1. Abridged Problem Statement
Given an undirected graph G with v vertices and e edges (v≤100), decide if G can arise from the following 3-SAT→clique reduction:
- Start with a 3-CNF formula of n clauses (3 literals per clause).
- Create 3n vertices, one per literal tji in clause i.
- For any two vertices from different clauses, add an edge if and only if the two literals are not contradictory (i.e. they are either the same sign of the same variable or they mention different variables).
- Finally, vertices are arbitrarily permuted.

Output "YES" if the given G matches a graph produced by this construction; otherwise "NO."

2. Detailed Editorial

Overview
We must detect whether the input graph G=(V,E) encodes some 3-CNF in the special reduction. Key facts of the reduction:
- |V| must be a multiple of 3 (there are exactly 3 literals per clause).
- Vertices are partitioned into n=|V|/3 independent triples (no edges inside each triple)—those represent the three literals of each clause.
- Between different triples (clauses), edges connect exactly those pairs of literals that are non-contradictory. Thus missing cross-clause edges signal a contradiction pair: same variable, opposite signs.
- Finally, an assignment of boolean values to variables corresponds to choosing one literal per clause forming an n-clique; its existence is not required by our test, only the structural pattern.

Step 1: Check |V|≡0 mod 3. If not, reject.
Step 2: Find all size-3 independent sets in G. For every triple {i,j,k} with no internal edges, record it as a candidate clause. We then require each vertex to belong to exactly one such triple. If any vertex lies in zero or multiple triples, output NO. This yields a unique partition of V into n clause-triples.

Step 3: "Complete" each triple by inserting the three missing edges among its vertices. After this augmentation, any remaining non-edges must lie strictly between different triples, each signaling a contradiction pair.

Step 4: Model literals and variable identities via 2-SAT style DSU over 2n nodes. Number your vertices 0…v−1. Build a DSU on 2v elements: for each literal i we interpret "i+n" as the negation of i. For every non-edge (u,v) in the augmented graph (u<v), we know u and v come from different clauses and must contradict. Thus u⇒¬v and v⇒¬u, which in DSU means unify(u, v+n) and unify(u+n, v).

Step 5: Check consistency of negations. If for any i, find(i)==find(i+n), there is a variable forced equal to its negation—reject.

Step 6: Validate that every contradiction class is complete: in the reduction, each pair of contradictory literals (across all clauses) are exactly the missing edges between the two opposite-sign classes. Concretely, for each vertex i, let cnt_a be the number of literals in the DSU-component of i, and cnt_b those in the component of i+n. Compute the sum of non-edge degrees over those vertices. All non-edges must go between the two components, and must form a complete bipartite graph K|cnt_a|,|cnt_b|. This requires sum_deg == 2·cnt_a·cnt_b.
If this holds for every i, accept ("YES"); else reject.

3. C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/graph/dsu.hpp>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

class DSU {
  public:
    int n;
    vector<int> par;
    vector<int> sz;

    DSU(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        par.assign(n + 1, 0);
        sz.assign(n + 1, 0);
        for(int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
    }

    int root(int u) { return par[u] = ((u == par[u]) ? u : root(par[u])); }
    bool connected(int x, int y) { return root(x) == root(y); }

    int unite(int x, int y) {
        x = root(x), y = root(y);
        if(x == y) {
            return x;
        }
        if(sz[x] > sz[y]) {
            swap(x, y);
        }
        par[x] = y;
        sz[y] += sz[x];
        return y;
    }

    vector<vector<int>> components() {
        vector<vector<int>> comp(n + 1);
        for(int i = 0; i <= n; i++) {
            comp[root(i)].push_back(i);
        }
        return comp;
    }
};

int n, m;
vector<vector<int>> adj;

void read() {
    cin >> n >> m;
    adj.assign(n, vector<int>(n, 0));
    for(int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        adj[u][v] = 1;
        adj[v][u] = 1;
    }
}

void solve() {
    // Decide whether the graph is the 3-SAT -> clique reduction graph. In
    // that reduction the 3n vertices split into n clauses of 3 vertices
    // each; vertices inside a clause are pairwise non-adjacent (a triple
    // anti-clique), and two vertices in different clauses are non-adjacent
    // exactly when they are contradictory literals of the same variable.
    //
    // First n must be divisible by 3. Then every vertex must belong to
    // exactly one triangle of mutual non-edges (its clause); we enumerate
    // all such anti-clique triples and require each vertex to appear in
    // precisely one. We add intra-clause edges so the remaining non-edges
    // are only the cross-clause contradictions.
    //
    // Those contradictions must pair literals as x / not-x: build a DSU on
    // 2n nodes (vertex i and its "negation" i+n). A non-edge (i, j) forces
    // i ~ j+n and i+n ~ j. The relation is consistent only if no i is in
    // the same class as i+n.
    //
    // Finally each variable class must contradict every literal of all the
    // other clauses exactly: for class of vertex i with cnt_a positive and
    // cnt_b negative occurrences, the number of cross non-edges it carries
    // (sum of complement-degrees) must equal 2 * cnt_a * cnt_b. If all
    // checks pass, print YES, else NO.

    if(n % 3 != 0) {
        cout << "NO\n";
        return;
    }

    vector<vector<tuple<int, int, int>>> three_anti_cliques_per_node(n);
    vector<int> clique_id(n, -1);
    int cnt_clique_id = 0;

    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            if(adj[i][j] == 0) {
                for(int k = j + 1; k < n; k++) {
                    if(adj[i][k] == 0 && adj[j][k] == 0) {
                        three_anti_cliques_per_node[i].emplace_back(j, k, i);
                        three_anti_cliques_per_node[j].emplace_back(i, k, j);
                        three_anti_cliques_per_node[k].emplace_back(i, j, k);
                        clique_id[i] = clique_id[j] = clique_id[k] =
                            cnt_clique_id++;
                    }
                }
            }
        }
    }

    for(int i = 0; i < n; i++) {
        if(three_anti_cliques_per_node[i].size() != 1) {
            cout << "NO\n";
            return;
        }
    }

    for(int i = 0; i < n; i++) {
        auto [j, k, l] = three_anti_cliques_per_node[i][0];
        adj[i][j] = 1;
        adj[j][i] = 1;
        adj[i][k] = 1;
        adj[k][i] = 1;
        adj[i][l] = 1;
        adj[l][i] = 1;
    }

    DSU dsu(2 * n);
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            if(adj[i][j] == 0) {
                dsu.unite(i, j + n);
                dsu.unite(i + n, j);
            }
        }
    }

    for(int i = 0; i < n; i++) {
        if(dsu.connected(i, i + n)) {
            cout << "NO\n";
            return;
        }
    }

    vector<int> deg(n, 0);
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(adj[i][j] == 0) {
                deg[i]++;
            }
        }
    }

    for(int i = 0; i < n; i++) {
        int cnt_a = 0, cnt_b = 0, sum = 0;
        for(int v = 0; v < n; v++) {
            if(dsu.connected(i, v)) {
                cnt_a++;
                sum += deg[v];
            }
        }

        for(int v = 0; v < n; v++) {
            if(dsu.connected(i, v + n)) {
                cnt_b++;
                sum += deg[v];
            }
        }

        if(cnt_a * cnt_b * 2 != sum) {
            cout << "NO\n";
            return;
        }
    }

    cout << "YES\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Comments

```python
import sys
sys.setrecursionlimit(10**7)

class DSU:
    def __init__(self, n):
        self.p = list(range(n))
        self.sz = [1]*n
    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]
    def union(self, a, b):
        a = self.find(a); b = self.find(b)
        if a == b: return
        if self.sz[a] > self.sz[b]:
            a, b = b, a
        self.p[a] = b
        self.sz[b] += self.sz[a]
    def same(self, a, b):
        return self.find(a) == self.find(b)

def main():
    data = sys.stdin.read().split()
    n, m = map(int, data[:2])
    edges = data[2:]
    # Build adjacency matrix
    adj = [[False]*n for _ in range(n)]
    ptr = 0
    for _ in range(m):
        u = int(edges[ptr]) - 1
        v = int(edges[ptr+1]) - 1
        ptr += 2
        adj[u][v] = adj[v][u] = True

    # Must have 3 literals per clause
    if n % 3 != 0:
        print("NO"); return

    # Find all independent triples
    triples_at = [[] for _ in range(n)]
    for i in range(n):
        for j in range(i+1, n):
            if not adj[i][j]:
                for k in range(j+1, n):
                    if not adj[i][k] and not adj[j][k]:
                        # record for i,j,k
                        triple = (i,j,k)
                        triples_at[i].append(triple)
                        triples_at[j].append(triple)
                        triples_at[k].append(triple)

    # Each vertex must appear in exactly one triple
    for lst in triples_at:
        if len(lst) != 1:
            print("NO"); return

    # Complete each triple by adding its 3 missing edges
    for i in range(n):
        i,j,k = triples_at[i][0]
        for a,b in [(i,j),(i,k),(j,k)]:
            adj[a][b] = adj[b][a] = True

    # Build 2-SAT style DSU on 2n nodes
    dsu = DSU(2*n)
    for u in range(n):
        for v in range(u+1, n):
            if not adj[u][v]:
                # u contradicts v => u=>¬v, v=>¬u
                dsu.union(u, v + n)
                dsu.union(u + n, v)

    # Check no literal is unified with its negation
    for i in range(n):
        if dsu.same(i, i+n):
            print("NO"); return

    # deg[x] = count of non-edges incident to x
    deg = [0]*n
    for i in range(n):
        deg[i] = adj[i].count(False)

    seen = [False]*(2*n)
    # For each variable (component pair), verify cross bipartiteness
    for i in range(n):
        ri, rni = dsu.find(i), dsu.find(i+n)
        if seen[ri]:
            continue
        seen[ri] = seen[rni] = True

        cntA = cntB = sumDeg = 0
        for x in range(n):
            rx = dsu.find(x)
            if rx == ri:
                cntA += 1
                sumDeg += deg[x]
            elif rx == rni:
                cntB += 1
                sumDeg += deg[x]
        # Missing edges should form a complete bipartite graph
        if sumDeg != 2*cntA*cntB:
            print("NO"); return

    print("YES")

if __name__ == "__main__":
    main()
```

5. Compressed Editorial
- Verify v%3==0.
- Enumerate all independent triples; demand each vertex belongs to exactly one → unique 3-partition.
- Fill in missing edges inside each triple (turn each into a clique).
- On the augmented graph, every remaining non-edge is across clauses and marks a contradiction.
- Build a 2n-element DSU (literal vs. negation). For each non-edge (u,v), `union(u, v+n)` and `union(u+n, v)`.
- Reject if any i is unified with i+n.
- Compute deg[i]=#non-edges at i; for each vertex i, count cnt_a (literals in i's component) and cnt_b (literals in negation component), and verify sum deg = 2·cnt_a·cnt_b, ensuring a complete bipartite pattern of contradictions.
- If all checks pass, answer YES; otherwise NO.
