Sunday Times Teaser 3163 – Letterboxes

by BRG on 6 May 2023

by Victor Bryant

Published Sunday May 07 2023 (link)

To enable me to spell out ONE, TWO, up to NINE one or more times, I bought large quantities of the letters E, F, G, H, I, N etc. Then in a box labelled “ONE” I put equal numbers of Os, Ns and Es; in a second box labelled “TWO” I put equal numbers of Ts, Ws and Os; in box “THREE” I put equal numbers of Ts, Hs and Rs, together with double that number of Es; etc. In this way I made nine boxes from which my grandson could take out complete sets to spell out the relevant digit. In total there was a prime number of each of the letters, with equal numbers of Ns and Vs, but more Ts. Furthermore, the grand total number of letters in the boxes was a two-figure prime.

In the order ONE to NINE, how many sets were in each box?

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BRG permalink


from itertools import product

# the box numbers as text
words = "ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE".split()

pr = {3, 5, 7}
prms = {2} | pr | set(n for n in range(11, 100, 2) if all(n % p for p in pr))

# there are a total of 36 letters so the maximum multiplier M(max) 
# is set by 3 x M(max) + 33 x 1 <= 99, giving M(max) less than 23
ltr_prms = [p for p in prms if p < 23]
mp_range = range(1, 23)

# the letters G, U, W and X occur only in EIGHT, FOUR, TWO and SIX 
# respectively so n8, n4, n2 and n6 must all be prime
for G, U, W, X in product(ltr_prms, repeat=4):
  n8, n4, n2, n6 = G, U, W, X
  
  for n3 in mp_range:
    H = n3 + n8           # H is in THREE and EIGHT
    R = n3 + n4           # R is in THREE and FOUR
    T = n2 + n3 + n8      # T is in TWO, THREE and EIGHT
    if not {H, R, T} < prms:
      continue
    
    # O is in ONE, TWO and FOUR
    for n1 in mp_range:
      if (O := n1 + n2 + n4) not in prms:
        continue
      
      # F is in FOUR and FIVE 
      for n5 in mp_range:
        F = n4 + n5
        # N = n1 + n7 + 2.n9 = V = n5 + n7 ==> n9 = (n5 - n1) / 2
        n9, r = divmod(n5 - n1, 2)
        # I is in FIVE, SIX, EIGHT and NINE
        I = n5 + n6 + n8 + n9        
        if r or not (n9 in mp_range) or not {F, I} < prms:
          continue

        # V is in FIVE and SEVEN and less than T
        for n7 in mp_range:
          N = V = n5 + n7
          if V >= T or V not in prms:
            continue
          
          # S is in SIX and SEVEN
          # E is in ONE, THREE, FIVE, SEVEN, EIGHT and NINE
          S = n6 + n7
          E = n1 + n5 + n8 + n9 + 2 * (n3 + n7)
          if not {E, S} < prms:
            continue
          
          # the multiples of the nine boxes
          ms = (n1, n2, n3, n4, n5, n6, n7, n8, n9)
          # the overall number of letters is prime 
          sm = sum(m * len(w) for m, w in zip(ms, words))
          if sm not in prms:
            continue
          
          ss = []
          for ms, w in zip(ms, words):
            ss.append(f"{ms}x {w}")
          print(f"{'; '.join(ss)}")
          print(f"total letters = {sm} ({E=} {F=} {G=} {H=} {I=} "
                f"{N=} {O=} {R=} {S=} {T=} {U=} {V=} {W=} {X=})")

from itertools import product

# the box numbers as text

words = "ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE".split()

pr = {3, 5, 7}

prms = {2} | pr | set(n for n in range(11, 100, 2) if all(n % p for p in pr))

# there are a total of 36 letters so the maximum multiplier M(max)

# is set by 3 x M(max) + 33 x 1 <= 99, giving M(max) less than 23

ltr_prms = [p for p in prms if p < 23]

mp_range = range(1, 23)

# the letters G, U, W and X occur only in EIGHT, FOUR, TWO and SIX

# respectively so n8, n4, n2 and n6 must all be prime

for G, U, W, X in product(ltr_prms, repeat=4):

n8, n4, n2, n6 = G, U, W, X

for n3 in mp_range:

H = n3 + n8 # H is in THREE and EIGHT

R = n3 + n4 # R is in THREE and FOUR

T = n2 + n3 + n8 # T is in TWO, THREE and EIGHT

if not {H, R, T} < prms:

continue

# O is in ONE, TWO and FOUR

for n1 in mp_range:

if (O := n1 + n2 + n4) not in prms:

continue

# F is in FOUR and FIVE

for n5 in mp_range:

F = n4 + n5

# N = n1 + n7 + 2.n9 = V = n5 + n7 ==> n9 = (n5 - n1) / 2

n9, r = divmod(n5 - n1, 2)

# I is in FIVE, SIX, EIGHT and NINE

I = n5 + n6 + n8 + n9

if r or not (n9 in mp_range) or not {F, I} < prms:

continue

# V is in FIVE and SEVEN and less than T

for n7 in mp_range:

N = V = n5 + n7

if V >= T or V not in prms:

continue

# S is in SIX and SEVEN

# E is in ONE, THREE, FIVE, SEVEN, EIGHT and NINE

S = n6 + n7

E = n1 + n5 + n8 + n9 + 2 * (n3 + n7)

if not {E, S} < prms:

continue

# the multiples of the nine boxes

ms = (n1, n2, n3, n4, n5, n6, n7, n8, n9)

# the overall number of letters is prime

sm = sum(m * len(w) for m, w in zip(ms, words))

if sm not in prms:

continue

ss = []

for ms, w in zip(ms, words):

ss.append(f"{ms}x {w}")

print(f"{'; '.join(ss)}")

print(f"total letters = {sm} ({E=} {F=} {G=} {H=} {I=} "

f"{N=} {O=} {R=} {S=} {T=} {U=} {V=} {W=} {X=})")

Frits permalink

# why is this necessary?

–> if you remove the n6 loop it will work fine. n6 is already set at the value of X.

Reply
- BRG permalink
  
  Thanks Frits, I looked at this code for literally hours last night without seeing this!
  
  Reply

Frits permalink

Without making the assumption that there are no more than 9 sets of letters in any box we can use the following.


highP = sorted(prms)[-1]
lngths = sorted(len(w) for w in words)
maxnum = (highP - sum(lngths[1:])) // lngths[0]

# the following two settings exhaustively explore the solution space
ltr_prms = [p for p in prms if p <= maxnum]
mp_range = range(1, maxnum + 1)

highP = sorted(prms)[-1]

lngths = sorted(len(w) for w in words)

maxnum = (highP - sum(lngths[1:])) // lngths[0]

# the following two settings exhaustively explore the solution space

ltr_prms = [p for p in prms if p <= maxnum]

mp_range = range(1, maxnum + 1)

Variable maxnum can even be determined to a lower value considering we have already a prime number of sets of letters in four boxes .

BRG permalink

Jim Randell has produced a neat solution to this teaser here. Here is my version of Jim’s code:


from collections import Counter

# the labels for the boxes
words = "ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE".split()

# primes less than 100
pr = {3, 5, 7}
prms = {2} | pr | set(x for x in range(11, 100, 2) if all(x % p for p in pr))

# solve progressively for the words in <wds> 
def solve(wds, m=Counter(), d=dict()):
  if not wds:
    # the overall number of letters is prime
    if not m.total() in prms:
      return
    # the number of N's and V's used are equal
    # but T is used more
    if not (m['T'] > m['N'] == m['V']):
      return
    yield m, d
  else:
    # collect the words used by each letter
    l2wds = {}
    for w in wds:
      for s in set(w):
        l2wds[s] = l2wds.get(s, set()) | {w}
    # pick letters used in the fewest words
    ltr = min(l2wds.keys(), key=(lambda s: len(l2wds[s])))
    # consider the word(s)
    for w in l2wds[ltr]:
      # the remaining words
      wds_ = [x for x in wds if x != w]
      # and the letters in our word that don't appear in them
      xs = set(w).difference(*wds_)
      m_ = m.copy()
      d_ = d.copy()
      for i in range(1, 100):
        m_ = m_ + Counter(w)
        if m_.total() >= 100:
          break
        # all our letters that don't appear later must be prime
        if not all((m_[x]) in prms for x in xs):
          continue
        d_[w] = i
        # attempt to solve for the remaining words
        yield from solve(wds_, m_, d_)

for m, d in solve(words):
  print('; '.join((f"{d[w]}x {w}" for w in words)))
  cs = ' '.join(f'{k}={v}' for k, v in sorted(m.items()))
  print(f"total letters = {m.total()} ({cs})")
  print()

from collections import Counter

# the labels for the boxes

words = "ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE".split()

# primes less than 100

pr = {3, 5, 7}

prms = {2} | pr | set(x for x in range(11, 100, 2) if all(x % p for p in pr))

# solve progressively for the words in <wds>

def solve(wds, m=Counter(), d=dict()):

if not wds:

# the overall number of letters is prime

if not m.total() in prms:

return

# the number of N's and V's used are equal

# but T is used more

if not (m['T'] > m['N'] == m['V']):

return

yield m, d

else:

# collect the words used by each letter

l2wds = {}

for w in wds:

for s in set(w):

l2wds[s] = l2wds.get(s, set()) | {w}

# pick letters used in the fewest words

ltr = min(l2wds.keys(), key=(lambda s: len(l2wds[s])))

# consider the word(s)

for w in l2wds[ltr]:

# the remaining words

wds_ = [x for x in wds if x != w]

# and the letters in our word that don't appear in them

xs = set(w).difference(*wds_)

m_ = m.copy()

d_ = d.copy()

for i in range(1, 100):

m_ = m_ + Counter(w)

if m_.total() >= 100:

break

# all our letters that don't appear later must be prime

if not all((m_[x]) in prms for x in xs):

continue

d_[w] = i

# attempt to solve for the remaining words

yield from solve(wds_, m_, d_)

for m, d in solve(words):

print('; '.join((f"{d[w]}x {w}" for w in words)))

cs = ' '.join(f'{k}={v}' for k, v in sorted(m.items()))

print(f"total letters = {m.total()} ({cs})")

print()

Sunday Times Teaser 3163 – Letterboxes

by Victor Bryant

Published Sunday May 07 2023 (link)

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