Sunday Times Teaser 2814 – Today’s Teaser

by Andrew Skidmore

Published: 28 August 2016 (link)

Today’s Teaser concerns writing a date as a string of six digits: for example, today’s date can be written as 280816. Our puzzle concerns two numbers that add up to a date. I have written down two six-figure numbers and then consistently replaced digits by letters. In this way the two numbers have become

TODAYS TEASER.

The sum of these two numbers equals the six-digit date of a Sunday in July, August or September this year.

What is the value of my TEASER?

2 Comments Leave one →

Brian Gladman permalink

Straightforward using the AlphaSum alphametic sum solver.


from datetime import date
from alphasum import AlphaSum

# the year of puzzle publication
year = 2016 % 100
# try the months July, August and September
for month in range(7, 10):  
  # and the days in these months with two digits
  for day in range(10, (31, 31, 30)[month - 7] + 1):
    # consider only dates that are Sundays
    if date(year, month, day).weekday() == 6:
      # convert the date into a six digit number as a string
      dt = '{:02d}{:02d}{:02d}'.format(day, month, year)
      # use the AlphaSum solver with this as the sum value
      # to find solutions for the alphametic sum
      AlphaSum(['TODAYS', 'TEASER'], dt).go()

from datetime import date

from alphasum import AlphaSum

# the year of puzzle publication

year = 2016 % 100

# try the months July, August and September

for month in range(7, 10):

# and the days in these months with two digits

for day in range(10, (31, 31, 30)[month - 7] + 1):

# consider only dates that are Sundays

if date(year, month, day).weekday() == 6:

# convert the date into a six digit number as a string

dt = '{:02d}{:02d}{:02d}'.format(day, month, year)

# use the AlphaSum solver with this as the sum value

# to find solutions for the alphametic sum

AlphaSum(['TODAYS', 'TEASER'], dt).go()

A permutation based solution is, however, faster:


from itertools import permutations
from datetime import date

# T is 1 since the leading digit of the date is 3 or less
T = 1
# use YS + ER == 16 mod 100 to compute S from E, R and Y 
for E, R, Y in permutations(set(range(10)).difference([T]), 3):
  S = (16 - 10 * (E + Y) - R) % 100
  if S < 10 and S not in {E, R, Y}:
    
    # permute the remaining digits for the remaining letters
    for A, D, O in permutations(
          set(range(10)).difference([E, R, S, T, Y]), 3):
      
      # compose the two addends and their sum
      TODAYS = 100000 * T + 10000 * O + 1000 * D + 100 * A + 10 * Y + S
      TEASER = 100000 * T + 10000 * E + 1000 * A + 100 * S + 10 * E + R
      sm = TODAYS + TEASER
      
      # form the date and check that the month and day are in range
      y, m, d = dt = [int(str(sm)[i:i+2]) for i in (4, 2, 0)]
      # check that the month and day are in range and that the
      # date is a Sunday
      if (7 <= m <= 9 and 1 <= d <= (31, 31, 30)[m - 7] 
                              and date(*dt).weekday() == 6):
        fs = "TODAY'S = {}, TEASER = {}, DATE = {}."
        print(fs.format(TODAYS, TEASER, sm))

from itertools import permutations

from datetime import date

# T is 1 since the leading digit of the date is 3 or less

T = 1

# use YS + ER == 16 mod 100 to compute S from E, R and Y

for E, R, Y in permutations(set(range(10)).difference([T]), 3):

S = (16 - 10 * (E + Y) - R) % 100

if S < 10 and S not in {E, R, Y}:

# permute the remaining digits for the remaining letters

for A, D, O in permutations(

set(range(10)).difference([E, R, S, T, Y]), 3):

# compose the two addends and their sum

TODAYS = 100000 * T + 10000 * O + 1000 * D + 100 * A + 10 * Y + S

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

sm = TODAYS + TEASER

# form the date and check that the month and day are in range

y, m, d = dt = [int(str(sm)[i:i+2]) for i in (4, 2, 0)]

# check that the month and day are in range and that the

# date is a Sunday

if (7 <= m <= 9 and 1 <= d <= (31, 31, 30)[m - 7]

and date(*dt).weekday() == 6):

fs = "TODAY'S = {}, TEASER = {}, DATE = {}."

print(fs.format(TODAYS, TEASER, sm))

Jim Randell permalink

There is a very limited set of results to the sum, so my first attempt was just to feed them directly into the generic Alphametic solver in the enigma.py library:


% pypy enigma.py SubstitutedExpression --answer="TEASER" "TODAYS + TEASER in [100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916]" 
(TODAYS + TEASER in [100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916])
(153790 + 127026 in [100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916]) / A=7 D=3 E=2 O=5 R=6 S=0 T=1 Y=9 / 127026
TEASER = 127026 [1 solution]

% pypy enigma.py SubstitutedExpression --answer="TEASER" "TODAYS + TEASER in [100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916]"

(TODAYS + TEASER in [100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916])

(153790 + 127026 in [100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916]) / A=7 D=3 E=2 O=5 R=6 S=0 T=1 Y=9 / 127026

TEASER = 127026 [1 solution]

This runs in 590ms.

Pleasingly the result of the sum is the date the puzzle was published.

We can speed things up by applying the day, month and year restrictions in separate expressions:


% python enigma.py SubstitutedExpression \
    --invalid="0,T" \
    --answer="TEASER" \
    "(YS + ER) % 100 = 16" \
    "6 < ((DAYS + ASER) // 100) % 100 < 10" \
    "(TODAYS + TEASER) // 100 in (1007, 1707, 2407, 3107, 1408, 2108, 2808, 1109, 1809, 2509)"
((YS + ER) % 100 = 16) (6 < ((DAYS + ASER) // 100) % 100 < 10) ((TODAYS + TEASER) // 100 in (1007, 1707, 2407, 3107, 1408, 2108, 2808, 1109, 1809, 2509))
((90 + 26) % 100 = 16) (6 < ((3790 + 7026) // 100) % 100 < 10) ((153790 + 127026) // 100 in (1007, 1707, 2407, 3107, 1408, 2108, 2808, 1109, 1809, 2509)) / A=7 D=3 E=2 O=5 R=6 S=0 T=1 Y=9 / 127026
TEASER = 127026 [1 solution]

% python enigma.py SubstitutedExpression \

--invalid="0,T" \

--answer="TEASER" \

"(YS + ER) % 100 = 16" \

"6 < ((DAYS + ASER) // 100) % 100 < 10" \

"(TODAYS + TEASER) // 100 in (1007, 1707, 2407, 3107, 1408, 2108, 2808, 1109, 1809, 2509)"

((YS + ER) % 100 = 16) (6 < ((DAYS + ASER) // 100) % 100 < 10) ((TODAYS + TEASER) // 100 in (1007, 1707, 2407, 3107, 1408, 2108, 2808, 1109, 1809, 2509))

((90 + 26) % 100 = 16) (6 < ((3790 + 7026) // 100) % 100 < 10) ((153790 + 127026) // 100 in (1007, 1707, 2407, 3107, 1408, 2108, 2808, 1109, 1809, 2509)) / A=7 D=3 E=2 O=5 R=6 S=0 T=1 Y=9 / 127026

TEASER = 127026 [1 solution]

This runs in 80ms, which I think is sufficiently fast that I didn’t refine it further.

The enigma.py library is available from [ https://www.magwag.plus.com/jim/enigma.html ].

Sunday Times Teaser 2814 – Today’s Teaser

by Andrew Skidmore

Published: 28 August 2016 (link)

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