Question

我有一个3x3的数字拼图，如下所示：

3 | 5 | 2
7 | 8 | 9
1 | 6 | 4

成为解决方案：

1 | 2 | 3
4 | 5 | 6
7 | 8 | 9

规则是我只能移动附近的“零件”，直到获得解决方案为止。

我对此的想法是计算偏移量，然后将其运行到“有效的”算法中以获得有效的解决方案。但是，我只能想到使用bruteforce并检查程序为找到最有效的步骤所要做的步骤。

我的意思是：

(2,  0) | (0,  1) | (-1,  0)
(0,  1) | (0,  1) | ( 0,  1)
(0, -2) | (1, -1) | (-2, -1)

在笛卡尔平面中x和y的偏移量。我得到了以下内容，它可以计算偏移量，但是对“花式算法”没有任何想法。

https://ideone.com/0RP83x

是否有一种不使用蛮力就可以使解决方案运动最少的有效方法？

Answer 1

可以将网格视为一个节点（图形的一部分）。

让我们来写网格

abc
def
ghi

作为一行abcdefghi。

您从节点352789164开始，并且您想到达节点123456789。

节点的邻居是您通过应用交换可以到达的所有节点。例如123456789用于邻居

[
  213456789, 132456789,
  123546789, 123465789,
  123456879, 123456798,
  423156789, 153426789,
  126453789, 123756489,
  123486759, 123459786
]

然后您可以通过提供以下内容来应用A *：

d(nodeA, nodeB) = weight(nodeA, nodeB) = 1（所有交换费用均为1）
h(node) =达到解决方案所需的最少交换次数。

要获得最少的h，请考虑计算错位的数字。

如果您有偶数个错位数字，则需要最少“一半”的交换
如果您有奇数个错位数字，则一半+ 1（例如，目标123的312需要进行2次交换）

下面的js示例中，我从wiki复制粘贴的代码

function h (node) {
  const s = ''+node
  let misplaced = 0
  for(let i = 0; i < s.length; ++i) {
    if (parseInt(s[i]) != i+1) {
      misplaced++
    }
  }
  if (misplaced % 2 === 0) {
    return misplaced / 2
  }
  return Math.ceil(misplaced / 2)
}

function d (a, b) {
  return 1
}

const swaps = (_ => {
  const arr = [[1,2],[2,3],[4,5],[5,6],[7,8],[8,9],[1,4],[2,5],[3,6],[4,7],[5,8],[6,9]]
  function makePermFunc([a,b]) {
    a--
    b--
    return function (node) {
      const s = (''+node)
      const da = parseInt(s[a])
      const db = parseInt(s[b])
      const powa = 9 - a - 1
      const powb = 9 - b - 1
      node -= da * 10**powa
      node -= db * 10**powb
      node += da * 10**powb
      node += db * 10**powa
      return node
    }
  }
  const funcs = arr.map(makePermFunc)

  return node => funcs.map(f => f(node))
})();

//https://en.wikipedia.org/wiki/A*_search_algorithm
function reconstruct_path (cameFrom, current) {
  const total_path = [current]
  while(cameFrom.has(current)) {
    current = cameFrom.get(current)
    total_path.unshift(current)
  }
  return total_path
}


// A* finds a path from start to goal.
// h is the heuristic function. h(n) estimates the cost to reach goal from node n.
function A_Star(start, goal, h) {
  // The set of discovered nodes that may need to be (re-)expanded.
  // Initially, only the start node is known.
  const openSet = new Set([start])

  // For node n, cameFrom[n] is the node immediately preceding it on the cheapest path from start to n currently known.
  const cameFrom = new Map()

  // For node n, gScore[n] is the cost of the cheapest path from start to n currently known.
  const gScore = new Map()
  gScore.set(start, 0)

  // For node n, fScore[n] := gScore[n] + h(n).
  const fScore = new Map()
  fScore.set(start, h(start))

  while (openSet.size) {
    //current := the node in openSet having the lowest fScore[] value
    let current
    let bestScore = Number.MAX_SAFE_INTEGER
    for (let node of openSet) {
      const score = fScore.get(node)
      if (score < bestScore) {
        bestScore = score
        current = node
      }
    }
    
    if (current == goal) {
      return reconstruct_path(cameFrom, current)
    }
    openSet.delete(current)
    swaps(current).forEach(neighbor => {
      // d(current,neighbor) is the weight of the edge from current to neighbor
      // tentative_gScore is the distance from start to the neighbor through current
      const tentative_gScore = gScore.get(current) + d(current, neighbor)
      if (!gScore.has(neighbor) || tentative_gScore < gScore.get(neighbor)) {
        // This path to neighbor is better than any previous one. Record it!
        cameFrom.set(neighbor, current)
        gScore.set(neighbor, tentative_gScore)
        fScore.set(neighbor, gScore.get(neighbor) + h(neighbor))
        if (!openSet.has(neighbor)){
          openSet.add(neighbor)
        }
      }
    })
  }
  // Open set is empty but goal was never reached
  return false
}

console.log(A_Star(352789164, 123456789, h).map(x=>(''+x).split(/(...)/).filter(x=>x).join('\n')).join('\n----\n'))
console.log('a more basic one: ', A_Star(123654987, 123456789, h))

解决难题（最佳解决方案）

1 个答案: