我有一个困惑的3x3数字，如下所示：
3 | 5 | 2
7 | 8 | 9
1 | 6 | 4
正在解决：
1 | 2 | 3
4 | 5 | 6
7 | 8 | 9
规则是我只能移动附近的“零件”，直到找到解决方案为止。
我对此的看法是计算偏移量，然后将其运行到“有效的”算法中以寻求有效的解决方案。但是，我只能考虑使用bruteforce并检查程序为找到最有效的步骤而执行的步骤数。
以偏移量表示：
(2, 0) | (0, 1) | (-1, 0)
(0, 1) | (0, 1) | ( 0, 1)
(0, -2) | (1, -1) | (-2, -1)
哪个是笛卡尔平面中x和y的偏移量。我得到了以下内容，该内容的确计算了偏移量，但对“花式算法”没有任何想法。
是否有一种有效的方法可以在不使用蛮力的情况下获得最少的运动？
可以将网格视为节点（图形的一部分）。
让我们写网格
abc
def
ghi
作为单行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有邻居
[
213456789, 132456789,
123546789, 123465789,
123456879, 123456798,
423156789, 153426789,
126453789, 123756489,
123486759, 123459786
]
然后您可以通过提供：来申请A *：
d(nodeA, nodeB) = weight(nodeA, nodeB) = 1（所有掉期成本为1）h(node) =达到解决方案所需的最小交换次数。要获得最小的h，请考虑计算错位的数字。
下面的示例中，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))