不用勉强,以后慢慢就会懂了”,后来也真的在不断印证这句话;但我很担心随着年纪的增长,记忆力和学习能力不断退化之后,恐怕这个 flag 会逐渐倒下。所以趁着眼下尚能理解进去,尽量用自己的语言做一下记录。

精确覆盖问题

有一些由整数 1 \sim n 中的数字组成的集合 S_1, S_2, \cdots, S_m,要求选择若干个集合 S_i,使得 1 \sim n 中每个整数都在选出的集合中的某个出现且恰好仅出现一次。举个栗子:

不妨假设 n=7, m=6,集合为:

\begin{aligned} S_1&=\lbrace 1, 4, 7 \rbrace\\ S_2&=\lbrace 1, 4 \rbrace\\ S_3&=\lbrace 4,5,7 \rbrace\\ S_4&=\lbrace 3, 5, 6 \rbrace\\ S_5&=\lbrace 2, 3, 6, 7 \rbrace\\ S_6&=\lbrace 2, 7 \rbrace\\ \end{aligned}

则一个精确覆盖为 \lbrace S_2, S_4, S_6 \rbrace,因为 \lbrace 1, 4 \rbrace, \lbrace 3, 5, 6 \rbrace, \lbrace 2, 7 \rbrace 无重复、无遗漏地包含了 1 \sim 7 中的所有整数。

我们可以用一个 m \times n01 矩阵来表示集合,其中,0 表示不包含,1 表示包含。比如第 (i, j) 个位置若为 0,则说明 S_i 中不包含 j。上文中的栗子用矩阵表示如下所示:

1234567
S_11001001
S_21001000
S_30001101
S_40010110
S_50110011
S_60100001

则精确覆盖问题可重新表述为:在一个 m \times n01 矩阵中,选择若干行,对这些行做向量加法,得到的结果为 (1, 1, \cdots, 1)。即:

  • 选出的行里,不存在某列同时在两行中值均为 1
  • 所有选出的行叠加在一起的结果覆盖所有列(每一列的值都不为 0)。

算法 X

算法 X(Algorithm X),其实就是回溯,可能是专门针对覆盖问题提出的算法吧。算法描述如下:

每次选择一个没有被删除列,然后枚举该列为 1 的所有行,尝试删除这些行,递归搜索后再恢复这些行。尝试删除行时,还要将该行中所有值为 1 的列也一并删除[1],恢复时也同样。

若没有可以选择的列了(即所有列都被删除了),说明已经找到一个精确覆盖的解了[2];若还有列但是没有行了[3]说明原问题无解。

舞蹈链

舞蹈链(Dancing Link),一种支持快速删除、恢复列和行的数据结构。

舞蹈链是一个十字型双向链表结构,链表中的每个节点对应上述 01 矩阵中的一个 1。另外还有 n+1 个虚拟节点,其中每列最上方一个虚拟节点作为该列链表的头指针,而所有虚拟节点的最前方有一个虚拟节点,作为虚拟节点的头指针,它也是舞蹈链的头指针。如下图所示(图片来源于网络):

dancing-link.png

使用四个数组 L, R, U, D 分别表示舞蹈链中节点的左、右、上、下四个方向的指针,下标为节点的编号;同时列编号作为该列的虚拟节点,0 表示舞蹈链的头指针对应的虚拟节点。

  • 删除行:只需要修改将该行中所有节点的上下行的下、上指针互指就好了。

    dancing-link.01.ts 
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    // 设要删除的某行中某个节点编号为 R[i]
    for (let j = R[i]; j !== i; j = R[j]) {
    U[D[j]] = U[j]
    D[U[j]] = D[j]
    }
  • 恢复行:只需要将该行中所有节点的上下行的下、上指针分别指向该该节点就好了。

    dancing-link.01.ts 
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    // 设要删除的某行中某个节点编号为 R[i]
    for (let j = R[i]; j !== i; j = R[j]) {
    U[D[j]] = j
    D[U[j]] = j
    }
  • 删除列和恢复列也类似,此处略去。

DLX 算法

使用了舞蹈链的算法 X 通常被称为 DLX 算法。此处给出 DLX 算法的 Typescript 实现[4]

dlx.ts  | 240 lines.
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/**
* The algorithm X that applied the dancing-link, it is also called as "DLX".
* It is used to solve the exact-cover problem.
*
* Dancing-link: A cross doubly linked list, each column has a virtual node as
* the head pointer, and at the top of all virtual nodes there is an additional
* virtual node as the head pointer of the virtual node, which is also the head
* pointer of the entire dancing-link. In the implementation of using an array
* to simulate a linked list, the virtual node is represented by a column
* number, and the head pointer of the dancing-link can be represented by 0.
*
* @see https://me.guanghechen.com/post/algorithm/dlx/
*/
export interface DLX {
/**
* Initialize the dancing-link.
* @param totalColumns number of columns
*/
init(totalColumns: number): void
/**
* Release memory variables.
*/
destroy(): void
/**
* Add a row to the dancing-link.
*
* It should be noted that after solving the exact-cover problem, the
* result is a list of selected row numbers, so the row number should be
* specified as a value that can carry information.
*
* @param r the row number
* @param columns columns on the row
*/
addRow(rowNo: number, columns: ReadonlyArray<number>): void
/**
* Try to find a precise coverage.
*
* When a solution is found, return the row numbers of all selected rows,
* otherwise return null.
*/
solve(): number[] | null
}
/**
* Generate an object that encapsulates the DLX algorithm.
*
* @param MAX_N maximum number of nodes in the dancing-link
* @returns
*/
export function createDLX(MAX_N: number): DLX {
// The number of nodes in the dancing-link (including the virtual nodes on
// the column).
let sz: number
// the number of columns in the dancing-link.
let totalColumns: number
const selectedRowNos: number[] = new Array(MAX_N) // list of row numbers of selected rows
let countOfSelectedRows: number // the number of selected rows
const count: number[] = new Array(MAX_N) // lhe number of nodes of a column in the dancing-link
const row: number[] = new Array(MAX_N) // the row number of a node in the dancing-link
const col: number[] = new Array(MAX_N) // the column number of a node in the dancing-link
const L: number[] = new Array(MAX_N) // left pointer of cross-link list
const R: number[] = new Array(MAX_N) // right pointer of cross-link list
const U: number[] = new Array(MAX_N) // up pointer of cross-link list
const D: number[] = new Array(MAX_N) // down pointer of cross-link list
return { init, destroy, addRow, solve }
/**
* @see DLX#init
* @public
*/
function init(_totalColumns: number): void {
totalColumns = _totalColumns
sz = _totalColumns + 1
// Resize arrays.
if (selectedRowNos.length < sz) {
selectedRowNos.length = sz
count.length = sz
row.length = sz
col.length = sz
L.length = sz
R.length = sz
U.length = sz
D.length = sz
}
for (let i = 0; i < sz; ++i) {
L[i] = i - 1
R[i] = i + 1
U[i] = i
D[i] = i
}
R[_totalColumns] = 0
L[0] = _totalColumns
count.fill(0, 0, sz)
}
/**
* @see DLX#destroy
* @public
*/
function destroy(): void {
selectedRowNos.length = 0
count.length = 0
row.length = 0
col.length = 0
L.length = 0
R.length = 0
U.length = 0
D.length = 0
}
/**
* @see DLX#addRow
* @public
*/
function addRow(r: number, columns: ReadonlyArray<number>): void {
const first = sz
for (let i = 0; i < columns.length; ++i, ++sz) {
const c = columns[i]
row[sz] = r
col[sz] = c
count[c] += 1
// Connect left and right nodes
L[sz] = sz - 1
R[sz] = sz + 1
// Connect top and bottom nodes,
// c is the virtual node on the c-th column, and is also the head pointer
// of the linked list of the column, so at this time U[c] is the last
// element of the column
D[sz] = c
D[U[c]] = sz
U[sz] = U[c]
U[c] = sz
}
// Since this is a circular linked list, the first and last columns of the
// current row are connected to each other.
R[sz - 1] = first
L[first] = sz - 1
}
/**
* @see DLX#solve
* @public
*/
function solve(): number[] | null {
if (!algorithmX(0)) return null
return selectedRowNos.slice(0, countOfSelectedRows)
}
/**
* Remove a column from the dancing-link.
* @param c column number
* @private
*/
function removeColumn(c: number): void {
L[R[c]] = L[c]
R[L[c]] = R[c]
for (let i = D[c]; i !== c; i = D[i]) {
for (let j = R[i]; j !== i; j = R[j]) {
U[D[j]] = U[j]
D[U[j]] = D[j]
count[col[j]] -= 1
}
}
}
/**
* Restore a previously deleted column
* @param c column number
* @private
*/
function restoreColumn(c: number): void {
for (let i = U[c]; i !== c; i = U[i]) {
for (let j = L[i]; j !== i; j = L[j]) {
count[col[j]] += 1
U[D[j]] = j
D[U[j]] = j
}
}
L[R[c]] = c
R[L[c]] = c
}
/**
* Algorithm X.
*
* Recursively solve the problem of precise coverage, enumerate which rows are
* selected in the recursive process, remove the selected rows and all the
* columns on the rows, and restore these rows and columns during the
* backtrack.
*
* @param dep recursion depth
* @private
*/
function algorithmX(dep: number): boolean {
// Find a solution when the dancing-link is empty.
if (R[0] === 0) {
// Record the length of the solution.
countOfSelectedRows = dep
return true
}
/**
* Optimization: Find the column with the least number of nodes, and try to
* cover from this column.
*/
let c = R[0]
for (let i = R[0]; i !== 0; i = R[i]) {
if (count[i] < count[c]) c = i
}
// Remove this column.
removeColumn(c)
for (let i = D[c]; i !== c; i = D[i]) {
selectedRowNos[dep] = row[i]
for (let j = R[i]; j !== i; j = R[j]) removeColumn(col[j])
// Recursively processing.
if (algorithmX(dep + 1)) return true
// Backtrack.
for (let j = L[i]; j !== i; j = L[j]) restoreColumn(col[j])
}
// Backtrack.
restoreColumn(c)
return false
}
}

我把它封装在了 @algorithm.ts/dlx 中,你可以通过 npm 包导入它:

 
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import { createDLX } from 'algorithm.ts/dlx'
// 创建一个基于至多有 1000 个节点的舞蹈链的 dlx 算法
const dlx = createDLX(1000)
// 初始化 dlx 算法,总共有 1000 列
dlx.init(1000)
// 添加行,此处为伪代码
dlx.addRow(...)
// 尝试找到一个精确覆盖
dlx.solve()

求解数独问题

数独是精确覆盖问题的一个特例。为了套用 DLX 算法框架,首先需要弄清楚如何构建 01 矩阵。一般来说,可以将列对应成约束,而将行对应到策略,即选择某个策略时能够满足哪些约束。比如考虑经典 x^2 \times x^2 数独游戏,其有如下类型的约束:

  • Slot(a,b): 第 a 行第 b 列格子要有数字;
  • Row(a,b): 第 a 行要有数字 b
  • Col(a,b): 第 a 列要有数字 b
  • Sub(a,b): 第 a 个子方阵中要有方阵 c

一共有 x^2 \times x^2 \times 4 中约束。再考虑可选择的策略,即在第 r 行第 c 列填入数字 v,且可以满足约束 Slot(r,c)Row(r,v)Col(c,v)\displaystyle Sub\left(\left\lfloor \frac{r}{x} \right\rfloor \times x + \left\lfloor \frac{c}{x} \right\rfloor,v\right)

代码如下:

sudoku.ts  | 83 lines.
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import { createDLX } from './dlx'
/**
* @param SUDOKU_SIZE_SQRT 数独子方阵大小(即数独行数的平方根)
*/
export function createSudokuSolver(SUDOKU_SIZE_SQRT: number): SudokuSolver {
const ebs = 1e-6
const SUDOKU_SIZE = SUDOKU_SIZE_SQRT * SUDOKU_SIZE_SQRT
const SUDOKU_SIZE_SQUARE = SUDOKU_SIZE * SUDOKU_SIZE
const MAX_NODES = SUDOKU_SIZE_SQUARE * 4
let codeA: number, codeB: number, codeC: number
const columns: number[] = new Array<number>(4)
const solver = createDLX(MAX_NODES)
return { solve }
function solve(puzzle: number[][]): boolean {
solver.init(MAX_NODES)
for (let r = 0; r < SUDOKU_SIZE; ++r) {
for (let c = 0; c < SUDOKU_SIZE; ++c) {
const w = puzzle[r][c]
// (r,c) 所属的子方阵编号
const s =
Math.floor(r / SUDOKU_SIZE_SQRT + ebs) * SUDOKU_SIZE_SQRT +
Math.floor(c / SUDOKU_SIZE_SQRT + ebs)
for (let v = 0; v < SUDOKU_SIZE; ++v) {
if (w === -1 || w === v) {
columns[0] = encode(SudokuConstraint.SLOT, r, c)
columns[1] = encode(SudokuConstraint.ROW, r, v)
columns[2] = encode(SudokuConstraint.COL, c, v)
columns[3] = encode(SudokuConstraint.SUB, s, v)
solver.addRow(encode(r, c, v), columns)
}
}
}
}
const answer: number[] | null = solver.solve()
if (answer === null) return false
for (const code of answer) {
decode(code)
// eslint-disable-next-line no-param-reassign
puzzle[codeA][codeB] = codeC
}
return true
}
function encode(a: number, b: number, c: number): number {
return a * SUDOKU_SIZE_SQUARE + b * SUDOKU_SIZE + c + 1
}
function decode(code: number): void {
let c = code - 1
codeC = c % SUDOKU_SIZE
c = Math.floor(c / SUDOKU_SIZE + ebs)
codeB = c % SUDOKU_SIZE
c = Math.floor(c / SUDOKU_SIZE + ebs)
codeA = c
}
}
/**
* Sudoku constraints.
*/
export enum SudokuConstraint {
SLOT = 0, // Slot(a, b) 表示第 a 行 b 列个格子上要有数字
ROW = 1, // Row(a, b) 表示第 a 行要有数字 b
COL = 2, // Col(a, b) 表示第 a 列要有数字 b
SUB = 3, // Sub(a, b) 表示第 a 个子方阵要有数字 b
}
export interface SudokuSolver {
/**
* 数独的谜题格子,从 0 开始填,若某个格子未被填,则将其置为 -1
* @param puzzle
*/
solve(puzzle: number[][]): boolean
}

我把求解数独的算法封装在了 @algorithm.ts/sudoku 中,下面是求解 3^2 \times 3^2 数独谜题的示例[5]

sudoku live demo 
  • 刘汝佳《算法竞赛入门经典──训练指南》 P406 6.3.3 精确覆盖问题和 DLX 算法
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