Sunday Times Teaser 3204 – Pressing Problem

by BRG on February 16, 2024

by Howard Williams

Published Friday February 16 2024 (link)

The Mint’s machine can press circular coins from square plates of precious metal in straight rows with no gaps between coins. Currently, the coins are pressed with the same number of coins in each column and row, with their centres lying on the same vertical and horizontal straight lines.

As newly appointed Warden of the Mint, Newton set about reviewing its operational efficiency. He found that more rows can be squeezed into the same plate by shifting some rows so that each coin in it touches two coins in the row above; each of these rows will have one fewer coin in it. With this method, the best that can be done is to produce exactly 8 per cent more coins per plate.

How many more coins per plate can be produced in this way?

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sep = 3 ** (1 / 2) / 2

# consider the number of columns
for nc in range(1, 50):
  # the number of coins in the first arrangement 
  ncns_1 = nc * nc
  # the maximum number of rows in a hexagonal arrangement
  max_r = int((nc - 1) / sep)
  
  # find the arrangement that yields the most additional coins
  ncns_2, sex = 0, (0, 0)
  # consider <s> shifted rows, each following an unshifted row, 
  # giving an overall vertical height of 1 + (2s - 1).sep
  for s in range(max_r // 2 + 1):
    # the remaining space for more unshifted rows
    sp = nc - 1 - (2 * s - 1) * sep
    # the following row requires <sep> space, further rows
    # require unit space
    ex = 0 if sp < sep else 1 + int(sp - sep)
    # the number of coins in the second arrangement
    ncns = s * (2 * nc - 1) + ex * nc
    # store the vaues for the most extra coins
    if ncns > ncns_2:
      ncns_2 = ncns
      sex = (s, ex)
  # have we found the specified percentage increase?
  if 100 * ncns_2 == 108 * ncns_1:
    s, ex = sex
    d, r = ncns_2 - ncns_1, 2 * s + ex 
    print(f"{d} coins more [{nc} columns, {r} rows "
          f"({s} shifted), coins: ({ncns_1}, {ncns_2})].")
    break

sep = 3 ** (1 / 2) / 2

# consider the number of columns

for nc in range(1, 50):

# the number of coins in the first arrangement

ncns_1 = nc * nc

# the maximum number of rows in a hexagonal arrangement

max_r = int((nc - 1) / sep)

# find the arrangement that yields the most additional coins

ncns_2, sex = 0, (0, 0)

# consider <s> shifted rows, each following an unshifted row,

# giving an overall vertical height of 1 + (2s - 1).sep

for s in range(max_r // 2 + 1):

# the remaining space for more unshifted rows

sp = nc - 1 - (2 * s - 1) * sep

# the following row requires <sep> space, further rows

# require unit space

ex = 0 if sp < sep else 1 + int(sp - sep)

# the number of coins in the second arrangement

ncns = s * (2 * nc - 1) + ex * nc

# store the vaues for the most extra coins

if ncns > ncns_2:

ncns_2 = ncns

sex = (s, ex)

# have we found the specified percentage increase?

if 100 * ncns_2 == 108 * ncns_1:

s, ex = sex

d, r = ncns_2 - ncns_1, 2 * s + ex

print(f"{d} coins more [{nc} columns, {r} rows "

f"({s} shifted), coins: ({ncns_1}, {ncns_2})].")

break

This code varies the plate size to see what other solutions are possible:


sep = 3 ** (1 / 2) / 2

# consider the plate size <ps> and size increment <pi>
ps, pi = 1.0, 0.1
while True:
  ps += pi
  if ps >= 25.0:
    break
  nc = int(ps)
  # the number of coins in the first arrangement 
  ncns_1 = nc * nc
  # the maximum number of rows in a hexagonal arrangement
  max_r = int((ps - 1) / sep) + 1
  ncns_2, sex = (0, 0)
  # consider s shifted rows, each following an unshifted row, 
  # and giving an overall vertical height of 1 + (2s - 1).sep
  for s in range(max_r // 2 + 1):
    # the remaining space for more unshifted rows
    sp = ps - 1 - (2 * s - 1) * sep
    # the following row requires <sep> space, further rows
    # require unit space
    ex = 0 if sp < sep else 1 + int(sp - sep)
    # the number of coins in a hexagonal arrangement
    ncns = s * (2 * nc - 1) + ex * nc
    if ncns > ncns_2:
      ncns_2 = ncns
      sex = (s, ex)
  if 100 * ncns_2 == 108 * ncns_1:
    s, ex = sex
    d, r = ncns_2 - ncns_1, 2 * s + ex 
    print(f"{d:>2} coins more [plate: {ps:>4.1f}; columns: {nc:>2}; rows: {r:>2} "
          f"(shifted {s:>2}); coins: ({ncns_1:>3}, {ncns_2:>3})].")

sep = 3 ** (1 / 2) / 2

# consider the plate size <ps> and size increment <pi>

ps, pi = 1.0, 0.1

while True:

ps += pi

if ps >= 25.0:

break

nc = int(ps)

# the number of coins in the first arrangement

ncns_1 = nc * nc

# the maximum number of rows in a hexagonal arrangement

max_r = int((ps - 1) / sep) + 1

ncns_2, sex = (0, 0)

# consider s shifted rows, each following an unshifted row,

# and giving an overall vertical height of 1 + (2s - 1).sep

for s in range(max_r // 2 + 1):

# the remaining space for more unshifted rows

sp = ps - 1 - (2 * s - 1) * sep

# the following row requires <sep> space, further rows

# require unit space

ex = 0 if sp < sep else 1 + int(sp - sep)

# the number of coins in a hexagonal arrangement

ncns = s * (2 * nc - 1) + ex * nc

if ncns > ncns_2:

ncns_2 = ncns

sex = (s, ex)

if 100 * ncns_2 == 108 * ncns_1:

s, ex = sex

d, r = ncns_2 - ncns_1, 2 * s + ex

print(f"{d:>2} coins more [plate: {ps:>4.1f}; columns: {nc:>2}; rows: {r:>2} "

f"(shifted {s:>2}); coins: ({ncns_1:>3}, {ncns_2:>3})].")

Sunday Times Teaser 3204 – Pressing Problem

by Howard Williams

Published Friday February 16 2024 (link)

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