Sunday Times Teaser 3265 – Easter Prayer

by BRG on April 20, 2025

by Andrew Skidmore

Published Sunday April 20 2025 (link)

Millions of people will today turn their thoughts to those throughout the world whose lives are made miserable by the ravages of war and bitter conflict. I have taken a selection of letters from the alphabet and given each a different single-digit value. In this way the two words that form the title of this teaser represent two six-figure numbers, one of which is a factor of the other.

There are two letters in EASTER PRAYER for which the following is true. If I told you the value of that letter alone, you wouldn’t be able to work out all the others, but if I told you both their values then you could work out all the values.

What is the value of PRAYER?

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BRG permalink


from itertools import permutations, combinations
from collections import defaultdict

# translate the letter sequence <s> using the dictionary <d>
s2n = lambda d, s: int(''.join(str(d[c]) for c in s))

# find solutions for the alphametic puzzle in which 'easter' and 'prayer'
# are six digit numbers such that one is a factor of the other
ltrs = 'aeprsty'

# collect solutions indexed on letter/digit pairs  
ld2s = defaultdict(set)
# consider all arrangements of five different digits 
for p5 in permutations(range(10), 5):
  d = dict(zip('aerstpy', p5 + (0, 0)))
  if d['e'] == 0:
    continue 
  easter, _ra_er = (s2n(d, p) for p in ('easter', 'prayer'))
  # collect possible 'prayer' values by considering the multiplier/divisor
  pyr = ([p for m in range(2, 10) if 1e5 <= (p := easter * m) < 1e6] + 
         [p for m in range(2, 10) if (p := easter // m) >= 1e5 and not easter % m])
  for p in pyr:
    # check the known digits in prayer and extract values for 'p' and 'y'
    if p - 100000 - p % 1000 + p % 100 != _ra_er:
      continue
    d['p'], d['y'] = p // 100000, (p // 100) % 10
    easter, prayer = (s2n(d, w) for w in ('easter', 'prayer'))
    # map letter/digit pairs to solutions given by these assignments
    for l, d in d.items():
      ld2s[l, d] |= {(easter, prayer)}

# find letter/digit pairs that do not identify unique solutions
ldps = [ld for ld, s in ld2s.items() if len(s) > 1]

# look for a pair of letter/digit pairs that identifies a unique solution ...
# (credit John Z)
for (l1, d1), (l2, d2) in combinations(ldps, 2):
  if len(ld2s[l1, d1] & ld2s[l2, d2]) == 1:
    Easter, Prayer = (ld2s[l1, d1] & ld2s[l2, d2]).pop()
    print(f"{Prayer = } [{Easter = }, {l1} -> {d1}, {l2} -> {d2}].")

from itertools import permutations, combinations

from collections import defaultdict

# translate the letter sequence <s> using the dictionary <d>

s2n = lambda d, s: int(''.join(str(d[c]) for c in s))

# find solutions for the alphametic puzzle in which 'easter' and 'prayer'

# are six digit numbers such that one is a factor of the other

ltrs = 'aeprsty'

# collect solutions indexed on letter/digit pairs

ld2s = defaultdict(set)

# consider all arrangements of five different digits

for p5 in permutations(range(10), 5):

d = dict(zip('aerstpy', p5 + (0, 0)))

if d['e'] == 0:

continue

easter, _ra_er = (s2n(d, p) for p in ('easter', 'prayer'))

# collect possible 'prayer' values by considering the multiplier/divisor

pyr = ([p for m in range(2, 10) if 1e5 <= (p := easter * m) < 1e6] +

[p for m in range(2, 10) if (p := easter // m) >= 1e5 and not easter % m])

for p in pyr:

# check the known digits in prayer and extract values for 'p' and 'y'

if p - 100000 - p % 1000 + p % 100 != _ra_er:

continue

d['p'], d['y'] = p // 100000, (p // 100) % 10

easter, prayer = (s2n(d, w) for w in ('easter', 'prayer'))

# map letter/digit pairs to solutions given by these assignments

for l, d in d.items():

ld2s[l, d] |= {(easter, prayer)}

# find letter/digit pairs that do not identify unique solutions

ldps = [ld for ld, s in ld2s.items() if len(s) > 1]

# look for a pair of letter/digit pairs that identifies a unique solution ...

# (credit John Z)

for (l1, d1), (l2, d2) in combinations(ldps, 2):

if len(ld2s[l1, d1] & ld2s[l2, d2]) == 1:

Easter, Prayer = (ld2s[l1, d1] & ld2s[l2, d2]).pop()

print(f"{Prayer = } [{Easter = }, {l1} -> {d1}, {l2} -> {d2}].")

My initial solution had a more complicated final section which I replaced with the much better one provided by John Z below.

John Z permalink

Coding the s2n function on line 6 as:

Python

s2n = lambda d, s: sum(d[c] * 10 ** n for n, c in enumerate(s[::-1]))

1
2
3

s2n = lambda d, s: sum(d[c] * 10 ** n for n, c in enumerate(s[::-1]))

may result in a slight speed improvement. My ‘longhand’ calculation below is probably faster still, though.

Reply
- BRG permalink
  
  Yes, a bit faster but the direct arithmetic calculation with numeric constants
  will I think be much faster.
  
  Reply

John Z permalink


# dictionary of possible solutions indexed by letter and value
dLV = {}

sr10 = set(range(10))
for E in range(1, 10): # first letter cannot be 0
  for A in sr10 - {E}:
    for S in sr10 - {E, A}:
      for T in sr10 - {E, A, S}:
        for R in sr10 - {E, A, S, T}:
          for P in set(range(1, 10)) - {E, A, S, T, R}:
            for Y in sr10 - {E, A, S, T, R, P}:

              EASTER = E * 100000 + A * 10000 + S * 1000 + T * 100 + E * 10 + R

              PRAYER = P * 100000 + R * 10000 + A * 1000 + Y * 100 + E * 10 + R

              if not EASTER % PRAYER or not PRAYER % EASTER:

                print(EASTER, PRAYER)

                for i, v in zip('EASTRPY', (E, A, S, T, R, P, Y)):
                  
                  # insert EASTER PRAYER values in dictionary
                  dLV[(i, v)] = dLV.get((i, v), set()) | {(EASTER, PRAYER)}

# find letters that appear in two different EASTER PRAYER values
lv = [x[0] for x in dLV.items() if len(x[1]) == 2]

# solution = intersection of their EASTER PRAYER values
sol =(dLV[lv[0]] & dLV[lv[1]]).pop()

print ('EASTER =', sol[0], 'PRAYER =', sol[1])

# dictionary of possible solutions indexed by letter and value

dLV = {}

sr10 = set(range(10))

for E in range(1, 10): # first letter cannot be 0

for A in sr10 - {E}:

for S in sr10 - {E, A}:

for T in sr10 - {E, A, S}:

for R in sr10 - {E, A, S, T}:

for P in set(range(1, 10)) - {E, A, S, T, R}:

for Y in sr10 - {E, A, S, T, R, P}:

EASTER = E * 100000 + A * 10000 + S * 1000 + T * 100 + E * 10 + R

PRAYER = P * 100000 + R * 10000 + A * 1000 + Y * 100 + E * 10 + R

if not EASTER % PRAYER or not PRAYER % EASTER:

print(EASTER, PRAYER)

for i, v in zip('EASTRPY', (E, A, S, T, R, P, Y)):

# insert EASTER PRAYER values in dictionary

dLV[(i, v)] = dLV.get((i, v), set()) | {(EASTER, PRAYER)}

# find letters that appear in two different EASTER PRAYER values

lv = [x[0] for x in dLV.items() if len(x[1]) == 2]

# solution = intersection of their EASTER PRAYER values

sol =(dLV[lv[0]] & dLV[lv[1]]).pop()

print ('EASTER =', sol[0], 'PRAYER =', sol[1])

John Z permalink

A cleaner version using permutations instead of nested for loops is in the works.

Reply
- Frits permalink
  
  @John, your len(x[1]) == 2 in line 28 is too restrictive. E.g. adding 2 incorrect
  (E, A, S, T, R, P, Y) entries to dLV ([7, 9, 6, 3, 5, 1, 0] and [7, 9, 3, 8, 5, 1, 6])
  should not result in only one solution.
  
  Reply

John Z permalink


from itertools import permutations

# dictionary of possible solutions indexed by letter and value
dLV = {}

for E, A, S, T, R, P, Y in permutations(range(10), 7):
  if E > 0 and P > 0: # first letters cannot be zero

    EASTER = E * 100000 + A * 10000 + S * 1000 + T * 100 + E * 10 + R

    PRAYER = P * 100000 + R * 10000 + A * 1000 + Y * 100 + E * 10 + R

    if not EASTER % PRAYER or not PRAYER % EASTER:

      for L, v in zip('EASTRPY', (E, A, S, T, R, P, Y)):
        
        # insert EASTER PRAYER values in dictionary
        dLV[(L, v)] = dLV.get((L, v), set()) | {(EASTER, PRAYER)}

# find letter / value pairs that appear in two different EASTER PRAYER values
lv = [x[0] for x in dLV.items() if len(x[1]) == 2]

# solution = intersection of their EASTER PRAYER values
sol =(dLV[lv[0]] & dLV[lv[1]]).pop()

print ('EASTER =', sol[0], 'PRAYER =', sol[1])

from itertools import permutations

# dictionary of possible solutions indexed by letter and value

dLV = {}

for E, A, S, T, R, P, Y in permutations(range(10), 7):

if E > 0 and P > 0: # first letters cannot be zero

EASTER = E * 100000 + A * 10000 + S * 1000 + T * 100 + E * 10 + R

PRAYER = P * 100000 + R * 10000 + A * 1000 + Y * 100 + E * 10 + R

if not EASTER % PRAYER or not PRAYER % EASTER:

for L, v in zip('EASTRPY', (E, A, S, T, R, P, Y)):

# insert EASTER PRAYER values in dictionary

dLV[(L, v)] = dLV.get((L, v), set()) | {(EASTER, PRAYER)}

# find letter / value pairs that appear in two different EASTER PRAYER values

lv = [x[0] for x in dLV.items() if len(x[1]) == 2]

# solution = intersection of their EASTER PRAYER values

sol =(dLV[lv[0]] & dLV[lv[1]]).pop()

print ('EASTER =', sol[0], 'PRAYER =', sol[1])

John Z permalink

Here is a ‘less restrictive’


from itertools import permutations, combinations

# dictionary of possible solutions indexed by letter and value
dLV = {}

for E, A, S, T, R, P, Y in permutations(range(10), 7):
  if E > 0 and P > 0: # first letters cannot be zero

    EASTER = E * 100000 + A * 10000 + S * 1000 + T * 100 + E * 10 + R

    PRAYER = P * 100000 + R * 10000 + A * 1000 + Y * 100 + E * 10 + R

    if not EASTER % PRAYER or not PRAYER % EASTER:

      for L, v in zip('EASTRPY', (E, A, S, T, R, P, Y)):
        
        # insert EASTER PRAYER values in dictionary
        dLV[(L, v)] = dLV.get((L, v), set()) | {(EASTER, PRAYER)}

# two find letter / value pairs that satisfy criteria
for a, b in combinations(dLV.items(), 2):
  
  if len(a[1]) > 1 and len(b[1]) > 1 and len(a[1] & b[1]) == 1:

    print('Letters / values:', a[0], b[0])
    sol = (a[1] & b[1]).pop()
    print ('EASTER =', sol[0], 'PRAYER =', sol[1])

from itertools import permutations, combinations

# dictionary of possible solutions indexed by letter and value

dLV = {}

for E, A, S, T, R, P, Y in permutations(range(10), 7):

if E > 0 and P > 0: # first letters cannot be zero

EASTER = E * 100000 + A * 10000 + S * 1000 + T * 100 + E * 10 + R

PRAYER = P * 100000 + R * 10000 + A * 1000 + Y * 100 + E * 10 + R

if not EASTER % PRAYER or not PRAYER % EASTER:

for L, v in zip('EASTRPY', (E, A, S, T, R, P, Y)):

# insert EASTER PRAYER values in dictionary

dLV[(L, v)] = dLV.get((L, v), set()) | {(EASTER, PRAYER)}

# two find letter / value pairs that satisfy criteria

for a, b in combinations(dLV.items(), 2):

if len(a[1]) > 1 and len(b[1]) > 1 and len(a[1] & b[1]) == 1:

print('Letters / values:', a[0], b[0])

sol = (a[1] & b[1]).pop()

print ('EASTER =', sol[0], 'PRAYER =', sol[1])

approach

Sunday Times Teaser 3265 – Easter Prayer

by Andrew Skidmore

Published Sunday April 20 2025 (link)

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