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Jim Randell 8:08 am on 1 July 2026 Permalink | Reply
Tags: by: A K Austin ( 5 )
Brain-Teaser 943: Order in football
From The Sunday Times, 17th August 1980 [link]

There are three teams In the Midchester football league: Albion, United and Victoria. During the season they play each other twice; once at home and once away.

A team gets two points for a win and one point tor a draw. After each match the local paper publishes the current league table. In this table the teams are listed according to points and if two or more teams have equal points then they are listed according to goal difference, (i.e. “goals for” minus “goals against”). If two or more teams have equal points and equal goal difference then they are listed alphabetically.

At the end of last season the sports editor published the results of all the matches:

Albion 4 – United 0
Albion 0 – Victoria 0
United 0 – Albion 1
United 2 – Victoria 0
Victoria 3 – Albion 0
Victoria 1 – United 2

He also said that each of the six league tables published during the season had put the teams in a different order. Thus each of the six possible orders of the teams (AUV, AVU, UAV, UVA, VAU, VUA) had occurred in one of the six league tables. Finally, he said that the third match of the season had been the draw, A vs. V.

List the six matches in the order in which they were played during the season.

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

[teaser943]
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Ruud and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:09 am on 1 July 2026 Permalink | Reply
  I thought about writing a recursive solver that checks the orders of the teams are all different as each match is chosen, but it is more straightforward to just generate the possible orderings of the matches, calculate the orders of the teams after each match, and see if they are all different.
  
  This Python program runs in 74ms. (Internal runtime is 1.9ms).
```
from enigma import (compare, subsets, join, seq_all_different, printf)

teams = "AUV"  # labels for the teams
points = [0, 1, 2]  # point for l, d, w

# produce the orders for the teams after each of the matches is played
def orders(ms):
  # go through the matches, and accumulate points, goal difference for each team
  (pts, gd, tss) = (dict.fromkeys(teams, 0), dict.fromkeys(teams, 0), list())
  for ((X, gX), (Y, gY)) in ms:
    # assign points
    pts[X] += compare(gX, gY, vs=points)
    pts[Y] += compare(gY, gX, vs=points)
    # count goals
    gd[X] += (gX - gY)
    gd[Y] += (gY - gX)
    # calculate order
    ts = join(sorted(teams, key=(lambda k: (pts[k], gd[k], -ord(k))), reverse=1))
    tss.append(ts)
  return tss

# the matches, and the outcomes:
# this match was 3rd
match3 = (("A", 0), ("V", 0))
# and these are the remaining matches
matches = [
  (("A", 4), ("U", 0)),
  (("U", 0), ("A", 1)),
  (("U", 2), ("V", 0)),
  (("V", 3), ("A", 0)),
  (("V", 1), ("U", 2)),
]

# choose an order for the remaining matches
for ms in subsets(matches, size=len, select='P', fn=list):
  # insert match3 into the 3rd position (= index 2)
  ms.insert(2, match3)
  # calculate the team orders after each match
  tss = orders(ms)
  # check they are all different
  if seq_all_different(tss):
    for (i, (((X, x), (Y, y)), ts)) in enumerate(zip(ms, tss), start=1):
      printf("{i}: {X} vs. {Y}  ({x} - {y});  order = {ts}")
    printf()
```
  Solution: The order of the matches was: U vs. V; V vs. A; A vs. V; U vs. A; A vs. U; V vs. U.
  
  Giving the following orderings of the teams after each match:
  
  match 1: U vs. V = 2 – 0
  order: U (2 points, +2 goal diff); A (0 points, 0 goal diff); V (0 points, −2 goal diff)
  
  match 2: V vs. A = 3 – 0
  order: U (2 points, +2 goal diff); V (2 points, +1 goal diff); A (0 points, −3 goal diff)
  
  match 3: A vs. V = 0 – 0
  order: V (3 points, +1 goal diff); U (2 points, +2 goal diff); A (1 points, −3 goal diff)
  
  match 4: U vs. A = 0 – 1
  order: V (3 points, +1 goal diff); A (3 points, −2 goal diff); U (2 points, +1 goal diff)
  
  match 5: A vs. U = 4 – 0
  order: A (5 points, +2 goal diff); V (3 points, +1 goal diff); U (2 points, −3 goal diff)
  
  match 6: V vs. U = 1 – 2
  order: A (5 points, +2 goal diff); U (4 points, −2 goal diff); V (3 points, 0 goal diff)
  
  LikeLike
- Ruud 8:18 pm on 1 July 2026 Permalink | Reply
  Correction:
```
import peek
import itertools

for matches in itertools.permutations("AU40 AV00 UA01 UV20 VA30 VU12".split()):
    if matches[2] != "AV00":
        continue
    points = dict.fromkeys("AUV", 0)
    goals_difference = dict.fromkeys("AUV", 0)
    score = {"A": (0, 0, "A"), "U": (0, 0, "U"), "V": (0, 0, "V")}
    seen = set()
    for match in matches:
        team0, team1 = match[:2]
        goals0, goals1 = map(int, match[2:])
        points[team0] += (goals0 > goals1) + (goals0 >= goals1)
        points[team1] += (goals1 > goals0) + (goals1 >= goals0)
        goals_difference[team0] += goals0 - goals1
        goals_difference[team1] += goals1 - goals0
        score[team0] = (points[team0], goals_difference[team0], team0)
        score[team1] = (points[team1], goals_difference[team1], team1)
        order = tuple(x[2] for x in sorted(score.values(), key=lambda x: (-x[0], -x[1], ord(x[2]))))
        if order in seen:
            break
        seen.add(order)
    else:
        peek(matches)
```
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Jim Randell 8:21 am on 15 May 2026 Permalink | Reply
Tags: by: Rupert Segar
Brain teaser 975: Ambiguous birthdays
From The Sunday Times, 29th March 1981 [link]

Recently I attended the Numerical Astrologers Conference, held annually in Brighton. At this meeting, members wore badges on which were written not only their names, but also their dates of birth. These dates were written numerically in the usual English manner of first putting the day, then the month, then the year (e.g. 7th March 1946 would be represented by 7/3/46, and the 31st December 1954 by 31/12/54). The year always uses two digits.

I had just entered the meeting when I was accosted by an internationally acclaimed cabalist, who, I noticed, shared with me the month of his birthday, though he was born in a year eight years prior to the year of my birth. He had been looking at my badge, and now told me that the digits of my date of birth were, in sequence, in exactly the reverse order to his own.

We were eagerly discussing this coincidence when an old acquaintance of mine introduced himself to the cabalist, who then made the same claim to my friend about their dates of birth. At first this puzzled me, as I knew my friend to be younger than myself, but, upon inspection, both the claims proved to be correct.

Strangely, both of us had dates of birth whose digits, when taken in completely the reverse order, were exactly the same as those of the cabalist.

Numerically, what was the cabalist’s date of birth?

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

[teaser975]
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ruudvanderham and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:22 am on 15 May 2026 Permalink | Reply
  The cabalist’s DOB cannot be 6 digits or 4 digits, as there is only one way to reverse these:
  
  UV/WX/YZ → ZY/XW/VU
  W/X/YZ → Z/Y/XW
  
  But if it had 5 digits, there are 2 ways it can be reversed:
  
  V/WX/YZ or VW/X/YZ → Z/YX/VW or ZY/X/VW
  
  As the friend is younger than the setter they must be born later in the year, so:
  
  setter = ZY/X/VW
  friend = Z/YX/VW
  
  And the cabalist and the setter share their birth month, so:
  
  cabalist = VW/X/YZ
  setter = ZY/X/VW
  friend = Z/YX/VW
  
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  It runs in 86ms. (Internal runtime of the generated code is 2.2ms).
```
#! python3 -m enigma -rr

SubstitutedExpression

# convert a 2-digit year
--code="year = lambda y: (1900 + y if y < 70 else 1800 + y)"
# check for a valid date
--code="from datetime import date"
--code="valid = lambda d, m, y: catch(date, year(y), m, d) is not None"

# the cabalist's DOB is VW/X/YZ
# and can be reversed as setter = ZY/X/WV, friend = Z/YX/WV
--distinct=""
--invalid="0,VYZ" # dates do not have components with leading zeros

# check the three dates are valid
"valid(VW, X, YZ)"
"valid(ZY, X, WV)"
"valid(Z, YX, WV)"

# the [2-digit] years are 8 apart
"year(YZ) + 8 == year(WV)"

--template="VW/X/YZ -> ZY/X/WV and Z/YX/WV"
--solution=""
```
  Solution: The cabalist’s date of birth was 12/1/13 (12th January 1913).
  
  And so the setter’s date of birth is 31/1/21 (31st January 1921) and the friend’s date of birth is 3/11/21 (3rd November 1921).
  
  LikeLike
- ruudvanderham 2:01 pm on 15 May 2026 Permalink | Reply
```
import datetime

date0 = datetime.datetime(1880, 1, 1)
date1 = datetime.datetime(1980, 1, 1)

day = {}
month = {}
date = date0
while date < date1:
    date = date + datetime.timedelta(1)
    as_str = f"{date.day}{date.month}{date.year % 100:02d}"[::-1]
    if len(as_str) == 5 and "0" not in as_str[:3]:
        year = int(as_str[3:])
        year += 1800 if year > 80 else 1900
        if year - date.year == 8:
            for i in range(2):
                month[i] = int(as_str[i + 1 : 3])
                day[i] = int(as_str[: i + 1])
                if month == date.month:
                    i_equal = i
                try:
                    datetime.datetime(year, month[i], day[i])
                except ValueError:
                    break
            else:
                for i in range(2):
                    if month[i] == date.month:
                        if datetime.datetime(year, month[i], day[i]) < datetime.datetime(year, month[1 - i], day[1 - i]):
                            print(
                                f"Cabalist: {date:%Y-%m-%d} / Me: {datetime.datetime(year, month[i], day[i]):%Y-%m-%d} / Friend:{datetime.datetime(year, month[1 - i], day[1 - i]):%Y-%m-%d}"
                            )
```
  LikeLike
Jim Randell 9:04 am on 28 April 2026 Permalink | Reply
Tags: by: Alfred Moritz ( 2 )
Brainteaser 1049: A question of colour
From The Sunday Times, 5th September 1982 [link]

A frame at snooker, as every player — and every watcher of “Pot Black” — will know, consists of fifteen red balls, worth one point each, and six “colours”: yellow = 2, green = 3, brown = 4, blue = 5, pink = 6 and black = 7.

The reds must all be potted first: after each red the player chooses a single colour to pot (immediately replaced on the table) before addressing the next red. Any miss (or potting of the wrong colour) causes the other player to have his turn. After the last red has been potted, with an attempt at a colour, all the colours are on the table, and they must be potted in ascending order of value.

The game ends when all the colours have been potted, or earlier if the current player has reached a score that cannot be equalled or exceeded by his opponent. For our present purposes, there are no penalty points or snookers.

Unlike the champions we see on television, Arthur and Bertie recently played a frame which was as undistinguished in quality (though neither actually incurred any penalty points) as it was remarkable in other ways. Each of them potted the same number of balls; the scores were level immediately before Bertie potted the last of the reds; and the winner — Arthur by a single point —was not decided until the table was finally cleared. They then discovered that Arthur’s score was the lowest possible for such a close win.

What was Arthur’s score? Which player potted each of the final six “colours” (e.g. B, B, B, B, B, A).

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

[teaser1049]
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Jim Randell is discussing. Toggle Comments
- Jim Randell 9:04 am on 28 April 2026 Permalink | Reply
  We are looking for the lowest possible scores.
  
  The scores are level with 1 red remaining. So 14 of the reds have been potted. For the lowest possible scores the associated colours must have been missed. So each player has potted 7 reds (and no colours) for a total score of 7.
  
  The remaining balls are: the final red, a possible additional colour, and the 6 colours in order.
  
  So, if the table is cleared and each player pots the same number of balls: B pots the final red, and a colour, to put him on 8 + x points (and has now potted 9 balls). So of the 6 remaining colours B pots 2 and A pots 4.
  
  This Python program examines possible ends to the match.
  
  It runs in 70ms. (Internal runtime is 143µs).
```
from enigma import (Enumerator, printf)

colours = [2, 3, 4, 5, 6, 7]

# play the remaining colours
def play(A, B, cols, As=[], Bs=[]):
  # are we done?
  if not cols:
    yield (A, B, cols, As, Bs)
  else:
    rest = sum(cols)
    # has someone won?
    if A > B + rest or B > A + rest:
      yield (A, B, cols, As, Bs)
    else:
      # someone pots the next colour
      x = cols[0]
      yield from play(A + x, B, cols[1:], As + [x], Bs)
      yield from play(A, B + x, cols[1:], As, Bs + [x])

# solve the puzzle, where B pots the final red with colour <x>
def solve(x):
  for (A, B, cols, As, Bs) in play(7, 8 + x, colours):
    # A wins by 1 point
    if not (A == B + 1): continue
    # the table is cleared
    if cols: continue
    # A and B potted the same number of balls
    if not (len(As) == 4 and len(Bs) == 2): continue
    # return the final scores and colours potted
    yield (A, B, As, Bs)

# choose a colour for A to pot with the final red
for x in colours:
  ss = Enumerator(solve(x))
  for (A, B, As, Bs) in ss:
    # output solution
    printf("A pots colours {As}; total = {A} pts")
    printf("B pots final red + {x}; pots colours {Bs}; total = {B} pts")
    printf()
  # we only need the smallest solution
  if ss.count > 0: break
```
  Solution: Arthur’s score was 23. The final six colours were potted as: A, A, A, B, B, A.
  
  The match started with A and B each potting 7 reds and no colours, and then:
  
  → A=7, B=7, [1 red remaining]
  B pots the final red and the green (3)
  → A=7, B=11, [27 points remaining]
  A pots the yellow (2)
  → A=9, B=11, [25 points remaining]
  A pots the green (3)
  → A=12, B=11, [22 points remaining]
  A pots the brown (4)
  → A=16, B=11, [18 points remaining]
  B pots the blue (5)
  → A=16, B=16, [13 points remaining]
  B pots the pink (6)
  → A=16, B=22, [7 points remaining]
  A pots the black (7)
  → A=23, B=22 [final scores]
  
  B has potted 7 + 2 + 2 = 11 balls. And A has potted 7 + 4 = 11 balls.
  
  LikeLike
Jim Randell 7:33 am on 24 March 2026 Permalink | Reply
Tags: by: Angela Newing ( 40 )
Brain-Teaser 968: Friends and neighbours
From The Sunday Times, 8th February 1981 [link]

My old school friends, Rose and May, both live on the sunny side of Woodland Grove in a row of five houses. I know that May lives with her parents, Mr and Mrs Joiner, in the house called “The Oaks”, but Rose had never told me her surname nor the name of her house.

I wanted to send both of them invitations to my 21st birthday party, so I asked the local paper boy if he knew which house Rose lived in or the name of her parents. He was not sure, but came up with the following facts:

1. Each of the five houses is occupied by a married couple with one unmarried daughter.
2. The last of the five houses on the right is called “Fir Trees”.
3. The Turner family lives in the house immediately to the right of the house called “The Willows”.
4. The Carpenters live next door but one to the house called “Silver Birches”.
5. The second house from the left is the home of the Sawyer family.
6. Cherry is the daughter of the couple in the middle house.
7. Ivy lives with her parents at The Elms”.
8. Hazel is the daughter of Mr and Mrs Woodman.

From this information I was able to deduce Rose’s surname and the name of her house.

What are they?

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

[teaser968]
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Jim Randell is discussing. Toggle Comments
- Jim Randell 7:34 am on 24 March 2026 Permalink | Reply
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  It runs in 80ms. (Internal runtime of the generated code is 85µs).
```
#! python3 -m enigma -rr

SubstitutedExpression

# allocate house numbers 1-5 to:
--digits="1-5"
# house names:
--macro="@Elm = A"
--macro="@Fir = B"
--macro="@Oak = C"
--macro="@Bir = D"
--macro="@Wil = E"
# family name:
--macro="@Car = F"
--macro="@Joi = G"
--macro="@Saw = H"
--macro="@Tur = I"
--macro="@Woo = J"
# daughter:
--macro="@Che = K"
--macro="@Haz = L"
--macro="@Ivy = M"
--macro="@May = N"
--macro="@Ros = P"
--distinct="ABCDE,FGHIJ,KLMNP"

# "May Joiner lives in Oak"
"@May = @Joi"
"@May = @Oak"

# 2: "House 5 is Fir"
"@Fir = 5"

# 3: "Turners live in the house immediately right of Willow"
"@Wil + 1 = @Tur"

# 4: "Carpenters live next door but one to Birch"
"abs(@Car - @Bir) = 2"

# 5: "Sawyers are house 2"
"@Saw = 2"

# 6: "Cherry is in house 3"
"@Che = 3"

# 7: "Ivy is in Elm"
"@Ivy = @Elm"

# 8: "Hazel Woodman"
"@Haz = @Woo"

--solution="ABCDEFGHIJKLMNP"
--template=""
--verbose="1-W"
```
  And here is a Python wrapper to prettify the output:
```
from enigma import (SubstitutedExpression, irange, printf)

# load the puzzle
p = SubstitutedExpression.from_file(["{dir}/teaser968.run"])

# link symbols to values
s2h = dict(A="The Elms", B="Fir Trees", C="The Oaks", D="Silver Birches", E="The Willows")
s2f = dict(F="Carpenter", G="Joiner", H="Sawyer", I="Turner", J="Woodman")
s2n = dict(K="Cherry", L="Hazel", M="Ivy", N="May", P="Rose")

# solve the puzzle
for s in p.solve(verbose=0):
  # fill out the slots
  (house, family, name) = (dict((s[k], v) for (k, v) in m.items()) for m in (s2h, s2f, s2n))
  # output the solution
  for i in irange(1, 5):
    (h, f, n) = (m.get(i, '???') for m in (house, family, name))
    printf("{i}: {n} {f}, \"{h}\"")
  printf()
```
  Solution: Rose is Rose Sawyer, she lives at “The Willows”.
  
  The complete solution is:
  
  1: Ivy Carpenter, “The Elms”
  2: Rose Sawyer, “The Willows”
  3: Cherry Turner, “Silver Birches”
  4: May Joiner, “The Oaks”
  5: Hazel Woodman, “Fir Trees”
  
  LikeLike
Jim Randell 6:51 am on 20 March 2026 Permalink | Reply
Tags: by: Sir Brian Young ( 3 )
Brain teaser 985: Sanders explores the river
From The Sunday Times, 7th June 1981 [link]

Sanders decided one day to explore the river upstream from his camp. The current flowed, and Sanders paddled, at constant rates; and when he stopped paddling, he changed the boat’s speed so that it immediately (or so let us assume) moved at the speed of the current.

Sanders’ companion, equipped with a watch and notebook, did his best to keep a log of the great voyage, but later Sanders found that the log read as follows:

08:30 — We leave camp. Sanders paddles upstream.
10:10 — We pass a blue jetty, and a shirt which is drifting down with the stream.
10:15 — Sanders stops paddling and rests.
??:?? — We are again abreast of the blue jetty. Sanders resumes his paddling upstream.
11:16 — Sanders stops paddling, turns the boat in mid-stream, and rests.
11:34 — The paddling downstream begins.
??:?? — Once again abreast of the blue jetty.
12:30 — We pass the shirt.
??:?? — We reach the camp.
??:?? — The shirt reaches the camp.

What are the four missing times?

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

[teaser985]
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John Crabtree and Jim Randell are discussing. Toggle Comments
- Jim Randell 6:52 am on 20 March 2026 Permalink | Reply
  
  There are no distance units given, so we shall say the jetty is 100 units distant from the camp.
  
  Suppose the river flows at a rate of r units/min, and Sanders rows (in still water) at a rate of s units/min.
  
  Then Sanders effective speed against the flow of the river is (s − r), and his effective speed with the flow of the river is (s + r).
  
  Between 8:30 and 10:10 = 100 minutes, Sanders rows 100 units against the flow of the river to the jetty:
  
  s − r = 100 / 100 = 1
  ⇒
  s = r + 1
  
  So, rowing against the river, Sanders proceeds at a rate of 1 unit/min. And rowing with the river he proceeds at a rate of (2r + 1).
  
  He then rows for another 5 minutes (until 10:15) against the river, so travels an additional 5 units distance beyond the jetty.
  
  And takes time t to drift back to the jetty (at 10:15 + t = the first unknown time):
  
  t = 5/r
  
  The shirt drifts down river for 140 minutes (between 10:10 and 12:30), at which point it has travelled a distance from the jetty towards the camp of: (140r).
  
  Sanders then rows against the river for (61 − t) minutes (between 10:15 + t and 11:16), and so achieves a distance from the jetty of: (61 − t).
  
  He drifts back for 18 minutes (until 11:34), and is nearer to the jetty by (18r) units.
  
  So his distance to the shirt intercept point (at 12:30) is:
  
  (61 − t) − (18r) + (140r)
  = (61 − t + 122r)
  
  Which he travels (rowing with the flow of the river) in 56 minutes (between 11:34 and 12:30).
  
  Hence:
  
  (61 − t + 122r) = 56(2r + 1)
  ⇒
  61 − t + 122r = 112r + 56
  t = 10r + 5
  
  Equating our equations for t:
  
  10r + 5 = 5/r
  ⇒
  10r² + 5r − 5 = 0
  5(r + 1)(2r − 1) = 0
  ⇒
  r = 1/2
  
  Hence t = 10.
  
  So the first unknown time (drift back to the jetty) is 10:25.
  
  And the shirt has drifted 140r = 70 units from the jetty, and so the intercept points is 30 units before the camp.
  
  Sanders is rowing with the river, at a rate of (2r + 1) = 2 units/min, so arrives at the camp 15 minutes after passing the shirt.
  
  Hence, Sanders reaches the camp at 12:30 + 15m = 12:45 (= third unknown time).
  
  And the shirt, drifting at a rate of 1/2 units/min reaches the camp 60 minutes after being passed.
  
  So the fourth unknown time is 12:30 + 60m = 13:30 (= fourth unknown time).
  
  The second unknown time is how long after 11:34 does it take Sanders to row with the river a distance of:
  
  (61 − t) − 18r
  = 51 − 9
  = 42 units
  
  The time taken is:
  
  42 / 2
  = 21 minutes
  
  And so the second unknown time (rowing downstream past the jetty) is 11:34 + 21m = 11:55.
  
  Solution: The missing times are: 10:25, 11:55, 12:45, 13:30.
  
  LikeLike
  - Jim Randell 11:02 am on 20 March 2026 Permalink | Reply
    Here is a numerical solution using the same approach:
```
from enigma import (find_zero, fdiv, intr, sprintf, printf)

# suppose the rate of flow of the river is <r>
def solve(r):
  s = r + 1  # Sanders rate of rowing

  # the first interception occurs at 10:10, and then ...

  # the shirt travels downstream for 140 minutes (10:10 - 12:30)
  # and so travels a distance of d1 = 140.r
  d1 = 140 * r

  # Sanders rows for another 5 minutes (to 10:15), moving a distance
  # of 5 beyond the jetty and is then carried back by the river to the
  # jetty at a time of (10:15 + t)
  # r.t = 5
  t = fdiv(5, r)

  # Sanders then rows for (61 - t) minutes to a distance d2 beyond the jetty
  d2 = 61 - t
  # and drifts back towards the jetty for 18 minutes (distance d3)
  d3 = d2 - 18 * r

  # and then rows for 56 minutes (11:34 - 12:30) with the river to make the
  # second interception
  # return the distance between Sanders and the shirt at 12:30
  return d1 - (56 * (s + r) - d3)

# this is equivalent to:
# solve = lambda r: 10*r + 5 - fdiv(5, r)

# find the value of r when the distance is zero
r = find_zero(solve, 0, 100).v
s = r + 1
t = fdiv(5, r)
printf("[r={r:.02f}, s={s:.02f}, t={t:.02f}]")

# express time hh:mm + xx minutes (to the nearest minute)
def time(hh, mm, xx=0):
  (h, m) = divmod(mm + xx, 60)
  return sprintf("{h:02d}:{m:02d}", h=intr(hh + h), m=intr(m))

# compute the unknown times:
# when Sanders drifts back to the jetty
t1 = time(10, 15, t)
printf("unknown time 1 = {t1}")

# when Sanders rows back to the jetty
t2 = time(11, 34, fdiv(61 - t - 18*r, r + s))
printf("unknown time 2 = {t2}")

# remaining distance to camp from second interception
rd = 100 - 140*r

# when Sanders reaches camp
t3 = time(12, 30, fdiv(rd, r + s))
printf("unknown time 3 = {t3}")

# when the shirt reaches camp
t4 = time(12, 30, fdiv(rd, r))
printf("unknown time 4 = {t4}")
```
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- John Crabtree 9:53 pm on 25 March 2026 Permalink | Reply
  
  Sanders sees the shirt at 10:10 and 12:30. During that time, the upstream distance that he travels through the water must equal the downstream distance through the water. Therefore he spends the same time padding upstream (10:10 to 11:16 minus t) and downstream (11:34 to 12:30) Therefore t = 10 minutes.
  Between 10:10 and 10:25 Sanders paddles for 5 of the 15 minutes. Therefore s = 3r. etc.
  I have not seen the published solution.
  
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Jim Randell 9:29 am on 3 March 2026 Permalink | Reply
Tags: by: E. R. J. Benson ( 2 )
Brain teaser 1006: Seafood dinner
From The Sunday Times, 8th November 1981 [link]

The exclusive Ichthyophage Club was invited to hold its annual dinner at a resort on the Costa Fortuna last summer. Several members accepted the invitation, and each was asked to bring one of the local seafood specialities as their contribution to the feast.

The fare consisted of:

• the local edible starfish, which has five legs,
• the squid, which has ten,
• and octopus.

Each guest provided one such creature, and in fact more people provided octopus than provided starfish.

A chef was hired locally. He arranged the food so that each guest received a plateful consisting of the same number of legs — at least one leg from each species — but no guest’s plate was made up in the same way as any other guest’s plate. In other words, no guest had the same combination of legs on his plate as any other guest did.

When the food had been arranged in this way, the chef was grieved to find that all the legs had been used, leaving no edible fragments for him to take home to his family.

How many guests were there?

How many brought starfish, how many brought squid, and how many brought octopus?

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

[teaser1006]
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- Jim Randell 9:29 am on 3 March 2026 Permalink | Reply
  Suppose there were n guests, and x of them brought a starfish, y of them brought a squid, z of them brought an octopus (each with their full complement of legs).
  
  And they are each served a different arrangement of k legs, then:
  
  n = x + y + z
  k = (5x + 10y + 8z) / n
  
  This Python program considers increasing numbers of guests, and looks for viable decompositions that allow a different arrangement of legs to be served to each of them.
  
  It runs in 74ms. (Internal runtime is 334µs).
```
from enigma import (enigma, Enumerator, irange, inf, div, subsets, unzip, printf)

# set defaults for decompose
decompose = enigma.partial(enigma.decompose, increasing=0, sep=0, min_v=1)

# solve the puzzle for <n> guests
def solve(n):
  # each guest brings one of:
  # starfish (5 legs) = x; squid (10 legs) = y; octopus (8 legs) = z
  for (x, y, z) in decompose(n, 3):
    # "more people provided octopus than starfish"
    if not (z > x): continue

    # calculate the number of legs per guest
    ts = (5*x, 10*y, 8*z)
    k = div(sum(ts), n)
    if k is None: continue

    # look for a set of n decompositions
    for ss in subsets(decompose(k, 3), size=n):
      # that give the correct number of leg types
      ns = tuple(sum(vs) for vs in unzip(ss))
      if not (ns == ts): continue
      # return solution
      yield ((x, y, z), k, ss)
      break  # one decomposition is enough

# consider the number of guests
for n in irange(3, inf):
  # look for a viable scenario with <n> guests
  sols = Enumerator(solve(n))
  for ((x, y, z), k, ss) in sols:
    printf("{n} guests ({x} starfish, {y} squid, {z} octopus) -> {k} legs per plate")
    printf("-> plates = {ss}")
  # are we done?
  if sols.count > 0: break
```
  Solution: There were 8 guests. 2 brought starfish, 3 brought squid, 3 brought octopus.
  
  The total number of legs available was: 2×5 + 3×10 + 3×8 = 64 legs.
  
  And so each of the guests was served 8 legs.
  
  The distinct served plates were (starfish legs, squid legs, octopus legs):
  
  (1, 1, 6)
  (1, 2, 5)
  (1, 3, 4)
  (1, 4, 3)
  (1, 5, 2)
  (1, 6, 1)
  (2, 4, 2)
  (2, 5, 1)
  total = (10, 30, 24)
  
  Although the program only looks for one way to make up the plates, this is in fact the only way for n = 8.
  
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Jim Randell 3:34 pm on 14 January 2026 Permalink | Reply
Tags: by: Mary Connor ( 4 )

Brain teaser 983: Superchamp

From The Sunday Times, 24th May 1981 [link]

The five champion athletes competed against one another in their five special events to decide who was the supreme champion (the one with the highest number of points). Three points were awarded for the winner in each event. two points for second place, and one for third.

The Superchamp won more events than anyone else. All scored a different number of points, and all scored in a different number of events.

Ben and Colin together scored twice as many points as Fred. Eddie was the only one to win his own special event, and in it the Superchamp came second. In another event Eddie gained one point less than the hurdler. The hurdler only came third in the hurdles, which was won by Denis, who was last in the long jump but scored in the sprint. Fred won the long jump and was second in the walk, but came nowhere in the hurdles.

Ben scored in the long jump and Colin scored in the sprint. The thousand-metre man was third in two events. The long jumper scored more points than the sprinter.

Who was the [long] jumper, and what was his score?

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

[teaser983]

Jim Randell 3:35 pm on 14 January 2026 Permalink | Reply

There are a lot of conditions to this puzzle, so I used the minizinc.py library to allow me to construct a declarative model for it in MiniZinc, which is then solved, and the Python program used to format the solutions.

It runs in 545ms.

from enigma import (require, printf)
from minizinc import MiniZinc

require("sys.version", "3.6") # needed to support [ use_embed=1 ]

# indices for names and events (1..5)
(B, C, D, E, F) = (1, 2, 3, 4, 5)
(hurdles, long_jump, sprint, m1000, walk) = (1, 2, 3, 4, 5)

# construct the MiniZinc model
m = MiniZinc([
  'include "globals.mzn"',

  # scores for the competitors in each event: pts[event, competitor] = score
  "array [1..5, 1..5] of var 0..3: pts",

  # scores in each event are: 3, 2, 1, 0, 0
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 3) = 1)", # 1st = 3pts
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 2) = 1)", # 2nd = 2pts
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 1) = 1)", # 3rd = 1pts
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 0) = 2)", # unplaced = 0pts

  # functions:
  "function var int: points(var int: j) = sum (i in 1..5) (pts[i, j])",
  "function var int: wins(var int: j) = sum (i in 1..5) (pts[i, j] = 3)",
  "function var int: scored(var int: j) = sum (i in 1..5) (pts[i, j] > 0)",

  # "the superchamp scored more points than anyone else"
  "var 1..5: S",
  "constraint forall (j in 1..5 where j != S) (points(j) < points(S))",

  # "the superchamp won more events than anyone else"
  "constraint forall (j in 1..5 where j != S) (wins(j) < wins(S))",

  # each competitor scored a different number of points
  "constraint all_different (j in 1..5) (points(j))",

  # each competitor scored in a different number of events
  "constraint all_different (j in 1..5) (scored(j))",

  # "B and C together scored twice as many points as F"
  "constraint points({B}) + points({C}) = 2 * points({F})",

  # "E was the only one to win his own event"
  # "and in that event the superchamp came second"
  "array [1..5] of var 1..5: spe",
  "constraint all_different(spe)",
  "constraint forall (j in 1..5) ((j == {E}) <-> (pts[spe[j], j] = 3))",
  "constraint pts[spe[{E}], S] = 2",

  # "in another event E gained 1 point less than the hurdler"
  "var 1..5: hurdler",
  "constraint spe[hurdler] = {hurdles}",
  "constraint exists (i in 1..5) (pts[i, {E}] + 1 = pts[i, hurdler])",

  # "the hurdler only came third in hurdles"
  "constraint pts[{hurdles}, hurdler] = 1",

  # "hurdles was won by D"
  "constraint pts[{hurdles}, {D}] = 3",

  # "D came last in long jump"
  "constraint pts[{long_jump}, {D}] = 0",

  # "D scored in the sprint"
  "constraint pts[{sprint}, {D}] > 0",

  # "F won the long jump"
  "constraint pts[{long_jump}, {F}] = 3",

  # "F was 2nd in the walk"
  "constraint pts[{walk}, {F}] = 2",

  # "F did not score in hurdles"
  "constraint pts[{hurdles}, {F}] = 0",

  # "B scored in the long jump"
  "constraint pts[{long_jump}, {B}] > 0",

  # "C scored in the sprint"
  "constraint pts[{sprint}, {C}] > 0",

  # "the 1000m man was 3rd in 2 events"
  "var 1..5: m1000er",
  "constraint spe[m1000er] = {m1000}",
  "constraint sum (i in 1..5) (pts[i, m1000er] = 1) = 2",

  # "the long jumper scored more points than the sprinter"
  "var 1..5: long_jumper",
  "constraint spe[long_jumper] = {long_jump}",
  "var 1..5: sprinter",
  "constraint spe[sprinter] = {sprint}",
  "constraint points(long_jumper) > points(sprinter)",

], use_embed=1)


# map indices back to names / events (for output)
name = {1: "Ben", 2: "Colin", 3: "Denis", 4: "Eddie", 5: "Fred"}
event = {1: "hurdles", 2: "long jump", 3: "sprint", 4: "1000m", 5: "walk"}

# solve the model and find:
# pts = results in each event
# spe = special event for each competitor
for (pts, spe) in m.solve(solver="minizinc -a", result="pts spe"):
  # collect points
  t = dict((k, 0) for k in name.keys())

  # output the results for each event
  for (i, rs) in enumerate(pts, start=1):
    (p1, p2, p3) = (rs.index(p) + 1 for p in (3, 2, 1))
    printf("{e}: 1st = {p1}, 2nd = {p2}, 3rd = {p3}", e=event[i], p1=name[p1], p2=name[p2], p3=name[p3])
    # collect points
    t[p1] += 3
    t[p2] += 2
    t[p3] += 1
  printf()

  # determine superchamp
  S = max(t.keys(), key=t.get)
  assert all(k == S or v < t[S] for (k, v) in t.items())

  # output the results for each competitor
  for j in sorted(event.keys()):
    printf("{j} [{e}]: {t} points{S}", j=name[j], e=event[spe[j - 1]], t=t[j], S=(' [SUPERCHAMP]' if j == S else ''))
  printf("--")
  printf()

Solution: Denis is the long jumper. He scored 5 points.

The program finds 2 possible scenarios:

hurdles: 1st = Denis, 2nd = Eddie, 3rd = Ben
long jump: 1st = Fred, 2nd = Ben, 3rd = Eddie
sprint: 1st = Ben, 2nd = Denis, 3rd = Colin
1000m: 1st = Eddie, 2nd = Ben, 3rd = Fred
walk: 1st = Ben, 2nd = Fred, 3rd = Eddie

Ben [hurdles]: 11 points [SUPERCHAMP]
Colin [sprint]: 1 points
Denis [long jump]: 5 points
Eddie [1000m]: 7 points
Fred [walk]: 6 points

hurdles: 1st = Denis, 2nd = Eddie, 3rd = Colin
long jump: 1st = Fred, 2nd = Ben, 3rd = Colin
sprint: 1st = Colin, 2nd = Denis, 3rd = Eddie
1000m: 1st = Eddie, 2nd = Colin, 3rd = Fred
walk: 1st = Colin, 2nd = Fred, 3rd = Eddie

Ben [sprint]: 2 points
Colin [hurdles]: 10 points [SUPERCHAMP]
Denis [long jump]: 5 points
Eddie [1000m]: 7 points
Fred [walk]: 6 points

The solution published in the book (which runs to 3 pages) only gives the first of these solutions.

I think this is due to the interpretation of “In another event Eddie gained one point less than the hurdler”. The book solution seems to infer (hurdler = 3pts, Eddie = 2pts) or (hurdler = 2pts, Eddie = 1pt), and concludes it is not possible for C to be the champ.

And in the first scenario, in the long jump, we have: the hurdler (Ben) scores 2pts (2nd place), and Eddie scores 1pt (3rd place).

However, I think the wording also allows (hurdler = 1pt, Eddie = 0pts), and in the second scenario, in the long jump, we have: the hurdler (Colin) scores 1pt (3rd place), and Eddie scores 0pts (unplaced).

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