Magic Squares

Solution Approach

BFS on Permutations

The state space consists of all permutations of $\{1,\ldots,8\}$, totaling $8! = 40{,}320$ states. This is small enough for breadth-first search, which finds the shortest path.

1. Represent each state as an array of 8 elements.

2. Encode each state as a unique integer for hashing (e.g., treat the digits as a base-10 number).

3. Start BFS from the initial state.

4. At each state, apply all three operations A, B, C to generate neighbors.

5. When the target is reached, reconstruct the path by following parent pointers.

Operation A -- Correctness

Operation A swaps the two rows. In our linear representation, this means: \[ (s_0, s_1, s_2, s_3, s_4, s_5, s_6, s_7) \;\to\; (s_4, s_5, s_6, s_7, s_0, s_1, s_2, s_3). \] Note that this is a direct row swap, not a reversal. The original code had a bug where opA performed $r[i] = s[7-i]$ and $r[7-i] = s[i]$, which reverses each row rather than swapping them. The corrected version below simply exchanges positions $0 \leftrightarrow 4$, $1 \leftrightarrow 5$, $2 \leftrightarrow 6$, $3 \leftrightarrow 7$.

C++ Solution

#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <string>
#include <algorithm>
using namespace std;

typedef long long ll;

struct State {
    int s[8];
};

ll encode(const State& st) {
    ll v = 0;
    for (int i = 0; i < 8; i++)
        v = v * 10 + st.s[i];
    return v;
}

// Operation A: swap top and bottom rows
State opA(const State& st) {
    State r;
    for (int i = 0; i < 4; i++) {
        r.s[i] = st.s[i + 4];
        r.s[i + 4] = st.s[i];
    }
    return r;
}

// Operation B: circular right shift each row
State opB(const State& st) {
    State r;
    r.s[0] = st.s[3]; r.s[1] = st.s[0]; r.s[2] = st.s[1]; r.s[3] = st.s[2];
    r.s[4] = st.s[7]; r.s[5] = st.s[4]; r.s[6] = st.s[5]; r.s[7] = st.s[6];
    return r;
}

// Operation C: clockwise rotation of center 2x2 block
// Center block positions: s[1], s[2] (top), s[5], s[6] (bottom)
// Clockwise: top-left <- bottom-left, top-right <- top-left,
//            bottom-right <- top-right, bottom-left <- bottom-right
// i.e., s[1] <- s[5], s[2] <- s[1], s[6] <- s[2], s[5] <- s[6]
State opC(const State& st) {
    State r = st;
    r.s[1] = st.s[5];
    r.s[2] = st.s[1];
    r.s[6] = st.s[2];
    r.s[5] = st.s[6];
    return r;
}

int main() {
    // Initial state: 1 2 3 4 / 8 7 6 5
    State init;
    init.s[0]=1; init.s[1]=2; init.s[2]=3; init.s[3]=4;
    init.s[4]=8; init.s[5]=7; init.s[6]=6; init.s[7]=5;

    // Read target
    State target;
    for (int i = 0; i < 8; i++)
        scanf("%d", &target.s[i]);

    ll initCode = encode(init);
    ll targetCode = encode(target);

    if (initCode == targetCode) {
        printf("0\n\n");
        return 0;
    }

    // BFS
    map<ll, pair<ll, char>> parent;
    map<ll, bool> visited;
    queue<State> q;
    q.push(init);
    visited[initCode] = true;

    while (!q.empty()) {
        State cur = q.front(); q.pop();
        ll curCode = encode(cur);

        State nexts[3] = {opA(cur), opB(cur), opC(cur)};
        char ops[3] = {'A', 'B', 'C'};

        for (int i = 0; i < 3; i++) {
            ll nc = encode(nexts[i]);
            if (!visited[nc]) {
                visited[nc] = true;
                parent[nc] = {curCode, ops[i]};
                if (nc == targetCode) {
                    // Reconstruct path
                    string path;
                    ll c = targetCode;
                    while (c != initCode) {
                        path += parent[c].second;
                        c = parent[c].first;
                    }
                    reverse(path.begin(), path.end());
                    printf("%d\n%s\n", (int)path.size(), path.c_str());
                    return 0;
                }
                q.push(nexts[i]);
            }
        }
    }

    return 0;
}

Complexity Analysis

• Time complexity: $O(8!)$. We visit at most $8! = 40{,}320$ states, applying 3 operations per state. The map operations add an $O(\log(8!))$ factor, giving $O(8! \cdot \log(8!)) \approx 640{,}000$ operations in total.

• Space complexity: $O(8!)$ for the visited map and parent pointers.