Sunday Times Teaser 2650 – Squares of Oblongs

by Des MacHale

I gave a set of nine rectangular cards of dimensions 1×2, 2×3, 3×4, 4×5, 5×6, 6×7, 7×8, 8×9 and 9×10 to each of two friends.

I asked each of them to arrange some of their cards to form a complete square and it turned out that those they made were the smallest and the largest that were possible.

What, in increasing order, were the sizes of all the unused cards?

3 Comments Leave one →

brian gladman permalink

This solution uses a collection of routines that I have put together in jigsaw.py


from jigsaw import orientate, print_rect, select_pieces, solve_rect

# compile a dictionary of the pieces
pcs = dict()
for n in range(1, 10):
  p = tuple((x, y) for x in range(1, n + 1) for y in range(1, n + 2))
  pcs[chr(ord('A') + n - 1)] = orientate(p, two_sided=True)

# for saving smallest and largest solutions
s_min, s_max, g_min, g_max = 19, 0, None, None

# the maximum square possible with all tiles is 18 x 18
for s in range(2, 19):
  # selct pieces that match the area of a square of this side length
  for spc in select_pieces(s ** 2, pcs.keys(), pcs, m=s, n=s):
    # compile a list of the unused pieces
    nu = ''.join(p for p in pcs.keys() if p not in spc)
    
    # look for an arrangement of pieces in the square 
    for g in solve_rect(spc, s, s, pcs):
      # save details for smallest and largest squares found
      if s < s_min:
        s_min, g_min = s, (nu, g)
        break
      if s > s_max:
        s_max, g_max = s, (nu, g)
        break

# output the list of unused pieces
f, s, l = ', '.join((f"{(n:=ord(c) - ord('A') + 1)} x {n + 1}" 
            for c in sorted(g_min[0] + g_max[0]))).rpartition(', ')
print(f"The unused pieces are {' and '.join((f, l))}.\n")

# print the smallest and largest arrangements of pieces into squares
print(f"Minimum square {s_min} x {s_min}:")
print_rect(g_min[1], s_min, s_min)
print(f"Maximum square {s_max} x {s_max}:")
print_rect(g_max[1], s_max, s_max)

from jigsaw import orientate, print_rect, select_pieces, solve_rect

# compile a dictionary of the pieces

pcs = dict()

for n in range(1, 10):

p = tuple((x, y) for x in range(1, n + 1) for y in range(1, n + 2))

pcs[chr(ord('A') + n - 1)] = orientate(p, two_sided=True)

# for saving smallest and largest solutions

s_min, s_max, g_min, g_max = 19, 0, None, None

# the maximum square possible with all tiles is 18 x 18

for s in range(2, 19):

# selct pieces that match the area of a square of this side length

for spc in select_pieces(s ** 2, pcs.keys(), pcs, m=s, n=s):

# compile a list of the unused pieces

nu = ''.join(p for p in pcs.keys() if p not in spc)

# look for an arrangement of pieces in the square

for g in solve_rect(spc, s, s, pcs):

# save details for smallest and largest squares found

if s < s_min:

s_min, g_min = s, (nu, g)

break

if s > s_max:

s_max, g_max = s, (nu, g)

break

# output the list of unused pieces

f, s, l = ', '.join((f"{(n:=ord(c) - ord('A') + 1)} x {n + 1}"

for c in sorted(g_min[0] + g_max[0]))).rpartition(', ')

print(f"The unused pieces are {' and '.join((f, l))}.\n")

# print the smallest and largest arrangements of pieces into squares

print(f"Minimum square {s_min} x {s_min}:")

print_rect(g_min[1], s_min, s_min)

print(f"Maximum square {s_max} x {s_max}:")

print_rect(g_max[1], s_max, s_max)

It runs in 280 milliseconds.

ahmet cetinbudaklar permalink

Dear Brian:

After checking the number of teasers you are providing I have noticed thast this teaser number shoul read 2650 instead of 2560.

Reply
brian gladman permalink

Thankyou Ahmet – now corrected.

Reply

Sunday Times Teaser 2650 – Squares of Oblongs

by Des MacHale

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