Sunday Times Teaser 2875 – Easy as ABC

by Graham Smithers

Published October 29 2017 (link)

I have ten cards and on each is one of the letters A, B, C, E, L, N, T, V, W and Y. On the back of each card is a different digit.

The digits on T, E, N add to 10. The digits on T, W, E, L, V, E add to 12.

The digits on T, W, E, N, T, Y add to 20.

The digits on A, B, C add to ….

If I told you that last total, then you should be able to answer the following question:

What are the digits on T, E and N respectively?

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Brian Gladman permalink


from collections import defaultdict
from itertools import permutations

# index solutions on A + B + C
d = defaultdict(set)

# consider possible digits for T, E and N
s0 = set(range(10))
for T, E, N in permutations(s0, 3):
  if sum((T, E, N)) == 10:
    
    # consider possible digits for W, L and V  
    s1 = s0.difference([T, E, N])
    for W, L, V in permutations(s1, 3):
      if sum((T, W, E, L, V, E)) == 12:
    
        # conside the possible digit for Y
        s2 = s1.difference([W, L, V])
        for Y in s2:
          if sum((T, W, E, N, T, Y)) == 20:
            
            # the remaining digits are A, B and C
            d[sum(s2.difference([Y]))].add((T, E, N))

for sm, vset in d.items():
  if len(vset) == 1:
    (T, E, N), = vset    
    print(f'T = {T}, E = {E}, N = {N} (A + B + C = {sm})')

from collections import defaultdict

from itertools import permutations

# index solutions on A + B + C

d = defaultdict(set)

# consider possible digits for T, E and N

s0 = set(range(10))

for T, E, N in permutations(s0, 3):

if sum((T, E, N)) == 10:

# consider possible digits for W, L and V

s1 = s0.difference([T, E, N])

for W, L, V in permutations(s1, 3):

if sum((T, W, E, L, V, E)) == 12:

# conside the possible digit for Y

s2 = s1.difference([W, L, V])

for Y in s2:

if sum((T, W, E, N, T, Y)) == 20:

# the remaining digits are A, B and C

d[sum(s2.difference([Y]))].add((T, E, N))

for sm, vset in d.items():

if len(vset) == 1:

(T, E, N), = vset

print(f'T = {T}, E = {E}, N = {N} (A + B + C = {sm})')

This can also be solved using constraint programming. Here is a solution using Google’s Python interface to their Operational Research Tools:


from ortools.constraint_solver import pywrapcp
from collections import defaultdict

# 1. Set up the solver
solver = pywrapcp.Solver('Sunday Times Teaser 2875')

# 2. Create model variables
A, B, C, E, L, N, T, V, W, Y = dgts = [solver.IntVar(0, 9, x) for x in 'ABCELNTVWY']

# 3. Set up the constraints
solver.Add(solver.AllDifferent(dgts))
solver.Add(T + E + N == 10)
solver.Add(T + W + E + L + V + E == 12)
solver.Add(T + W + E + N + T + Y == 20)

# 4. Initialise the solver
solution = solver.Assignment()
solution.Add(dgts)
solver.NewSearch(solver.Phase(dgts, solver.CHOOSE_FIRST_UNBOUND, 
                                    solver.ASSIGN_MIN_VALUE))
# index solutions on the sum A + B + C
abc_to_ten = defaultdict(set)

# 5. Find possible solutions
while solver.NextSolution():
  for x in dgts:
    a, b, c, e, l, n, t, v, w, y = vals = [x.Value() for x in dgts]
    abc_to_ten[a + b + c].add((t, e, n))
solver.EndSearch()

# output solutions that are unique for the sum A + B + C
for sm, vset in abc_to_ten.items():
  if len(vset) == 1:
    (T, E, N), = vset
    print(f'T = {T}, E = {E}, N = {N} (A + B + C = {sm})')

from ortools.constraint_solver import pywrapcp

from collections import defaultdict

# 1. Set up the solver

solver = pywrapcp.Solver('Sunday Times Teaser 2875')

# 2. Create model variables

A, B, C, E, L, N, T, V, W, Y = dgts = [solver.IntVar(0, 9, x) for x in 'ABCELNTVWY']

# 3. Set up the constraints

solver.Add(solver.AllDifferent(dgts))

solver.Add(T + E + N == 10)

solver.Add(T + W + E + L + V + E == 12)

solver.Add(T + W + E + N + T + Y == 20)

# 4. Initialise the solver

solution = solver.Assignment()

solution.Add(dgts)

solver.NewSearch(solver.Phase(dgts, solver.CHOOSE_FIRST_UNBOUND,

solver.ASSIGN_MIN_VALUE))

# index solutions on the sum A + B + C

abc_to_ten = defaultdict(set)

# 5. Find possible solutions

while solver.NextSolution():

for x in dgts:

a, b, c, e, l, n, t, v, w, y = vals = [x.Value() for x in dgts]

abc_to_ten[a + b + c].add((t, e, n))

solver.EndSearch()

# output solutions that are unique for the sum A + B + C

for sm, vset in abc_to_ten.items():

if len(vset) == 1:

(T, E, N), = vset

print(f'T = {T}, E = {E}, N = {N} (A + B + C = {sm})')

Sunday Times Teaser 2875 – Easy as ABC

by Graham Smithers

Published October 29 2017 (link)

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