Sunday Times Teaser 3163 – Letterboxes

by BRG on May 6, 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?

From → Uncategorized

5 Comments Leave one →

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)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta