[THOUGHT] We need to interpret the problem carefully. We have a regular cube (a cube graph with 12 edges). There are 12 different colors available, each edge colored with exactly one color. The condition: any two edges that share a vertex must receive different colors. That means edges incident to the same vertex must have distinct colors. Since each vertex has degree 3, at each vertex the three edges meeting there must be three different colors.

We are asked: "Express the probability that all 12 edges are colored with pairwise distinct colors as a reduced fraction q/p and calculate p+q." Wait, but there are exactly 12 edges and 12 different colors. "All 12 edges are colored with pairwise distinct colors" means each edge gets a distinct color, i.e., a bijection from edges to colors. But we already have 12 different colors, and each edge gets exactly one color, so "pairwise distinct colors" means we use all 12 colors exactly once. That's equivalent to a permutation assignment of colors to edges.

But there is also the condition that edges sharing a vertex must have different colors. If we assign colors bijectively, then automatically edges sharing a vertex will have different colors because all colors are distinct. So the condition "any two edges that share a vertex must receive different colors" is automatically satisfied if we assign distinct colors to all edges. However, the problem likely means: We randomly color each edge with one of the 12 colors (each edge independently? with equal probability? but there are 12 different colors, so each edge gets a distinct color? That's contradictory: if we randomly assign colors from 12 colors to 12 edges, the total number of assignments is 12! (if we assign colors without repetition). But if we allow repetitions, then total assignments would be 12^12. But the phrase "12 different colors" suggests we have a palette of 12 distinct colors, but edges could receive the same color? The condition says "any two edges that share a vertex must receive different colors." That's a constraint. And we want probability that all 12 edges are colored with pairwise distinct colors (i.e., all colors used exactly once). So we need to compute: (# of colorings satisfying the vertex condition AND using all colors exactly once) / (# of colorings satisfying the vertex condition). But the denominator might be total colorings satisfying the vertex condition (with possible repetitions). However, the problem statement: "A regular cube has its 12 edges colored with 12 different colors. Each edge is colored with exactly one color, and any two edges that share a vertex must receive different colors. Express the probability that all 12 edges are colored with pairwise distinct colors as a reduced fraction q/p and calculate p+q."

Interpretation: We have 12 colors available. We assign each edge a color uniformly at random from the 12 colors, independently? But with the constraint that edges sharing a vertex must have different colors. That constraint must be satisfied for the random assignment to be valid. So we are considering uniform distribution over all colorings that satisfy the vertex condition. Among those, we want the probability that all edges have distinct colors (i.e., a permutation). So we need to count:

Let S = set of colorings of the 12 edges with colors from {1,...,12} such that at each vertex, the three incident edges have distinct colors. Note: edges not sharing a vertex may have same color.

We want |T| / |S| where T = subset of S where all 12 colors are distinct (bijection).

Then probability = |T| / |S| = q/p reduced fraction. Compute p+q.

We need to compute |S| and |T|.

First, compute total colorings without constraints: 12^12. But we have constraints at each vertex. The cube has 8 vertices, each vertex imposes that its 3 incident edges have distinct colors. However, constraints are overlapping because edges are shared among vertices.

We can think of edge coloring of cube graph with 12 colors (colors can repeat) such that at each vertex, incident edges have distinct colors. That's equivalent to a proper edge-coloring of the cube graph with at most 12 colors (but we allow fewer colors). Actually, proper edge-coloring means edges sharing a vertex have different colors. That's exactly our condition. So S is the set of proper edge-colorings of the cube graph using colors from {1,...,12}. We want number of such colorings.

The cube graph is bipartite (since cube is bipartite). Edge chromatic number of a bipartite graph equals maximum degree Δ. Here Δ=3. So it's class 1 graph: 3 colors suffice for proper edge-coloring. But we have 12 colors available, so we can use more colors.

Counting proper edge-colorings with exactly 12 colors available (colors can repeat) is not trivial. We can use inclusion-exclusion or polynomial method.

Alternatively, we can compute |S| by considering assignments edge by edge with constraints. But constraints are overlapping.

Maybe we can compute |S| by using the fact that we can first color the edges incident to each vertex independently? Not independent.

Better approach: Use the fact that cube graph is 3-regular bipartite. There is known formula for number of proper edge-colorings with k colors? For bipartite graphs, number of proper edge-colorings with k colors equals number of proper vertex colorings of line graph? Not simple.

We can brute-force compute |S| by enumerating all 12^12 assignments? That's huge (12^12 ≈ 8.9e12). Too large.

We need smarter counting.

We can think of coloring edges such that at each vertex, colors are distinct. Equivalent to assigning each edge a color such that for each vertex, the three edges get three distinct colors. Since each edge is incident to two vertices, the color of an edge must be distinct from colors of other edges incident to those two vertices.

We can try to count using inclusion-exclusion on violating vertices? Not straightforward.

Maybe we can use the fact that cube graph can be decomposed into 3 perfect matchings (since it's 3-regular bipartite and class 1). Indeed, cube graph has a 3-edge-coloring (like RGB). So we can color edges with 3 colors such that each vertex sees three colors. That's a proper edge-coloring with 3 colors. For k colors, we can think of assigning each edge a color from {1,...,k} such that at each vertex, colors are distinct. This is equivalent to a homomorphism from cube graph to K_k (complete graph) as edge-coloring? Actually edge-coloring corresponds to graph homomorphism from line graph to K_k.

But maybe we can compute |S| using polynomial: For each vertex, we require that the three incident edges have distinct colors. For a given assignment of colors to edges, each vertex imposes that the three colors are distinct. So total number of colorings satisfying all vertex constraints = sum over all colorings of indicator that each vertex's edges are distinct. We can use inclusion-exclusion over vertices: For each subset of vertices, count colorings where edges incident to those vertices are allowed to have repeats (i.e., no restriction). Actually inclusion-exclusion: Let V be set of 8 vertices. For each subset A of vertices, we count colorings where for each vertex in A, the incident edges have distinct colors? That's not standard.

Better: Let's define for each vertex v, the bad event B_v that two of its incident edges have same color. We want number of colorings with no bad events. Use inclusion-exclusion over bad events. But events are not independent.

We can try to compute using Burnside's lemma? Not.

Given complexity, maybe the problem expects us to compute probability that all edges distinct given random proper edge-coloring? Wait, maybe the intended interpretation: We randomly assign distinct colors to edges (permutation uniformly). Condition: edges sharing a vertex must have different colors. But if we assign permutation, condition automatically satisfied. Then probability = 1? That seems trivial. So that's not it.

Maybe: We have 12 different colors, we randomly color each edge with one of the 12 colors (independent, uniform). Then we condition on the event that the coloring satisfies the vertex condition. Among those that satisfy, what's the probability that all edges are distinct? That's what we thought.

But counting |S| might be doable via dynamic programming on cube graph. Cube graph is small: 12 edges, 8 vertices. We can brute-force over all 12^12 assignments? That's 8.9e12, too many. But we can use symmetry and reduce.

We can use Python to enumerate all proper edge-colorings with 12 colors? Even with 12 colors, number of possibilities huge. But we can use inclusion-exclusion to count |S| exactly.

Let's denote edges as e1,...,e12. Colors from 1 to 12. For each vertex, we have constraint that colors on its 3 edges are all different. There are 8 vertices. Let's list edges per vertex.

We need adjacency of edges: each edge belongs to 2 vertices. So constraints are pairwise between edges sharing a vertex.

We can think of graph G where vertices are edges of cube, and two edges are adjacent if they share a vertex. That's the line graph of cube. The constraint is that adjacent edges in line graph must have different colors. So S is the set of proper vertex colorings of line graph with 12 colors. Line graph of cube has 12 vertices, each vertex degree? Let's compute: each edge shares a vertex with other edges: at each vertex, 3 edges mutually adjacent. So in line graph, each edge is adjacent to (3-1)*2 = 4? Actually, each edge is incident to 2 vertices, each vertex has 2 other edges, so total 4 edges adjacent to it (except if edges are opposite? Let's check: In cube, each edge shares a vertex with 4 other edges? Example: an edge of cube: it has two endpoints, each endpoint has two other edges (excluding the given edge), total 4 edges. But some of those may be the same? No, they are distinct. So degree of each vertex in line graph is 4. So line graph is 4-regular graph on 12 vertices. So S is number of proper vertex colorings of this line graph with 12 colors. That's easier: count proper colorings of a graph with 12 colors. That's chromatic polynomial evaluated at 12. So |S| = P_G(12) where P_G(k) is chromatic polynomial of line graph of cube.

Similarly, |T| = number of proper colorings using all 12 colors exactly once, i.e., bijective colorings. That's equal to number of proper colorings that are also permutations. But if we have 12 colors and 12 vertices, a bijection is just a labeling of vertices with colors 1..12. So |T| = number of proper colorings that use all colors exactly once = number of proper colorings that are also bijections = number of proper colorings with exactly 12 colors (no repetition). That's equal to number of proper colorings with at most 12 colors minus those with at most 11 colors? Actually, number of colorings using exactly 12 colors = total proper colorings with 12 colors minus those that use at most 11 colors. But that's messy.

Alternatively, |T| is the number of proper colorings that are also permutations, which is equal to the number of automorphisms? Not.

Wait: If we color vertices of line graph (which correspond to edges of cube) with colors 1..12 bijectively, then we are essentially assigning a permutation of colors to edges. The condition is that adjacent vertices (edges sharing a vertex) must have different colors. Since all colors are distinct, adjacency condition is automatically satisfied if no two adjacent edges have same color. But with bijection, they are all different anyway, so adjacency condition is automatically satisfied? No: adjacency condition requires that adjacent edges have different colors. With bijection, all colors are different, so adjacent edges will have different colors because they have different colors. So any bijection is automatically a proper coloring. Therefore, |T| = number of bijections from edges to colors that satisfy adjacency condition. But adjacency condition is automatically satisfied because all colors distinct. So |T| = number of bijections = 12!.

But wait: Is it possible that two adjacent edges get same color in a bijection? No, because bijection assigns distinct colors to all edges. So indeed any bijection is proper. So |T| = 12!.

Now |S| = number of proper colorings of line graph with 12 colors (colors can repeat). That's chromatic polynomial P_G(12). So probability = 12! / P_G(12).

We need chromatic polynomial of line graph of cube. Let's compute line graph of cube. First, need graph of cube edges adjacency.

Let's label cube vertices as coordinates (0/1)^3. Edges connect vertices differing in one coordinate. There are 12 edges: 4 edges in x-direction, 4 in y-direction, 4 in z-direction.

Let's label edges: Ex: edges along x-axis: (0,0,0)-(1,0,0) etc. We'll list.

Better: We can generate cube graph and its line graph in Python using networkx or manually.

We'll compute chromatic polynomial of line graph. There are known formulas for chromatic polynomial of line graph of cube? Might be computed via deletion-contraction.

We can compute using Python with sympy or networkx? sympy has chromatic_polynomial function for graphs. Let's try.

We'll create graph of cube (8 vertices, 12 edges), then compute line graph, then compute chromatic polynomial.

But chromatic polynomial for graph with 12 vertices might be large polynomial. We just need value at 12.

Alternatively, we can compute number of proper colorings directly via inclusion-exclusion using formula for chromatic polynomial: P_G(k) = sum_{i=0}^{n} (-1)^i * a_i * k^{n-i}, where a_i are coefficients. But we can compute P_G(12) by brute force enumeration of colorings? With 12 colors, total colorings 12^12 ≈ 8.9e12, too many. But we can use inclusion-exclusion over edges? Not.

We can use algorithm for counting proper colorings using inclusion-exclusion over independent sets? There is formula using partition of vertices.

Given small graph (12 vertices), we can compute chromatic polynomial via deletion-contraction recursively. We'll implement using sympy's Graph.chromatic_polynomial method.

Let's try.

[/THOUGHT] [PYTHON] import sympy from sympy import symbols

Create cube graph

vertices: tuples (x,y,z) with x,y,z in {0,1}

vertices = [(x,y,z) for x in range(2) for y in range(2) for z in range(2)]

edges: connect if Hamming distance 1

edges = [] for v in vertices: x,y,z = v # neighbor in x if x == 0: edges.append((v, (1,y,z))) # neighbor in y if y == 0: edges.append((v, (x,1,z))) # neighbor in z if z == 0: edges.append((v, (x,y,1)))

edges list contains each edge twice (undirected). Remove duplicates.

edges_set = set() for e in edges: if e[0] != e[1]: edges_set.add(tuple(sorted(e))) edges = list(edges_set) print("Cube graph: vertices", len(vertices), "edges", len(edges))

Create line graph: vertices correspond to edges of cube, adjacency if two edges share a vertex.

line_vertices = edges # each edge is a vertex in line graph

Build adjacency: two edges share a vertex if they have a common endpoint.

line_adj = {} for e1 in edges: line_adj[e1] = [] for e2 in edges: if e1 == e2: continue # check if share a vertex if (e1[0] in e2) or (e1[1] in e2): line_adj[e1].append(e2)

Create sympy Graph for line graph

G = sympy.Graph() for v in line_vertices: G.add_node(v) for v, neighbors in line_adj.items(): for w in neighbors: if v < w: # add edge once G.add_edge(v, w)

print("Line graph: vertices", G.order(), "edges", G.size())

Compute chromatic polynomial

k = symbols('k') P = G.chromatic_polynomial() print("Chromatic polynomial (first few terms):", P.expand()[:200])

Evaluate at k=12

P12 = P.subs(k, 12) print("P(12) =", P12)

Compute 12!

fact12 = sympy.factorial(12) print("12! =", fact12)

Probability = 12! / P(12)

prob = fact12 / P12 print("Probability =", prob)

Convert to reduced fraction

prob_frac = sympy.nsimplify(prob, [sympy.Rational]) print("Probability as fraction:", prob_frac) if isinstance(prob_frac, sympy.Rational): p = prob_frac.denominator q = prob_frac.numerator print("q/p =", q, "/", p) print("p+q =", p+q) else: print("Not rational?") [/PYTHON]

[THOUGHT] The sympy.Graph class might not exist in the version available. We need an alternative approach. We can compute the chromatic polynomial using other methods or compute the number of proper colorings directly via inclusion-exclusion using the principle of counting colorings of a graph with k colors, which is given by the chromatic polynomial. Since the line graph has 12 vertices, we can compute the chromatic polynomial via recursion manually, but that's tedious. Instead, we can compute the number of proper colorings with 12 colors using a combinatorial enumeration algorithm. Since 12^12 is too large, we can use the inclusion-exclusion formula over edges of the line graph? Actually, the number of proper colorings of a graph G with k colors is given by:

[ P_G(k) = \sum_{i=0}^{|E|} (-1)^i \cdot c_i \cdot k^{n-i} ]

where (c_i) is the number of ways to choose i edges such that after contracting those edges, the resulting graph has n-i connected components? Not exactly. There is a formula using the Tutte polynomial, but easier: we can compute the chromatic polynomial using deletion-contraction recursion. We can implement it recursively for small graph (12 vertices, ? edges). Let's compute the line graph's adjacency matrix.

First, let's construct the line graph of the cube. We already have edges list. We'll assign indices to edges for easier handling.

We'll compute adjacency matrix of line graph: size 12x12. Then compute chromatic polynomial via recursion using the fact that for a graph G, P(G,k) = P(G-e, k) - P(G/e, k) where e is an edge. We can implement recursively, but recursion depth might be large. However, graph has 12 vertices and each vertex degree 4, so number of edges in line graph: each vertex degree 4, so total edges = (12*4)/2 = 24. So line graph has 24 edges.

We can compute chromatic polynomial using sympy's graph theory maybe from sympy.combinatorics? Or we can use networkx? But networkx might not be available. We can implement our own recursion.

Alternatively, we can compute |S| directly using inclusion-exclusion over the constraints. Since constraints are per vertex of cube (not per edge of line graph). Let's go back to original problem: S is set of colorings of edges of cube with colors 1..12 such that at each vertex, the three incident edges have distinct colors. There are 8 vertices, each imposes 3 edges distinct. This is equivalent to proper edge-coloring of cube with 12 colors. Counting proper edge-colorings of a graph with k colors is related to the chromatic polynomial of the line graph, as we said.

Given the difficulty, maybe there is known result: The number of proper edge-colorings of a cube with k colors is k^12 - 83k^11 + ...? We can derive using inclusion-exclusion over vertices. Let's try inclusion-exclusion on vertices where bad event is that two edges incident to that vertex have same color. But events are overlapping.

Let edges be labeled. For each vertex v, let A_v be the set of colorings where the three edges incident to v have distinct colors. We want number of colorings in intersection of all A_v. Total colorings without restrictions: 12^12.

Number of colorings satisfying A_v: For vertex v with edges e1,e2,e3, we need them to be distinct. Number of colorings for these three edges: 121110. For the other 9 edges, any colors: 12^9. So |A_v| = 121110 * 12^9 = 12^10 * 1320? Actually 121110 = 1320. So |A_v| = 1320 * 12^9.

But intersections of multiple A_v are tricky because edges are shared.

We can try to compute using inclusion-exclusion over vertices:

|S| = sum_{X subset of V} (-1)^{|X|} * N(X)

where N(X) is number of colorings where for each vertex in X, the condition A_v is violated (i.e., at least two edges share color). But that's complicated.

Given time, maybe we can compute via brute force with symmetry reduction. Since 12^12 is huge, but we can use dynamic programming over cube's structure. Cube has symmetry group. We can use Burnside's lemma to count proper colorings up to symmetry? But we need exact count.

Alternatively, we can compute chromatic polynomial of line graph by using deletion-contraction algorithm implemented in Python without external libraries. Since graph has 12 vertices, we can represent graph as adjacency list. We'll implement recursive function chromatic_polynomial(G, k) that returns number of proper colorings with k colors. For small graph, we can compute directly using recursion with memoization.

We can compute P(G,12) by using the recurrence: P(G,k) = P(G-e, k) - P(G/e, k). Base cases: empty graph with n vertices: k^n. Complete graph K_n: k*(k-1)...(k-n+1).

We need to implement graph contraction: when contracting edge e=(u,v), we merge u and v, remove loops, keep multiple edges? For chromatic polynomial, contraction merges vertices and removes loops but keeps multiple edges? Actually, chromatic polynomial for multigraphs: contraction removes loops but keeps multiple edges. So we need to handle multigraphs.

We'll implement graph as adjacency list with edge counts? Since line graph is simple (no multiple edges), contraction may create multiple edges.

We'll implement using networkx? Not available. Let's implement ourselves.

We'll label vertices 0..11 for edges of cube. Build adjacency matrix.

First, list cube edges with indices:

Let's define cube vertices as 0..7 with coordinates (x,y,z) binary.

Edges: 0: (0,0,0)-(1,0,0) 1: (0,0,0)-(0,1,0) 2: (0,0,0)-(0,0,1) 3: (0,1,0)-(1,1,0) 4: (0,1,0)-(0,1,1) 5: (0,0,1)-(1,0,1) 6: (0,0,1)-(0,1,1) 7: (1,0,0)-(1,1,0) 8: (1,0,0)-(1,0,1) 9: (1,1,0)-(1,1,1) 10: (0,1,1)-(1,1,1) 11: (1,0,1)-(1,1,1)

But need to ensure each edge appears once. Let's generate systematically.

We'll generate all edges as tuples (v1,v2) where v1<v2 and Hamming distance 1.

We'll map each edge to index.

Then build line graph adjacency: two edges adjacent if they share a vertex.

We'll compute adjacency list.

Then implement chromatic polynomial counting via recursion with memoization using graph canonical representation (like sorted tuple of edges). Since graph has 24 edges, recursion tree might be large but manageable with memoization.

We'll implement function chrom_poly(graph_edges, n_vertices) where graph_edges is frozenset of edges (unordered pairs). n_vertices is number of vertices (fixed). We'll compute number of proper colorings with k colors as function of k? Actually we need value at k=12. So we can compute directly for k=12 using recursion that returns integer.

We can compute number of proper colorings with k colors using inclusion-exclusion formula based on partitions (using Bell numbers). For graph with 12 vertices, number of partitions is B_12=4213597, too large.

Better to use deletion-contraction recursion with caching. We'll implement:

def count_colorings(edges, n, k): # edges: set of tuples (u,v) with u<v, n: number of vertices # base: if no edges, return k**n # choose an edge e # G-e: edges without e # G/e: contract edge e: merge u,v into u, rename vertices, remove loops, keep multiple edges. # count = count_colorings(G-e) - count_colorings(G/e)

But contraction reduces number of vertices by 1.

We need to handle multiple edges: after contraction, if there are multiple edges between same pair, they are still edges (but chromatic polynomial for multigraph treats multiple edges as single edge? Actually, chromatic polynomial is same for multigraph as for simple graph because multiple edges don't affect proper coloring condition (they just require colors different anyway). So we can merge duplicate edges into one.

We'll implement contraction by relabeling vertices: new vertex set 0..n-2, map old vertices: vertices < min(u,v) keep same, vertices > max(u,v) shift down by 1, and u and v map to new vertex index = min(u,v). Then generate new edges set with renamed vertices, remove loops, and merge duplicates.

We'll use recursion with memoization keyed by (frozenset of edges, n). Since n is fixed 12 initially, but reduces.

Let's implement.

[/THOUGHT] [PYTHON] import itertools

Label cube vertices 0..7 corresponding to binary (x,y,z)

vertices = list(range(8))

Map index to coordinates

coord = {i: ((i>>2)&1, (i>>1)&1, i&1) for i in range(8)} # actually careful: let's define x,y,z

Better: i = 4x + 2y + z? Let's define i = x4 + y2 + z, where x,y,z in {0,1}

def idx(x,y,z): return x4 + y2 + z

Generate edges: for each vertex, connect to neighbors with higher index to avoid duplicates

edges_cube = [] for x in range(2): for y in range(2): for z in range(2): v = idx(x,y,z) if x == 0: v2 = idx(1,y,z) if v < v2: edges_cube.append((v, v2)) if y == 0: v2 = idx(x,1,z) if v < v2: edges_cube.append((v, v2)) if z == 0: v2 = idx(x,y,1) if v < v2: edges_cube.append((v, v2)) print("Cube edges:", edges_cube) print("Number of cube edges:", len(edges_cube))

Assign index to each cube edge

edge_index = {e:i for i,e in enumerate(edges_cube)} print("Edge indices:", edge_index)

Build line graph adjacency: vertices are cube edge indices

n_line = len(edges_cube) # 12 adj_line = [[] for _ in range(n_line)] for i in range(n_line): for j in range(i+1, n_line): e1 = edges_cube[i] e2 = edges_cube[j] # check if share a vertex if (e1[0] in e2) or (e1[1] in e2): adj_line[i].append(j) adj_line[j].append(i)

Create edge list for line graph (undirected, simple)

edges_line = [] for i in range(n_line): for j in adj_line[i]: if i < j: edges_line.append((i,j)) print("Line graph edges:", len(edges_line))

Should be 24? Let's check

print("Number of edges in line graph:", len(edges_line))

Now compute number of proper colorings with k colors using deletion-contraction

k = 12 # number of colors

We'll implement recursive function with memoization

from functools import lru_cache

def graph_key(edges_set, n): # edges_set: frozenset of tuples (u,v) with u<v # n: number of vertices (0..n-1) # canonical representation: sort edges and tuple of sorted edges # but frozenset is hashable return (edges_set, n)

@lru_cache(maxsize=None) def count_colorings(edges_tuple, n): # edges_tuple: tuple of edges sorted for consistency edges = set(edges_tuple) if not edges: return k**n # Choose an edge e = (u,v) e = next(iter(edges)) u,v = e # G-e edges_minus = edges.copy() edges_minus.remove(e) # Convert to tuple for caching edges_minus_t = tuple(sorted(edges_minus)) count1 = count_colorings(edges_minus_t, n) # Contract edge e # New vertex mapping: vertices 0..n-1, merge u and v into min(u,v) new_vertices = n-1 # mapping old -> new: for old vertex w: # if w == u or w == v -> new_u = min(u,v) # else if w > max(u,v) -> w-1 # else -> w (since w < min(u,v) or between min and max) new_u = min(u,v) new_v = max(u,v) def map_vertex(w): if w == u or w == v: return new_u elif w > new_v: return w-1 else: return w # Build new edges set after contraction new_edges = set() for (a,b) in edges: if (a,b) == e: continue na = map_vertex(a) nb = map_vertex(b) if na == nb: continue # remove loop # ensure na < nb if na > nb: na, nb = nb, na new_edges.add((na, nb)) # Merge multiple edges: set already handles duplicates new_edges_t = tuple(sorted(new_edges)) count2 = count_colorings(new_edges_t, new_vertices) return count1 - count2

Convert edges_line to tuple of sorted edges

edges_line_sorted = tuple(sorted(edges_line)) print("Computing number of proper colorings with k=", k) S = count_colorings(edges_line_sorted, n_line) print("|S| =", S)

Compute 12!

import math T = math.factorial(12) print("|T| =", T) prob = T / S print("Probability =", prob)

Convert to reduced fraction

from fractions import Fraction prob_frac = Fraction(T, S) print("Fraction:", prob_frac) p = prob_frac.denominator q = prob_frac.numerator print("q/p =", q, "/", p) print("p+q =", p+q) [/PYTHON]

[THOUGHT] The computation gives |S| = 1038542714880, |T| = 12! = 479001600, probability = 2835/6146678. Then p+q = 6146678 + 2835 = 6149513. However, we should double-check if the interpretation is correct. The probability fraction is reduced? Let's verify: 2835/6146678 simplifies? We used Fraction from Python, which reduces automatically. So it's reduced. Then p+q = 6146678 + 2835 = 6149513.

But wait: The probability seems small (~0.00046). That might be plausible. However, we must ensure that our counting of |S| is correct. We computed chromatic polynomial of line graph at k=12. But is |S| indeed the number of proper vertex colorings of line graph with 12 colors? Yes, because proper vertex coloring of line graph corresponds to edge-coloring of cube where edges sharing a vertex have different colors. So that matches.

We also assumed |T| = 12! because any bijection from edges to colors automatically satisfies the adjacency condition. That is correct because all colors are distinct, so adjacent edges necessarily have different colors. So |T| = 12!.

Thus probability = 12! / P_G(12) where G is line graph of cube. Our computation gave P_G(12) = 1038542714880. Let's verify quickly: 12! = 479001600. Multiply probability fraction: 479001600 / 1038542714880 = simplify dividing numerator and denominator by? Compute gcd: We can compute gcd(479001600, 1038542714880). But Python gave fraction 2835/6146678. That implies 1038542714880 / 479001600 = 6146678/2835? Let's compute: 1038542714880 / 479001600 = 2168.0? Actually compute: 479001600 * 2168 = 4790016002000=958003200000, 479001600168=80472268800, sum = 1.0384754688e12, close but not exactly. Let's trust Python's Fraction reduction.

Thus p+q = 6146678 + 2835 = 6149513.

We should ensure no off-by-one errors. Let's quickly recalc with a small sanity check: For k=12, number of proper colorings of line graph should be less than total colorings 12^12 = 8916100448256. Our computed |S| = 1038542714880 is about 1/8.6 of total, which seems plausible.

Thus final answer p+q = 6149513.

[/THOUGHT] \boxed{6149513}