我有一个 3x3 数字的谜题如下：
3 | 5 | 2
7 | 8 | 9
1 | 6 | 4
作为解决方案：
1 | 2 | 3
4 | 5 | 6
7 | 8 | 9
规则是我只能移动附近的"碎片"，直到我得到解决方案。
我对此的看法是计算偏移量，然后将其运行到"花哨"算法中以获得有效的解决方案。但是，我只能想到使用蛮力并检查程序为找到最有效的步骤所做的步骤数量。
在偏移中，我的意思是：
(2, 0) | (0, 1) | (-1, 0)
(0, 1) | (0, 1) | ( 0, 1)
(0, -2) | (1, -1) | (-2, -1)
哪些是笛卡尔计划中 x 和 y 的偏移量。我得到了以下内容，它确实计算了偏移量，但没有关于"花哨算法"的想法。
https://ideone.com/0RP83x
有没有一种有效的方法可以在不使用蛮力的情况下获得解决方案的最小动作？
可以将网格视为节点(图形的一部分(。
让我们编写网格
abc
def
ghi
作为单行abcdefghi.
您从节点352789164开始，想要到达节点123456789。
节点的邻居是您可以通过应用交换到达的所有节点。 例如123456789对邻居有
[
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申请 A*：
d(nodeA, nodeB) = weight(nodeA, nodeB) = 1(所有掉期费用为1(h(node)= 获得解决方案所需的最少交换次数。
要获得最小的h，请考虑计算错位的数字。
- 如果您有偶数个错位的数字，则至少需要"一半"的交换
- 如果您有奇数个错位的数字，那么一半 + 1(例如，目标 123 的 312 需要 2 次交换(
下面的js示例，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))