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  • Unknown's avatar

    Jim Randell 10:30 am on 6 September 2019 Permalink | Reply
    Tags:   

    Teaser 2804: Spots before the eyes 

    From The Sunday Times, 19th June 2016 [link] [link]

    At the opticians I was shown a sequence of six screens. On each there was a different number of spots ranging from 0 to 5 and I had to say how many I thought I could see. This was done with the left eye and then repeated with the right. The optician always used the same sequence and my answers were:

    (5 2 3 4 4 3)
    (4 1 4 5 2 3)

    In each case I got two correct. When I asked if this was the worst he’d seen, the optician showed me these three earlier sets of answers in which just one was correct in each:

    (2 2 1 2 1 4)
    (3 3 2 3 5 1)
    (0 4 5 1 3 2)

    What was the correct sequence?

    [teaser2804]

     
    • Jim Randell's avatar

      Jim Randell 10:33 am on 6 September 2019 Permalink | Reply

      Programatically we can consider all possible orderings of the numbers from 0 to 5, and look for a sequence that gives the appropriate number of matches in the 5 cases.

      This Python program runs in 39ms.

      Run: [ @repl.it ]

      from enigma import subsets, irange, printf

# sequences, values to check
ss = (
  ((5, 2, 3, 4, 4, 3), 2),
  ((4, 1, 4, 5, 2, 3), 2),
  ((2, 2, 1, 2, 1, 4), 1),
  ((3, 3, 2, 3, 5, 1), 1),
  ((0, 4, 5, 1, 3, 2), 1),
)

# count matches
def check(s1, s2):
  return sum(a == b for (a, b) in zip(s1, s2))

# consider possible sequences
for s in subsets(irange(0, 5), size=len, select="P"):
  if all(check(s, t) == n for (t, n) in ss):
    printf("{s}")

      Solution: The correct sequence is (4 2 0 1 5 3).

      Like

    • GeoffR's avatar

      GeoffR 11:00 am on 6 October 2020 Permalink | Reply 

      
% A Solution in MiniZinc 
include "globals.mzn";

% Correct sequence is A,B,C,D,E,F
var 0..5: A; var 0..5: B; var 0..5: C; var 0..5: D;
var 0..5: E; var 0..5: F;

constraint all_different ([A,B,C,D,E,F]);

% Two sequences with two correct answers
% Sequence 1 - 5 2 3 4 4 3
constraint sum ([A==5,B==2,C==3,D==4,E==4,F==3]) == 2;

% Sequence 2 - 4 1 4 5 2 3
constraint sum ([A==4,B==1,C==4,D==5,E==2,F==3]) == 2;

% Three sequences with one correct answers
% Sequence 3 - 2 2 1 2 1 4)
constraint sum ([A==2,B==2,C==1,D==2,E==1,F==4]) == 1;

% Sequence 4 - (3 3 2 3 5 1)
constraint sum ([A==3,B==3,C==2,D==3,E==5,F==1]) == 1;

% Sequence 5 - (0 4 5 1 3 2)
constraint sum ([A==0,B==4,C==5,D==1,E==3,F==2]) == 1;

solve satisfy;

output ["Correct sequence = " ++ show([A,B,C,D,E,F]) ];

% Correct sequence = [4, 2, 0, 1, 5, 3]
% ----------
% ==========

      Like

  • Unknown's avatar

    Jim Randell 9:43 am on 5 September 2019 Permalink | Reply
    Tags: by: K E Mawardy   

    Brainteaser 1675: The A-to-Z of sport 

    From The Sunday Times, 16th October 1994 [link]

    The individual sport or sports of the 26 members of the Venerable Sports Club means that they are just enough to form:

    a football team (11); or
    a hockey team (11); or
    a rugby team (15).

    The number of members playing only two of these sports is the same as the number of members playing rugby only, which is less than one third of the total membership.

    How many members play all three sports?

    This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

    [teaser1675]

     
    • Jim Randell's avatar

      Jim Randell 9:46 am on 5 September 2019 Permalink | Reply

      We can label the 7 non-empty parts of the Venn diagram, F, H, R, FH, FR, HR, FHR, according to which sports each member plays.

      We can then express the constraints in terms of these 7 integer variables.

      I expressed the constraints as a MiniZinc model, and then used the minizinc.py wrapper to collect the solutions and count them using Python.

      This program runs in 263ms.

      from minizinc import MiniZinc, var
from enigma import multiset, printf

# decision variables, the 7 non-empty areas of the venn diagram
vs = "F H R FH FR HR FHR"

# the MiniZinc model
model = f"""

  % declare the variables
  {var('0..26', vs.split())};

  % there are 26 members in total
  constraint F + H + R + FH + FR + HR + FHR = 26;

  % F's form a football team (11)
  constraint F + FH + FR + FHR = 11;

  % H's form a hockey team (11)
  constraint H + FH + HR + FHR = 11;

  % R's form a rugby team (15)
  constraint R + FR + HR + FHR = 15;

  % total number of two sports = number of only rugby
  constraint FH + FR + HR = R;

  % R is less than 1/3 of the total membership (= 26)
  constraint 3 * R < 26;

  solve satisfy;
"""

# create the model
m = MiniZinc(model)

# record FHR values
r = multiset()

# solve it
for (F, H, R, FH, FR, HR, FHR) in m.solve(result=vs):
  printf("[F={F} H={H} R={R} FH={FH} FR={FR} HR={HR} FHR={FHR}]")
  r.add(FHR)

# output solutions
for (k, v) in r.most_common():
  printf("FHR = {k} [{v} solutions]")

      Solution: 2 members play all three sports.

      There are 7 possible solutions:

      F=2 H=8 R=7 FH=1 FR=6 HR=0 FHR=2
      F=3 H=7 R=7 FH=1 FR=5 HR=1 FHR=2
      F=4 H=6 R=7 FH=1 FR=4 HR=2 FHR=2
      F=5 H=5 R=7 FH=1 FR=3 HR=3 FHR=2
      F=6 H=4 R=7 FH=1 FR=2 HR=4 FHR=2
      F=7 H=3 R=7 FH=1 FR=1 HR=5 FHR=2
      F=8 H=2 R=7 FH=1 FR=0 HR=6 FHR=2

      Like

      • Jim Randell's avatar

        Jim Randell 12:32 pm on 5 September 2019 Permalink | Reply

        And here’s a solution in Python. It runs in 93ms.

        Run: [ @repl.it ]

        from enigma import (irange, multiset, printf)

# decompose t into k numbers (min value m)
def decompose(t, k, m=0, s=[]):
  if k == 1:
    yield s + [t]
  else:
    for x in irange(m, t - k * m):
      yield from decompose(t - x, k - 1, m, s + [x])

# record FHR values
r = multiset()

# can form a football team of 11
for (F, FH, FR, FHR) in decompose(11, 4):
  # can form a hockey team of 11
  for (H, HR) in decompose(11 - FH - FHR, 2):
    # number of rugby only = total number of two sports
    R = FH + FR + HR
    # can form a rugby team of 15
    if not (R + FR + HR + FHR == 15): continue
    # R is less than 1/3 of the total membership (26)
    if not (3 * R < 26): continue
    # total membership is 26
    if not (F + H + R + FH + FR + HR + FHR == 26): continue

    printf("[F={F} H={H} R={R} FH={FH} FR={FR} HR={HR} FHR={FHR}]")
    r.add(FHR)

# output solutions
for (k, v) in r.most_common():
  printf("FHR = {k} [{v} solutions]")

        Like

        • John Crabtree's avatar

          John Crabtree 12:41 am on 6 September 2019 Permalink | Reply

          Let R members play rugby only and let A members play all three sports.
          Considering that there are 26 members and 37 players leads to 2A + R = 11,
          R ≤ 7 and so A ≥ 2.

          15 players play rugby, ie A + 2R ≥ 15, which leads to 3A ≤ 7.

          And so 2 members play all three sports.

          Liked by 1 person

  • Unknown's avatar

    Jim Randell 9:50 am on 3 September 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 497: [Alphabets Cricket Club] 

    From The Sunday Times, 6th December 1970 [link]

    All 26 members of the Alphabets Cricket Club (all surnames starting with a different letter of the alphabet) turned up for the annual general meeting at the Ugly Duck. Afterwards they divided into 5 teams of 5 and played a quick game of skittles, the lowest-scoring team to buy beer for the rest. (P. L. Zeigler sportingly stood down).

    Each player had just one ball at the 9 skittles, scoring, of course, a point for each skittle knocked down. The team captains varied in performance: Thomson got 9, Knight 5, Oldwhistle 3, C. F. Arrowroot 2, and the luckless Fisher 0. (Fisher’s team moved to have all captains’ scores ignored, but the plea was rejected; it was pointed out that anyhow Fisher’s weren’t due to buy the beer).

    The team totals were all different, as indeed they would have been had captains’ scores been ignored. No other player made the same score as any Captain, and no two team members of the same team made the same score. Oldwhistle’s team won.

    What was the order of finishing, and what were the individual scores of the team buying the beer?

    This puzzle was originally published with no title.

    [teaser497]

     
    • Jim Randell's avatar

      Jim Randell 9:50 am on 3 September 2019 Permalink | Reply

      This Python program considers the set of scores for the winning team, and then looks for possible scores for the remaining teams that gives the teams different scores (both including and excluding the captains scores).

      from enigma import (subsets, irange, printf)

# we can identify the teams by the captains scores (winning captain first)
caps = [ 3, 9, 5, 2, 0 ]

# team names (by captains score)
team = dict(zip(caps, "OTKAF"))

# possible remaining scores
scores = set(irange(0, 9)).difference(caps)

# solve for captains score in caps
# t1 = winning teams total
# t0 = winning teams total without captains score
# ss = scores so far
# t1s = team totals (with captains score)
# t0s = team totals (without captains score)
def solve(caps, t1, t0, ss=[], t1s=[], t0s=[]):
  # are we done?
  if not caps:
    yield ss
  else:
    # consider scores for the next team
    x = caps[0]
    for s in subsets(scores, size=4):
      s0 = sum(s)
      s1 = s0 + x
      # check scores are less than the winner
      if not (s1 < t1 and s0 < t0): continue
      # and also not repeated
      if s1 in t1s or s0 in t0s: continue
      # solve for the remaining captains
      yield from solve(caps[1:], t1, t0, ss + [s], t1s + [s1], t0s + [s0])
      

# find scores for the winning team
for s in subsets(scores, size=4):
  t = sum(s)
  # solve for the remaining teams
  for ss in solve(caps[1:], caps[0] + t, t):

    # record (0=team, 1=captain score, 2=other scores, 3=total score (with cap), 4=total score (without cap))
    rs = [ ('O', caps[0], s, caps[0] + t, t) ]
    for (c, x) in zip(caps[1:], ss):
      rs.append((team[c], c, x, c + sum(x), sum(x)))
    # sort according to total score (with cap)
    rs.sort(key=lambda t: t[3], reverse=1)

    # team F did not come last
    if rs[-1][0] == 'F': continue

    # output the table
    for (x0, x1, x2, x3, x4) in rs:
      printf("{x0}: {x1} + {x2} = {x3} ({x4})")
    printf()

      Solution: The finishing order was: 1st = Oldwhistle, 2nd = Thomson, 3rd = Knight, 4th = Fisher, 5th = Arrowroot. The individual scores of the team buying the beer were: 1, 2 (Arrowroot), 4, 6, 8.

      The complete set of results is:

      O: 3 + (4, 6, 7, 8) = 28 (25) = 1st (1st)
      T: 9 + (1, 4, 6, 7) = 27 (18) = 2nd (5th)
      K: 5 + (1, 4, 7, 8) = 25 (20) = 3rd (3rd)
      F: 0 + (1, 6, 7, 8) = 22 (22) = 4th (2nd)
      A: 2 + (1, 4, 6, 8) = 21 (19) = 5th (4th)

      There is also a “near miss” solution:

      O: 3 + (4, 6, 7, 8) = 28 (25) = 1st (1st)
      T: 9 + (1, 4, 6, 7) = 27 (18) = 2nd (5th)
      K: 5 + (1, 4, 7, 8) = 25 (20) = 3rd (3rd)
      A: 2 + (1, 6, 7, 8) = 24 (22) = 4th (2nd)
      F: 0 + (1, 4, 6, 8) = 19 (19) = 5th (4th)

      But this is disallowed as team F is last (and wouldn’t be last if the captains scores are ignored).

      Like

  • Unknown's avatar

    Jim Randell 10:41 am on 2 September 2019 Permalink | Reply
    Tags:   

    Teaser 2810: Almost a year apart 

    From The Sunday Times, 31st July 2016 [link] [link]

    Tomorrow’s date is 1st August 2016, which in the usual six-digit format is 01 08 16. This can be regarded as a “square” date since 10816 is a perfect square. No square date is followed by another square date exactly one year later, although some pairs of square dates are very nearly one year apart.

    Which two square dates (in their six-digit format) come closest to being a year apart?

    [teaser2810]

     
    • Jim Randell's avatar

      Jim Randell 10:43 am on 2 September 2019 Permalink | Reply

      Richard England likes puzzles involving square dates. See also: Enigma 1547, Enigma 1574, Enigma 1668, Enigma 1771.

      This Python program looks at possible squares and checked to see if they represent a viable date. It then looks for the two of these dates that are closest to the specified number of days apart.

      Run: [ @repl.it ]

      import datetime
from enigma import irange, sqrt, intc, intf, nsplit, Accumulator, subsets, printf

# accumulate results closest to N days apart
N = 365

# find square dates from years (20)00 - (20)99
dates = list()
for k in irange(intc(sqrt(10100)), intf(sqrt(311299))):
  (dd, mm, yy) = nsplit(k * k, base=100)
  try:
    d = datetime.date(2000 + yy, mm, dd)
  except ValueError:
    continue
  dates.append((d.toordinal(), dd, mm, yy))
dates.sort()

# find the two dates closest to N days apart
r = Accumulator(fn=min)
for p in subsets(dates, size=2):
  d = p[1][0] - p[0][0]
  r.accumulate_data(abs(N - d), p)

# output the solution
((i1, dd1, mm1, yy1), (i2, dd2, mm2, yy2)) = r.data
printf("{dd1:02d}.{mm1:02d}.{yy1:02} - {dd2:02d}.{mm2:02d}.{yy2:02d} = {d} days", d=i2 - i1)

      Solution: The two square dates closest to a year apart are 11.02.24 and 7.02.25.

      These dates are 362 days apart.

      The next closest dates are:

      26.01.00 – 01.02.01 = 372 days
      24.01.00 – 01.02.01 = 374 days

      Like

  • Unknown's avatar

    Jim Randell 7:29 am on 30 August 2019 Permalink | Reply
    Tags:   

    Teaser 2971: Six sisters on the ski lift 

    From The Sunday Times, 1st September 2019 [link] [link]

    The sum of the ages of six sisters known to me is 92. Though there is no single whole number greater than 1 that simultaneously divides the ages of any three of them, I did notice this morning, while they lined up for the ski lift, that the ages of any two of them standing one behind the other, had a common divisor greater than 1.

    In increasing order, how old are the six sisters?

    [teaser2971]

     
    • Jim Randell's avatar

      Jim Randell 8:06 am on 30 August 2019 Permalink | Reply

      This program builds up a list of the ages in the order they appear in the queue, such that each consecutive pair share a common factor (greater than 1), but no 3 of them have a common factor (greater than 1).

      It runs in 287ms.

      Run: [ @replit ]

      from enigma import (irange, mgcd, cached, subsets, printf)

# use a caching version of gcd
gcd = cached(mgcd)

# look for k numbers that sum to t
# each has a common factor with the previous number
# but no 3-set has a common factor > 1
def solve(t, k, m=1, ns=[]):
  if k == 0:
    yield ns
  else:
    # choose a value from the remaining total
    for n in (irange(m, t - (k - 1) * m) if k > 1 else [t]):
      # any new number must have a gcd > 1 with the previous number
      if len(ns) > 0 and gcd(n, ns[-1]) == 1: continue
      # but must have a gcd of 1 combined with all other pairs
      if any(gcd(n, a, b) > 1 for (a, b) in subsets(ns, size=2)): continue
      # solve for the remaining numbers
      yield from solve(t - n, k - 1, m, ns + [n])

# solve the puzzle
for ns in solve(92, 6, 2):
  printf("{s} -> {ns}", s=sorted(ns))

      Solution: The ages of the sisters are 7, 10, 13, 15, 21, 26.

      They are queuing in the order (7, 21, 15, 10, 26, 13), or the reverse of this.

      Which we can factorise as: (7, 7×3, 3×5, 5×2, 2×13, 13). So the common factors are 7, 3, 5, 2, 13.

      And, in fact, picking a sequence of different prime numbers and then multiplying them together in pairs will give us a sequence where consecutive pairs share a factor, but no three numbers share a common factor.

      However this procedure will not necessary find all possible solutions. For example, if the required total had been 96, there is a possible queue of (7, 7×3, 3×5, 5×2×2, 2×11, 11) that cannot be constructed in this way (note the 5×2×2 term), but it is found by the program given above.

      Like

      • Jim Randell's avatar

        Jim Randell 2:17 pm on 31 August 2019 Permalink | Reply

        Here is an alternative approach that uses a bit of analysis.

        Each consecutive pair of numbers in the queue must have some prime as a common factor. So we can start with a prime and then look at pairs that are multiples of that prime. We can then choose a different prime (if a prime is repeated we will have 3 numbers that are divisible by it) that divides the second number, and choose the third number that is a multiple of the new prime, and so we continue until we have all six numbers in the queue.

        This recursive Python 3 program is a little longer that my previous approach, but it runs quicker, and can be used to explore solutions for different totals and number of elements in the queue. (It also removes the solutions that that are reverse of another solution).

        For the parameters of the puzzle it runs in 80ms. (Internal runtime is 23ms).

        Run: [ @replit ]

        from enigma import (primes, irange, mgcd as gcd, subsets, arg, printf)

# T = total sum of numbers, K = number of numbers to consider
T = arg(92, 0, int)
K = arg(6, 1, int)
printf("[T={T} K={K}]")

# possible common prime factors
primes = set(primes.between(1, T // 3))

# return viable multiples of p
multiples = lambda p, t, k: irange(p, t - (k - 1) * 2, step=p)

# find k more numbers, that sum to t
# the next number should be a multiple of prime p
# ns = numbers found so far
# ps = primes used so far
def solve(k, t, p, ns=[], ps=set()):
  # are we done?
  if k == 1:
    if t % p == 0:
      yield ns + [t]
  else:
    # the next age n is some multiple of p and also some other prime q
    for q in primes.difference(ps):
      for n in multiples(p * q, t, k):
        yield from solve(k - 1, t - n, q, ns + [n], ps.union([q]))

# choose the first prime
for p in primes:
  # the first age is some multiple of p
  for n in multiples(p, T, K):
    # solve for the remaining 5 numbers
    for ns in solve(K - 1, T - n, p, [n], set([p])):
      if ns[0] > ns[-1]: continue # [optional] only consider queues one way round
      # check no 3-set has a gcd > 1
      if any(gcd(*s) > 1 for s in subsets(ns, size=3)): continue
      # output solution
      printf("{s} -> {ns}", s=sorted(ns))

        Like

    • Robert Brown's avatar

      Robert Brown 5:18 pm on 30 August 2019 Permalink | Reply

      That’s odd. My QuickBasic program (using a different algorithm!) runs in < 10ms.

      Like

  • Unknown's avatar

    Jim Randell 2:28 pm on 29 August 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 496: [Operatics] 

    From The Sunday Times, 29th November 1970 [link]

    The soprano, the tenor, and the manager are all well aware that the five operas in the repertoire will be sung on the first five evenings of next week.

    The tenor is expecting Ernani to be two nights after Don Giovanni and one night before Carmen, and he is expecting Bohème later in the week than Aida. The soprano is expecting Aida to be two nights before Ernani.

    The manager, who knows all their expectations and knows the true state of affairs, tells me that for every night next week the soprano and tenor had different operas on their lists — and neither is right; he adds that the true list shows Don Giovanni two nights before Bohème.

    I asked him what the tenor expected to sing on the evening booked for Aida; he told me, but I could still not be sure about next week’s true programme till he also told me what the soprano expected to sing on the evening booked for Aida. Then I knew the true programme.

    What do the soprano and tenor expect on the evening of Aida?

    This puzzle was originally published with no title.

    [teaser496]

     
    • Jim Randell's avatar

      Jim Randell 2:29 pm on 29 August 2019 Permalink | Reply

      This Python program selects possible sets of orderings for the manager, soprano and tenor, and then uses the [[ filter_unique() ]] function from the enigma.py library to reduce these to sets where knowledge of the tenor’s selection for the actual night of A is ambiguous, but the addition of the soprano’s selection gives a unique answer.

      This Python program runs in 90ms.

      from itertools import product
from enigma import (subsets, irange, filter_unique, unpack, join, printf)

# the five operas
operas = list("ABCDE")

# record possibilities for the manager, soprano, tenor
(ms, ss, ts) = (list(), list(), list())

# consider possible orders
for s in subsets(irange(1, 5), size=len, select="P"):
  (A, B, C, D, E) = s

  # tenor is expecting...
  # E then C then D, A before B
  if D + 2 == E and E + 1 == C and A < B:
    ts.append(s)

  # soprano is expecting...
  # A two nights before E
  if A + 2 == E:
    ss.append(s)

  # manager knows...
  # D two nights before B
  if D + 2 == B:
    ms.append(s)

# collect viable (m, s, t) orders
rs = list()

# choose a pair of orders for the soprano and tenor
for (s, t) in product(ss, ts):
  # they must differ in every position
  if any(a == b for (a, b) in zip(s, t)): continue

  # choose an order for the manager
  for m in ms:
    # it must differ from s and t in every position
    if any(a == b or a == c for (a, b, c) in zip(m, s, t)): continue

    rs.append((m, s, t))

# indices into the orders for A, B, C, D, E
(A, B, C, D, E) = (0, 1, 2, 3, 4)

# knowing what the tenor expects on the day of A is not enough to determine the correct order
(_, rs) = filter_unique(rs, unpack(lambda m, s, t: t.index(m[A])), unpack(lambda m, s, t: m))

# ... but knowing what the soprano expects as well is enough
(rs, _) = filter_unique(rs, unpack(lambda m, s, t: (t.index(m[A]), s.index(m[A]))), unpack(lambda m, s, t: m))

# output solutions
fmt = lambda x: join((n for (i, n) in sorted(zip(x, "ABCDE"))), sep=" ")
for (m, s, t) in rs:
  printf("m = {m}; s = {s}; t = {t}", m=fmt(m), s=fmt(s), t=fmt(t))

      Solution: On the night of Aida, the soprano is expecting Bohème, and the tenor is expecting Ernani.

      The complete orders are:

      C E D A B = manager
      D C A B E = soprano
      A D B E C = tenor

      Before the setter has asked any questions there is enough information that he can narrow the correct order down to one of four options.

      The setter asks the manager what the tenor expects to sing on the night booked for A. This is the 4th night, and so will get the answer E.

      This allows the setter to narrow the correct order down to one of two options.

      On also being told that the soprano expects to sing B on that night, only one of those options remains, giving the solution.

      Like

    • Pete Good's avatar

      Pete Good 9:43 am on 13 August 2023 Permalink | Reply

      Jim,

      I think that this teaser has two solutions. I found 10 sets of three schedules that appear to meet the constraints and not just 4 sets:-

      C E D A B , A D B E C, B C A D E
      E A D C B, A D B E C, C B A D E
      C E D A B, A D B E C, D B A C E
      C E D A B, A D B E C, D C A B E
      E D C B A, D A E C B, B C A D E
      E A D C B, A D B E C, B C A D E
      E C D A B, A D B E C, C B A D E
      E C D A B, A D B E C, D B A C E
      E A D C B, A D B E C, D C A B E
      E D C B A, D A E C B, C B A D E

      On the evening of Aida, the tenor and soprano are expecting ED, DB, EC, EB, BE, DC, ED, EC, DC or BE. There are two sets of schedules containing ED, EC, BE and DC so these can be discounted but DB and EB are both unique so it appears that there are solutions.

      Have I overlooked something?

      Like

      • Jim Randell's avatar

        Jim Randell 11:31 am on 13 August 2023 Permalink | Reply

        @Pete: You have found the right list of 10 possibilities.

        We are told that if we knew what the tenor was expecting on the day that A is actually to be performed, then that would not be enough to determine the correct order.

        Looking at the list, the tenor is expecting one of B, D, E to be performed on the day of A.

        Looking at the corresponding correct orders:

        B → (E D C B A)
        D → (E A D C B)
        E → (C E D A B) (E C D A B)

        The first 2 lead to a unique correct ordering, so we eliminate those cases and we are left with the 5 possibilities (where the tenor expects E to be performed on the day of A).

        From those 5 possibilities we look at what the soprano is expecting on the day of A.

        It is one of: B, C, D.

        Looking at the corresponding correct orders:

        B → (C E D A B)
        C → (C E D A B) (E C D A B)
        D → (C E D A B) (E C D A B)

        So only B will allow us to determine the correct ordering.

        Hence the soprano expects B, and the tenor expects E. And the correct ordering is (C E D A B).

        Like

    • Pete Good's avatar

      Pete Good 11:55 am on 13 August 2023 Permalink | Reply

      Thank you for explaining this.

      Like

  • Unknown's avatar

    Jim Randell 3:06 pm on 28 August 2019 Permalink | Reply
    Tags:   

    Teaser 2814: Today’s Teaser 

    From The Sunday Times, 28th August 2016 [link] [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?

    [teaser2814]

     
    • Jim Randell's avatar

      Jim Randell 3:07 pm on 28 August 2019 Permalink | Reply

      There are four potential dates in each of the given months, but removing those with a leading zero leaves 10 candidates.

      We can check if the alphametic sum corresponds to one of those dates in a single expression using the [[ SubstitutedExpression() ]] solver from the enigma.py library, but we can speed things up considerably by considering the year part separately, which must give 16 in the final two columns, and the month part which must give 07, 08, 09 in the next two columns.

      This run file executes in 168ms.

      Run: [ @replit ]

      #! python3 -m enigma -rr

SubstitutedExpression

--invalid="0,T"
--answer="TEASER"

# day/month must be:
# 10/07, 17/07, 24/07, 31/07 (Jul)
# 07/08, 14/08, 21/08, 28/08 (Aug)
# 04/09, 11/09, 18/09, 25/09 (Sep)
--code="dates = set([100716, 170716, 240716, 310716, 140816, 210816, 280816, 110916, 180916, 250916])"

# the result of the alphametic sum is one of the dates
"(TODAYS + TEASER) in dates"

# [optional] year
--code="yy = set(d % 100 for d in dates)"
"(YS + ER) % 100 in yy"

# [optional] month + year
--code="mmyy = set(d % 10000 for d in dates)"
"(DAYS + ASER) % 10000 in mmyy"

      Solution: TEASER = 127026.

      And so the sum is:

      153790 + 127026 = 280816

      So the date is 28th August 2016, the publication date of the puzzle.

      Like

  • Unknown's avatar

    Jim Randell 10:38 am on 27 August 2019 Permalink | Reply
    Tags:   

    Brainteaser 1661: Goals galore 

    From The Sunday Times, 10th July 1994 [link]

    Three football teams played each other once in a tournament. Athletic beat City, City beat Borough, and Borough beat Athletic.

    They could not decide which team should receive the trophy since:

    Athletic had scored the most goals;

    Borough had the best goal average (goals for ÷ goals against)

    City had the best goal difference (goals for − goals against)

    Fewer than 40 goals were scored in the tournament.

    What were the scores in the three games?

    This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

    [teaser1661]

     
    • Jim Randell's avatar

      Jim Randell 10:39 am on 27 August 2019 Permalink | Reply

      There are six bounded variables (the number of goals scored by each team in each of the three matches), and the conditions put further constraints on the variables.

      I used the MiniZinc system to solve the set of constraints, and the minizinc.py library to construct the constraints using Python.

      This Python 3 program runs in 228ms.

      from minizinc import MiniZinc, var
from enigma import join

# teams
(A, B, C) = teams = list('ABC')

# decision variables: XY = goals for X against Y in X vs Y match
vs = "AB BA AC CA BC CB"

# goals for / against team x
gf = lambda x: "(" + join((x + t for t in teams if t != x), sep=' + ') + ")"
ga = lambda x: "(" + join((t + x for t in teams if t != x), sep=' + ') + ")"

model = f"""

  % declare the variables
  {var('0..39', vs.split())};

  % the match outcomes
  constraint BA > AB; % B beat A
  constraint AC > CA; % A beat C
  constraint CB > BC; % C beat B

  % A scored the most goals
  constraint {gf(A)} > {gf(B)};
  constraint {gf(A)} > {gf(C)};

  % B had the best goal average
  constraint {gf(B)} * {ga(A)} > {gf(A)} * {ga(B)};
  constraint {gf(B)} * {ga(C)} > {gf(C)} * {ga(B)};

  % C had the best goal difference
  constraint {gf(C)} - {ga(C)} > {gf(A)} - {ga(A)};
  constraint {gf(C)} - {ga(C)} > {gf(B)} - {ga(B)};

  % fewer than 40 goals were scored in the tournament
  constraint AB + BA + AC + CA + BC + CB < 40;

  solve satisfy;
"""

# create the model
m = MiniZinc(model)

# solve the constraints
for (AB, BA, AC, CA, BC, CB) in m.solve(result=vs):
  print(f"A vs B = {AB}-{BA}; A vs C = {AC}-{CA}; B vs C = {BC}-{CB}")

      Solution: The scores in the matches are: A vs B = 3-7; A vs C = 13-12; B vs C = 0-3.

      In total there were 38 goals scored in the tournament.

      The different measures are:

      goals scored: A = 16, C = 15, B = 7
      goal average: B = 1.17, C = 1.15, A = 0.84
      goal difference: C = +2, B = +1, A = −3

      Like

    • GeoffR's avatar

      GeoffR 3:37 pm on 28 August 2019 Permalink | Reply

      Minizinc found two solutions for me, the first solution looks incorrect as the averages for B and C were the same(7/6 and 14/12, which does not agree with my constraint for goal averages.

      The second solution is a correct solution and agrees with the published solution.

      % A Solution in MiniZinc
include "globals.mzn";

% Teams A, B and C goal scores
% eg AB = goals for Team A playing Team B
% and BA = goals against Team A playing Team B
var 0..39:AB; var 0..39:BA;  
var 0..39:AC; var 0..39:CA; 
var 0..39:BC; var 0..39:CB; 

% goals for and against Teams A, B and C
var 0..39:gfA; var 0..39:gfB; var 0..39:gfC; 
var 0..39:gaA; var 0..39:gaB; var 0..39:gaC; 

% goal difference variables
var -10..39: gdifA; 
var -10..39: gdifB; 
var -10..39: gdifC; 

% goal average variables
var 0.01..10.1: gavA; 
var 0.01..10.1: gavB; 
var 0.01..10.1: gavC;

% total of all goals < 40
constraint AB + BA + AC + CA + BC + CB < 40;

% Team B beat Team A, Team A beat TeamC and Team C beat Team B
constraint BA > AB; 
constraint AC > CA; 
constraint CB > BC; 

% Goals for Teams A, B and C
constraint gfA = AB + AC;
constraint gfB = BA + BC;
constraint gfC = CA + CB;

% goals against Teams A, B and C
constraint gaA = BA + CA;
constraint gaB = AB + CB;
constraint gaC = AC + BC;

% A had the most goals of Teams A, B and C
constraint (AC + AB) > (BA + BC) /\ (AC + AB) > (CA + CB);

% C had the highest goal difference
constraint gdifA = AB + AC - BA - CA;
constraint gdifB = BA + BC - AB - CB;
constraint gdifC = CA + CB - AC - BC;

constraint gdifC > gdifA /\ gdifC > gdifB;

% B had the best goal average
constraint gavA = (AB + AC) / (BA + CA);
constraint gavB = (BA + BC) / (AB + CB);
constraint gavC = (CA + CB) / (AC + BC);

constraint gavB > gavC /\ gavB > gavA;

solve satisfy;

% show team scores in the three games
output [   "A v B = " ++ show(AB) ++ "-" ++ show(BA)
++ "\n" ++ "A v C = " ++ show(AC) ++ "-" ++ show(CA)
++ "\n" ++ "B v C = " ++ show(BC) ++ "-" ++ show(CB) ];

% A v B = 3-7
% A v C = 12-11
% B v C = 0-3
% ----------
% A v B = 3-7
% A v C = 13-12
% B v C = 0-3
% ----------
% ==========
% Finished in 433msec

% Detailed outputs
% AB = 3; BA = 7; AC = 12; CA = 11; BC = 0; CB = 3;
% gfA = 15; gfB = 7; gfC = 14;0
% gaA = 18; gaB = 6; gaC = 12;
% gdifA = -3; gdifB = 1; gdifC = 2;
% gavA = 0.833333333333333; gavB = 1.16666666666667; gavC = 1.16666666666667;
% ----------
% AB = 3; BA = 7; AC = 13; CA = 12; BC = 0; CB = 3;
% gfA = 16; gfB = 7; gfC = 15;
% gaA = 19; gaB = 6; gaC = 13;
% gdifA = -3; gdifB = 1; gdifC = 2;
% gavA = 0.842105263157895; gavB = 1.16666666666667; gavC = 1.15384615384615;
%----------
%==========

      Like

      • Jim Randell's avatar

        Jim Randell 9:18 am on 29 August 2019 Permalink | Reply

        @GeoffR: I suspect the second solution comes about because the calculations for “goal average” will be done using floating point numbers. floats are only approximations to real numbers, so if we compare two floats that should be equal, but have been arrived at by different calculations, it is likely that one will compare as larger than the other.

        I would change the model to keep things in the integer domain, by replacing constraints of the form:

        constraint a / b > c / d;

        with the equivalent:

        constraint a * d > c * b;

        Like

    • GeoffR's avatar

      GeoffR 3:12 pm on 29 August 2019 Permalink | Reply

      @Jim: Yes, I think you are right as I got a similar second incorrect solution in Python, with a similar approach. For me, it was an unforeseen consequence of using floats instead of integers.
      Thanks for the explanation.

      Like

      • Jim Randell's avatar

        Jim Randell 3:17 pm on 29 August 2019 Permalink | Reply

        Although in Python you can avoid floats by using the [[ fractions ]] module to give an exact representation of rational numbers for the goal averages.

        Like

  • Unknown's avatar

    Jim Randell 12:01 pm on 26 August 2019 Permalink | Reply
    Tags:   

    Teaser 2818: Accountability 

    From The Sunday Times, 25th September 2016 [link] [link]

    George and Martha’s bank account number is an 8-digit number consisting of the digits 1 to 8 in some order.

    Martha commented that, if you split the number into two 4-figure numbers, then the number formed by the last four digits is a multiple of the number formed by the first four digits.

    Then George listed all the different prime numbers that divided exactly into the account number. When he added together all those primes the total was less than a hundred.

    What is their account number?

    [teaser2818]

     
    • Jim Randell's avatar

      Jim Randell 12:01 pm on 26 August 2019 Permalink | Reply

      We can solve this puzzle using the general alphametic solver [[ SubstitutedExpression() ]] from the enigma.py library.

      The following run file executes in 218ms.

      #! python -m enigma -rr

SubstitutedExpression

# solver paraemters
--digits="1-8"
--answer="ABCDEFGH"

# expressions to solve
"EFGH % ABCD = 0"
"sum(p for (p, e) in prime_factor(ABCDEFGH)) < 100"

      Solution: The account number is 12876435.

      We have:

      6435 = 5 × 1287

      The number factorises as:

      12876435 = 3³ . 5 . 11 . 13 . 23 . 29

      So the sum of the prime divisors is 84.

      Like

  • Unknown's avatar

    Jim Randell 7:17 am on 25 August 2019 Permalink | Reply
    Tags: by: V W Lewis   

    Brain-Teaser 495: [Goal scorers] 

    From The Sunday Times, 22nd November 1970 [link]

    All five forwards (numbers 7-11) scored in their team’s eight-nil win. Their names, alphabetically, were Alan, Bob, Craig, Dave and Eric.

    Five goals were kicked (number 10 scoring the last of just three consecutive ones). Dave headed his second goal. He and two other forwards scored two goals each — heading one and kicking the other. Successive goals were scored by the other two forwards. No player scored two consecutive goals.

    The scores, in scoring order, were:

    1st goal: Number 11
    2nd goal: Alan
    3rd goal: Eric
    4th goal: Number 8
    5th goal: Bob
    6th goal: Number 7
    7th goal: Dave
    8th goal: Number 9

    What was each forward’s number?

    This puzzle was originally published with no title.

    [teaser495]

     
    • Jim Randell's avatar

      Jim Randell 7:18 am on 25 August 2019 Permalink | Reply

      This Python program runs in 95ms.

      from enigma import subsets, irange, tuples, diff, update, unpack, join, printf

names = "ABCDE"

# m: maps number (7-11) to name A-E
for s in subsets(irange(7, 11), size=len, select="P"):
  m = dict(zip(s, names))

  # use the map to fill out the list of goal scorers
  ns = [ m[11], 'A', 'E', m[8], 'B', m[7], 'D', m[9] ]

  # no-one scored consecutive goals
  if any(a == b for (a, b) in tuples(ns, 2)): continue

  # count the number of goals scored by each
  n = dict((x, ns.count(x)) for x in names)

  # D and 2 others scored 2 goals
  if n['D'] != 2: continue
  g2s = list(x for x in names if n[x] == 2)
  if len(g2s) != 3: continue

  # the 1 goal scorers goals were consecutive
  g1s = diff(names, g2s)
  if abs(ns.index(g1s[0]) - ns.index(g1s[1])) != 1: continue

  # assign the kicked goals we know (1 = kick, 0 = head)
  k = [ None ] * 8
  # D's goals are kick then head
  (j0, j1) = list(i for (i, x) in enumerate(ns) if x == 'D')
  (k[j0], k[j1]) = (1, 0)

  # number 10 scored the final goal of just three consecutive kicks
  for (i, x) in enumerate(ns):
    if x != m[10]: continue
    if i < 2: continue
    if k[i - 1] == 0 or k[i - 2] == 0: continue
    k[i] = k[i - 1] = k[i - 2] = 1
    if i > 2:
      if k[i - 3] == 1: continue
      k[i - 3] = 0
    if i < 7:
      if k[i + 1] == 1: continue
      k[i + 1] = 0

    # fill out any remaining 5 kicked goals
    j = k.count(1)
    if j > 5: continue
    ms = list(i for (i, x) in enumerate(k) if x is None)
    for js in subsets(ms, size=5 - j, select="P"):
      k2 = update(k, ms, list((1 if x in js else 0) for x in ms))

      # check the other scorers of 2 goals are a kick and a head
      if any(len(set(k2[i] for (i, x) in enumerate(ns) if x == y)) == 1 for y in diff(g2s, 'D')): continue

      # output solution
      for (k, v) in sorted(m.items(), key=unpack(lambda k, v: v)):
        printf("{v} = #{k}")
      printf("[goals = {gs}]", gs=join((n + "hk"[i] for (n, i) in zip(ns, k2)), sep=","))

      Solution: Alan is number 9. Bob is number 10. Craig is number 8. Dave is number 11. Eric is number 7.

      The goals scored are:

      1. Dave (#11), kicked
      2. Alan (#9), headed
      3. Eric (#7), kicked
      4. Craig (#8), kicked
      5. Bob (#10), kicked
      6. Eric (#7), headed
      7. Dave (#11), headed
      8. Alan (#9), kicked

      Like

  • Unknown's avatar

    Jim Randell 1:56 am on 24 August 2019 Permalink | Reply
    Tags:   

    Teaser 2970: Puzzling powers 

    From The Sunday Times, 25th August 2019 [link] [link]

    I came across a most remarkable number recently and wrote it down. All its digits were different and it was divisible by 11. In itself, that wasn’t particularly interesting, but I wrote down the number of digits in the number and then wrote down the sum of the digits in the number.

    I therefore had three numbers written down. What surprised me was that of the three numbers, one was a square and two were cubes.

    What is the remarkable number?

    [teaser2970]

     
    • Jim Randell's avatar

      Jim Randell 7:42 am on 24 August 2019 Permalink | Reply

      I started without making the assumption that the numbers are separately a square, a cube and a cube (in some order), i.e. one of them could be both a square and a cube.

      The smallest numbers that are both squares and cubes are 0, 1, 64, 729, … (i.e. the powers of 6).

      So we can write down 0 (a multiple of 11 using all different digits), which has 1 digit and a digit sum of 0. And we can say one of them is a square and two of them are cubes (as they are all both squares and cubes). If however, we require exactly 1 of them to be a square and exactly 2 of them to be cubes we can eliminate this solution.

      The number of digits is going to be between 1 and 10, and the digits sum is going to be between 0 and 45, so neither of those can be both a square and a cube, which means that the multiple of 11 must be either a square or a cube (possibly both), so can be written as either (11.m)² or (11.m)³.

      This Python program considers numbers of this form, and finds only one solution. It runs in 157ms.

      Run: [ @repl.it ]

      from enigma import (irange, iroot, nsplit, icount_exactly as count, is_square, is_cube, printf)

# consider multiples of 11 both squared and cubed
for p in (2, 3):
  for m in irange(11, iroot(9876543210, p), step=11):
    n = m**p
    # digits
    ds = nsplit(n)
    # number of digits
    k = len(ds)
    # all digits different
    if len(set(ds)) != k: continue
    # digit sum
    s = sum(ds)

    # one is a square, two are cubes
    ns = (n, k, s)
    if count(ns, is_square, 1) and count(ns, is_cube, 2):
      printf("n={n} k={k} s={s}")

      Solution: The number is 26198073.

      Changing [[ count() ]] to use [[ icount_at_least() ]] does not yield any further solutions.

      The number is (27×11)³.

      The digit length is 8 (= 2³).

      The digit sum is 36 (= 6²).

      If we look at the 6th power of multiples of 11 we find that the powers of 11, 22, 33, 44 all have repeated digits, and the power of 55 is longer than 10 digits. So none of the numbers are going to be both a square and a cube. We can use this analysis to reduce the run time of the program, for example we only have to check numbers with 1, 4, 8, 9 digits. So we can reduce the maximum possible number (at line 5) to 987654321.

      For k = 1 the only possible value is n = 0, and we have already eliminated this as a solution.

      For k = 4 (a square), the only multiple of 11 cubed with 4 digits is n = 11³ = 1331, which has repeated digits.

      For k = 8 (a cube) and k = 9 (a square) the only possible digit sum is s = 36 (a square), so we see the solution must be a multiple of 11 cubed, with 8 different digits and a digit sum of 36. So we only have to check values n = (11.m)³ for m = 20 … 42 to find a value with 8 different digits. There are 2 possible values, and only one of these has a digit sum of 36.

      Like

  • Unknown's avatar

    Jim Randell 9:54 am on 21 August 2019 Permalink | Reply
    Tags:   

    Teaser 2824: New extension 

    From The Sunday Times, 6th November 2016 [link] [link]

    George has just moved departments and he has a new four-figure telephone extension number. He shows it to Martha and proudly points out that, if you delete any two of the digits of the extension number, then the remaining two-figure number is prime. Martha then lists the resulting six different primes and comments delightedly that more than one of them divides exactly into the extension number.

    What is George’s new extension number?

    [teaser2824]

     
    • Jim Randell's avatar

      Jim Randell 9:55 am on 21 August 2019 Permalink | Reply

      We can express the puzzle as a set of alphametic constraints and use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve it.

      The following run file executes in 81ms.

      #! python -m enigma -rr

# solver to use
SubstitutedExpression

# solver parameters
--distinct=""
--invalid=""
--answer="ABCD"

# 2-digit subsequences are prime
"is_prime(AB)"
"is_prime(AC)"
"is_prime(AD)"
"is_prime(BC)"
"is_prime(BD)"
"is_prime(CD)"

# and all different
"all_different(AB, AC, AD, BC, BD, CD)"

# and more than 1 of them divides ABCD
"sum(ABCD % x == 0 for x in (AB, AC, AD, BC, BD, CD)) > 1"

      Solution: George’s new extension number is 4371.

      4371 is divisible by 47 and by 31.

      Like

    • GeoffR's avatar

      GeoffR 8:43 am on 22 August 2019 Permalink | Reply 

      
% A Solution in MiniZinc
include "globals.mzn";

var 1..9:A; var 1..9:B; var 1..9:C;  var 1..9:D;

predicate is_prime(var int: x) = 
x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x))))) 
((i < x) -> (x mod i > 0));

% the four-figure telephone extension number
var 1000..9999: ABCD = 1000*A + 100*B + 10*C + D;

% six different two-digit numbers
constraint all_different([AB, AC, AD, BC, BD, CD]);

var 10..99: AB = 10*A + B;
var 10..99: AC = 10*A + C; 
var 10..99: AD = 10*A + D;
var 10..99: BC = 10*B + C;
var 10..99: BD = 10*B + D;
var 10..99: CD = 10*C + D;

constraint is_prime(AB) /\ is_prime(AC) /\ is_prime(AD)
/\ is_prime(BC) /\ is_prime(BD) /\ is_prime(CD);

% more than one two-digit number divides exactly into the extension number
constraint sum ([ABCD mod AB == 0, ABCD mod AC == 0,
ABCD mod AD == 0, ABCD mod BC == 0, ABCD mod BD == 0,
ABCD mod CD == 0]) > 1;

solve satisfy;
output ["George's Ext No. = " ++ show(ABCD)];

% George's Ext No. = 4371
% ----------
% ==========

      Like

  • Unknown's avatar

    Jim Randell 9:47 am on 20 August 2019 Permalink | Reply
    Tags:   

    Brainteaser 1659: Prime cuts 

    From The Sunday Times, 26th June 1994 [link]

    My new season ticket number is a prime obtained by rearranging five consecutive digits. By cutting off successive single digits from alternate ends I can steadily reduce this to a four-figure prime, a three-figure prime, a two-figure prime and finally a one-figure prime.

    You too can use some short cuts to find my season ticket number.

    This puzzle is included in the book Brainteasers (2002).

    [teaser1659]

     
    • Jim Randell's avatar

      Jim Randell 9:48 am on 20 August 2019 Permalink | Reply

      If we denote the 5-digit number ABCDE, then we can see that all of (C, BCD, ABCDE) must be prime.

      And depending on whether we start truncating on the left or the right one of (CD, BCDE) or (BC, ABCD) must be prime.

      We can write these conditions as a set of alphametic constraints and use the general alphametic solver [[ SubstitutedExpression() ]] from the enigma.py library to solve them.

      This run file executes in 146ms.

      Run: [ @repl.it ]

      #! python -m enigma -rr

SubstitutedExpression

--answer="ABCDE"

# the digits are consecutive
"max(A, B, C, D, E) - min(A, B, C, D, E) = 4"

# odd length truncations
"is_prime(ABCDE)"
"is_prime(BCD)"
"is_prime(C)"

# even length truncations
"(is_prime(BCDE) and is_prime(CD)) or (is_prime(ABCD) and is_prime(BC))"

      Solution: The ticket number is 68597.

      It is composed from the digits 5, 6, 7, 8, 9, and can be truncated to give the following primes: 68597, 8597, 859, 59, 5.

      Like

    • GeoffR's avatar

      GeoffR 12:47 pm on 20 August 2019 Permalink | Reply 

      % A Solution in MiniZinc
include "globals.mzn"; 

% Pattern of Primes
% C
% CD
% BCD
% BCDE
% ABCDE

var 1..9:A; var 1..9:B; var 1..9:C; var 1..9:D; var 1..9:E;

constraint all_different([A,B,C,D,E]);

% the five digit prime number contain consecutive digits
var set of int : s5 = {A,B,C,D,E};
constraint max ([A,B,C,D,E]) - min([A,B,C,D,E]) == card(s5) - 1;
 
predicate is_prime(var int: x) = 
x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x))))) 
((i < x) -> (x mod i > 0));

var 10..99: CD = 10*C + D;
var 100..999: BCD = 100*B + 10*C + D;
var 1000..9999: BCDE = 1000*B + 100*C + 10*D + E;
var 10000..99999: ABCDE = 10000*A + 1000*B + 100*C + 10*D + E;

constraint is_prime(C) /\ is_prime(CD) /\ is_prime(BCD)
/\ is_prime (BCDE) /\ is_prime(ABCDE);

solve satisfy;

output["Primes numbers are " ++ show(ABCDE) ++ ", " 
++ show(BCDE) ++ ", " ++ show(BCD) ++ ", " ++ show(CD) ++ 
" and " ++ show(C) ];

% Primes numbers are 68597, 8597, 859, 59 and 5
% ----------
% ==========

      Like

  • Unknown's avatar

    Jim Randell 4:31 pm on 19 August 2019 Permalink | Reply
    Tags:   

    Teaser 2884: Farewell 

    From The Sunday Times, 31st December 2017 [link] [link]

    Today I am retiring after editing this column for 40 very pleasurable years. My heartfelt thanks go out to all the setters and solvers of puzzles with whom I have corresponded.

    To celebrate the 40 years I threw a party for the puzzle-setters. At the party we assigned a different whole number from 1 to 26 to each letter of the alphabet; for example, we had A=13 and Z=3. We did this in such a way that, for each person present (including me), the values of the letters of their Christian name added to 40.

    Bob, Graham, Hugh, Nick and Richard were there, as were two of Andrew, Angela, Danny, Des, Ian, John, Mike, Peter, Robin, Steve and Tom.

    Which two?

    Victor Bryant also contributed many Enigma puzzles for New Scientist under the pseudonym Susan Denham.

    [teaser2884]

     
    • Jim Randell's avatar

      Jim Randell 4:32 pm on 19 August 2019 Permalink | Reply

      I treated this puzzle as a pair of linked alphametic problems in base 27 (so the symbols have values from 1 to 26), using the [[ SubstitutedExpression() ]] solver from the enigma.py library. It’s not especially quick, but it is moderately succinct. It runs in 2.60s.

      Run: [ @replit ]

      from enigma import (SubstitutedExpression, irange, join, subsets, printf)

# initial words
ws0 = [ "VICTOR", "BOB", "GRAHAM", "HUGH", "NICK", "RICHARD" ]

# additional words
ws1 = [ "ANDREW", "ANGELA", "DANNY", "DES", "IAN", "JOHN", "MIKE", "PETER", "ROBIN", "STEVE", "TOM" ]

# make a puzzle where words in <ws> add to <n>
def puzzle(ws, n, **args):
  kw = dict(base=27, digits=irange(1, 26), verbose=0) # default args
  kw.update(args)
  return SubstitutedExpression(list((join(w, sep=' + '), n) for w in ws), **kw)

# format a puzzle solution <s>
fmt = lambda s: join((join((k, "=", s[k])) for k in sorted(s.keys())), sep=" ")

# make a puzzle with where words in <ws0> add up to 40, with initial conditions
p0 = puzzle(ws0, 40, s2d=dict(A=13, Z=3))

# find solutions to the first puzzle
for s0 in p0.solve():

  # choose 2 words from <ws1>
  for ws2 in subsets(ws1, size=2):

    # make a puzzle where words in <ws2> add to 40, using the letters to numbers assignment in <s0>
    p2 = puzzle(ws2, 40, s2d=s0)

    # find a solution to the second puzzle
    for s2 in p2.solve(first=1):
      # output the solution <s2>
      printf("+ {ws2} [{s2}]", ws2=join(ws2, sep=", "), s2=fmt(s2))

      Solution: DES and MIKE also add up to 40.

      There are four ways the letters can be assigned to make the required set of names add up to 40.

      The following letters are fixed:

      A=13 Z=3
      B=16 D=10 E=18 G=7 H=4 M=2 O=8 R=1 S=12 U=25

      Then C, I, K, N, T, V have one of the following assignments:

      C=5 I=6 K=14 N=15 T=9 V=11
      C=5 I=6 K=14 N=15 T=11 V=9
      C=6 I=5 K=15 N=14 T=9 V=11
      C=6 I=5 K=15 N=14 T=11 V=9

      The remaining letters: J, L, P, W, Y (and the unused letters: F, Q, X) take on the remaining values: 17, 19, 20, 21, 22, 23, 24, 26 (in any order).

      Like

    • GeoffR's avatar

      GeoffR 7:38 pm on 19 August 2019 Permalink | Reply

      I had to change to the Chuffed Solver after the Geocode solver froze – this has happened previously.

      I found multiple solutions, all with MIKE and DES adding up to 40.
      I noticed the names which seemed to vary most in value were ANGELA, JOHN and PETER.

      % A Solution in MiniZinc
include "globals.mzn";

var 1..26:A;   var 1..26:B;   var 1..26:C;   var 1..26:D;
var 1..26:E;   var 1..26:F;   var 1..26:G;   var 1..26:H;   
var 1..26:I;   var 1..26:J;   var 1..26:K;   var 1..26:L;
var 1..26:M;   var 1..26:N;   var 1..26: O;  var 1..26:P;   
var 1..26:Q;   var 1..26:R;   var 1..26:S;   var 1..26:T;   
var 1..26:U;   var 1..26:V;   var 1..26:W;   var 1..26:X;   
var 1..26:Y;   var 1..26:Z;

constraint A == 13 /\ Z == 3;

constraint all_different ([A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,
P,Q,R,S,T,U,V,W,X,Y,Z]);

% Bob, Graham, Hugh, Nick and Richard were there, plus Victor
% - all have letters adding up to 40
constraint B+O+B == 40 /\ G+R+A+H+A+M == 40 /\ H+U+G+H == 40
/\ N+I+C+K == 40 /\ R+I+C+H+A+R+D == 40 /\ V+I+C+T+O+R == 40;

% Two of Andrew, Angela, Danny, Des, Ian, John, Mike, 
% Peter, Robin, Steve and Tom have letters summing to 40
constraint sum( [A+N+D+R+E+W == 40, A+N+G+E+L+A == 40, 
D+A+N+N+Y == 40, D+E+S == 40, I+A+N == 40, J+O+H+N == 40, 
M+I+K+E == 40, P+E+T+E+R == 40, R+O+B+I+N == 40, 
S+T+E+V+E == 40, T+O+M == 40]) == 2;  

solve satisfy;

output [ 
    "ANDREW = " ++ show (A+N+D+R+E+W) 
++ " ANGELA = " ++ show(A+N+G+E+L+A)
++ " DANNY = " ++ show(D+A+N+N+Y) 
++ "\nDES = " ++ show(D+E+S)
++ " IAN = " ++ show(I+A+N)
++ " JOHN = " ++ show(J+O+H+N)
++ "\nMIKE = " ++ show(M+I+K+E)
++ " PETER = " ++ show(P+E+T+E+R)
++ " ROBIN = " ++ show(R+O+B+I+N)
++ "\nSTEVE = " ++ show(S+T+E+V+E)
++ " TOM = " ++ show(T+O+M) ];

% Multiple solutions found
% DES = 40 and MIKE = 40 (20 solutions checked)

% Typical solution
% ANDREW = 81 ANGELA = 87 DANNY = 79
% DES = 40 IAN = 34 JOHN = 46
% MIKE = 40 PETER = 68 ROBIN = 46
% STEVE = 68 TOM = 19
% ----------
% Finished in 396msec

      Like

  • Unknown's avatar

    Jim Randell 3:23 pm on 18 August 2019 Permalink | Reply
    Tags: by: Len Wood   

    Brain-Teaser 494: [Squares from rectangles] 

    From The Sunday Times, 15th November 1970 [link]

    Tina and Cindy each started with a rectangular sheet of cardboard. The two rectangles had equal dimensions.

    Each girl cut her sheet into 5 squares using 4 straight cuts; the first 3 cuts producing one square each and the fourth cut producing two squares. The first square obtained by Tina had twice the area of the first square obtained by Cindy.

    If Tina’s fifth square had an area of one square inch, what was the area of the rectangles?

    This puzzle was originally published with no title.

    [teaser494]

     
    • Jim Randell's avatar

      Jim Randell 3:24 pm on 18 August 2019 Permalink | Reply

      Instead of considering cutting up the rectangle we consider adding squares to an existing rectangle.

      Tina’s 4th and 5th squares are 1 inch squares, so come from a 2 × 1 rectangle. A square can be added to any rectangle by extending it horizontally or vertically, and we need to add three more squares, so there are 8 possible starting rectangles.

      Cindy’s rectangle is also one of these 8 appropriately scaled to make the rectangles have the same area. If Cindy’s rectangle is scaled by f along each side then the area is scaled by f².

      Also the area of Tina’s first square is twice the area of Cindy’s first square.

      This Python 3 program finds the 8 possible rectangles and their divisions and finds a pair that give an appropriate scale factor.

      It runs in 87ms.

      from enigma import (subsets, printf)

# extend the rectangle (x, y) with k more squares
def solve(x, y, k, rs=[]):
  if k == 0:
    yield (x, y, rs)
  else:
    # grow in the x direction by adding a y square
    yield from solve(x + y, y, k - 1, rs + [y])
    # grow in the y direction by adding an x square
    yield from solve(x, x + y, k - 1, rs + [x])

# choose rectangles for T and C
for ((Tx, Ty, Tss), (Cx, Cy, Css)) in subsets(solve(1, 2, 3, [1, 1]), size=2, select="P"):
  (A, f2A) = (Tx * Ty, Cx * Cy)
  # check scale factors
  if f2A * Tss[-1] ** 2 == 2 * A * Css[-1] ** 2:
    # output solution
    printf("area = {A} [T: {Tx}x{Ty} = {Tss}, fC: {Cx}x{Cy} = {Css}]", A=Tx * Ty)

      Solution: The area of the rectangles was 28 square inches.

      Tina’s rectangle was 4×7, and her squares had area: 16, 9, 1, 1, 1.

      Cindy’s rectangle was (7√2)×(2√2), and her squares had area: 8, 8, 8, 2, 2.

      Here is a diagram of the two rectangles:

      Like

  • Unknown's avatar

    Jim Randell 8:43 pm on 15 August 2019 Permalink | Reply
    Tags:   

    Teaser 2969: Slide rules 

    From The Sunday Times, 18th August 2019 [link] [link]

    Using her ordinary 15cm ruler and Zak’s left-handed version (numbers 1 to 15 reading right to left) Kaz could display various fractions. For instance, putting 5 on one ruler above 1 on the other ruler, the following set of fractions would be displayed: 5/1, 4/2, 3/3, 2/4 and 1/5. Zak listed the fifteen sets starting from “1 above 1” up to “15 above 1”.

    Kaz chose some fractions with values less than one from Zak’s sets (using just the digits 1 to 9, each once only in her selection). Of these, two were in simplest form, one of which had consecutive numerator and denominator. Zak correctly totalled Kaz’s selection, giving the answer as a fraction in simplest form. Curiously, the answer’s numerator and denominator were both palindromic.

    Give Zak’s answer.

    [teaser2969]

     
    • Jim Randell's avatar

      Jim Randell 10:45 pm on 15 August 2019 Permalink | Reply

      This Python program collects possible fractions, and then uses a recursive function to select viable sets that Kaz could have chosen. These sets are then examined to find candidates where the sum is expressed as a fraction consisting of two palindromes.

      Run: [ @repl.it ]

      from fractions import Fraction as F
from enigma import (irange, nsplit, gcd, join, printf)

# collect fractions less than 1
fs = list()

# consider overlaps from 1 to 15
for n in irange(1, 15):
  for (a, b) in zip(irange(n, 1, step=-1), irange(1, n, step=1)):
    if a < b:
      fs.append((a, b))

# generate sets of fractions, using the digits in ds, and:
# one of the fractions consists of consecutive numbers
# two of the fractions are in lowest terms
def choose(fs, ds=set(), ss=[]):
  # are we done?
  if not ds:
    # find fractions in their lowest terms
    rs = list((a, b) for (a, b) in ss if gcd(a, b) == 1)
    # there should be exactly 2 of them
    if len(rs) != 2: return
    # and exactly one of them must consist of consecutive numbers
    if sum(b == a + 1 for (a, b) in rs) != 1: return
    yield ss
  else:
    for (i, (a, b)) in enumerate(fs):
      d = nsplit(a) + nsplit(b)
      if len(set(d)) < len(d) or not ds.issuperset(d): continue
      yield from choose(fs[i + 1:], ds.difference(d), ss + [(a, b)])

# check for palindromes
def is_palindromic(n):
  ds = nsplit(n)
  return ds == ds[::-1]

# choose the set of fractions
for ss in choose(fs, set(irange(1, 9))):
  # calculate the sum of the fractions
  t = sum(F(a, b) for (a, b) in ss)
  if is_palindromic(t.numerator) and is_palindromic(t.denominator):
    printf("{ss} = {t}", ss=join((join((a, "/", b)) for (a, b) in ss), sep=" + "))

      Solution: Zak’s answer is 545/252.

      The fractions Kaz has chosen are:

      3/12 + 4/8 + 5/9 + 6/7 = 545/252

      There are 10 sets of fractions that Kaz could have chosen, but only one of these sets has a total consisting of palindromic numerator and denominator when expressed in lowest terms.

      3/5 + 7/8 + 6/9 + 4/12 = 99/40
      3/5 + 7/8 + 6/9 + 2/14 = 1919/840
      4/5 + 6/8 + 3/12 + 7/9 = 116/45
      3/6 + 5/9 + 7/8 + 4/12 = 163/72
      3/6 + 5/9 + 7/8 + 2/14 = 1045/504
      4/6 + 5/9 + 7/8 + 3/12 = 169/72
      5/6 + 4/8 + 3/12 + 7/9 = 85/36
      4/8 + 6/7 + 5/9 + 3/12 = 545/252 [*** SOLUTION ***]
      3/9 + 6/7 + 5/8 + 4/12 = 361/168
      3/9 + 6/7 + 5/8 + 2/14 = 47/24

      Like

  • Unknown's avatar

    Jim Randell 10:03 am on 15 August 2019 Permalink | Reply
    Tags: by: Ben Tarlow   

    Brainteaser 1657: Not a black and white case 

    From The Sunday Times, 12th June 1994 [link]

    Marvo has been up to his tricks again. He shuffled a normal pack of 52 cards and then looked at each one in turn without letting Alice see them. In each case she had to predict whether the card was red or black and Marvo gave no indication of whether Alice’s guesses were right or wrong. But after 51 guesses Alice had guessed “black” 26 times and “red” 25 times and Marvo announced that Alice had exactly 10 predictions wrong so far.

    (1) Which colour (if either) is more likely for Alice’s last card?

    Marvo repeated the game with Bob. After 51 guesses Bob too had guessed “black” 26 times and “red” 25 times and Marvo announced that Bob had exactly 11 predictions wrong so far.

    (2) Which colour (if either) is more likely for Bob’s last card?

    Marvo then repeated the game once more with Carol. After 51 guesses Carol too had guessed “black” 26 times and “red” 25 times and Marvo announced that Carol had got at least 10 predictions wrong so far.

    (3) Which colour (if either) is more likely for Carol’s last card?

    This puzzle is included in the book Brainteasers (2002). The puzzle text was, and the answers required were changed slightly, but the idea of the puzzle remains the same.

    [teaser1657]

     
    • Jim Randell's avatar

      Jim Randell 10:04 am on 15 August 2019 Permalink | Reply

      Alice has guessed “black” 26 times and “red” 25 times.

      If she has got n guesses wrong, then we can suppose b of the “black” guesses were wrong (i.e. they were actually red cards), and so (n − b) of the “red” guesses were wrong (i.e. they were actually black cards).

      So at this point the total number of actual black cards is:

      the (26 − b) cards that Alice correctly guessed as black; plus
      the (n − b) cards that were guessed as red, but were actually black.

      And similarly for red, giving:

      black = 26 + n − 2b
      red = 25 − n + 2b

      There can’t be more than 26 actual black cards, so:

      26 + n − 2b ≤ 26
      n ≤ 2b
      b ≥ n/2

      And there can’t be more than 26 actual red cards:

      25 − n + 2b ≤ 26
      2b ≤ n + 1
      b ≤ (n + 1)/2

      So:

      n/2 ≤ b ≤ (n + 1)/2

      Which fixes the value of b, depending on whether n is odd or even.

      In Alice’s case n = 10, so b = 5. Hence there have actually been 26 black cards and 25 red cards, so the remaining card has to be red.

      In Bob’s case n = 11, so b = 6. And there have actually been 25 black cards and 26 red cards, so the remaining card has to be black.

      And we see that if n is even then the remaining card must be red, and if it is odd the remaining card must be black.

      Solution: (1) Alice’s final card is certainly red. (2) Bob’s final card is certainly black.

      Carol has made at least 10 incorrect predictions. So she has made 10, 11, 12, …, 51, and depending on the exact number the remaining card is red, black, red, … black. There are 24 cases where the card must be red and 24 cases where the card must be black.

      We need to count the number of scenarios for each outcome.

      When n = 10 then b = 5 and there have been 26 black guesses (5 are guessed incorrectly) and 25 red guesses (5 are guessed incorrectly). So the number of ways of choosing the incorrect guesses are:

      C(26, 5) × C(25, 5)

      each one giving rise to the remaining card being red.

      When n = 11 and b = 6 we get:

      C(26, 6) × C(25, 5)

      each one giving rise to the remaining card being black.

      (I think if we are counting the total number of possible scenarios each of these numbers would be multiplied by an additional factor to account for the order of the cards and the guesses, but we don’t need to do that to compare the relative likelihood of red and black outcomes for the final card).

      This Python program counts up the number of possibilities. It runs in 85ms.

      Run: [ @repl.it ]

      from enigma import irange, C, fdiv, compare, printf

r = b = 0
for n in irange(10, 51):
  (k, p) = divmod(n, 2)
  if p == 0:
    r += C(26, k) * C(25, k)
  else:
    b += C(26, k + 1) * C(25, k)

f = lambda x: fdiv(x, r + b)
x = compare(r, b, ("black", "equal", "red"))
printf("{x} [red = {r} ({fr:.4%}), black = {b} ({fb:.4%})]", fr=f(r), fb=f(b))

      Solution: (3) It is slightly more likely that Carol’s final card is red.

      The calculated figures give that Carol’s final card is red 50.0001% of the time and black 49.9999% of the time. So it is close.

      The puzzle as originally published in The Sunday Times asked for the outcome with the following numbers of wrong predictions: (a) exactly 10; (b) least 10; (c) at least 15.

      We have already covered (a) and (b), and for (c) the program can easily be modified to start counting from 15, rather than 10. The result is that it is slightly more likely that the card is black (50.02%) than red (49.98%).

      Liked by 1 person

  • Unknown's avatar

    Jim Randell 8:14 am on 13 August 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 493: Absorbers 

    From The Sunday Times, 8th November 1970 [link]

    Three men from the Absorbers’ Association, three from Bacchante’s Bank and three from the Convivial Corporation, took a drink together.

    From each concern, one man drank sherry, one gin and one whisky. Of the three choosing sherry, one chose medium, one dry and one extra dry. With gin, one took tonic, one lemon and one vermouth. With whisky, one plain water, one soda water and one dry ginger.

    The dry sherry drinker did not belong to the same concern as the plain water man, nor to the same concern as the vermouth man. The drinkers of dry ginger, lemon and medium sherry were from three different concerns.

    Either the tonic man or plain water man (but not both) was from the Absorbers’ Association. The selectors of soda and vermouth were not colleagues and neither represented Bacchante’s Bank.

    The tonic man belonged either to the same concern as the dry sherry drinker or to the same concern as the medium sherry man.

    Which type of sherry was chosen by a Convivial Corporation man and what did [the] other men from that same concern take with gin and take with whisky?

    This puzzle is included in the book Sunday Times Brain Teasers (1974).

    [teaser493]

     
    • Jim Randell's avatar

      Jim Randell 8:14 am on 13 August 2019 Permalink | Reply

      The solution space is quite small (there are (3!)³ = 216 possible arrangements), so we can just examine them all and eliminate those which don’t satisfy the conditions.

      This Python program runs in 83ms.

      Run: [ @repl.it ]

      from itertools import product
from enigma import (subsets, printf)

for ((M, D, X), (T, L, V), (W, S, G)) in product(subsets("ABC", size=len, select='P'), repeat=3):

  # "The dry sherry drinker did not belong to the same concern as the
  # plain water man, nor to the same concern as the vermouth man"
  if D == W or D == V: continue

  # "The drinkers of dry ginger, lemon and medium sherry were from
  # three different concerns"
  if len(set([G, L, M])) != 3: continue

  # "Either the tonic man or plain water man (but not both) was from
  # the Absorbers' Association"
  if not ((T == 'A') ^ (W == 'A')): continue

  # "The selectors of soda and vermouth were not colleagues and
  # neither represented Bacchante's Bank"
  if S == V or S == 'B' or V == 'B': continue

  # "The tonic man belonged either to the same concern as the dry
  # sherry drinker or to the same concern as the medium sherry man"
  if not (T == D or T == M): continue
  
  printf("Sherry: M={M} D={D} X={X}; Gin: T={T} L={L} V={V}; Whisky: W={W} S={S} G={G}")

      Solution: The CC men drank extra dry sherry, gin and lemon, whisky and soda.

      The other drinks are:

      AA = medium sherry, gin and vermouth, whisky and water
      BB = dry sherry, gin and tonic, whisky and ginger

      Like

  • Unknown's avatar

    Jim Randell 11:28 am on 12 August 2019 Permalink | Reply
    Tags:   

    Teaser 2885: Croquet mallet 

    From The Sunday Times, 7th January 2018 [link] [link]

    My croquet mallet is less than a metre high and consists of a cylindrical shaft attached to a heavier cuboid head, both of uniform density. Knowing the total weight of the mallet, I wanted to work out the weight of the shaft. I found the point of balance along the shaft and measured the distances from there to each end of the shaft, the smaller of which was less than the height of the head. Each of these three distances was a whole prime number of cm, and taking the three distances together with the height of the mallet, no digit was repeated. I worked out that the weight of the head was a single digit multiple of the weight of the shaft.

    What was the height of my mallet?

    [teaser2885]

     
    • Jim Randell's avatar

      Jim Randell 11:29 am on 12 August 2019 Permalink | Reply

      When the mallet is balanced we have:

      where a, b, c are all primes and b < c < a.

      The total length of the mallet t = a + b + c < 100.

      And between them, the numbers a, b, c, t have no repeated digits.

      The shaft has a weight w acting at a distance of (a + b)/2 − b = (a − b)/2 from the balance point.

      The head has a weight kw acting at a distance of b + c/2.

      So for these to balance we have:

      w(a − b)/2 = kw(b + c/2)
      (a − b) = k(2b + c)
      k = (a − b) / (2b + c)

      where k is a single digit integer.

      This Python program finds the solution in 78ms.

      Run: [ @replit ]

      from enigma import (primes, subsets, is_duplicate, div, printf)

for (b, c, a) in subsets(primes.between(1, 99), size=3, select='C'):
  t = a + b + c
  if not (t < 100): continue
  if is_duplicate(a, b, c, t): continue

  k = div(a - b, 2 * b + c)
  if k is None or k < 2 or k > 9: continue

  printf("t={t} [a={a} b={b} c={c}, k={k}]")

      Solution: The mallet is 90cm high.

      The values are: t=90, a=83 b=2 c=5, k=9.

      Like

  • Unknown's avatar

    Jim Randell 8:08 am on 11 August 2019 Permalink | Reply
    Tags:   

    Brainteaser 1649: Ages and ages! 

    From The Sunday Times, 17th April 1994 [link]

    Today is the birthday of both Mary and Nelly, so it is a good time to have a conundrum about their ages.

    When Mary was one third of the age she is today, Nelly was half the age that Mary was when Nelly was the age that Mary was when Nelly was the age that Mary was when Nelly was a quarter of the age she is today.

    If neither of them is yet eligible for an old-age pension, how old are they both today?

    This puzzle is included in the book Brainteasers (2002).

    [teaser1649]

     
    • Jim Randell's avatar

      Jim Randell 8:09 am on 11 August 2019 Permalink | Reply

      A lot of the puzzle text is concerned with the difference between the ages, so if we suppose N’s current age is n and M’s current age is (n + d), and then look at the expression:

      When Mary was one third of the age she is today, Nelly was half the age that Mary was when Nelly was the age that Mary was when Nelly was the age that Mary was when Nelly was a quarter of the age she is today

      We can successively refine it:

      → When Mary was (1/3)n + (1/3)d, Nelly was half the age that Mary was when Nelly was the age that Mary was when Nelly was the age that Mary was when Nelly was (1/4)n
      → When Mary was (1/3)n + (1/3)d, Nelly was half the age that Mary was when Nelly was the age that Mary was when Nelly was (1/4)n + d
      → When Mary was (1/3)n + (1/3)d, Nelly was half the age that Mary was when Nelly was (1/4)n + 2d
      → When Mary was (1/3)n + (1/3)d, Nelly was half (1/4)n + 3d
      → When Mary was (1/3)n + (1/3)d, Nelly was (1/8)n + (3/2)d
      → (1/3)n + (1/3)d = (1/8)n + (3/2)d + d
      → (5/24)n = (13/6)d
      → 5n = 52d

      So d must be a multiple of 5, and n must be a multiple of 52.

      For (n + d) < 65 we must have n = 52, d = 5.

      Solution: Nelly is 52. Mary is 57.

      Liked by 1 person

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