Distinct GCDs (Easy Version)

Let $m=n-1$. We need $m$ adjacent GCDs, all distinct, while using as few distinct array values as possible.

The clean way in is to stop staring at the numbers and look at the pattern of equal values.

Graph Model

Suppose the array uses exactly $k$ distinct values. Make one vertex for each distinct value.

For every adjacent pair $(a_i,a_{i+1})$:

• if they are different values, add the edge between those two vertices;
• if they are equal, add a self-loop on that vertex.

If the same unordered pair of values appeared twice, the GCD would be the same both times. So a valid array gives a walk that never reuses an unordered pair. In graph words, it is a trail in a graph with:

• at most one edge between every pair of different vertices;
• at most one loop on each vertex.

So the best possible sequence using $k$ values is exactly the longest possible trail in the complete graph with loops.

Call this maximum number of edges $g(k)$.

Computing $g(k)$

Start with the full graph on $k$ vertices: all ordinary edges plus one loop on every vertex. It has

$$\binom{k}{2}+k=\frac{k(k+1)}2$$

edges.

A loop contributes $2$ to degree, so every vertex has degree

$$(k-1)+2=k+1.$$

A connected graph has an Euler trail using all its edges iff it has either $0$ or $2$ odd-degree vertices.

If $k$ is odd, then $k+1$ is even. Every vertex is even, so the full graph has an Euler circuit. Therefore

$$g(k)=\frac{k(k+1)}2.$$

If $k$ is even, then every vertex is odd in the full graph. That is too many odd vertices; Euler says no. To get an Euler trail, we need at most two odd vertices.

Deleting an ordinary edge flips the parity of its two endpoints. Deleting a loop flips nothing, so deleting loops is just throwing away edges like an idiot. We need to flip at least $k-2$ vertices, so we must delete at least

$$\frac{k-2}{2}$$

ordinary edges.

This is achievable: delete a matching of size $\frac{k-2}{2}$, for example $(0,1),(2,3),\ldots$. Then exactly two vertices remain odd, the graph is still connected, and it has an Euler trail. Hence

$$g(k)=\frac{k(k+1)}2-\frac{k-2}{2}$$

for even $k$.

So the minimum possible number of distinct values is the smallest $k$ such that

$$g(k)\ge n-1.$$

That proves the lower bound. Now we need actually build the damn thing.

Building The Value Pattern

For that minimum $k$, construct the graph described above:

• if $k$ is odd, use the full looped complete graph;
• if $k$ is even, delete the matching $(0,1),(2,3),\ldots,(k-4,k-3)$.

This graph has an Euler trail with exactly $g(k)$ edges. Run Hierholzer's algorithm and get a vertex sequence of length $g(k)+1$. Since $g(k)\ge n-1$, take the first $n$ vertices.

Every adjacent pair in this prefix is a different graph edge, so no unordered pair repeats.

Making GCDs Unique

Now the number theory part: assign a real integer to each vertex so that every unordered pair of vertices has a different GCD.

Use a $10\times 10$ grid. For $0\le i,j<10$, define

$$X_{i,j}=2^{i+j}\cdot 3^i\cdot 5^j\cdot 7^{9-i}\cdot 11^{9-j}.$$

There are $100$ values. In this easy version $n\le 700$, and the needed $k$ is at most $37$, so this is plenty.

These values fit easily:

$$X_{i,j}=7^9 11^9\left(\frac67\right)^i\left(\frac{10}{11}\right)^j\le 7^9 11^9<10^{18}.$$

Now prove the important property. Take two grid points $(i_1,j_1)$ and $(i_2,j_2)$, and let

$$G=\gcd(X_{i_1,j_1},X_{i_2,j_2}).$$

GCD takes minimum prime exponents, so

$$v_3(G)=\min(i_1,i_2),\qquad v_7(G)=9-\max(i_1,i_2).$$

From those two exponents, we recover the unordered pair $\{i_1,i_2\}$.

Similarly,

$$v_5(G)=\min(j_1,j_2),\qquad v_{11}(G)=9-\max(j_1,j_2),$$

so we recover $\{j_1,j_2\}$.

There is only one possible ambiguity: maybe the pair is

$$(i_{lo},j_{lo}), (i_{hi},j_{hi})$$

or maybe it is crossed:

$$(i_{lo},j_{hi}), (i_{hi},j_{lo}).$$

The exponent of $2$ kills that ambiguity:

$$v_2(G)=\min(i_1+j_1,i_2+j_2).$$

For the parallel pairing, this minimum is $i_{lo}+j_{lo}$. For the crossed pairing, if both coordinates actually differ, the minimum is strictly larger. If one coordinate does not differ, then there was no real ambiguity anyway.

So the GCD uniquely determines the unordered pair of grid points. That includes self-pairs, so loops are covered too.

Putting It Together

For each test case:

1. Find the smallest $k$ with $g(k)\ge n-1$.
2. Build the Euler trail on the corresponding $k$-vertex graph.
3. Output the first $n$ grid-values indexed by that trail.

The trail guarantees no unordered pair repeats. The grid construction guarantees different unordered pairs have different GCDs. The definition of $k$ proves the number of distinct elements is minimum.

Complexity is tiny: the graph has at most about $700$ edges for this version, so precomputation is basically free.