Sunday Times Teaser 2621 – Come Dancing

by Angela Newing

Published: 16 December 2012 (link)

Angie, Bianca, Cindy and Dot were given dance partners and performed in front of four judges, Juno, Kraig, Len and Marcia. Each judge placed the performances in order and gave 4 marks to the best, then 3, 2 and 1 point to the others. The dancers’ marks were then added up and they finished in alphabetical order with no ties. Angie’s winning margin over Bianca was larger than Bianca’s over Cindy, and Bianca’s winning margin over Cindy was larger than Cindy’s over Dot. Juno’s ordering of the four was different from the final order, and Kraig’s 4 points and Len’s 3 points went to dancers who were not in the top two.

How many points did Cindy get from Juno, Kraig, Len and Marcia respectively?

3 Comments Leave one →

Brian Gladman permalink

Here is my solution


from itertools import permutations

points = (4, 3, 2, 1)
for K in permutations(points):
  # K's 4 not awarded to top two
  if 4 in K[:2]:
    continue
  for L in permutations(points):
    # L's 3 not awarded to top two
    if 3 in L[:2]:
      continue
    for J in permutations(points):
      # J's order different to the result
      if J == points:
        continue
      for M in permutations(points):
        scores = list(zip(J, K, L, M))
        A, B, C, D = list(sum(x) for x in scores)
        # result and margins in alphabetical order 
        if A > B > C > D and A - B > B - C and B - C > C - D:
          print(scores[2])

from itertools import permutations

points = (4, 3, 2, 1)

for K in permutations(points):

# K's 4 not awarded to top two

if 4 in K[:2]:

continue

for L in permutations(points):

# L's 3 not awarded to top two

if 3 in L[:2]:

continue

for J in permutations(points):

# J's order different to the result

if J == points:

continue

for M in permutations(points):

scores = list(zip(J, K, L, M))

A, B, C, D = list(sum(x) for x in scores)

# result and margins in alphabetical order

if A > B > C > D and A - B > B - C and B - C > C - D:

print(scores[2])

Frits permalink

More efficient to loop over points for Angie, Bianca, Cindy and Dot.
This puzzle I would normally solve with the Enigma.py SubstitutedExpression() solver.


from itertools import product

# A > 13 as sum(13, 10, 8, 7) = 38 (below 40)
minA = next(i for i in range(11, 17) 
        if i + sum(i - j for j in [3, 5, 6]) > 39)

# B > 9 as sum(9, 7, 6) = 23 leaving 17 for A
minB = next(i for i in range(8, 17) 
        if 40 - (i + sum(i - j for j in [2, 3])) < 17)

points = (4, 3, 2, 1)
for A in product(points, repeat=4):
  if (sumA := sum(A)) < minA: continue
  # K's 4 not awarded to top two
  # L's 3 not awarded to top two
  if A[1] == 4 or A[2] == 3: continue
  for B in product(points, repeat=4):
    if (sumB := sum(B)) < minB: continue
    # K's 4 not awarded to top two
    # L's 3 not awarded to top two
    if B[1] == 4 or B[2] == 3: continue
    if sumA - sumB < 3: continue 
    # different scores from A
    if any(B[i] == A[i] for i in range(4)): continue
    for C in product(points, repeat=4):
      if any(C[i] in {A[i], B[i]} for i in range(4)): continue
      if sumB - (sumC := sum(C)) < 2 or sumA - sumB <= sumB - sumC: continue
      D = tuple(set(range(1, 5)).difference([A[i], B[i], C[i]]).pop() for i in range(4))
      if sumC - (sumD := sum(D)) < 1 or sumB - sumC <= sumC - sumD: continue
      # J's order different to the result
      if (A[0], B[0], C[0], D[0]) == points: continue
      print(f"answer: {C}")

from itertools import product

# A > 13 as sum(13, 10, 8, 7) = 38 (below 40)

minA = next(i for i in range(11, 17)

if i + sum(i - j for j in [3, 5, 6]) > 39)

# B > 9 as sum(9, 7, 6) = 23 leaving 17 for A

minB = next(i for i in range(8, 17)

if 40 - (i + sum(i - j for j in [2, 3])) < 17)

points = (4, 3, 2, 1)

for A in product(points, repeat=4):

if (sumA := sum(A)) < minA: continue

# K's 4 not awarded to top two

# L's 3 not awarded to top two

if A[1] == 4 or A[2] == 3: continue

for B in product(points, repeat=4):

if (sumB := sum(B)) < minB: continue

# K's 4 not awarded to top two

# L's 3 not awarded to top two

if B[1] == 4 or B[2] == 3: continue

if sumA - sumB < 3: continue

# different scores from A

if any(B[i] == A[i] for i in range(4)): continue

for C in product(points, repeat=4):

if any(C[i] in {A[i], B[i]} for i in range(4)): continue

if sumB - (sumC := sum(C)) < 2 or sumA - sumB <= sumB - sumC: continue

D = tuple(set(range(1, 5)).difference([A[i], B[i], C[i]]).pop() for i in range(4))

if sumC - (sumD := sum(D)) < 1 or sumB - sumC <= sumC - sumD: continue

# J's order different to the result

if (A[0], B[0], C[0], D[0]) == points: continue

print(f"answer: {C}")

Frits permalink

This subprogram will print only one possible solution for total points for Angie, Bianca, Cindy and Dot. After this a manual solution is easier than another program.


from math import prod

tri = lambda n: (n * (n + 1)) // 2

# decompose <t> into <k> numbers from <ns> with increasing gaps 
def decompose(t, k, ns, s=[]):
  if not k:
    # K's 4 not awarded to top two
    if not t and s[-1] < 16:
      yield s[::-1]
  else:
    for i, n in enumerate(ns):
      if len(s) < 2:
        gap = 0 
      else:
        gap = s[-1] - s[-2]
        # K's 4 and L's 3 not awarded to top two
        if s[0] + s[1] < 15:  # min bottom two: 4 * 1 + 2 * 2 + 3 + 4
          break   
      if s and n - s[-1] <= gap: continue
      # after n and gap <n - s[-1]> at least a specific number must still 
      # follow      
      #if s and t - n < sum([n  + tri(n - s[-1] + i) - tri(n - s[-1])
      #                     for i in range(1, k)]): continue
      
      # points left <= maximum points in next round
      if t - n > 4 * prod([4, 3, 2, 1][:k - 1]): continue
      yield from decompose(t - n, k - 1, ns[i + 1:], s + [n])  

for a, b, c, d in decompose(40, 4, range(4, 17)):
  print((a, b, c, d))

from math import prod

tri = lambda n: (n * (n + 1)) // 2

# decompose <t> into <k> numbers from <ns> with increasing gaps

def decompose(t, k, ns, s=[]):

if not k:

# K's 4 not awarded to top two

if not t and s[-1] < 16:

yield s[::-1]

else:

for i, n in enumerate(ns):

if len(s) < 2:

gap = 0

else:

gap = s[-1] - s[-2]

# K's 4 and L's 3 not awarded to top two

if s[0] + s[1] < 15: # min bottom two: 4 * 1 + 2 * 2 + 3 + 4

break

if s and n - s[-1] <= gap: continue

# after n and gap <n - s[-1]> at least a specific number must still

# follow

#if s and t - n < sum([n + tri(n - s[-1] + i) - tri(n - s[-1])

# for i in range(1, k)]): continue

# points left <= maximum points in next round

if t - n > 4 * prod([4, 3, 2, 1][:k - 1]): continue

yield from decompose(t - n, k - 1, ns[i + 1:], s + [n])

for a, b, c, d in decompose(40, 4, range(4, 17)):

print((a, b, c, d))

Sunday Times Teaser 2621 – Come Dancing

by Angela Newing

Published: 16 December 2012 (link)

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