New Scientist Enigma 957 – Open The Box!

1. 
   Brian Gladman permalink

   Python

   from itertools import count, combinations, permutations # form ordered rows of <m> numbers from those in <nbrs> with # each row summing to <sm> (<m> divides the length of <nbrs>) def make_rows(nbrs, m, sm, rows=()): if not nbrs: yield rows else: # select the middle m - 2 numbers for cmm2 in combinations(nbrs[1:], m - 2): # compute and check the last number t = sm - nbrs[0] - sum(cmm2) if t in nbrs and t > cmm2[-1]: t = tuple([nbrs[0], *cmm2, t]) nsn = tuple(x for x in nbrs if x not in t) yield from make_rows(nsn, m, sm, rows + (t,)) # check that five columns with equal sums can be made by # reordering the values in the given set of four rows def make_cols(rows): sm = sum(sum(r) for r in rows) // 5 # the set of possible fourth rows spr4 = set(permutations(rows[3])) # leave the first row as it is and permute # the second and third rows for r2 in permutations(rows[1]): for r3 in permutations(rows[2]): # form five columns of three values and # compute the fourth value in each column c123 = tuple(zip(rows[0], r2, r3)) # form the last values in each column and # check that it is a possible fourth row c4 = tuple(sm - sum(c) for c in c123) if c4 in spr4: # return the five columns return tuple(x + (y,) for x, y in zip(c123, c4)) # consider increasing numbers of coins held by each # executive (n even, maximum stolen = (5.n / 2 - 3) # + (n - 1) >= 111 ==> n >= 34) for n in count(34, 2): # the numbers of coins in the 20 boxes ns = list(range(1, 11)) + list(range(n - 10, n)) # for saving the burglar's take(s) haul = dict() # make four rows of five boxes with the same # total number of coins in each row for rows in make_rows(ns, 5, 5 * n // 2): # select the two rows that are burgled for r1, r2 in permutations(rows, 2): # select three boxes from one row for c3 in combinations(r1, 3): s3 = sum(c3) # ... and one from another for c in r2: # the amount has three non zero digits if s3 + c >= 111 and '0' not in str(s3 + c): cols = make_cols(rows) if cols and s3 + c not in haul: haul[s3 + c] = (rows, cols) if haul: for h, (r, c) in haul.items(): print(f"Each had {n} coins, burglar's haul {h}:") print(f"rows = {r}") print(f"cols = {c}") print() exit()

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   from itertools import count, combinations, permutations   # form ordered rows of <m> numbers from those in <nbrs> with # each row summing to <sm> (<m> divides the length of <nbrs>) def make_rows(nbrs, m, sm, rows=()):   if not nbrs:     yield rows   else:     # select the middle m - 2 numbers     for cmm2 in combinations(nbrs[1:], m - 2):       # compute and check the last number       t = sm - nbrs[0] - sum(cmm2)       if t in nbrs and t > cmm2[-1]:         t = tuple([nbrs[0], *cmm2, t])         nsn = tuple(x for x in nbrs if x not in t)         yield from make_rows(nsn, m, sm, rows + (t,))   # check that five columns with equal sums can be made by # reordering the values in the given set of four rows def make_cols(rows):   sm = sum(sum(r) for r in rows) // 5   # the set of possible fourth rows   spr4 = set(permutations(rows[3]))   # leave the first row as it is and permute   # the second and third rows   for r2  in permutations(rows[1]):     for r3 in permutations(rows[2]):       # form five columns of three values and       # compute the fourth value in each column       c123 = tuple(zip(rows[0], r2, r3))       # form the last values in each column and       # check that it is a possible fourth row       c4 = tuple(sm - sum(c) for c in c123)       if c4 in spr4:         # return the five columns         return tuple(x + (y,) for x, y in zip(c123, c4))   # consider increasing numbers of coins held by each # executive (n even, maximum stolen = (5.n / 2 - 3) # + (n - 1) >= 111 ==> n >= 34) for n in count(34, 2):   # the numbers of coins in the 20 boxes   ns = list(range(1, 11)) + list(range(n - 10, n))     # for saving the burglar's take(s)   haul = dict()   # make four rows of five boxes with the same   # total number of coins in each row   for rows in make_rows(ns, 5, 5 * n // 2):        # select the two rows that are burgled     for r1, r2 in permutations(rows, 2):       # select three boxes from one row       for c3 in combinations(r1, 3):         s3 = sum(c3)         # ... and one from another         for c in r2:           # the amount has three non zero digits           if s3 + c >= 111 and '0' not in str(s3 + c):             cols = make_cols(rows)             if cols and s3 + c not in haul:               haul[s3 + c] = (rows, cols)                  if haul:     for h, (r, c) in haul.items():       print(f"Each had {n} coins, burglar's haul {h}:")       print(f"rows = {r}")       print(f"cols = {c}")       print()     exit()

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