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Jim Randell 6:21 am on 25 October 2026 Permalink | Reply
Tags: by: Henry C Warburton ( 2 )
Teaser 3344: Wining and dying
From The Sunday Times, 25th October 2026 [link] [link]

You are a king with 15 wine barrels in your cellar. You know that exactly one of the barrels contains a lethal poison, but you do not know which. On day one you give some wine (or none) to each of your four wine tasters. Any who took poison will be dead by the next day. The trouble is that you cannot tell the tasters apart, so if any of them do die you cannot be certain what wine it was you gave them. On day two you observe how many are still alive and can repeat the process with them, carrying on until you have identified the poisoned barrel.

On which day can you be absolutely certain which is the poisoned wine?

[teaser3344]
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Jim Randell is discussing. Toggle Comments
- Jim Randell 6:22 am on 25 October 2026 Permalink | Reply
  If we cannot distinguish the tasters, then the outcome of a tasting is how many of the tasters survive. So with 4 tasters the possible outcomes are that 0, 1, 2, 3, 4 of them survive, which is only 5 possible outcomes. And there are 15 barrels, so we are going to need more than 1 round of tastings.
  
  Let B(k, r) be the maximum number of barrels that can be resolved given k tasters and r rounds of tasting.
  
  In 0 rounds we can only determine the poison barrel if there is only 1 barrel to choose from.
  
  B(k, 0) = 1
  
  In 1 round of tasting there can be (k + 1) different outcomes (i.e. 0 .. k of the tasters die). So, if we label the barrels 0 .. k, and have the same number of tasters taste each barrel as the number on the label, then we can determine the poison barrel directly from the number of tasters that die.
  
  B(k, 1) = k + 1
  
  For r > 1 we can have groups of barrels that are tasted by 0 .. k of the tasters, and the number of barrels that we allocate to each group must be the number that can be resolved by the surviving tasters (who were not part of that group).
  
  B(k, r) = sum (j = 0 .. k) B(j, r − 1)
  
  This gives a straightforward recursive implementation of B(k, r).
  
  This Python program looks at increasing numbers of rounds with 4 tasters, until we are able to resolve 15 barrels.
```
from enigma import (irange, inf, cache, arg, printf)

# B(k, r) = max number of barrels that can be resolved
# with <r> rounds of tasting and <k> tasters (initially)
@cache
def B(k, r):
  # in 0 rounds we can resolve 1 barrel
  if r == 0: return 1
  # in 1 round we can resolve (k + 1) barrels
  if r == 1: return k + 1
  # otherwise allocate barrels to tasters according to how many
  # barrels can be resolved by j surviving tasters in (r - 1) rounds
  return sum(B(j, r - 1) for j in irange(0, k))

n = arg(15, 0, int, prompt="number of barrels")
k = arg(4, 1, int, prompt="number of tasters")

# find minimum number of rounds to resolve <n> barrels
for r in irange(0, inf):
  b = B(k, r)
  printf("[B({k}, {r}) = {b}]")
  if b >= n:
    printf("{n} barrels, {k} tasters -> {r} rounds")
    break
```
  Solution: [To Be Revealed]
  
  [Details – To Be Revealed]
  
  See also: Teaser 2784.
  
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Jim Randell 11:29 am on 23 October 2026 Permalink | Reply
Tags: by: Adrian Somerfield ( 15 )
Teaser 2343: [Video recording]
From The Sunday Times, 19th August 2007 [link]

In the standard three-hour video cassette, the tape moves past the playing head at a constant speed from the stock spool to the take-up spool, which are the same size. The empty spool has a diameter of 2cm and the full spool has external diameter 8cm.

When I started to record an item recently, the speed of rotation of the take-up spool was 40% less than that of the stock spool and the recording ended exactly at the end of the tape.

How long was the recording?

This puzzle was originally published with no title.

[teaser2343]
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Jim Randell is discussing. Toggle Comments
- Jim Randell 11:30 am on 23 October 2026 Permalink | Reply
  
  If the tape moves across the head at a constant linear speed of v.
  
  And at the start of recording the stock spool has a radius of s and the take up spool has a radius of t.
  
  Using the angular velocities 𝛚s (stock spool) 𝛚t (take-up spool), the constant velocity of the tape is given by:
  
  v = s . 𝛚s = t . 𝛚t
  
  But:
  
  𝛚t = 0.6 𝛚s
  
  Hence:
  
  s = 0.6 t
  ⇒
  s² = 0.36 t²
  
  And the total amount of tape on a full spool is:
  
  (4² − 1²) = 15
  
  And this is distributed between the spools at the start of the recording, so:
  
  (s² − 1²) + (t² − 1²) = 15
  ⇒
  s² + t² = 17
  ⇒
  1.36 t² = 17
  t² = 17/1.36 = 12.5
  
  So the proportion of tape on the take up spool at the start of the recording is:
  
  (t² − 1²) / 15
  = (12.5 − 1) / 15
  = 11.5 / 15
  = 138 / 180
  
  i.e. there is 42/180 of the tape remaining on the stock spool, and this is the tape that is used for the recording.
  
  The entire tape runs for 3 hours = 180 minutes, so:
  
  Solution: The length of the recording is 42 minutes.
  
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Jim Randell 2:46 pm on 21 October 2026 Permalink | Reply
Tags: by: Robert Gray ( 3 )
Brainteaser 1343: Mecca bound
From The Sunday Times, 29th May 1988 [link]

When Sheik Abdul heard that some pilgrims were leaving the district and travelling south to Mecca he immediately offered his well-appointed bus so that they could travel in comfort. Unfortunately, there were just too many pilgrims to travel comfortably. Abdul asked them to form a straight line facing south towards Mecca. He explained that he would carry out an operation that would result in some pilgrims facing north. These pilgrims would not go on the bus. Just then, Abdul noticed that his nephews were in the line-up. They occupied the end positions and the centre position.

Abdul went along the line asking each man to make a half-turn. Then he went along the line asking every second man to make a half-turn and so face south again. Next he asked every third man to make a half-turn; if he was facing north, he was to face south, and if he was facing south, he was to face north. And so on, When Abdul went along the line for the last time only the man at the end was asked to make a half-turn. Abdul found that his nephews were all facing north.

“That is only half of the operation”, he announced hastily. We continue and do the same again”. When it was pointed out that all the pilgrims would then be facing south, he said that this time he would start at the other end.

How many of the pilgrims went on the bus to Mecca?

[teaser1343]
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Ruud van der Ham and Jim Randell are discussing. Toggle Comments
- Jim Randell 2:48 pm on 21 October 2026 Permalink | Reply
  See also: Teaser 3131, Puzzle #08.
  
  This Python program runs in 66ms. (Internal runtime is 379µs).
```
from enigma import (irange, inf, rev, printf)

# apply the process to <ss>
def process(n, ss):
  for k in irange(1, n):
    # flip every k'th item
    for i in irange(k, n, step=k):
      ss[i - 1] ^= 1
  return ss

# consider odd numbers of participants (so there is a "centre")
for n in irange(3, inf, step=2):
  # start with everyone facing south (= 1)
  ss = [1] * n
  # run the process in the forward direction
  ss = process(n, ss)
  # all the nephews are now facing north (= 0)
  if not all(ss[i - 1] == 0 for i in [1, n, (n + 1) // 2]): continue
  # run the process in the reverse direction
  ss = rev(process(n, rev(ss)))
  # count number facing south
  printf("n={n}: S={S}\n-> {ss}\n", S=sum(ss))
  break
```
  Solution: 41 pilgrims went on the bus to Mecca.
  
  There were 49 pilgrims in the queue.
  
  At the end of both operations the pilgrims in positions: 4th, 9th, 14th, 16th, 34th, 36th, 41st, 46th positions are facing north.
  
  These can be paired as follows to give sums of 50 (= 49 + 1):
  
  4th & 46th
  9th & 41st
  14th & 36th
  16th & 34th
  
  Where one position (underlined) in each pair is a square number (counting from the start). The other position corresponds to a square number position counting from the end.
  
  The 25th (centre) position, is paired with itself, and so ends up facing south.
  
  The next solutions occur with 1681 pilgrims, and 57121 pilgrims:
  
  n = 49 = 7²; 49 + 1 = 2 × 5² (with 8 facing north)
  n = 1681 = 41²; 1681 + 1 = 2 × 29² (with 76 facing north)
  n = 57121 = 239²; 57121 + 1 = 2 × 169² (with 468 facing north)
  
  If we consider solutions to the Pell’s equation: X² − 2Y² = −1, the we get solutions of the form:
  
  n = X²; X² + 1 = 2 × Y²
```
% run pells.py 1 -2 -1  
solving: 1.X^2 + -2.Y^2 = -1
[1] X=1 Y=1 -> -1
[2] X=7 Y=5 -> -1
[3] X=41 Y=29 -> -1
[4] X=239 Y=169 -> -1
[5] X=1393 Y=985 -> -1
[6] X=8119 Y=5741 -> -1
[7] X=47321 Y=33461 -> -1
[8] X=275807 Y=195025 -> -1
[9] X=1607521 Y=1136689 -> -1
[10] X=9369319 Y=6625109 -> -1
[11] X=54608393 Y=38613965 -> -1
[12] X=318281039 Y=225058681 -> -1
[13] X=1855077841 Y=1311738121 -> -1
[14] X=10812186007 Y=7645370045 -> -1
[15] X=63018038201 Y=44560482149 -> -1
[16] X=367296043199 Y=259717522849 -> -1
[17] X=2140758220993 Y=1513744654945 -> -1
[18] X=12477253282759 Y=8822750406821 -> -1
[19] X=72722761475561 Y=51422757785981 -> -1
[20] X=423859315570607 Y=299713796309065 -> -1
[first 20 solutions]
```
  See: OEIS A002315 [@oeis], OEIS A001653 [@oeis].
  
  LikeLike
- Ruud van der Ham 3:29 pm on 21 October 2026 Permalink | Reply
```
import itertools

for n in itertools.count(5, 2):
    p = [True for _ in range(n)]
    for i in range(n):
        for j in range(i, n, i + 1):
            p[j] = not p[j]
    if p[0] == p[n // 2] == p[n - 1]:
        for i in range(n):
            for j in range(i, n, i + 1):
                p[n - j - 1] = not p[n - j - 1]
        print("number in line:", n)
        print("number of pilgrims on the bus:", sum(p))
        print("line:", "".join(map(lambda x: "NS"[x], p)))
        break
```
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Jim Randell 6:43 am on 18 October 2026 Permalink | Reply
Tags: by: Mark Valentine ( 19 )
Teaser 3343: Cartesian darts
From The Sunday Times, 18th October 2026 [link] [link]

An AI app is learning to play a game of virtual darts. The board is a square, subdivided into nine squares. The resulting central square is divided in the same way, and the process continues ad infinitum.

The value of each outer square is 1. Each smaller square has the same constant prime multiple of the value of the next larger square. A dart landing in a given square scores the average value of all squares of the same size or smaller, weighted by area. Thus, the closer the dart lands to the centre, the higher the score.

At first, the app sometimes got a non-integer score, but after improving always got an integer score.

What was the non-integer score (as a fraction)?

[teaser3343]
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- Jim Randell 6:44 am on 18 October 2026 Permalink | Reply
  Let the score for a dart in level k be S[k]. The outer squares (value 1) are level 0.
  
  Then:
  
  S[0] = 8/9 + S[1]/9
  S[1] = (8/9)p + S[2]/9
  S[2] = (8/9)p^2 + S[3]/9
  …
  S[k] = (8/9)p^k + S[k + 1]/9
  
  So:
  
  S[0] = (8/9)(1 + p/9 + p^2/9^2 + p^3/9^3 + …)
  
  The infinite geometric sum has a common ratio of p/9, so will converge if p < 9, and will converge to: 9/(9 − p).
  
  Hence:
  
  S[0] = (8/9)(9/(9 − p)) = 8/(9 − p)
  S[k + 1] = 9 S[k] − 8 p^k
  
  Where p is a prime less than 9, i.e. p = 2, 3, 5, 7.
  
  This Python program considers possible primes for p, and calculates the first 11 terms of the sequence S[]. It then looks for sequences that start off with non-integer scores, but after a certain point give only integer scores. (The puzzle can be read to require only S[0] to give a non-integer score, but my code does not require that).
  
  It runs in 81ms. (Internal runtime is 246µs).
```
from enigma import (Rational, primes, first, clump, seq2str, printf)

Q = Rational()
N = 11  # number of terms to consider

# calculate terms in the sequence
# S[k + 1] = 9 S[k] - 8 p^k
def seq(p, s0):
  (s, d) = (s0, 8)
  while True:
    yield s
    s = 9*s - d
    d *= p

# consider primes < 9
for p in primes.between(2, 8):
  # calculate S(0)
  S0 = Q(8, 9 - p)
  # calculate N terms of S[]
  S = list(first(seq(p, S0), N))
  # look for a sequence that starts with non-integer values
  # but then settles down to integer values
  zs = list(clump(x % 1 == 0 for x in S))
  sol = (len(zs) == 2 and zs[0][0] is False and zs[1][0] is True)
  # output solution
  printf(
    "p = {p}: S = {S} {x}" ,
    S=seq2str(S, enc=("[", ", ...]")),
    x=("-> [*** SOLUTION ***]" if sol else "")
  )
```
  Solution: The initial non-integer score was 4/3.
  
  The sequence is: [4/3, 4, 12, 36, 108, 324, 972, 2916, 8748, 26244, 78732, …].
  
  So it starts off with a non-integer S[0] value, but then each subsequent S[k] value is an integer (common ratio is 3).
  
  In general: S[k] = 8 . p^k / (9 − p).
  
  For p = 2 the value is always a non-integer (S[k] = 2^(k + 3) / 7).
  
  For p = 3 the first value is non-integer (S[0] = 4/3), but subsequent values are integer. (S[k] = 4 . 3^(k − 1)).
  
  For p = 5 the values are always integer. (S[k] = 2 . 5^k).
  
  For p = 7 the values are always integer. (S[k] = 4 . 7^k).
  
  If the p value is not restricted to primes, there is also a solution at p = 6:
  
  S[k] = 2^(k + 3) . 3^(k − 1)
  S = [8/3, 16, 96, 576, 3456, 20736, 124416, 746496, 4478976, 26873856, 161243136, …]
  
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Jim Randell 11:15 am on 15 October 2026 Permalink | Reply
Tags: by: R. Buchanan-Dunlop ( 3 )

Brainteaser 1342: Balancing trick

From The Sunday Times, 22nd May 1988 [link]

The other day Harry visited me and he produced eight boxes each of which contained five new golf balls, and, on producing them, he picked out one ball from one of the boxes, and he said that he knew that that ball was of regulation weight, and he added that he also knew that 38 of the other balls were also of regulation weight, but there was one ball amongst the other 39 that was either heavier or lighter than the regulation ball. On saying this he handed me a simple set of scales, one square tray on one side balanced against a similar tray on the other side, each tray being large enough to hold up to 14 of the balls in a pyramid. The scales had no weights attached to them, and Harry asked me how many weighings of the balls it would take me to discover for certain, in all eventualities, which was the maverick ball, and also to tell whether it was lighter or heavier than the others.

He told me that he would give me one box of balls for each weighing less than eight that it took me in order to achieve this.

How many boxes of balls was I able to win from this wager?

[teaser1342]

Jim Randell 11:17 am on 15 October 2026 Permalink | Reply

See also: Teaser 1952-04-13, Teaser 2500, Enigma 1589.

This is a “type 2” problem – “how to determine the counterfeit coin from 39 with an extra known good coin”.

I have adapted the program given in my comment for Teaser 2500 that generates decision trees for several types of problems as below (also [link]):

#! python3

# this program can output the decision tree for 4 kinds of coin
# weighing puzzle, where we have a collection of <n> coins, one of
# which is counterfeit and is either lighter or heavier than the
# remaining coins (which all weigh the same), and we must determine by
# comparative weighings which of the coins is counterfeit, and whether
# it is heavier or lighter.
#
# type 1 = find the suspect coin from a collection of <n> coins.
#
# type 2 = find the suspect coin from a collection of <n> coins, where
# an additional known good reference coin is provided.
#
# type 3 = find the suspect coin from a collection of <n> coins, where
# some of the coins are red, and some of the coins are black. there is
# either 1 or 2 more red coins than black coins. if the suspect coin
# is red it is heavy, and if it is black it is light.
#
# type 4 = find the suspect lighter coin from a collection of <n> coins.
#
# additionally we can also solve the following problem:
#
# type 1* = find the suspect coin from a collection of <n> coins, but
# we don't need to know if it is heavier or lighter.
#
# this is solved by putting one of the coins aside and treating the
# remaining <n-1> coins as a "type 1" problem. if all weighings
# balance (and we reach the "impossible" leaf of the decision tree),
# then all the coins are good, apart from the one set aside, so it
# must be counterfeit. and so this problem is the same as solving a
# "type 1" problem with <n-1> coins.
#
# the program is run as:
#
#  % python3 coin-weighing.py <n> <p>
#
# where <n> is the number of coins/weighings, and <p> is the problem number (1-4)
#
# if the <n> parameter is given as "w<k>" then the maximum number of
# coins for <k> weighings is used to construct the decision tree.
#
# similarly the <n> parameter can be specified as "n<n>" for the number of coins

from enigma import (irange, divc, arg, nl, printf)

IMP = "(impossible)"

# divide a list into thirds
# such that the first two thirds are equal size
def thirds(ss):
  (i, r) = divmod(len(ss), 3)
  if r == 2: i += 1
  j = 2*i
  return (ss[:i], ss[i:j], ss[j:])

def output_weighing(p, w, A, B):
  printf("{p}[{w}] weighing {w}:")
  printf("{p}[{w}] A = {A}", A=sorted(A))
  printf("{p}[{w}] B = {B}", B=sorted(B))

# find the light suspect coin
def solve_lighter(coins, w=0, p=''):
  n = len(coins)
  if n < 1:
    printf("{p}-> {IMP}")
  elif n == 1:
    printf("{p}[{w}] -> {coins[0]}L")
  elif n == 2 or n == 3:
    w += 1
    (A, B) = ([coins[0]], [coins[1]])
    output_weighing(p, w, A, B)
    printf("{p}[{w}] A < B -> {A[0]}L")
    printf("{p}[{w}] A > B -> {B[0]}L")
    printf("{p}[{w}] A = B -> \\")
    printf("{IMP}" if n == 2 else "{coins[2]}L")
  else:
    # divide the coins into three piles
    (A, B, C) = thirds(coins)
    w += 1
    output_weighing(p, w, A, B)
    printf("{p}[{w}] A < B:")
    w1 = solve_lighter(A, w, p + "  ")
    printf("{p}[{w}] A > B:")
    w2 = solve_lighter(B, w, p + "  ")
    printf("{p}[{w}] A = B:")
    w3 = solve_lighter(C, w, p + "  ")
    w = max(w1, w2, w3)
  return w

# find suspect coin in reds (if heavier), or in blacks (if lighter)
def solve_red_black(reds, blacks, red='H', black='L', good=-1, w=0, p=''):
  ns = (nr, nb) = (len(reds), len(blacks))
  assert nr - nb in {1, 2}
  if ns == (1, 0):
    printf("{p}[{w}] -> \\")
    x = reds[0]
    printf("{x}{red}" if x != good else IMP)
  elif nr == 2:
    # weigh the two reds
    w += 1
    (r0, r1) = (reds[0], reds[1])
    output_weighing(p, w, [r0], [r1])
    printf("{p}[{w}] A < B -> \\")
    x = (r1 if red == 'H' else r0)
    printf("{x}{red}" if x != good else IMP)
    printf("{p}[{w}] A > B -> \\")
    x = (r0 if red == 'H' else r1)
    printf("{x}{red}" if x != good else IMP)
    printf("{p}[{w}] A = B -> \\")
    printf("{blacks[0]}{black}" if nb == 1 and blacks[0] != good else IMP)
  else:
    # divide the reds into 3 piles
    k = divc(nr, 3)
    (r1, r2, r3) = (reds[:k], reds[k:2*k], reds[2*k:])
    # divide the blacks into 3 corresponding piles
    i = len(r1) - 1
    j = i + len(r2) - 1
    (b1, b2, b3) = (blacks[:i], blacks[i:j], blacks[j:])
    w += 1
    output_weighing(p, w, r1 + b1, r2 + b2)
    printf("{p}[{w}] A < B:")
    (rs, bs) = ((r2, b1) if red == 'H' else (r1, b2))
    w1 = solve_red_black(rs, bs, red, black, good, w, p + "  ")
    printf("{p}[{w}] A > B:")
    (rs, bs) = ((r1, b2) if red == 'H' else (r2, b1))
    w2 = solve_red_black(rs, bs, red, black, good, w, p + "  ")
    printf("{p}[{w}] A = B:")
    if len(b3) > len(r3):
      w3 = solve_red_black(b3, r3 , black, red, good, w, p + "  ")
    else:
      # increase the reds with a known good red
      gd = r1[0]
      w3 = solve_red_black([gd] + r3, b3, red, black, gd, w, p + "  ")
    w = max(w1, w2, w3)
  return w

# find suspect coin with extra known good coin
def solve_test_coin(coins, good, w=0, p=''):
  n = len(coins)
  if n < 1:
    printf("{p}-> {IMP}")
  elif n == 1:
    # weigh the coin against the known good coin
    w += 1
    x = coins[0]
    output_weighing(p, w, [good], [x])
    printf("{p}[{w}] A < B -> {x}H")
    printf("{p}[{w}] A > B -> {x}L")
    printf("{p}[{w}] A = B -> {IMP}")
  else:
    # divide the coins into 3 piles
    i = n // 3
    j = 2*i + 1
    (c1, c2, c3) = (coins[:i], coins[i:j], coins[j:])
    w += 1
    output_weighing(p, w, [good] + c1, c2)
    printf("{p}[{w}] A < B:")
    w1 = solve_red_black(c2, c1, 'H', 'L', -1, w, p + "  ")
    printf("{p}[{w}] A > B:")
    w2 = solve_red_black(c2, c1, 'L', 'H', -1, w, p + "  ")
    printf("{p}[{w}] A = B:")
    w3 = solve_test_coin(c3, good, w, p + "  ")
    w = max(w1, w2, w3)
  return w

# find suspect coin
def solve_general(coins, w=0, p=''):
  n = len(coins)
  if n < 3:
    printf("{p}-> {IMP}")
  else:
    # divide the coins into 3 piles
    (c1, c2, c3) = thirds(coins)
    w += 1
    output_weighing(p, w, c1, c2)
    printf("{p}[{w}] A < B:")
    w1 = solve_red_black(c1 + [c3[0]], c2, 'L', 'H', -1, w, p + "  ")
    printf("{p}[{w}] A > B:")
    w2 = solve_red_black(c1 + [c3[0]], c2, 'H', 'L', -1, w, p + "  ")
    printf("{p}[{w}] A = B:")
    w3 = solve_test_coin(c3, c1[0], w, p + "  ")
    w = max(w1, w2, w3)
  return w

# number of coins (or weighings) and problem type
n = arg("n13", 0, prompt="number of coins (n<int>; or weighings w<int>)")
p = arg("1", 1, prompt="problem type (1-4)")
printf("[arg = {n}; problem = {p} (of 1-4)]")

# parse the n parameter
def fn(n, f):
  # n<n> = <n> coins; w<n> = <n> weighings
  if n.startswith('n'): return int(n[1:])
  if n.startswith('w'): return f(int(n[1:]))
  # old style: <n> = n<n>; -<n> = w<n>
  n = int(n)
  return (f(-n) if n < 0 else n)

if p == "1":
  # solve the general case
  n = fn(n, (lambda k: (3**k - 3) // 2))
  printf("---{nl}find suspect coin from {n} coins{nl}---")
  coins = list(irange(1, n))
  printf("coins = {coins}")
  w = solve_general(coins)
  printf("---{nl}{n} coins -> {w} weighings")

if p == "2":
  # solve with a known good coin (0)
  n = fn(n, (lambda k: (3**k - 1) // 2))
  printf("---{nl}find suspect coin from {n} coins, with an extra known good coin (0){nl}---")
  coins = list(irange(1, n))
  printf("coins = {coins}")
  w = solve_test_coin(coins, 0)
  printf("---{nl}{n} coins -> {w} weighings")

if p == "3":
  # solve red/black
  n = fn(n, (lambda k: 3**k))
  r = ((n + 2) // 2 if n > 0 else 0)
  b = n - r
  printf("---{nl}find suspect coin from {n} coins, where {r} coins are red and {b} coins are black{nl}if the suspect coin is red it is heavy, if it is black it is light{nl}---")
  (reds, blacks) = (list(irange(1, r)), list(irange(r + 1, n)))
  printf("reds = {reds}{nl}blacks = {blacks}")
  w = solve_red_black(reds, blacks)
  printf("---{nl}{n} coins -> {w} weighings")

if p == "4":
  # solve lighter
  n = fn(n, (lambda k: 3**k))
  printf("---{nl}find suspect lighter coin from {n} coins{nl}---")
  coins = list(irange(1, n))
  printf("coins = {coins}")
  w = solve_lighter(coins)
  printf("---{nl}{n} coins -> {w} weighings")

In this case we have a “type 2” problem with 39 coins/balls:

% run coin-weighing.py n39 2        
[arg = n39; problem = 2 (of 1-4)]
---
find suspect coin from 39 coins, with an extra known good coin (0)
---
coins = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39]
[1] weighing 1:
[1] A = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
[1] B = [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27]
...
---
39 coins -> 4 weighings

We see that the maverick ball can always be determined in just 4 weighings, and the first weighing requires 14 balls in each pan of the balance, which is permitted. (And subsequent weighings use fewer balls).

Solution: You can guarantee to win 4 boxes of balls.

Although this is not the largest number of balls possible with 4 weighings:

For a “type 2” problem with k weighings we can find the counterfeit among n = (3^k − 1) / 2 objects.

k = 4 → n = (3^4 − 1) / 2 = 40

So, we could solve the same puzzle if all 40 balls in the boxes are suspect, and Harry provided a separate regulation-weight ball from his pocket.

% run coin-weighing.py w4 2 
[arg = w4; problem = 2 (of 1-4)]
---
find suspect coin from 40 coins, with an extra known good coin (0)
---
coins = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40]
[1] weighing 1:
[1] A = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
[1] B = [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27]
...
---
40 coins -> 4 weighings

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