В каждой клетке таблицы 3 × 5 стоит натуральное чи...

python
import itertools

nums=list(range(1,16))
triples=[t for t in itertools.combinations(nums,3) if sum(t)==24]
len(triples), triples[:10]

(25,
 [(1, 8, 15),
  (1, 9, 14),
  (1, 10, 13),
  (1, 11, 12),
  (2, 7, 15),
  (2, 8, 14),
  (2, 9, 13),
  (2, 10, 12),
  (3, 6, 15),
  (3, 7, 14)])

python
triples_set=[set(t) for t in triples]

solutions=[]
def search(remaining, start, current):
    if not remaining:
        solutions.append(current.copy()); 
        return
    # select smallest remaining element to reduce branching
    s=min(remaining)
    # choose triples containing s
    for i, t in enumerate(triples_set):
        if s in t and t<=remaining:
            current.append(t)
            search(remaining - t, i+1, current)
            current.pop()

remaining=set(nums)
search(remaining, 0, [])
len(solutions)

11

python
solutions[:2]

[[{1, 8, 15}, {2, 9, 13}, {3, 10, 11}, {4, 6, 14}, {5, 7, 12}],
 [{1, 8, 15}, {2, 10, 12}, {3, 7, 14}, {4, 9, 11}, {5, 6, 13}]]

python
import itertools, math

def find_row_assignment(partition):
    part=[tuple(sorted(list(s))) for s in partition] # triples
    # Try all permutations per column (3!^5=243) manageable
    for perms in itertools.product(itertools.permutations((0,1,2)), repeat=5):
        rows=[0,0,0]
        ok=True
        for col, triple in enumerate(part):
            perm=perms[col]
            for idx,row in enumerate(perm):
                rows[row]+=triple[idx]
        if rows==[40,40,40]:
            # build grid
            grid=[[None]*5 for _ in range(3)]
            for col,triple in enumerate(part):
                perm=perms[col]
                for idx,row in enumerate(perm):
                    grid[row][col]=triple[idx]