[THOUGHT] I will:

1. Build the cube graph (Q3) with 8 vertices labeled 0..7, edges between vertices with Hamming distance 1.
2. Construct its line graph L(Q3) whose vertices correspond to cube edges, and adjacency corresponds to sharing a vertex.
3. Use sympy’s Graph.chromatic_polynomial() to compute the chromatic polynomial P(k) of L(Q3).
4. Evaluate P(12) to get the total number of proper edge-colorings with 12 colors (denominator).
5. The number of colorings where all 12 edges are pairwise distinct is 12! (any injective assignment is automatically proper).
6. Form the reduced fraction q/p = 12! / P(12), reduce it, and output p+q.

Now I will implement this in Python. [/THOUGHT] [PYTHON] from sympy import symbols, factorial, Rational from sympy.combinatorics.graph import Graph

Step 1: Build cube graph Q3 with vertices 0..7

Vertices: 3-bit numbers 0..7; edge if Hamming distance is 1.

cube_vertices = list(range(8)) cube_edges = []

def hamming_distance(a, b): return bin(a ^ b).count("1")

for i in cube_vertices: for j in range(i+1, 8): if hamming_distance(i, j) == 1: cube_edges.append((i, j))

Sanity check: cube should have 12 edges

print("Cube edges:", cube_edges, "Count:", len(cube_edges))

Step 2: Construct line graph L(Q3)

Vertices of line graph are indices of cube_edges: 0..11

line_vertices = list(range(len(cube_edges))) line_edges = []

for i in range(len(cube_edges)): for j in range(i+1, len(cube_edges)): e1 = cube_edges[i] e2 = cube_edges[j] # edges adjacent in cube if they share a vertex if len(set(e1).intersection(e2)) > 0: line_edges.append((i, j))

Sanity check: properties of line graph

print("Line graph vertex count:", len(line_vertices)) print("Line graph edge count:", len(line_edges))

Step 3: Build sympy Graph and compute chromatic polynomial

G_line = Graph() G_line.add_vertices(*line_vertices) for u, v in line_edges: G_line.add_edge(u, v)

k = symbols('k') chrom_poly = G_line.chromatic_polynomial(k) print("Chromatic polynomial P(k) of line graph:") print(chrom_poly)

Step 4: Evaluate at k=12 to get total number of proper colorings

P_12 = chrom_poly.subs(k, 12) P_12_simplified = int(P_12) print("P(12) =", P_12_simplified)

Step 5: Number of colorings with all 12 edges distinct is 12!

num_distinct = int(factorial(12)) print("12! =", num_distinct)

Step 6: Probability = 12! / P(12), reduce fraction and compute p+q

from sympy import fraction, simplify prob = Rational(num_distinct, P_12_simplified) prob_simplified = simplify(prob) q, p = prob_simplified.numerator, prob_simplified.denominator print("Probability simplified q/p =", prob_simplified) print("q =", q, "p =", p) print("p+q =", p + q) [/PYTHON]

[THOUGHT] I need to recompute the total number of proper edge colorings of the cube with 12 colors without using the missing sympy.graph module. I’ll:

1. Build the cube graph Q3 (8 vertices, 12 edges).
2. Build its line graph L(Q3) with 12 vertices (one per cube edge), edges joining vertices whose cube edges share an endpoint.
3. Count proper colorings of L(Q3) with 12 colors without enumerating all 12^12 assignments:
   • Use abstract colors (1..m) and backtracking over vertices 0..11, building color patterns up to relabeling symmetry.
   • For each completed pattern using m abstract colors, it corresponds to (12)_m = 12·11·...·(12−m+1) actual colorings.
   • Sum (12)_m over all patterns to get P(12), the total number of proper edge colorings with 12 actual colors.
4. The favorable colorings where all 12 edges have distinct colors are exactly 12! (any bijection edge→color is automatically proper).
5. Compute the probability = 12! / P(12), reduce it, and output p+q.

Now I'll implement this backtracking and counting in Python. [/THOUGHT] [PYTHON] from math import factorial from fractions import Fraction

Step 1: Build cube graph Q3 (vertices 0..7, edges between vertices with Hamming distance 1)

cube_vertices = list(range(8)) cube_edges = []

def hamming_distance(a, b): return bin(a ^ b).count("1")

for i in cube_vertices: for j in range(i+1, 8): if hamming_distance(i, j) == 1: cube_edges.append((i, j))

print("Cube edges:", cube_edges, "Count:", len(cube_edges))

Step 2: Build line graph L(Q3)

Vertices: 0..11 each representing an edge in cube_edges

n = len(cube_edges) adj = [set() for _ in range(n)]

for i in range(n): for j in range(i+1, n): e1 = cube_edges[i] e2 = cube_edges[j] if len(set(e1).intersection(e2)) > 0: adj[i].add(j) adj[j].add(i)

print("Line graph vertex count:", n) print("Line graph edge count:", sum(len(s) for s in adj) // 2)

Step 3: Precompute falling factorials (12)_m

k_colors = 12 ffall = [1] * (n + 1) for m in range(1, n + 1): f = 1 for j in range(m): f *= (k_colors - j) ffall[m] = f

print("Falling factorials (12)_m for m=0..12:", ffall)

Step 4: Backtracking over color patterns (abstract colors)

total_P12 = 0

def dfs(i, colors, maxc): """ i: current vertex index to color (0..n) colors: list of assigned abstract colors for vertices 0..i-1 maxc: maximum abstract color label used so far """ global total_P12 if i == n: # Completed a color pattern using 'maxc' abstract colors total_P12 += ffall[maxc] return

# Colors used by neighbors of vertex i among already-colored vertices
used_neighbor_colors = set()
for nb in adj[i]:
    if nb < i:
        used_neighbor_colors.add(colors[nb])

# Try existing colors 1..maxc that are not used by neighbors
for c in range(1, maxc + 1):
    if c not in used_neighbor_colors:
        colors.append(c)
        dfs(i + 1, colors, maxc)
        colors.pop()

# Try introducing a new color maxc+1, if we have not exceeded k_colors
newc = maxc + 1
if newc <= k_colors:
    colors.append(newc)
    dfs(i + 1, colors, newc)
    colors.pop()

dfs(0, [], 0)

print("Total proper colorings P(12) =", total_P12)

Step 5: Number of colorings with all 12 edges distinct: 12!

num_distinct = factorial(12) print("12! =", num_distinct)

Step 6: Probability and reduced fraction q/p, then p+q

prob = Fraction(num_distinct, total_P12) q, p = prob.numerator, prob.denominator print("Probability q/p =", prob) print("q =", q, "p =", p) print("p+q =", p + q) [/PYTHON]

[THOUGHT] Using the line graph of the cube and counting proper colorings with 12 colors via backtracking (weighted by falling factorials for actual color choices), we found:

• Total proper edge-colorings with 12 colors: P(12) = 1038542714880.
• Colorings with all 12 edges distinct: 12! = 479001600. Thus the desired probability is 12! / P(12) = 2835/6146678, so p = 6146678, q = 2835, and p+q = 6149513. [/THOUGHT] \boxed{6149513}