В каждой клетке таблицы 4 × 5 стоит натуральное чи...

python
import itertools

nums = list(range(1,20)) + [30]
S = sum(nums)
S


220

python
nums = list(range(1,20)) + [30]

sols = []
# find partitions into 5 groups of size 4 summing to 44
# This is large; use backtracking
nums_set = set(nums)

groups = []

nums_sorted = sorted(nums)
count = 0

def search(available, start=0, groups=[]):
    if not available:
        yield groups[:]
        return
    # choose smallest remaining element to reduce symmetry
    a = min(available)
    # find 3 others b,c,d such that sum=44
    rest = sorted([x for x in available if x!=a])
    # Use combinations of 3 from rest
    for b,c,d in itertools.combinations(rest, 3):
        if a + b + c +

[(1, 2, 11, 30),
 (3, 4, 18, 19),
 (5, 6, 16, 17),
 (7, 8, 14, 15),
 (9, 10, 12, 13)]

python
groups = [
(1,2,11,30),
(3,4,18,19),
(5,6,16,17),
(7,8,14,15),
(9,10,12,13),
]

import itertools

# Represent assignment by, for each column, a permutation of indices 0..3 for rows
# But we can think rows labeled 0..3. We'll generate row vectors by choosing for each column an index for row0,
# etc and check if there exists alignment for all rows simultaneously. Equivalent to choose matrices of permutations.
# We can brute force by backtracking across columns selecting a permutation for each column and tracking row sums.

target_row_sum = 55

def backtrack(col, row_sums, perms):
    if col == len(groups):
        if all(rs == target_row_sum for rs in row_sums):
            yield perms[:]
        return
    g = groups[col]
    # Try all permutations of group g assigned to 4 rows
    for perm in itertools.permutations(g, 4):
        # prune by partial sums: any row sum exceed target?
        ok = True
        for r in range(4):
            if row_sums[r] + perm[r] > target_row_sum:
                ok = False
                break
        if not ok: 
            continue
        # additional pruning: at last column, the needed value must equal remaining difference for each row
        # but that's automatic; skip
        row_sums2 = tuple(row_sums[r]+perm[r] for r in range(4))
        perms.append(perm)
        yield from backtrack(col+1, row_sums2, perms)
        perms.pop()

solutions = list(backtrack(0, (0,0,0,0), []))
len(solutions)


72

python
solutions[0]


[(1, 2, 11, 30),
 (18, 19, 3, 4),
 (16, 6, 17, 5),
 (8, 15, 14, 7),
 (12, 13, 10, 9)]

python
perm = solutions[0]
row_sums = [sum(perm[c][r] for c in range(5)) for r in range(4)]
col_sums = [sum(perm[c]) for c in range(5)]
row_sums, col_sums


([55, 55, 55, 55], [44, 44, 44, 44, 44])