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  • Unknown's avatar

    Jim Randell 12:50 pm on 24 September 2023 Permalink | Reply
    Tags:   

    Brainteaser 1095: “Seven anagrams” 

    From The Sunday Times, 31st July 1983 [link]

    Members of TOPIQ (the society with membership limited to those with a tested I.Q. above a specified figure), are required to test one another’s mental quickness with suitable problems from time to time. Once, when I was asked to set a problem, I recalled an old verse which holds several anagrams of the same six letters. The verse went something like this:

    Hushed was the RUSTLE of leaves in the breeze,
    Faded, the LUSTRE of light on the trees,
    As the SUTLER sat in his ULSTER grey,
    Quaffing his ale at the close of the day.
    Then he came to the end of his drinking song:
    “Thou RULEST the weak; thou LUREST the strong,
    John Barleycorn, my Jo!”

    In my problem the six anagrams were omitted. Members firstly had to find these. They were told that, if each of the six letters in the anagrams represented a different one of digits 1 to 6, then the sum of all the six-digit numbers of the anagrams was a 7-digit figure starting with LS and ending with UT. They were then asked the following question, solved by most members in under 10 minutes:

    Having, from the sum, solved the number-letter relationship, what number do you get for RESULT?

    What do you get?

    [teaser1095]

     
    • Jim Randell's avatar

      Jim Randell 12:51 pm on 24 September 2023 Permalink | Reply

      I used the [[ SubstitutedExpression ]] solver from the enigma.py library to solve this puzzle.

      I introduced additional symbols x, y, z, which can take on any value (not just 1-6) for the result of the sum.

      This run file executes in 71ms. (Internal runtime of the generated program is 1.4ms – or 467µs if you use [[ SubstitutedExpression.split_sum() ]]).

      #! python3 -m enigma -rr

SubstitutedExpression

--symbols="ELRSTUxyz"
--distinct="ELRSTU"
--invalid="0|7|8|9,ELRSTU"

"RUSTLE + LUSTRE + SUTLER + ULSTER + RULEST + LUREST = LSxyzUT"

--answer="RESULT"

      Solution: RESULT = 254316.

      The numbers (in the order given) give the following sum:

      234615 + 134625 + 436152 + 314652 + 231546 + 132546 = 1484136

      Like

  • Unknown's avatar

    Jim Randell 4:27 pm on 22 September 2023 Permalink | Reply
    Tags:   

    Teaser 3183: Partial eclipse 

    From The Sunday Times, 24th September 2023 [link] [link]

    Alf, Bert and Charlie were comparing their experiences of the eclipse from different locations. The moon appeared to be exactly the same size as the sun as it passed in front of part of the sun’s disc. The magnitude of an eclipse measures the maximum part of the diameter of the sun’s disc that is obscured. Taking the apparent diameter of the sun as 100, they noted the following:

    Alf: “My magnitude was a whole number”.
    Bert: “If I took the part of the sun’s circumference that was obscured, multiplied it by the radius and then subtracted the area of the sun’s disc that was obscured, I also got a whole number”.
    Charlie: “Both of those statements are true for all of us, but we all saw different magnitudes greater than 10”.

    In increasing order, what were those three magnitudes?

    As well as this puzzle being Teaser 3183 there are now 3183 puzzles posted in total between the Enigmatic Code and S2T2 sites.

    [teaser3183]

     
    • Jim Randell's avatar

      Jim Randell 4:55 pm on 22 September 2023 Permalink | Reply

      If we suppose the apparent radius of the two discs is R, and we consider the part of the circumference of the sun that is obscured, then it forms the arc of a circular sector that subtends an angle of 2θ at the centre of the sun’s disc. The length of the obscured circumference is then: 2Rθ.

      The area of the part of the sun’s disc that is obscured A can be calculated as the sum of two equal obscured segments, each being the area of the circular sector less the area of the triangle formed by replacing the arc of the sector with a chord.

      A = 4 [(𝛑R²)(θ/2𝛑) − (R cos(θ))(R sin(θ))/2)
      A = 2R²θ − 2R²sin(θ)cos(θ)
      A = R²(2θ − sin(2θ))

      (See also: [@wikipedia])

      And so B’s measure is:

      B = 2R²θ − A
      ⇒
      B = 2R²sin(θ)cos(θ)
      B = R²sin(2θ)

      And A’s measure (the magnitude) m is given by:

      m = 2R(1 − cos(θ))

      So given an integer magnitude, we can calculate a rational value for cos(θ), and using the identity cos²(θ) + sin²(θ) = 1 we can calculate a value for sin(θ), which must also be rational if B is to have an integer value.

      This Python program runs in 61ms. (Internal runtime is 413µs).

      from enigma import (Rational, irange, is_square_q, sq, as_int, printf)

Q = Rational()

# apparent radius
R = 50

# consider possible magnitudes from 11 to 99
for m in irange(11, 99):
  # calculate cos(theta), sin(theta)
  cost = 1 - Q(m, 2 * R)
  sint = is_square_q(1 - sq(cost), F=Q)
  if sint is None: continue
  # calculate B's measure
  B = 2 * sq(R) * sint * cost
  # is it an integer?
  B = as_int(B, default=None)
  if B is None: continue
  # output solution
  printf("m={m} -> B={B}")

      Solution: The magnitudes are: 20, 40, 72.

      And we have:

      m = 20 → B = 2400
      m = 40 → B = 2400
      m = 72 → B = 1344

      There are only 3 values because the “partner” value for m = 72 (which also gives B = 1344) is m = 4.

      Like

      • Frits's avatar

        Frits 8:47 pm on 22 September 2023 Permalink | Reply

        I had to look up some formula and came to the same formula for B.

           
# apparent radius
R = 50

# consider possible magnitudes from 11 to 99
for m in range(11, 100):
  # calculate cos(theta)
  c = 1.0 - m / (2 * R)
  # calculate sin(theta), c^2 + s^2 = 1
  s = (1 - c**2)**.5
  # B = r^2.sin(2 * theta), sin(2x) = 2.sin(x).cos(x) 
  B = 2 * R * R * s * c
  
  # is it close to an integer?
  if abs(B - round(B)) > 1e-6: continue
 
  # output solution
  print(f"m={m} -> B={B:.2f}")

        Like

      • Jim Randell's avatar

        Jim Randell 2:38 pm on 23 September 2023 Permalink | Reply

        With a bit more work we can simplify the calculation of B to:

        B = (1/2)(2R − m)√(m(4R − m))

        Which gives a simpler program, that only uses integers:

        from enigma import (irange, is_square, div, printf)

# apparent radius
R = 50

# consider possible magnitudes from 11 to 99
for m in irange(11, 99):
  r = is_square(m * (4 * R - m))
  if r is None: continue
  B = div((2 * R - m) * r, 2)
  if B is None: continue
  # output solution
  printf("m={m} -> B={B}")

        Like

  • Unknown's avatar

    Jim Randell 9:32 am on 21 September 2023 Permalink | Reply
    Tags: by: A. H. R. Grimsey   

    Brainteaser 1485: Safe depository 

    From The Sunday Times, 24th February 1991 [link]

    The number of my safe deposit box in the Norsex Bank has four different digits, decreasing from left to right. Curiously, that of my fiancée has the same figures but in the reverse order. The bank manager, Mr K. Aprekar, has said that when we are married next month, he will give us a presently-vacant box number which not only contains the four digits we now share, but is in fact the difference between our two numbers.

    What is it?

    [teaser1485]

     
    • Jim Randell's avatar

      Jim Randell 9:32 am on 21 September 2023 Permalink | Reply

      Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

      The following run file executes in 71ms. (Internal runtime of the generated program is 496µs).

      #! python3 -m enigma -rr

SubstitutedExpression

# suppose the current numbers are: ABCD and DCBA

# numbers are decreasing
"A > B" "B > C" "C > D"

# the new number (= the difference between the original two numbers)
--macro="@new = ABCD - DCBA"
# is composed of the same digits
"set(nsplit(@new, 4)) == {A, B, C, D}"

--answer="@new"

      Solution: The new box number is 6174.

      We have:

      7641 − 1467 = 6174

      Like

  • Unknown's avatar

    Jim Randell 12:33 pm on 19 September 2023 Permalink | Reply
    Tags: ,   

    Teaser 2642: New world order 

    From The Sunday Times, 12th May 2013 [link] [link]

    In maths class Pat has been learning about progressions from his “New World” text book. The book has three sections, namely “Arithmetic”, “Algebra” and “Geometry” which overall run from page 1 to page 500 inclusive — with each section starting on a new page. The geometry section is over twice as long as the arithmetic section.

    As an exercise Pat calculated the sum of the page numbers in each section and he was surprised to find that the sum of the page numbers in “Algebra” was double that of the sum of the page numbers in “Geometry”.

    How many pages are there in the “Arithmetic” section?

    [teaser2642]

     
    • Jim Randell's avatar

      Jim Randell 12:34 pm on 19 September 2023 Permalink | Reply

      The total sum of all the page numbers = T = tri(500).

      If the Arithmetic section is pages 1 .. x, the Algebra section pages (x + 1) .. y, and the Geometry section pages (y + 1) .. 500, then the sums of the page numbers in each section are:

      Ar = tri(x)
      Al = tri(y) − tri(x)
      Ge = T − tri(y)
      ⇒
      tri(y) = T − Ge

      and:

      Al = 2 Ge
      ⇒
      Ge = (T − Ar) / 3

      So, by choosing a value for x, we can calculate the values of Ar, Al, Ge and hence y.

      This Python program runs in 55ms. (Internal runtime is 267µs).

      from enigma import (irange, tri, div, is_triangular, printf)

# Ar: sum(1 .. x) = tri(x)
# Al: sum(x + 1 .. y) = tri(y) - tri(x)
# Ge: sum(y + 1 .. 500) = tri(500) - tri(y)
#
# Al = 2 * Ge

# total sum of all pages
T = tri(500)

# consider the end page for arithmetic = x
for x in irange(2, 165):
  Ar = tri(x)
  Ge = div(T - Ar, 3)
  if Ge is None: continue
  # calculate y
  y = is_triangular(T - Ge)
  if y is None or not x < y < 500: continue
  if not (500 - y > 2 * x): continue
  # output solution
  printf("Ar = 1..{x}; Al = {x1}..{y}; Ge = {y1}..500", x1=x + 1, y1=y + 1)

      Solution: There are 45 pages in the Arithmetic section.

      So the sections are:

      Arithmetic: pages 1 – 45 (= 45 pages); sum = 1035
      Algebra: pages 46 – 409 (= 364 pages); sum = 82810 (= Geometry × 2)
      Geometry: pages 410 – 500 (= 91 pages); sum = 41405

      Like

  • Unknown's avatar

    Jim Randell 9:32 am on 17 September 2023 Permalink | Reply
    Tags:   

    Brain-Teaser 919: Golden age of steam 

    From The Sunday Times, 2nd March 1980 [link]

    Ever since the dawn of history, mankind has looked back to a Golden Age when everybody and everything was so much better.

    As far as our railway system is concerned, the Golden Age was the prewar era of the great steam locomotives. In spite of their noise, steam, and smoke pollution, (or perhaps even because of these characteristics), the older generation look back with nostalgia to that golden era. Let us therefore pose a not-too-difficult problem concerning those wonderful days.

    Two steam locomotives, each travelling at 50 mph, are driven by Tom and Dick. They are approaching Harry’s signal box from opposite directions.

    Tom sounds his whistle. Immediately upon hearing it, Dick replies. Harry hears both whistles with an interval of seven seconds between them. The two engines later pass the signal box at the same instant.

    You may assume that the speed of sound was 1100 feet per second relative to the ground. One mile has 5280 feet.

    How far (in feet) was Tom from the box when he sounded his whistle?

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994). The puzzle text above is taken from the book.

    [teaser919]

     
    • Jim Randell's avatar

      Jim Randell 9:33 am on 17 September 2023 Permalink | Reply

      The trains are travelling at the same speed, and (presumably on different tracks) pass the signal box at the same time, so they must start the same distance from the signal box.

      Working in feet per second, a speed of 50 mph = 220/3 ft/s.

      Suppose at time t = 0 both T and D are a distance x feet from H, and T sounds his whistle.

      At a time t = x / 1100 H hears T’s whistle.

      And if D hears the whistle when he is a distance y from H, then he hears T’s whistle at time t = (x + y) / 1100.

      D instantaneously responds (still at distance y), and H hears D’s whistle at time t = (x + 2y) / 1100.

      H hears the whistles 7 seconds apart, so:

      (x + 2y) / 1100 = 7 + x / 1100
      y = (7/2)1100 = 3850

      So D sounds his whistle when he is 3850 ft from H at time t = (x + 3850) / 1100.

      And at this time T must also be 3850 ft from H.

      Hence:

      x = 3850 + (220/3)(x + 3850) / 1100
      ⇒ x = 4400

      Solution: Tom was 4400 ft from the signal box when he sounded his whistle.

      Like

    • Hugo's avatar

      Hugo 9:49 am on 17 September 2023 Permalink | Reply

      It must have been a cold day: the speed of sound is about 1100 ft/s at 5°C and 60% relative humidity
      (about 3 ft/s less in dry air at that temperature).

      Like

  • Unknown's avatar

    Jim Randell 4:45 pm on 15 September 2023 Permalink | Reply
    Tags:   

    Teaser 3182: Stand and be counted 

    From The Sunday Times, 17th September 2023 [link] [link]

    When bad light ended the first day’s play, the total score was 182 with three batters having been “out” during the day.

    In the successive stands, the lower-numbered and the higher-numbered batters scored in the ratios 7:3, 10:7, 1:11 and 11:19, in that order. The total score was the sum of the scores of the five batters.

    The third batter was the highest scorer.

    [In cricket the batting team starts with batters 1 and 2 together, and each is able to score. When one of them is “out” the third batter comes “in” to replace them, when one of that pair is out the fourth batter comes in, and so on. The amount by which the team score has increased while a particular pair is together is called the “stand”].

    What were the scores of all the batters, in their batting order?

    [teaser3182]

     
    • Jim Randell's avatar

      Jim Randell 5:14 pm on 15 September 2023 Permalink | Reply

      This Python program works through each partnership, considering stands of the specified ratios.

      It runs in 60ms. (Internal runtime is 1.5ms).

      from enigma import (irange, inf, div, map2str, printf)

# total score
T = 182

# add deltas <ds> to keys <ks> of dict <d>
def add(d, ks, ds):
  d = dict(d)
  for (k, i) in zip(ks, ds):
    d[k] += i
  return d

# n = number of next bat
# A = current lower numbered bat
# B = current higher numbered bat
# rs = remaining stand ratios
# ss = current scores per bat
# t = total runs scored
def solve(n, A, B, rs, ss, t):
  # choose scores for this stand
  (a, b) = rs[0]
  # remaining ratios
  rs_ = rs[1:]
  if rs_:
    # minimum remaining total
    rt = sum(a + b for (a, b) in rs_)
    # consider possible multipliers for this stand
    for k in irange(1, inf):
      (rA, rB) = (a * k, b * k)
      t_ = t + rA + rB
      if t_ + rt > T: break
      ss_ = add(ss, [A, B], [rA, rB])
      # each batsman may be dismissed
      yield from solve(n + 1, A, n, rs_, ss_, t_)
      yield from solve(n + 1, B, n, rs_, ss_, t_)
  else:
    # can we achieve the total?
    k = div(T - t, a + b)
    if k is not None:
      yield add(ss, [A, B], [a * k, b * k])

rs = [(7, 3), (10, 7), (1, 11), (11, 19)]  # stand ratios
ss = dict((k, 0) for k in irange(1, 5))  # scores

# find totals for each batsman
for ss in solve(3, 1, 2, rs, ss, 0):
  # batsman 3 scored the most
  if max(ss.values()) != ss[3]: continue
  # output solution
  printf("{ss}", ss=map2str(ss, sep=" ", enc=""))

      Solution: The totals are: bat 1 = 21; bat 2 = 49; bat 3 = 52; bat 4 = 22; bat 5 = 38.

      The stands are:

      1: bat 1 = 21; bat 2 = 9; ratio = 7:3; bat 1 dismissed
      2: bat 2 = 40; bat 3 = 28; ratio = 10:7; bat 2 dismissed
      3: bat 3 = 2; bat 4 = 22; ratio = 1:11; bat 4 dismissed
      4: bat 3 = 22; bat 5 = 38; ratio = 11:19; total = 182 runs

      Like

      • Jim Randell's avatar

        Jim Randell 7:18 pm on 15 September 2023 Permalink | Reply

        Here’s an alternative (faster) version.

        We can generate possible batting pairings for each stand. And the denominators of the ratios are all different, so we can use the [[ express() ]] function to determine the multipliers for each stand.

        We then use the ratios and the multiplies to calculate the scores for each batsman.

        This Python program runs in 58ms, and has an internal runtime of 386µs.

        from enigma import (irange, subsets, express, map2str, printf)

# generate possible pairings
def _pairs(k, A, B, n, ds=[]):
  ds = ds + [(A, B)]
  # are we done?
  if k < 1:
    yield ds
  else:
    # one of A and B is dismissed
    yield from _pairs(k - 1, B, n, n + 1, ds)
    yield from _pairs(k - 1, A, n, n + 1, ds)

pairs = list(_pairs(3, 1, 2, 3))

# calculate scores with a sequence of multipliers and a sequence of pairs
def scores(ks, ps):
  ss = dict((i, 0) for i in irange(1, 5))
  for (k, (a, b), (x, y)) in zip(ks, ps, [(7, 3), (10, 7), (1, 11), (11, 19)]):
    ss[a] += k * x
    ss[b] += k * y
  return ss

# generate possible multipliers for each stand
for (k1, k3, k2, k4) in express(182, [10, 12, 17, 30], min_q=1):
  for ps in pairs:
    ss = scores((k1, k2, k3, k4), ps)
    if ss[3] == max(ss.values()):
      printf("{ss}", ss=map2str(ss, sep=" ", enc=""))

        Like

    • Frits's avatar

      Frits 6:01 pm on 15 September 2023 Permalink | Reply 

      from enigma import SubstitutedExpression

# the alphametic puzzle
p = SubstitutedExpression(
  [
    # 10 * W + 17 * X + 12 * Y + 30 * Z = total score 182
    "div(182 - (10 * W + 17 * X + 12 * Y), 30) == Z",
    
    # who whon the first session
    "F < 3",
    # who whon the third session
    "2 < T < 5",
    
    # third batter must have won in 2nd session
    
    # possible score for the first  batter 7 * W + (F == 1) * 10 * X
    # possible score for the second batter 3 * W + (F == 2) * 10 * X 
    # possible score for the third  batter 7 * X + Y + 11 * (T == 3) * Z
    # possible score for the fourth batter 11 * Y + (T == 4) * 11 * Z
    #          score for the fifth  batter 19 * Z
    
    # the third batter was the highest scorer
    "max(sc := [7 * W + (F == 1) * 10 * X, \
                3 * W + (F == 2) * 10 * X , \
                7 * X + Y + 11 * (T == 3) * Z, \
                11 * Y + (T == 4) * 11 * Z, \
                19 * Z]) == sc[2]", 
  ],
  answer="sc",
  distinct="",
  digits=range(1, 10),
  reorder=0,
  verbose=0,    # use 256 to see the generated code
)

# print answers
for ans in p.answers():
  print(f"answer: {ans}")

      Like

      • Frits's avatar

        Frits 6:17 pm on 15 September 2023 Permalink | Reply

        @Jim, Thank you for leaving the easy solution (SubstitutedExpression) for me.

        Like

      • Frits's avatar

        Frits 6:30 pm on 15 September 2023 Permalink | Reply

        Theoretically I should have used base=13 and digits=range(1, 13) to allow for high W and Y’s.

        Like

    • Frits's avatar

      Frits 11:49 am on 16 September 2023 Permalink | Reply

      A more general version without analysis.

      from enigma import SubstitutedExpression

# ratio sums
rs = [10, 17, 12, 30]
multipliers = "WXYZ"

# maxima per multiplier
d = {multipliers[i]: (182 - sum(rs) + rs[i]) // rs[i] for i in range(4)}    
mx = max(d.values())

# invalid digit / symbol assignments
d2i = {n: set(k for k, v in d.items() if n > v) for n in range(1, mx + 1)} 
# F/S/T were not "out" in the 1st/2nd/3rd play
for i, v in enumerate("FST"):
  for j in range(3 + i, mx + 1):
    d2i[j].add(v)

# the alphametic puzzle, use multipliers W, X, Y and Z
p = SubstitutedExpression(
  [ 
    # 10 * W + 17 * X + 12 * Y + 30 * Z = total score 182
    "div(182 - (17 * X + 12 * Y + 30 * Z), 10) = W",
    
    # S wasn't "out" in the second play
    "S in {F, 3}",
    # T wasn't "out" in the third play
    "T in {S, 4}",
    
    # score for the first  batter 7W + (F == 1).10X + (S == 1).Y + (T == 1).11Z
    # score for the second batter 3W + (F == 2).10X + (S == 2).Y + (T == 2).11Z
    # score for the third  batter                7X + (S == 3).Y + (T == 3).11Z
    # score for the fourth batter                            11Y + (T == 4).11Z
    # score for the fifth  batter                                           19Z
    
    # the third batter was the highest scorer
    "max(sc := [7 * W + (F == 1) * 10 * X + (S == 1) * Y + (T == 1) * 11 * Z, \
                3 * W + (F == 2) * 10 * X + (S == 2) * Y + (T == 2) * 11 * Z, \
                                    7 * X + (S == 3) * Y + (T == 3) * 11 * Z, \
                                                  11 * Y + (T == 4) * 11 * Z, \
                                                                      19 * Z])\
         == sc[2]", 
  ],
  answer="sc",
  d2i=d2i,
  distinct="",
  base=mx + 1,
  digits=range(1, mx + 1),
  reorder=0,
  verbose=0,    # use 256 to see the generated code
)

# print answers
for ans in p.answers():
  print(f"answer: {ans}")

      Like

    • Jim Olson's avatar

      Jim Olson 3:46 am on 26 September 2023 Permalink | Reply

      What is the correct answer?

      Like

  • Unknown's avatar

    Jim Randell 8:35 am on 14 September 2023 Permalink | Reply
    Tags: by: J David Jones   

    Teaser 2008: Divide and add 

    From The Sunday Times, 11th March 2001 [link]

    In the following arithmetic the fractions have four digits in the numerator and denominator, and the digits have been replaced consistently by letters. with different letters used for different digits.

    ETON/NOTE + TANG/GNAT

    Each fraction is a whole number.

    What is the total?

    [teaser2008]

     
    • Jim Randell's avatar

      Jim Randell 8:36 am on 14 September 2023 Permalink | Reply

      We can use the [[ SubstitutedExpression ]] solver from the enigma.py library to solve this puzzle.

      The following run file executes in 72ms. (Internal runtime of the generated program is 1.2ms).

      #! python3 -m enigma -rr

SubstitutedExpression

"ETON % NOTE = 0"
"TANG % GNAT = 0"

--answer="(ETON // NOTE) + (TANG // GNAT)"

      Solution: The total is: 13.

      The expression is:

      9801/1089 + 8712/2178 = 9 + 4 = 13

      Like

  • Unknown's avatar

    Jim Randell 10:12 am on 12 September 2023 Permalink | Reply
    Tags:   

    Teaser 2651: Parsnip Farm 

    From The Sunday Times, 14th July 2013 [link] [link]

    On Parsnip Farm there was a triangular field with one of its boundaries running north to south. From that boundary’s southernmost point Farmer Giles built a fence in an easterly direction, thus partitioning the field into two smaller triangular fields, with the area of the northern field being forty per cent of the area of the southern field. Each side of each field was a whole number of furlongs in length, one of the sides of the southern field being twenty-five furlongs in length.

    What are the lengths of the other two sides of the southern field?

    [teaser2651]

     
    • Jim Randell's avatar

      Jim Randell 10:12 am on 12 September 2023 Permalink | Reply

      If the area of the original field is A, then it is split such that the northern field has an area N = (2/7)A and the southern field S = (5/7)A.

      The boundaries of N are all integers and so form a Pythagorean triple (x, y, z).

      And we can expand the northern field by a factor of f until its hypotenuse exactly covers the boundary of the original field, and we get this:

      The we can then write the area of the original field in two ways:

      A = (7/2)N = 7xy / 4
      A = fxy / 2
      ⇒
      f = 7/2

      And we can calculate the unknown sides of S:

      z′ = (f − 1)z = 5z/2
      y′ = hypot(7x/2, 5y/2) = hypot(7x, 5y) / 2

      This Python program runs in 58ms. (Internal runtime is 334µs).

      Run: [ @replit ]

      from enigma import (pythagorean_triples, div, ihypot, ordered, printf)

for (a, b, z) in pythagorean_triples(100):
  z_ = div(5 * z, 2)
  if z_ is None: continue
  N = ordered(a, b, z)
  for (x, y) in [(a, b), (b, a)]:
    y_ = div(ihypot(7 * x, 5 * y), 2)
    if y_ is None: continue
    S = ordered(x, z_, y_)
    if 25 in S:
      # output solution
      printf("N = {N}; S = {S}")

      Solution: The other two sides of the southern field are 6 and 29 furlongs.

      N is a (6, 8, 10) triangle, (area = 24 square furlongs ≈ 0.97 sq km).

      S is a (6, 25, 29) triangle, (area = 60 square furlongs ≈ 2.43 sq km).

      So the original field was a (8, 29, 35) triangle, (area = 84 square furlongs ≈ 3.40 sq km).

      Like

    • Frits's avatar

      Frits 5:44 pm on 12 September 2023 Permalink | Reply

      Avoiding imports.

       
is_square = lambda n: int(rt := n**.5) == rt

# generate Pythagoren triples for a fixed hypothenuse
def triples(h):
  sqs = {i * i for i in range(1, h)}
  return [(int(s1**.5), int(s2**.5)) for s1 in sqs 
           if (s2 := h**2 - s1) >= s1 and s2 in sqs]
  
side = 25

# can x be 25?  
# then f.x = 87.5, y' can never be a whole number  (f = 3.5)
# if other side (2.5y) is k.0 or k.5, so x != 25

# can y' be 25?
for c, d in triples(side): 
   # are these lenghts multiples of 2.5 and 3.5?
  x, y = -1, -1
  if (c % 2.5 == 0 and d % 3.5 == 0):
    y, x = c // 2.5, d // 3.5    
  if ((c % 3.5 == 0 and d % 2.5 == 0)):
    x, y = c // 3.5, d // 2.5    
  if x == -1: continue
  
  # y' can be 25
  if is_square(z2 := x**2 + y**2):
    z = int(z2**.5)
    if z % 2 == 0:
      print(f"answer {int(x)} and {int(2.5 * z)} furlongs")
  
# can z' be 25?
if side % 5: exit(0)
z = int(side * 2 / 5)

ts = triples(z)
# determine z'
for x, y in ts + [x[::-1] for x in ts]:
  if is_square(y1 := 3.5**2 * x**2 + 2.5**2 * y**2):
    print(f"answer {int(x)} and {int(y1**.5)} furlongs")

      Like

  • Unknown's avatar

    Jim Randell 8:05 am on 10 September 2023 Permalink | Reply
    Tags:   

    Brain-Teaser 928: Confused contracts 

    From The Sunday Times, 4th May 1980 [link]

    The results of the hands in a bridge tournament were written like this: “Joe Smith, two ♥”. Before I could calculate the scores, my dog tore up the results of the last four players. Fortunately the writing remained legible. By manoeuvring the fragments, and knowing there had been one contract in each suit, and one at each of the one, two, three and four levels, I was able to construct various feasible combinations.

    I first tried putting ♦ opposite Ted, Pete and Reg in turn. Ted could have ♦ only when Mr Green had the two contract; Pete only when ♠ were four; and Reg only when Mr Black had the one. Similarly, if Reg or Mr Black had ♣, then Mr Green had four; If Mr White had ♥ or ♣ then Mr Black had the other; if Reg had ♥ then Pete had ♦; if Mr Brown had ♥ then Mr Black had four; if Mr Black had ♥ then Mr Green had two; and if Mr Black didn’t have ♥ then ♠ were in less than four.

    Mr Black could have one only when Pete had three ♠; Mr Green could have two only when ♥ were four; Mr Black had four only when Reg had ♦; Mr Green had four only when ♦ were in three; and ♣ could be in three or four only when Reg had ♥.

    Finally I noted that Ted and Vic had the same coloured suits whenever I tried putting ♠ with the three, but different coloured suits if Mr White did not have ♥.

    Luckily, I then remembered that ♦ had actually been bid at the four level.

    What was the suit of the three contract, and what was the full name of the player who bid it?

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994). The puzzle text above is taken from the book.

    [teaser928]

     
    • Jim Randell's avatar

      Jim Randell 8:06 am on 10 September 2023 Permalink | Reply

      I know nothing of Bridge, so at first the puzzle text made no sense to me, but there is a handy hint in the 1974 book:

      Ignoring the Bridge jargon, you have to match each of the numbers 1 to 4 with one of the suits, a forename, and a surname.

      I used the [[ SubstitutedExpression ]] solver to allocate the values among the three groups, so all we need to do is translate the text into a collection of constraints which we can represent by Python expressions.

      I translated: “If A had X or Y, then B had the other” into: “(A = X) ⇒ (B = Y)” and “(A = Y) ⇒ (B = X)”.

      Note that we might expect to deduce “Reg != Black” from the statement “if Reg or Mr Black had …”, but if we try this then there are no solutions.

      The following run file executes in 66ms. (Internal runtime of the generated code is 185µs).

      Run: [ @replit ]

      #! python3 -m enigma -rr

# allocate values 1-4 to the following:
#
#  suits    : C D S H [Clubs, Diamonds, Spades, Hearts]
#  forenames: T R P V [Ted, Reg, Pete, Vic]
#  surnames : B K G W [Brown, blacK, Green, White]

SubstitutedExpression

--digits="1-4"
--distinct="CDSH,TRPV,BKGW"

# specify the conditions
"implies(T == D, G == 2)"
"implies(P == D, S == 4)"
"implies(R == D, K == 1)"
"implies(R == C or K == C, G == 4)"
"implies(W == H, K == C)" "implies(W == C, K == H)"
"implies(R == H, P == D)"
"implies(B == H, K == 4)"
"implies(K == H, G == 2)"
"implies(K != H, S != 4)"
"implies(K == 1, P == 3 == S)"
"implies(G == 2, H == 4)"
"implies(K == 4, R == D)"
"implies(G == 4, D == 3)"
"implies(C == 3 or C == 4, R == H)"
"implies(S == 3, {T, V} in [{C, S}, {D, H}])"
"implies(W != H, {T, V} not in [{C, S}, {D, H}])"

# diamonds was the 4th contract
--assign="D,4"

# mention all the variables
"true(C, D, S, H, T, R, P, V, B, K, G, W)"

--template="(C D S H) (T R P V) (B K G W)"
--solution=""

      Solution: The 3 contract was in hearts. It was bid by Pete Green.

      The contracts are fully defined:

      1♣ = Ted Brown
      2♠ = Reg Black
      3♥ = Pete Green
      4♦ = Vic White

      And we see that Reg = Black.

      Like

    • Frits's avatar

      Frits 4:06 pm on 11 September 2023 Permalink | Reply

      More than 200 lines of code.

      from itertools import combinations 

# allocate values 1-4 to the following:
#
#  suits    : C D S H [Clubs, Diamonds, Spades, Hearts]
#  forenames: T R P V [Tom, Reg, Pete, Vic]
#  surnames : B K G W [Brown, blacK, Green, White]

# 0/1/2 for no/some/all details
detail = 1

rev = lambda x: "neq" if x == "eq" else "eq"

# a specific print format
p_frz = lambda x, y: f"{x:>3}-{''.join(sorted(y))}"

# add implication if <x> then <y> 
def implies(x, y):
  if "=" in x:
    op1 = "eq" if "==" in x else "neq"
    op2 = "eq" if "==" in y else "neq"
    # use frozenset to make it hashable
    var1, var2 = frozenset([x[0], x[-1]]), frozenset([y[0], y[-1]])
    imp[(op1, var1)] = imp.get((op1, var1), []) + [(op2, var2)]
    # if a implies b then not b implies not a
    imp[(rev(op2), var2)] = imp.get((rev(op2), var2), []) + [(rev(op1), var1)]
  else:
    if x in imp:
      if y not in imp[x]:
        # check if y results in a falsehood
        if y == (rev(x[0]), x[1]):
          print(f"ERROR {x}-{y} is false")
         
        if y[1] in fb[x[1]] and x[0] == y[0] == "eq":  
          if detail: 
            print(f"{p_frz(x[0], x[1])} is false as it leads to "
                  f"eq-{''.join(y[1])} which is not allowed")
          add_truth("neq", x[1])
        imp[x] += [y]
    else:
       imp[x] = [y]

# print implications
def print_imp(d):
  print()  
  for (a, b), vs in sorted(d.items()):
    print(f"{p_frz(a, b)}  : ", end=' ')
    for (c, d) in vs:
      print(f"{p_frz(c, d)}", end=', ')
    print()
  print()  
      
# add entries to truth list <t>
def add_truth(c, s):
  s_ = frozenset(s)
  if (c, s_) in t: return
  l_ = list(s)
  if detail: print("add to t:", p_frz(c, s_))
  t.add((c, s_))
  if c != "eq": return
  
  # add forbidden entries
  for f in fb[s_]:
    if detail > 1: print("add inherited rules to t:", p_frz("neq", f))
    t.add(("neq", f)) 

  toadd = []
  # propagate 'eq-ab': if 'neq-ac' exists then also 'neq-bc' must be true
  for d, f in t:
    if d != "neq": continue
    for i, x in enumerate(l_):
      if x not in f: continue
      o, = f.difference([x])
      if o not in bd[l_[1 - i]]:
        toadd.append(frozenset([l_[1 - i], o]))

  for x in toadd:
    add_truth("neq", x)
            
imp, fb, bd = dict(), dict(), dict()

# specify the conditions
implies("T == D", "G == 2")
implies("P == D", "S == 4")
implies("R == D", "K == 1")
implies("W == H", "K == C") 
implies("W == C", "K == H")
implies("R == H", "P == D")
implies("B == H", "K == 4")
implies("K == H", "G == 2")
implies("K != H", "S != 4")
implies("K == 1", "P == 3")
implies("K == 1", "P == S")
implies("K == 1", "S == 3")
implies("G == 2", "H == 4")
implies("K == 4", "R == D")
implies("G == 4", "D == 3")

#implies("R == C or K == C", "G == 4")
implies("R == C", "G == 4")
implies("K == C", "G == 4")
#implies("C == 3 or C == 4", "R == H")
implies("C == 3", "R == H")
implies("C == 4", "R == H")

#implies("S == 3", "{T, V} in [{C, S}, {D, H}]")
#implies("W != H", "{T, V} not in [{C, S}, {D, H}]")
implies("S == 3", "W == H")

types = [x.split() for x in ["C D S H", "T P R V", "B K G W", "1 2 3 4"]]
# buddies
bd = {x: [y for y in types[i] if y != x] 
          for i, tp in enumerate(types) for x in tp}
# forbidden values
for a, b in combinations(bd.keys(), 2):
  if b in bd[a]: continue
  fb[frozenset([a, b])] = [frozenset([a, x]) for x in bd[b]] + \
                          [frozenset([b, x]) for x in bd[a]] 

if detail > 1:
  print("forbidden\n---------")
  for k, vs in sorted(fb.items()):
    print(''.join(k), end="  :  ")
    for v in vs:
      print(''.join(v), end=", ")
    print()
  print()  

t = set()
# diamonds was the 4th contract
add_truth('eq', frozenset(['D', '4']))

loop, ln = 1, len(t)
while True:

  if detail:
    print(f"\nloop {loop}\n------")
    if loop < 3: print_imp(imp)

  if detail > 1:
    print(f"\ntruths\n------")
    for x in sorted(t, key = lambda x: (x[0], sorted(x[1]) ) ):
      print(p_frz(x[0], ''.join(sorted(x[1]))))
  
  # expand implications by chaining
  for k, vs in imp.items():
    for v in vs:
      # check if v is allowed ...
      if (rev(v[0]), v[1]) in t:
        # ... if not disallow k
        add_truth(rev(k[0]), k[1])
      
      if v in imp:
        # add implications of v to k
        for i in imp[v]:
          implies(k, i)

  # check for 3 non-equalities in a group
  toadd = []
  for c, f in t:
    if c != "neq": continue
    lst = list(f)
    # for both elements within f
    for i in range(2):
      y = [("neq", frozenset([x, lst[1-i]])) in t for x in bd[lst[i]]]
      if sum(y) != 2: continue
      # we have 3 neq's so the remaining entry in group must be eq
      toadd.append(frozenset([lst[1-i], bd[lst[i]][y.index(0)]]))
  
  for x in toadd:
    add_truth("eq", x)
  
  # only if normal logic has been exhausted
  if len(t) == ln:
    lookup = lambda d, x: [f for c, f in d if c == "eq" and x in f and 
                           len(f | set("1234")) == 5]
        
    # implies("S == 3", "{T, V} in [{C, S}, {D, H}]")
    # implies("W != H", "{T, V} not in [{C, S}, {D, H}]")
    s3 = ("eq", frozenset(["S", "3"])) in t
    ns3 = ("neq", frozenset(["S", "3"])) in t
    if not s3 and not ns3: break
    
    # D is 4 so try to find H
    tH = lookup(t, "H")      
    if not tH: break

    tT, tV = lookup(t, "T"), lookup(t, "V")    
    if not tV and not tT: break

    H, = tH[0].difference(["H"]) 
    DH = {'4', H}
    
    vr, ot = "V" if tV else "T", "T" if tV else "V"
   
    v, = (tV[0] if tV else tT[0]).difference([vr]) 

    n = DH.difference([v]).pop() if v in DH else (set("1234") - DH - {v}).pop()
    add_truth("neq" if ns3 else "eq", frozenset([ot, n]))
  
  r3 = [''.join(f.difference(["3"])) for c, f in t if c == "eq" and '3' in f]
  # is there enough data in row 3 for the answer?
  if len(r3) == 3: 
    print(f"\nanswer: {r3[0]}, {r3[1]} and {r3[2]}") 
    exit(0)
  
  ln = len(t)  
  loop += 1

      Like

  • Unknown's avatar

    Jim Randell 4:51 pm on 8 September 2023 Permalink | Reply
    Tags:   

    Teaser 3181: “No slack, Alice!” says Grace 

    From The Sunday Times, 10th September 2023 [link] [link]

    Grace mimicked Mr Dodgson: “Take two different positive odd numbers with no shared prime factor. Multiply the larger by their sum, giving the perimeter of a distinct right-angled triangle with whole-number sides. Only such values and their multiples work”.

    Alice created a loop from part of a 1m thread. She was able to pull it tightly on a pinboard into a right-angled triangle with whole-number cm sides in two ways (with different-shaped triangles).

    Alice then pulled the same loop tight over the thin polar spike of the classroom globe, down two meridians and along the equator. Thinking “Logic’s dead!”, she saw 90° equatorial angles and a non-zero polar angle, which obviously didn’t add to 180°.

    Grace calculated the polar angle to the nearest degree. Curiously, transposing its two digits gave the globe’s whole-number centimetre radius.

    Find this radius.

    [teaser3181]

     
    • Jim Randell's avatar

      Jim Randell 5:09 pm on 8 September 2023 Permalink | Reply

      If the string has length p and angle (in radians) made by the loop at the pole is 𝛂 on a globe of radius r, then we have:

      p = r(𝛑 + 𝛂)

      This Python program finds perimeters (that are whole numbers of centimetres, less than 100), such that (at least) two dissimilar Pythagorean triangles can be constructed, and then looks for a corresponding radius for a globe that fits the remaining conditions.

      It runs in 58ms. (Internal runtime is 296µs).

      from math import (pi, degrees)
from enigma import (pythagorean_triples, group, item, fdiv, irange, intr, nrev, printf)

# generate possible pythagorean triangles, return (<perimeter>, <triangles>)
def generate(M):
  for ss in pythagorean_triples(M // 2):
    p = sum(ss)
    if p < M:
      yield (p, ss)

# group pythagorean triangles by their perimeter
g = group(generate(100), by=item(0), f=item(1))

# consider possible perimeters
for (p, ts) in g.items():
  if len(ts) < 2: continue
  printf("[p={p} -> {ts}]")

  # consider possible globe radii
  for r in irange(10, 99):
    # calculate the polar angle (to the nearest degree)
    a = intr(degrees(fdiv(p, r) - pi))
    if a > 99: continue
    if a < 10: break
    if a == nrev(r):
      printf("-> r={r} a={a}")

      Solution: The globe has a radius of 19 cm.

      The loop used 90 cm of thread.

      Which allows (planar) right-angled triangles with sides of (15, 36, 39) cm and (9, 40, 41) cm to be formed.

      It can also be used on a 19cm radius globe to form a triangle on the globe with approx. 30.31 cm running along the equator and 29.85 cm running up meridians to the north pole, meeting at an angle of approx. 91.4°. So the three sides are each close to 30 cm, and the angles are all about 90°.

      Like

  • Unknown's avatar

    Jim Randell 9:28 am on 7 September 2023 Permalink | Reply
    Tags: ,   

    Teaser 2645: One square 

    From The Sunday Times, 2nd June 2013 [link] [link]

    I have consistently replaced digits by letters, using different letters for different digits, and in this way I have found that:

    (FIVE − FOUR)² = ONE

    Now, if I were to tell you whether or not THREE is divisible by 3, and whether or not FOUR is divisible by 4, then you could work out FIVE.

    What number is FIVE?

    Note: As originally published this puzzle has no solution (and was reproduced in this flawed form in the book The Sunday Times Brain Teasers Book 2 (2020)).

    However the additional information that “FIVE is greater than FOUR” allows the published answer to be found.

    [teaser2645]

     
    • Jim Randell's avatar

      Jim Randell 9:29 am on 7 September 2023 Permalink | Reply

      With the extra condition that “FIVE is greater than FOUR” we find the published answer.

      So maybe the setter intended to say: “… I have found that FIVE − FOUR is a positive number, whose square is ONE” (or: “the square root of ONE is equal to FIVE – FOUR“).

      This Python program runs in 147ms.

      Run: [ @replit ]

      from enigma import (SubstitutedExpression, group, unpack, item, peek, seq2str, printf)

# the alphametic problem (with an additional condition)
p = SubstitutedExpression(
  ["sq(FIVE - FOUR) = ONE", "FIVE > FOUR"],
  answer="(THREE, FOUR, FIVE)",
)

# collect the answers, and record values of FIVE (= item(2))
# by (THREE divisible by 3, FOUR divisible by 4).
fn = unpack(lambda n3, n4, n5: (int(n3 % 3 == 0), int(n4 % 4 == 0)))
d = group(p.answers(verbose=0), by=fn, f=item(2), fn=set)

# output the collections, and find the solution
for (k, vs) in d.items():
  if len(vs) == 1:
    printf("(d3, d4) = {k} -> FIVE = {v} [*** SOLUTION ***]", v=peek(vs))
  else:
    printf("(d3, d4) = {k} -> FIVE = {vs}", vs=seq2str(vs, sort=1, enc="{}"))

      Solution: FIVE = 5901.

      If we are told THREE is not divisible by 3, and FOUR is divisible by 4, then the only possible assignments are:

      FIVE = 5901, FOUR = 5872, ONE = 841, THREE = 36211
      FIVE = 5901, FOUR = 5872, ONE = 841, THREE = 63211

      Any of the other of the divisibility possibilities lead to multiple candidate values for FIVE, so this gives the answer.

      However if we do not require FIVE > FOUR, then we find many further possible solutions to the alphametic problem:

      FIVE = 1496, FOUR = 1520, ONE = 576, THREE = 38066
      FIVE = 1496, FOUR = 1520, ONE = 576, THREE = 83066

      FIVE = 3791, FOUR = 3820, ONE = 841, THREE = 56011
      FIVE = 3791, FOUR = 3820, ONE = 841, THREE = 65011

      FIVE = 5791, FOUR = 5820, ONE = 841, THREE = 36011
      FIVE = 5791, FOUR = 5820, ONE = 841, THREE = 63011

      FIVE = 6791, FOUR = 6820, ONE = 841, THREE = 35011
      FIVE = 6791, FOUR = 6820, ONE = 841, THREE = 53011

      FIVE = 8496, FOUR = 8520, ONE = 576, THREE = 13066
      FIVE = 8496, FOUR = 8520, ONE = 576, THREE = 31066

      And so we cannot work out the value of FIVE.

      We can see these candidates by removing the [[ "FIVE > FOUR" ]] on line 5 of the program.

      Like

      • Frits's avatar

        Frits 12:32 pm on 8 September 2023 Permalink | Reply

        The reported run time 147ms was probably with PyPy (I get 423ms with CPython).

        With this program CPython performs better than PyPY.

        from enigma import (SubstitutedExpression, group, unpack, item, peek, seq2str, printf)

# invalid digit / symbol assignments
d2i = dict()
for d in range(0, 10):
  vs = set()
  if d == 0: vs.update('OFT')
  # if E is odd then R is even and if E is even then R is even as well
  if d % 2: vs.update('R')
  # a number that ends in 2, 3, 7 or 8 is not a perfect square
  if d in {2, 3, 7, 8}: vs.update('E')
  d2i[d] = vs
 
# the alphametic problem (with an additional condition)
p = SubstitutedExpression(
  [ # FIVE > FOUR
    "I > O", 
    "sq(IVE - OUR) = ONE",],
  answer="(FOUR, FIVE, THREE)",
  #answer="(FOUR, FIVE, \
  #         R + 2*E + sum(i for i in range(10) if i not in {F,I,V,E,O,U,R,N}))",
  d2i=d2i,
  verbose=0
)
 
# collect the answers, and record values of FIVE (= item(1))
# by (THREE divisible by 3, FOUR divisible by 4).
fn = unpack(lambda n4, n5, n3: (int(n3 % 3 == 0), int(n4 % 4 == 0)))
d = group(p.answers(), by=fn, f=item(1), fn=set)
 
# output the collections, and find the solution
for (k, vs) in d.items():
  if len(vs) == 1:
    printf("(d3, d4) = {k} -> FIVE = {v} [*** SOLUTION ***]", v=peek(vs))
  else:
    printf("(d3, d4) = {k} -> FIVE = {vs}", vs=seq2str(vs, sort=1, enc="{}"))

        Like

        • Jim Randell's avatar

          Jim Randell 12:13 pm on 9 September 2023 Permalink | Reply

          @Frits: Yes, 147ms was with PyPy 7.3.12. With CPython 3.12 I get an overall runtime of 255ms.

          Although, if you are solving the rescued problem you can use the expression [[ "is_square(ONE) + FOUR = FIVE" ]] which is a bit faster, and it is even faster to use [[ "is_square(ONE) + OUR = IVE" ]] (93ms with CPython 3.12).

          Like

  • Unknown's avatar

    Jim Randell 9:09 am on 5 September 2023 Permalink | Reply
    Tags:   

    Teaser 2637: Powerful team 

    From The Sunday Times, 7th April 2013 [link] [link]

    George and Martha support the local football team. The squad is numbered from 1 to 22, with 11 playing at any one time and the remaining 11 sitting on the bench. At the start of this week’s match Martha commented that among their 11 players on the pitch there were no three consecutive numbers, and that the same was true of the 11 sitting on the bench. She also commented that the sum of the 11 players’ numbers on the pitch was a perfect square. At half-time one substitution was made and in the second half George noted that all that Martha had said was still true, but now the perfect square was higher.

    What (in increasing order) were the numbers of the 11 players in the first half?

    [teaser2637]

     
    • Jim Randell's avatar

      Jim Randell 9:10 am on 5 September 2023 Permalink | Reply

      This Python program runs in 349ms. (Internal runtime is 279ms).

      Run: [ @replit ]

      from enigma import (irange, subsets, tuples, is_square, diff, cproduct, update, printf)

# available players
players = list(irange(1, 22))

# check for <k> consecutive numbers in sorted list <ns>
is_consecutive = lambda ns, k: any(t[0] + k == t[-1] + 1 for t in tuples(ns, k))

# choose 11 of the players to form the team
for ps in subsets(players, size=11):
  # the sum is a perfect square
  s = sum(ps)
  if not is_square(s): continue
  # no 3 consecutive numbers are used
  if is_consecutive(ps, 3): continue

  # and the unused players
  xs = diff(players, ps)
  # no 3 consecutive numbers
  if is_consecutive(xs, 3): continue

  # substitute a player from ps with one from xs
  for ((i, p), (j, x)) in cproduct([enumerate(ps), enumerate(xs)]):
    s_ = s - p + x
    if not (s_ > s and is_square(s_)): continue
    # perform the substitution
    ps_ = sorted(update(ps, [(i, x)]))
    if is_consecutive(ps_, 3): continue
    xs_ = sorted(update(xs, [(j, p)]))
    if is_consecutive(xs_, 3): continue

    # output solution
    printf("{ps} = {s} -> {ps_} = {s_}")

      Solution: The players in the first half were: 1, 2, 4, 5, 7, 8, 10, 12, 14, 17, 20.

      The sum of the numbers in the first half is 100 (= 10²).

      In the second half player 1 was substituted by player 22, causing the sum to increase to 121 (= 11²).

      We can easily see that the smallest possible square is 100, and the largest possible square is 144.

      The difference in the squares is also the difference in the numbers of the substituting and substituted players.

      So the only possible difference is 21, going from 100 → 121.

      And the substitution must be that player 1 was substituted by player 22.

      Like

  • Unknown's avatar

    Jim Randell 8:54 am on 3 September 2023 Permalink | Reply
    Tags: by: Ian Adamson   

    Brain-Teaser 937: Number please? 

    From The Sunday Times, 6th July 1980 [link]

    It was so stifling that afternoon that I had been dozing on my bed with the curtains closed, the air conditioning full on, and a bottle of Vichy water on my bedside table.

    The harsh ring of the telephone brought me to my full consciousness. “This is X9 calling”, said the voice, “it’s been a hot afternoon”. This was the message I had been waiting for, so I replied, “Yes, but it’s cooler in Stockholm.”

    “Now listen carefully”, continued X9, “I want you to telephone me at eleven o’clock tonight. The first digit of my number is twice the first digit of your age; the second digit of my number is three times the second (and last) of your age. Have you got that?”

    “Yes”, I acknowledged, “but why this secrecy?”

    “There may be people listening in. My third digit is the difference of my first two digits — and the fourth is the sum of my first two digits. The final digit of my number is the sum of my third and fourth digits.”

    At first, I worked out an incorrect telephone number, as I had forgotten that I had just recently celebrated my birthday — the difference between the two telephone numbers being a multiple of my age.

    What number should I telephone?

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994).

    [teaser937]

     
    • Jim Randell's avatar

      Jim Randell 8:54 am on 3 September 2023 Permalink | Reply

      This Python program runs in 53ms. (Internal runtime is 95µs).

      from enigma import (cproduct, nconcat, tuples, printf)

# generate possible (<age>, <number>) pairs
def generate():
  # for a 2-digit age we have:
  # 1st digit = 1, 2, 3, 4
  # 2nd digit = 0, 1, 2, 3
  for (x, y) in cproduct([(1, 2, 3, 4), (0, 1, 2, 3)]):
    # calculate the 5-digit number <abcde>
    a = 2 * x
    b = 3 * y
    c = abs(a - b)
    d = a + b
    e = c + d
    if d > 9 or e > 9: continue
    # return viable pair
    yield (nconcat(x, y), nconcat(a, b, c, d, e))

# collect the pairs in a dict
d = dict(generate())

# look for consecutive ages
for (k1, k2) in tuples(sorted(d.keys()), 2):
  if k1 + 1 != k2: continue
  (v1, v2) = (d[k1], d[k2])
  # difference in numbers is a multiple of the current age
  if (v1 - v2) % k2 > 0: continue
  # output solution
  printf("{k2} -> {v2} [{k1} -> {v1}]")

      Solution: The telephone number is 43178.

      This is derived from an age of 21.

      21 → (4 = 2×2)(3 = 3×1)(1 = 4 − 3)(7 = 4 + 3)(8 = 1 + 7) = 43178

      So the incorrectly calculated number is 40448, derived from age 20:

      20 → (4 = 2×2)(0 = 3×0)(4 = 4 − 0)(4 = 4 + 0)(8 = 4 + 4) = 40448.

      And 43178 − 40448 = 2730 = 130×21.

      In fact the only other ages that give viable phone numbers are:

      10 → 20224
      11 → 23156

      But in this case 23156 − 20224 = 2932 is not divisible by 11.

      Like

    • Frits's avatar

      Frits 3:51 pm on 5 September 2023 Permalink | Reply 

       
# formula for X9's telephone number
x9 = lambda a, b: (20224 - (f := 2*a < 3*b) * 404) * a + (2730 + f * 606) * b

# Assume that for the incorrect telephone number the same rules must appply.
# This means that the 2nd digit of current age can't be 0, 
# also 2nd digit of current age can't be 3 (then 4th digit > 9)

for n in [11, 12, 21, 31]:
  d1, d2 = n // 10, n % 10 
  
  correct = x9(d1, d2)
  incorrect = x9(d1, d2 - 1)
  # sum of 2nd and 3rd digit must be less than 10
  if any(int(x[2]) + int(x[3]) > 9 for x in (str(correct), str(incorrect))):
    continue
  
  # difference in numbers is a multiple of the current age
  if (correct - incorrect) % n: continue
  print("answer:", correct)

      Like

  • Unknown's avatar

    Jim Randell 4:41 pm on 1 September 2023 Permalink | Reply
    Tags:   

    Teaser 3180: Taking the chair 

    From The Sunday Times, 3rd September 2023 [link] [link]

    There were eight people on the committee: four men — Jingo, King, Ling and Ming, and four women — Sheena, Tina, Una and Vina. They had to choose a chairperson from among themselves and so each of them voted for their choice, each person choosing someone of the opposite sex. Jingo’s choice’s choice was King. Also, Ling’s choice’s choice’s choice was Sheena. Furthermore, Tina’s choice’s choice’s choice’s choice was Una.

    Just two people got fewer votes than their choice did. After further discussion the woman with the most votes was  made chairperson.

    (a) Who was that?
    (b) Who voted for her?

    [teaser3180]

     
    • Jim Randell's avatar

      Jim Randell 4:58 pm on 1 September 2023 Permalink | Reply

      See also: Tantalizer 8.

      This Python program runs in 126ms. (Internal runtime is 72ms).

      Run: [ @replit ]

      from enigma import (irange, subsets, cproduct, multiset, map2str, seq2str, printf)

# multiple lookups on dict <d>
def nget(d, k, n):
  for _ in irange(1, n):
    k = d[k]
  return k

# male and female labels
(male, female) = ("JKLM", "STUV")

# each chooses a member of the opposite gender
choose = lambda vs, k: subsets(vs, size=k, select='M')
for (ms, fs) in cproduct([choose(female, len(male)), choose(male, len(female))]):
  # map label to choice
  m = dict(zip(male + female, ms + fs))

  # check the conditions:
  if not nget(m, 'J', 2) == 'K': continue
  if not nget(m, 'L', 3) == 'S': continue
  if not nget(m, 'T', 4) == 'U': continue

  # count the votes
  vote = multiset.from_seq(m.values())

  # exactly 2 people got fewer votes than their choice
  if not sum(vote.count(k) < vote.count(v) for (k, v) in m.items()) == 2: continue

  # output voting map
  printf("{m}", m=map2str(m, arr="->"))
  # find max votes for a female
  t = max(vote.count(k) for k in female)
  # output winner
  for w in female:
    if vote.count(w) == t:
      vs = list(k for (k, v) in m.items() if v == w)
      printf("-> winner = {w} [choice of {vs}]", vs=seq2str(vs, enc=""))
  printf()

      Solution: (a) Una was made chairperson. (b) Jingo and King voted for her.

      The votes are:

      J → U
      K → U
      L,M → S,V

      S → L
      T → K
      U → K
      V → M

      L and M voted for S and V in some order.

      K and U got 2 votes each. And L, M, S, V got 1 vote each.

      Like

    • Frits's avatar

      Frits 8:14 pm on 1 September 2023 Permalink | Reply

      Without a function for multiple lookup.

      from itertools import product
 
# male and female labels
(male, female) = ("JKLM", "STUV")

# all possible votes for men
ms = [m for m in product(female, repeat=4) if 'S' in m and 'U' in m]
# all possible votes for women
fs = [f for f in product(male, repeat=4) if 'K' in f]

# remove entries where 3 or more people nobody votes for
votes = [(m, f) for m, f in product(ms, fs) if len(set(m + f)) > 5]

# Jingo's choice's choice was King, J -> f? -> K
votes = [(m, f) for m, f in votes if f[female.index(m[0])] == 'K']

# Ling's choice's choice's choice was Sheena L -> f? -> m? -> S
votes = [(m, f) for m, f in votes 
         if m[male.index(f[female.index(m[2])])] == 'S']

# Tina's choice's choice's choice's choice was Una, T -> m? -> f? -> m? -> U
votes = [(m, f) for m, f in votes 
         if m[male.index(f[female.index(m[male.index(f[1])])])] == 'U']

# exactly 2 people got fewer votes than their choice
for m, f in votes:
  d = dict()
  # build frequency dictionary
  for i in range(4):
    d[male[i]] = f.count(male[i])
    d[female[i]] = m.count(female[i])

  # count the number of people who got fewer votes than their choice
  c = sum((d[male[i]] < d[m[i]]) + (d[female[i]] < d[f[i]]) for i in range(4))  
  
  if c == 2: 
    mx = max([(d[w], w) for w in female])
    print(f"That was {mx[1]}, voted for by "
          f"{' and '.join(male[i] for i, x in enumerate(m) if x == mx[1])}"
          f", {m}, {f}")

      Like

    • Frits's avatar

      Frits 11:15 pm on 1 September 2023 Permalink | Reply 

      from enigma import SubstitutedExpression

# male and female labels
(male, female) = ("JKLM", "STUV")

# the alphametic puzzle
p = SubstitutedExpression(
  [
   # Jingo's choice's choice was King, J -> f? -> K
   "(f := [S, T, U, V])[J] == 1",
   
   # Ling's choice's choice's choice was Sheena, L -> f? -> m? -> S
   "(m := [J, K, L, M])[f[L]] == 0",
  
   # Tina's choice's choice's choice's choice was Una, T -> m? -> f? -> m? -> U
   "m[f[m[T]]] == 2",
   
   # exactly 2 people got fewer votes than their choice
   "sum((f.count(i) < m.count(m[i])) if i < 4 else \
        (m.count(i - 4) < f.count(f[i - 4])) for i in range(8)) == 2",
  ],
  answer="m, f",
  distinct="",
  digits=range(4),
  reorder=0,
  verbose=0,    # use 256 to see the generated code
)

# print answers
for ans in p.answers():
  # build frequency dictionary
  d = {i: ans[0].count(i) for i in range(4)}
  mx = max([(d[i], i) for i in range(4)])  
 
  print(f"That was {female[mx[1]]}, voted for by "
        f"{' and '.join(male[i] for i, x in enumerate(ans[0]) if x == mx[1])}"
        f", {[female[x] for x in ans[0]]} {[male[x] for x in ans[1]]}")

      Like

  • Unknown's avatar

    Jim Randell 9:21 am on 31 August 2023 Permalink | Reply
    Tags:   

    Teaser 1952: Naturally enough 

    From The Sunday Times, 13th February 2000 [link]

    Imagine putting a digit into each of the 20 boxes of this grid so that, reading across, there are four five-figure numbers (labelled 1–4) and, reading down, five four-figure numbers (labelled 5–9). Do this in such a way that (naturally) the resulting numbers 1–9 have the following properties:

    1, 4 and 9 are squares;
    1 and 8 are cubes;
    2, 3, 5 and 7 are primes;
    6 is the product of two primes;
    the average of 1–9 is more than 4.

    You should then be able to answer the question:

    What is the five-figure number forming 4 across?

    Although this brings the total number of Teaser puzzles set by Victor Bryant to 94 there are a further 5 set by Dr V W Bryant and 1 set by V. Bryant, so it seems likely there are now 100 Teaser puzzles set by Victor Bryant on the S2T2 site (of the 931 Teaser puzzles posted). Which is just over 10.7% (about 1 in 9) of the puzzles on the site.

    If we include the 413 Enigma puzzles set by Victor Bryant under the pseudonym of Susan Denham (geddit?) that are posted on the Enigmatic Code site we get a total of 513 / 3160 puzzles, just over 16.2% (or about 1 in 6) of the puzzles set over both sites that originate from Victor Bryant.

    [teaser1952]

     
    • Jim Randell's avatar

      Jim Randell 9:21 am on 31 August 2023 Permalink | Reply

      Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

      It is not particularly fast, but squeaks in just under 1s with a runtime of 999ms. (Internal runtime of the generated program is 860ms).

      Run: [ @replit ]

      #! python3 -m enigma -rr

#     [5][6][7][8][9]
#  [1] A  B  C  D  E
#  [2] F  G  H  I  J
#  [3] K  L  M  N  P
#  [4] Q  R  S  T  U

SubstitutedExpression

--distinct=""

# [1] [4] [9] are squares
"is_square(ABCDE)"
"is_square(QRSTU)"
"is_square(EJPU)"

# [1] [8] are cubes
"is_cube(ABCDE)"
"is_cube(DINT)"

# [2] [3] [5] [7] are primes
"is_prime(FGHIJ)"
"is_prime(KLMNP)"
"is_prime(AFKQ)"
"is_prime(CHMS)"

# [6] is the product of two primes
"len(factor(BGLR)) = 2"

# the average (mean) of [1]-[9] is > [4]
"avg([ABCDE, FGHIJ, KLMNP, QRSTU, AFKQ, BGLR, CHMS, DINT, EJPU]) > QRSTU"

# answer is [4]
--answer="QRSTU"

--template="ABCDE FGHIJ KLMNP QRSTU / AFKQ BGLR CHMS DINT EJPU"
--solution=""
--denest=-1
#--verbose="A" # just show the answer

      Solution: [4] = 15376.

      There are 807 ways to fill out the grid. But [4] is the same in all of them.

      In fact the following values are determined:

      Like

  • Unknown's avatar

    Jim Randell 8:53 am on 29 August 2023 Permalink | Reply
    Tags:   

    Teaser 2638: Digilist 

    From The Sunday Times, 14th April 2013 [link] [link]

    I have written down a list of positive whole numbers in ascending order. Overall they use each of the digits 0 to 9 exactly once. There is more than one even number in the list, and the majority of the numbers are primes. The final number in the list is a perfect square and it is the sum of all the other numbers in the list.

    What is my list of numbers?

    [teaser2638]

     
    • Jim Randell's avatar

      Jim Randell 8:56 am on 29 August 2023 Permalink | Reply

      There is more than 1 even number on the list (i.e. at least 2), and the majority of numbers on the list are primes.

      There is only one even prime, so there must be at least 2 primes on the list, so the list must have at least 3 numbers.

      If the final square had 5 digits there would only be 5 digits remaining to form at least 2 numbers that sum to a 5 digit number, and this is not possible.

      So we only need to consider results that are 2, 3, or 4 digits.

      My first program was straightforward, but a bit slow (execution time was 1.9s).

      This Python program constructs an alphametic problem by allocating 2, 3 or 4 digits to the result of the sum, and then chops the remaining digits into numbers that make up the terms. It constructs a corresponding alphametic puzzle, and then uses the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve that.

      It runs in 128ms.

      Run: [ @replit ]

      from enigma import (
  SubstitutedExpression, irange, express, str_upper, tuples,
  sprintf as f, join, printf
)

# construct appropriate alphametic words
def words(qs, ds, k, syms=str_upper):
  for (q, d) in zip(qs, ds):
    # return <q> <d>-length words
    for _ in irange(1, q):
      yield syms[:d]
      syms = syms[d:]
  # and a <k>-length result
  yield syms[:k]

# the result is a <k> digit square
for k in [2, 3, 4]:
  # form numbers from the remaining digits
  ds = list(irange(1, k))
  for qs in express(10 - k, ds):
    if qs[-2:] == [0, 0]: continue
    # make appropriate length alphametic words for the numbers
    ss = list(words(qs, ds, k))

    # the alphametic puzzle sum
    sq = ss.pop(-1)
    expr = f("{ss} = {sq}", ss=join(ss, sep=" + "))
    # additional constraints
    exprs = [
      f("is_square({sq})"),  # result is a square
    ]
    # remove duplicates (same length symbols are ordered)
    for (x, y) in tuples(ss, 2):
      if len(x) == len(y):
        exprs.append(f("{x} < {y}", x=x[0], y=y[0]))
    # now add in expressions over the entire list
    ss.append(sq)
    ns = join(ss, sep=", ", enc="[]")
    exprs.extend([
      # more than 1 of the numbers is even
      f("sum(n % 2 == 0 for n in {ns}) > 1 "),
      # the majority of the numbers are prime
      f("sum(is_prime(n) for n in {ns}) > {t}", t=len(ss) // 2),
    ])

    # make a solver for the puzzle
    p = SubstitutedExpression.split_sum(
      expr, extra=exprs,
      d2i={ 0: list(s[0] for s in ss) }, # all numbers are positive
      template=expr,
    ).solver()

    # solve the puzzle
    for s in p.solve(verbose=0):
      # output the sum
      printf("{s}", s=p.substitute(s, expr))

      Solution: The numbers are: 2, 3, 80, 491, 576.

      And:

      2 + 3 + 80 + 491 = 576

      There is more than one even number (2, 80, 576), and 3 of the 5 numbers are prime (2, 3, 491).

      The wording of the puzzle excludes 0 from being on the list, but if it were allowed we would have an alternative solution of:

      0 + 2 + 83 + 491 = 576

      Which has the same square total.

      There are also two further solutions that have the same square total (although a different one to above) but have fewer than 2 even numbers on the list:

      3 + 5 + 41 + 680 = 729
      3 + 5 + 80 + 641 = 729

      Like

  • Unknown's avatar

    Jim Randell 6:57 am on 27 August 2023 Permalink | Reply
    Tags:   

    Brain-Teaser 930: All scores different 

    From The Sunday Times, 18th May 1980 [link]

    The great thing about the new football method is that goals count whoever wins the match. With luck, this will create more excitement, and reward skills better than at present, for even a side losing heavily will have some incentive to attack.

    In this new method 10 points are given for a win, five points to each side for a draw, and one point for each goal scored.

    In a recent competition between four teams — A, B, C and D — who played each other once, the points were as follows:

    A = 25
    B = 34
    C = 6
    D = 20

    Each side scored at least one goal in each match, and — rather interestingly — the score was different in every match.

    Find the score in each match.

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994). The puzzle text above is taken from the book.

    The setter is given as “the late Eric Emmet”, and it is noted in the 1994 book that Eric Emmet had died by the time this puzzle was published.

    Eric Emmet also contributed the Puzzle series of puzzles to New Scientist, as well as several Enigma puzzles (including puzzles published as late as 1991).

    [teaser930]

     
    • Jim Randell's avatar

      Jim Randell 6:57 am on 27 August 2023 Permalink | Reply

      See also: Enigma 599.

      The total number of points awarded are: 25 + 34 + 6 + 20 = 85.

      There are 6 matches in the tournament, and 10 points are awarded depending on the match outcomes, accounting for 60 points in total.

      So there are 25 goals scored. Each side scored at least one goal in each match, so there are only 13 goals unaccounted for.

      In order to get different scores in each match we need to allow at least 4, but no more than 6 goals to be scored (per team) in any match.

      And we find a solution with up to 4 goals scored (per team) in a match.

      The following run file executes in 79ms. (Internal runtime of the generated program is 19ms).

      Run: [ @replit ]

      #! python3 -m enigma -rr

# matches are:
#
#  A vs B = E - F
#  A vs C = G - H
#  A vs D = I - J
#  B vs C = K - L
#  B vs D = M - N
#  C vs D = P - Q

SubstitutedExpression

--distinct=""
--digits="1-4"

# points for X in match X vs Y, where the score is x - y
--code="pts = lambda x, y: compare(x, y, (0, 5, 10)) + x"

# points for A, B, C, D
"pts(E, F) + pts(G, H) + pts(I, J) == 25"
"pts(F, E) + pts(K, L) + pts(M, N) == 34"
"pts(H, G) + pts(L, K) + pts(P, Q) ==  6"
"pts(J, I) + pts(N, M) + pts(Q, P) == 20"

# all matches have different scores
"seq_all_different(ordered(x, y) for (x, y) in [(E, F), (G, H), (I, J), (K, L), (M, N), (P, Q)])"

# output match scores
--header=""
--template="AvB = {E}-{F}; AvC = {G}-{H}; AvD = {I}-{J}; BvC = {K}-{L}; BvD = {M}-{N}; CvD = {P}-{Q}"
--solution=""

      Solution: The scores are: A vs B = 2-2; A vs C = 2-1; A vs D = 1-1; B vs C = 4-3; B vs D = 3-1; C vs D = 2-3.

      Like

      • Frits's avatar

        Frits 1:01 pm on 28 August 2023 Permalink | Reply

        @Jim, please elaborate on “And we find a solution with up to 4 goals scored in a match”.

        Did you add it because –digits=”1-6″ was rather slow?

        Like

        • Jim Randell's avatar

          Jim Randell 7:51 am on 29 August 2023 Permalink | Reply

          Limiting the digits to a maximum of 4 is enough to find a solution, and it runs quickly.

          For an exhaustive search we can increase the maximum digit to 6. The runtime is increased, but it is still less than 1s.

          But we can limit the total number of goals in any match (it cannot be more than 7), to reduce the runtime again. It complicates the recipe a little, but gives a complete solution that still runs quickly.

          Like

    • Frits's avatar

      Frits 11:45 am on 30 August 2023 Permalink | Reply

      Based on Jim’s setup.

       
# points for X in match X vs Y, where the score is x - y
pts = lambda x, y: x + 1 if x < y else x + 6 if x == y else x + 11

# decompose <t> into <k> numbers from <ns> so that sum(<k> numbers) equals <t>
def solve(t, ns, k=12, s=[]):
  if k == 1:
    if t in ns:
      yield s + [t]
  else: # perform early checks
  
    # H, L, P are already known
    if k == 9 and sum(s) != 3: return  # C must have had three losses
    
    # H, L, P, F, K, M are already known
    if k == 6:                 
      if sum(s) != 9: return   # B didn't loose and had one draw
      if s[4] <= s[1]: return  # C lost from B
    
    # H, L, P, F, K, M, E, N are already known  
    if k == 4:                  
      # B: 2 wins and a draw so (F > E and M == N) or (F == E and M > N)
      if not((s[3] > s[6] and s[5] == s[7]) or 
             (s[3] == s[6] and s[5] > s[7])): return
      # G > 0 and Q > 0  (and G + Q > 2)
      if sum(s) > 10: return
   
    # G > H and Q > P (and G + Q > 2)
    if k == 3 and s[-1] <= s[0]:  return 
    if k == 2 and (s[-1] <= s[2] or s[-2] + s[-1] <= 2):  return 
       
    for n in ns:
      if n > t: break
      yield from solve(t - n, ns, k - 1, s + [n])
      
# there are 25 goals scored and only 13 goals unaccounted for   
# B must have had two wins and one draw as otherwise one win and two draws
# leads to two "1 - 2" losses for C (who had definitely had three losses)

# matches are:
#
#  A vs B = 1 + E  -  1 + F
#  A vs C = 1 + G  -  1 + H
#  A vs D = 1 + I  -  1 + J
#  B vs C = 1 + K  -  1 + L
#  B vs D = 1 + M  -  1 + N
#  C vs D = 1 + P  -  1 + Q

# first unaccounted goals for C followed by unaccounted goals for B
for H, L, P, F, K, M, E, N, G, Q, I, J in solve(13, range(6)):
  games = [(E, F), (G, H), (I, J), (K, L), (M, N), (P, Q)]
  # all matches have different scores
  if len(set(seq := [tuple(sorted(x)) for x in games])) != len(seq): continue
  
  # check number of points
  if pts(E, F) + pts(G, H) + pts(I, J) != 25: continue
  if pts(J, I) + pts(N, M) + pts(Q, P) != 20: continue
  if pts(J, I) + pts(N, M) + pts(Q, P) != 20: continue
  if pts(H, G) + pts(L, K) + pts(P, Q) !=  6: continue
  
  desc = "AvB AvC AvD BvC BvD CvD".split()
  # add 1 to scores
  scores = [desc[i] + " = " + str(games[i][0] + 1) + "-" + 
            str(games[i][1] + 1) for i in range(6)]
  print(', '.join(scores))

      Like

  • Unknown's avatar

    Jim Randell 4:37 pm on 25 August 2023 Permalink | Reply
    Tags:   

    Teaser 3179: Sums and products 

    From The Sunday Times, 27th August 2023 [link] [link]

    Mrs Green uses a dice game to improve the maths skills of her pupils. The children sit four to a table and each child throws five dice. Points are awarded according to the sum and product for each child; bonus points are awarded if the five scores contain three or more of the same number.

    At Liam’s table all four children had the same sum but all achieved that sum in different ways. Surprisingly, all of the twenty dice scored more than one and no bonus points were awarded. The distribution of scores was:

    sixes … 5;
    fives … 4;
    fours … 2;
    threes … 4;
    twos … 5.

    Liam had the highest product.

    What, in ascending order, were Liam’s five scores?

    [teaser3179]

     
    • Jim Randell's avatar

      Jim Randell 4:54 pm on 25 August 2023 Permalink | Reply

      A bit more straightforward than the last couple of puzzles.

      This Python program runs in 52ms. (Internal runtime is 457µs).

      from enigma import (subsets, multiset, multiply, ediv, printf)

# required distribution of values
dist = multiset.from_dict({ 6: 5, 5: 4, 4: 2, 3: 4, 2: 5 })

# find the common sum <k> for the 4 children
k = ediv(dist.sum(), 4)

# generate possible scores
def scores(dist, k):
  # no value occurs more than 2 times
  m = multiset.from_pairs((k, min(2, v)) for (k, v) in dist.items())
  # look for scores from 5 dice, with the required sum
  for ss in subsets(sorted(m), size=5, select='uC'):
    if sum(ss) == k:
      yield ss

# consider 4-sets of scores
for ss in subsets(scores(dist, k), size=4):
  # look for required distribution
  if multiset.from_seq(*ss) == dist:
    printf("sum = {k} -> scores = {ss}")
    # find the set with the max product
    ans = max(ss, key=multiply)
    # output solution
    printf("ans = {ans}")

      Solution: Liam’s scores were: 3, 3, 4, 5, 5.

      Each child on Liam’s table threw 5 dice with a sum of 20:

      3, 3, 4, 5, 5 → product = 900 (Liam)
      2, 3, 3, 6, 6 → product = 648
      2, 2, 5, 5, 6 → product = 600
      2, 2, 4, 6, 6 → product = 576

      Liked by 1 person

    • Frits's avatar

      Frits 9:25 pm on 25 August 2023 Permalink | Reply 

      from enigma import SubstitutedExpression

# (A, B, C, D, E), (F, G, H, I, J), (K, L, M, N, O), (P, Q, R ,S, T) 
# are four rows of frequencies (0-2)

# the alphametic puzzle
p = SubstitutedExpression(
  [ # check first 3 scores of 5 throws
    "5 - A - B - C - D = E",
    "2 * A + 3 * B + 4 * C + 5 * D + 6 * E == 20",
    
    "F >= A",
    "5 - F - G - H - I = J",
    "2 * F + 3 * G + 4 * H + 5 * I + 6 * J == 20",
    
    "K >= F",
    "5 - K - L - M - N = O",
    "2 * K + 3 * L + 4 * M + 5 * N + 6 * O == 20",
    
    # fill up to target frequency
    "5 - A - F - K = P", 
    "P >= K",
    
    "4 - B - G - L = Q",
    "2 - C - H - M = R",
    "4 - D - I - N = S",
    "5 - E - J - O = T",
    
    "2 * P + 3 * Q + 4 * R + 5 * S + 6 * T == 20",
    
    # different throws, avoid duplicates
    "sorted(set(@rows)) == @rows",
  ],
  answer="[(multiply(i**y for i, y in enumerate(x, 2) if y), x) \
                     for x in @rows]",
  base=3,
  d2i="",
  macro=dict([("rows", "[(A, B, C, D, E), (F, G, H, I, J), \
                         (K, L, M, N, O), (P, Q, R ,S, T)]")]),
  distinct="",
  digits=range(0, 3),
  reorder=0,
  verbose=0,    # use 256 to see the generated code
)

# print answers
for ans in p.answers():
  nums = [[i] * y for i, y in enumerate(ans[0][1], 2) if y]
  print(f"answer: {sorted(y for x in nums for y in x)}")

      Like

  • Unknown's avatar

    Jim Randell 2:09 pm on 24 August 2023 Permalink | Reply
    Tags:   

    Brain-Teaser 922: Additional letters 

    From The Sunday Times, 23rd March 1980 [link]

    It is true, of course, that there are a lot of letters in this puzzle, but in spite of that I thought that for once Uncle Bungle was going to write it out correctly.

    In fact there was no mistake until the third and last line across but, in that line one of the letters, I’m afraid, was incorrect.

    This is another addition sum with letters substituted for digits. Each letter consistently stands for the same digit whenever it appears, and different letters stand for different digits — or at least the should do — and they do, but for the mistake in the last line across:

    What is the correct numerical 10-figure answer to the addition sum?

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994). The puzzle text above is taken from the book.

    [teaser922]

     
    • Jim Randell's avatar

      Jim Randell 2:09 pm on 24 August 2023 Permalink | Reply

      We can use [[ bungled_sum() ]] solver (as seen in Puzzle 56 etc.) for this puzzle.

      The following Python command line executes in 146ms.

      Run: [ @replit ]

      >>> from bungled_sum import bungled_sum
>>> bungled_sum(['YTBBEDMKD', 'YHDBTYYDD', 'EDYTERTPTY'], [2])

                                T
YTBBEDMKD + YHDBTYYDD = EDYTERTPHY / @[2,8] T -> H
695513243 + 673596633 = 1369109876 / B=5 D=3 E=1 H=7 K=4 M=2 P=8 R=0 T=9 Y=6

      Solution: The answer to the addition sum is 1369109876.

      The complete sum is:

      695513243 + 673596633 = 1369109876

      which would have been correctly posed as:

      YTBBEDMKD + YHDBTYYDD = EDYTERTPHY

      i.e. the penultimate (tens) digit of the result should have been H (not T).

      Like

  • Unknown's avatar

    Jim Randell 8:46 am on 22 August 2023 Permalink | Reply
    Tags:   

    Teaser 2635: Three sisters 

    From The Sunday Times, 24th March 2013 [link] [link]

    In Shakespeare’s less well-known prequel, King Lear shared all of his estate amongst his three daughters, each daughter getting a fraction of the estate. The three fractions, each in their simplest form, had numerators less than 100 and had the same denominators. Cordelia got the least, with Regan getting more, and Goneril the most (but her share was less than three times Cordelia’s). Each of the three fractions gave a decimal which recurred after three places (as in 0.abcabc…) and each digit from 1 to 9 occurred in one of the three decimals.

    What were the three fractions?

    [teaser2635]

     
    • Jim Randell's avatar

      Jim Randell 8:46 am on 22 August 2023 Permalink | Reply

      The recurring decimal 0.(abc)… is a representation of the fraction abc/999.

      But the three fractions we are dealing with can all be reduced to fractions with a 2-digit numerator, and a common denominator.

      So the common denominator must be a divisor of 999.

      This Python program runs in 54ms. (Internal runtime is 2ms)

      Run: [ @replit ]

      from enigma import (divisors, decompose, gcd, recurring, join, printf)

# return viable repetends
def repetend(n, d):
  r = recurring(n, d)
  return (None if r.nr else r.rr)

# consider possible denominator values 3 < d < 100
for d in divisors(999):
  if d < 4: continue
  if d > 297: break

  # calculate ascending splits of the denominator
  for (C, R, G) in decompose(d, 3, increasing=1, sep=1, min_v=1, max_v=99):
    # G's share is less than 3 times C's
    if not (G < 3 * C): continue
    # check fractions are in lowest terms
    if any(gcd(n, d) != 1 for n in (C, R, G)): continue

    # find the repetends
    reps = list(repetend(n, d) for n in (C, R, G))
    if None in reps: continue
    # check the digits used
    ds = join(reps)
    if '0' in ds or len(set(ds)) < 9: continue

    # output solution
    (rC, rR, rG) = reps
    printf("C = {C}/{d} = 0.({rC})...; R = {R}/{d} = 0.({rR})...; G = {G}/{d} = 0.({rG})...")

      Solution: The fractions are: Cordelia = 6/37, Reagan = 14/37, Goneril = 17/37.

      As decimals:

      C = 0.(162)…
      R = 0.(378)…
      G = 0.(459)…

      Like

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