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Jim Randell 8:40 am on 30 July 2024 Permalink | Reply
Tags: by: R Postill ( 20 )
Brain-Teaser 542: Finger trouble
From The Sunday Times, 14th November 1971 [link]

The Wabu tribesmen, though invariably truthful, are inveterate thieves, and this despite a tribal law which requires any man convicted of theft to have one finger amputated. Few of them now have a full complement and, since they count on their fingers, their answers to numerical questions are confusing to the outsider. (For instance, an 8-fingered man gives his answers to base 8, so that, for example, our figure 100 is for him 144).

Despite these handicaps, they are keen cricketers, and I recently watched their First Eleven. The Chief told me: “Only little Dojo has 10 fingers; the worst-equipped are Abo, Bunto and Coco with 4 apiece”. (I need hardly add that the Wabu would never elect a Chief who had been convicted of theft).

Later I asked some members of the team how many fingers (thumbs are, of course, included) the Eleven totalled. Epo said “242”, and Foto agreed; Gobo said “132”, and Hingo agreed. Kinko added: “I have more fingers than Iffo”.

How many fingers each have Epo, Hingo, Iffo and Jabo (the Wicket-keeper)?

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

[teaser542]
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Frits, Ruud, and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:41 am on 30 July 2024 Permalink | Reply
  We can use the [[ SubstitutedExpression ]] solver from the enigma.py library to solve this puzzle.
  
  The following run file executes in 83ms. (Internal runtime of the generate code is 1.9ms).
  
  Run: [ @replit ]
```
#! python3 -m enigma -rr

SubstitutedExpression

--base=11  # values are between 4 and 10
--distinct=""

# we are given some values
--assign="A,4"
--assign="B,4"
--assign="C,4"
--assign="D,10"
# and the others are between 5 and 9
--digits="5-9"
# additional constraints
"E = F"
"G = H"
"K > I"

# total number of fingers
--macro="@total = (A + B + C + D + E + F + G + H + I + J + K)"

# E (and F) says "242"
"int2base(@total, base=E) == '242'"

# G (and H) says "132"
"int2base(@total, base=G) == '132'"

--answer="(E, H, I, J)"
--template=""
```
  Solution: Epo has 5 fingers; Hingo has 7 fingers; Iffo has 8 fingers; Jabo has 9 fingers.
  
  The number of fingers for each member of the team are:
  
  A = 4
  B = 4
  C = 4
  D = 10
  E = 5
  F = 5
  G = 7
  H = 7
  I = 8
  J = 9
  K = 9
  total = 72
  
  And the total given in the appropriate bases:
  
  A, B, C → 1020 [base 4]
  D → 72 [base 10]
  E, F → 242 [base 5]
  G, H → 132 [base 7]
  I → 110 [base 8]
  J, K → 80 [base 9]
  
  LikeLike
  - Ruud 4:14 pm on 30 July 2024 Permalink | Reply
    It is really simple to solve this one by hand.
    But, here’s my as brute as possible code:
```
import itertools

a = b = c = 4
d = 10
for e in range(5, 11):
    sum_e = int("242", e)

    for g in range(4, 11):
        sum_g = int("132", g)
        if sum_e == sum_g:
            f = e
            h = g
            ijk = sum_e - (a + b + c + d + e + f + g + h)
            for i, j, k in itertools.product(range(1, 10), repeat=3):
                if i + j + k == ijk and k > i:
                    print(f"Epo={e} Hingo={h} Iffo={i} Jabo={j}")
```
    LikeLike
- Ruud 7:30 pm on 30 July 2024 Permalink | Reply
  Even more brute force:
```
import itertools

a = b = c = 4
d = 10
for e, f, g, h, i, j, k in itertools.product(range(5, 10), repeat=7):
    if e == f and g == h and (total := int("242", e)) == int("132", g) and k > i and (a + b + c + d + e + f + g + h + i + j + k) == total:
        print(f"Epo={e} Hingo={h} Iffo={i} Jabo={j}")
```
  LikeLike
- Frits 1:44 pm on 31 July 2024 Permalink | Reply
```
from enigma import SubstitutedExpression

# the alphametic puzzle
p = SubstitutedExpression(
  [
    # G*G + 3*G - 2*(E*E + 2*E) = 0
    "is_square(1 + G * (G + 3) // 2) - 1 = E",
    
    "K > I",
    "(G * (G + 3) + 2) -  (A + B + C + D + E + E + G + G + I + J) = K",
  ],
  answer="(E, G, I, J)",
  distinct="",
  s2d=dict(A=4, B=4, C=4, D=10),
  digits=range(5, 10),
  verbose=0,    # use 256 to see the generated code
)

names= "Epo Hingo Iffo Jabo".split()

# print answers
for ans in p.answers():
  print(f"{', '.join(f'{x}: {str(y)}'for x, y in zip(names, ans))}")     
```
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Jim Randell 6:58 am on 28 July 2024 Permalink | Reply
Tags: by: Phil Jackson ( 2 )
Teaser 3227: Chess club cupboard
From The Sunday Times, 28th July 2024 [link] [link]

When installing the Chess Club cupboard in its new room, the members carelessly jammed it on the low ceiling and back wall, as shown from the side (not to scale).

Measurements were duly taken and the following were noted. The floor and ceiling were level and the back wall was vertical. The cupboard was a rectangular cuboid. The ceiling height was just 2cm greater than the cupboard height. Also, the depth of the cupboard from front to back was 148 cm less than the cupboard height. Finally the point on the ceiling where the cupboard jammed was a distance from the back wall that was 125 cm less than the cupboard height.

What is the cupboard height?

[teaser3227]
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Jim Randell, Frits, Ruud, and 1 other are discussing. Toggle Comments
- Jim Randell 7:11 am on 28 July 2024 Permalink | Reply
  Originally I assumed some of the dimensions were whole numbers of centimetres, which allows you to find the answer, but we do not need to make this assumption.
  
  Here is a solution using the [[ Polynomial ]] class from the enigma.py library to generate a polynomial equation, and we can then look for positive real roots of the equation to determine the necessary dimensions.
  
  (Also, today is the 15th anniversary of the start of the enigma.py library).
  
  It runs in 73ms. (Internal runtime is 4.2ms).
  
  Run: [ @replit ]
```
from enigma import (Polynomial, sq, printf)

# let x = cupboard depth
fx = Polynomial('x')

# y = cupboard height
fy = fx + 148

# h = room height
fh = fy + 2

# d = distance top jam to top corner
fd = fy - 125

# if:
#   a = height of upper triangle
#   b = height of lower triangle
# we have:
#   h = a + b
# and:
#   a^2 = y^2 - d^2
#   b = x.d/y
#   a = h - b = (y.h - x.d)/y
#   a^2 = (y.h - x.d)^2 / y^2 = y^2 - d^2
#   y^2(y^2 - d^2) = (y.h - x.d)^2  # for x > 0
f = (sq(fy) * (sq(fy) - sq(fd)) - sq(fy * fh - fx * fd)).scale()
printf("[f(x) = {f}]", f=f.print())

# look for positive real roots of f(x) = 0
for x in f.roots(domain='F', include='+'):
  # output solution
  printf("cupboard = {x:g} x {y:g}; room height = {h:g}", y=fy(x), h=fh(x))
```
  Solution: The cupboard was 185 cm high.
  
  So the cupboard has dimensions: 37 cm × 185 cm, and the room is 187 cm high.
  
  The polynomial to determine the depth of the cupboard x is:
  
  f(x) = x^3 + 79x^2 − 1628x − 98568
  f(x) = (x − 37)(x^2 + 116x + 2664)
  
  The linear component has a positive root at:
  
  x = 37
  
  which provides the required answer for the depth of the cupboard.
  
  The quadratic component has two negative roots at:
  
  x = 10√7 − 58 ≈ −31.542
  x = −10√7 − 58 ≈ −84.458
  
  but these do not provide viable answers to the puzzle.
  
  LikeLike
  - Jim Randell 9:09 am on 29 July 2024 Permalink | Reply
    Or the polynomial can be solved numerically, to give an approximate answer.
    
    And this is slightly faster. The following Python program has an internal runtime of 324µs).
    
    Run: [ @replit ]
```
from enigma import (Polynomial, sq, find_zero, printf)

# let x = cupboard depth
# y = cupboard height
# h = room height
# d = distance top jam to top corner
fx = Polynomial('x')
fy = fx + 148
fh = fy + 2
fd = fy - 125

# construct polynomial to solve
f = (sq(fy) * (sq(fy) - sq(fd)) - sq(fy * fh - fx * fd)).scale()
printf("[f(x) = {f}]", f=f.print())

# find a solution numerically
x = find_zero(f, 1, 100).v

# output solution
printf("cupboard = {x:.2f} x {y:.2f}; room height = {h:.2f}", y=fy(x), h=fh(x))
```
    LikeLike
- GeoffR 12:21 pm on 28 July 2024 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";

var 1..250:h; var 1..250:y1; var 1..100:b;

% h = cupboard height
% y1 = depth below ceiling of top cupboard jam point
% b = distance of bottom jam point from vertical wall

% Two triangles to solve by Pythagorus
constraint h * h == (h - 125) * (h - 125) + (y1 * y1);

constraint (h - 148 ) * (h - 148) == (b * b) + (h + 2 - y1) * (h + 2 - y1);

solve satisfy;

output ["Cupboard height = " ++ show(h) ++ "cm."];
```
  LikeLike
- GeoffR 12:52 pm on 28 July 2024 Permalink | Reply
  
  I don’t suppose the setter intended it, but I found a second solution with a cupboard height of more than 400 cm higher than the presumed solution!
  
  LikeLike
  - Jim Randell 2:32 pm on 28 July 2024 Permalink | Reply
    
    @GeoffR: Originally I thought there were further solutions involving cupboards with non-integer dimensions, or very large cupboards (and rooms). But now I have done the analysis I found that the depth of the cupboard is defined by a cubic equation that has one positive real root (which gives the required answer) and two negative real roots, so I don’t think there are any further solutions, even if the size of the cupboard/room are not restricted.
    
    LikeLike
    - Frits 2:54 pm on 28 July 2024 Permalink | Reply
      
      @Jim,
      
      The equations I noted down for this puzzle also had a solution 10 * (9 + sqrt(7)) but that was less than 148 cm.
      
      LikeLike
      - Jim Randell 3:23 pm on 28 July 2024 Permalink | Reply
        
        @Frits: Presumably you were working in terms of the height of the cupboard. I was working in terms of the depth, so there was only one positive root.
        
        LikeLike
    - GeoffR 4:11 pm on 28 July 2024 Permalink | Reply
      @Jim: My original program gave the same outputs as your original solution.
      I increased the upper bounds of the variables to show the second solution I obtained with the second program below. Seems reasonable to use integers . Program is slow with the extra solution, but does it look OK?
      
      % A Solution in MiniZinc include "globals.mzn"; var 1..1250:h; var 1..1250:y1; var 1..1000:b; % h = cupboard height, rht = room height (for output) % y1 = depth below ceiling of top cupboard jam point % b = distance of bottom jam point from vertical wall % Two triangles to solve by Pythagorus constraint h * h == (h - 125) * (h - 125) + (y1 * y1); constraint (h - 148 ) * (h - 148) == (b * b) + (h + 2 - y1) * (h + 2 - y1); solve satisfy; output ["Cupboard height = " ++ show(h) ++ "cm." ++ "\n" ++ "[h, w, rht, y1, b ] = " ++ show( [h, h-148, h+2, y1, b ] )];
      
      LikeLike
      - Jim Randell 6:03 pm on 28 July 2024 Permalink | Reply
        
        @GeoffR: My very first program did not include a check that the upper and lower triangles were similar, which lead me to believe that there potentially multiple solutions without further constraints. But this allows cupboards that have a non-rectangular cross-section.
        
        However my subsequent analysis gives a single solution without further constraints. So we don’t need to limit the size of cupboard/room, and we don’t require measurements to have integer values.
        
        LikeLiked by 1 person
- Ruud 6:16 pm on 28 July 2024 Permalink | Reply
  
  I worked out by hand that the problem can be defined as finding the roots of the expression
  math.sqrt(250 *c -125*125)+((c-148)*(c-125)/c) – (c+2)
  I just put that in Wolfram Alpha as
  find roots of sqrt(250 *c -125*125)+((c-148)*(c-125)/c) – (c+2)
  to find the real solution (not revealed) and 10 (9 + sqrt(7)). The latter does not work here.
  
  LikeLike
  - Frits 6:55 pm on 28 July 2024 Permalink | Reply
    I did something similar with the same results.
    
    Adding the following to your code prevent multiple solutions:
```
  
% same angles
constraint h * b = (h - 148) * y1;
```
    LikeLike
Jim Randell 10:49 am on 25 July 2024 Permalink | Reply
Tags: by: J S Rowley ( 9 )
Brain-Teaser 416: Sharing sweets
From The Sunday Times, 27th April 1969 [link]

The Binks family had a birthday party recently for Ann, the youngest child. A tin of toffees was shared among the seven youngest members of the family, whose ages were all different and less than 20.

It was decided that the younger children should have more than the older; so Mr Binks suggested that each should receive the number obtained by dividing the total number of toffees by his or her age. This was done and it so happened that all the divisions worked out exactly and no toffees were left over.

Mary received 18 sweets.

How much older was she than Ann?

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

[teaser416]
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Frits, Ruud, Jim Randell, and 1 other are discussing. Toggle Comments
- Jim Randell 10:51 am on 25 July 2024 Permalink | Reply
  Labelling the children in age order A, …, G, so A is Ann, and the total number of toffees is T, we have:
  
  T = T/A + T/B + T/C + T/D + T/E + T/F + T/G
  ⇒
  1 = 1/A + 1/B + 1/C + 1/D + 1/E + 1/F + 1/G
  
  So we need to find 7 different reciprocals (with denominators less than 20) that sum to 1. And we can use the [[ reciprocals() ]] function from the enigma.py library to do that (originally written for Enigma 348).
  
  We can then assign one of the values T/B, …, T/G to 18 (for Mary), and check that all the remaining T/X values give a whole number.
  
  The following Python program runs in 66ms. (Internal runtime is 753µs).
  
  Run: [ @replit ]
```
from enigma import (reciprocals, printf)

# find 7 different reciprocals that sum to 1
for rs in reciprocals(7, 1, 1, M=19, g=1):
  # Ann is the youngest
  A = rs.pop(0)
  # Mary received 18 sweets
  for M in rs:
    T = M * 18
    if not all(T % r == 0 for r in rs): continue
    # output solution
    printf("A={A} M={M} [T={T} rs={rs}]", rs=[A] + rs)
```
  Solution: Mary (10) is 7 years older than Ann (3).
  
  There is only one set of 7 reciprocals that works:
  
  1 = 1/3 + 1/4 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18
  
  There are 180 toffees, and the ages are:
  
  3 → 60 toffees (Ann)
  4 → 45 toffees
  9 → 20 toffees
  10 → 18 toffees (Mary)
  12 → 15 toffees
  15 → 12 toffees
  18 → 10 toffees
  total = 180 toffees
  
  LikeLike
- GeoffR 1:32 pm on 25 July 2024 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";

var 1..19:A; var 1..19:B; var 1..19:C; var 1..19:D;
var 1..19:E; var 1..19:F; var 1..19:G;

constraint all_different ([A, B, C, D, E, F, G]);
% Order childrens ages
constraint A < B /\ B < C /\ C < D /\ D < E /\ E < F /\ F < G;

var 28..200:T;  % total number of toffees

% Distribute toffees to seven children, with none left over
constraint T  mod A == 0 /\ T  mod B == 0 /\ T  mod C == 0 /\ T  mod D == 0 
/\ T  mod E == 0 /\ T  mod F == 0 /\ T  mod G == 0;

constraint T == T div A  + T div B + T div C + T div D 
+ T div E + T div F + T div G;

solve satisfy;

output ["Childrens ages = " ++ show ( [A,B,C,D,E,F,G] )
++ "\n" ++ "Number of toffees = " ++ show(T) ];

% Childrens ages = [3, 4, 9, 10, 12, 15, 18]
% Number of toffees = 180
% ----------
% ==========
```
  As Mary received 18 sweets, she was 10 years old, as 10 * 18 = 180.
  As the youngest was Ann, she was 3 years old, 7 years younger than Mary.
  
  LikeLike
- Ruud 6:40 am on 26 July 2024 Permalink | Reply
  
  [removed]
  
  LikeLike
  - Jim Randell 9:21 am on 26 July 2024 Permalink | Reply
    
    @Ruud: Can you explain the reasoning behind your code? I tried changing the 18 to 24 to solve a variation of the puzzle, but it did not give a correct answer. (I also added the missing imports to your program).
    
    LikeLike
    - ruudvanderham 12:02 pm on 26 July 2024 Permalink | Reply
      @Jim
      My solution was incorrect. Could you remove it?
      Here’s my revised one, which is not much different from yours:
      
      import itertools import math number_of_toffee_mary = 18 for ages in itertools.combinations(range(1, 20), 7): if math.isclose(sum(1 / age for age in ages), 1): for age in ages: if all(divmod(age * number_of_toffee_mary, try_age)[1] == 0 for try_age in ages): print(f"{ages=}. Number of toffees={number_of_toffee_mary*age}. Mary is {age-ages[0]} years older than Ann")
      
      LikeLike
- Frits 4:59 pm on 27 July 2024 Permalink | Reply
```
from itertools import combinations

M = 19
N = 7

# return divisors of <n> between 2 and <m>
def divs(n, m=M):
  lst = [d for d in range(2, int(n**.5) + 1) if n % d == 0 and d <= m]
  return lst + [d for x in lst[::-1] if x * x != n and (d := n // x) <= m]

# reciprocal sum
def rs(s): 
  i, t = 0, 0.0
  ln = len(s)
  while t < 1 and i < ln:
    t += 1 / s[i]
    i += 1
  return i == ln and round(t, 5) == 1

# Mary received 18 sweets
for m in range(2, M + 1):
  # pick <N> divisors
  for c in combinations(divs(18 * m), N):
    # reciprocal sum must be 1
    if rs(c) == 1: 
      print(f"answer: {c}")
```
  LikeLike
Jim Randell 9:44 am on 23 July 2024 Permalink | Reply
Tags: by: Adrian Somerfield ( 13 )
Teaser 2576: Latin cube
From The Sunday Times, 5th February 2012 [link] [link]

I have a cube and on each face of it there is a capital letter. I look at one face from the front, hold the top and bottom faces horizontal, and rotate the cube. In this way I can correctly read a four-letter Roman numeral. Secondly, I hold the cube with the original front face underneath and the original top face at the front, and I rotate the cube keeping the new top and bottom faces horizontal. In this way I correctly see another four-letter Roman numeral. Thirdly, I return the cube to its original position and rotate it with the two sides vertical to give another correct four-letter Roman numeral. The sum of the second and third numbers is less than a hundred.

What was the third four-letter Roman numeral seen?

[teaser2576]
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Lise Andreasen and Jim Randell are discussing. Toggle Comments
- Jim Randell 9:45 am on 23 July 2024 Permalink | Reply
  Some clarification of the puzzle text is required to find a single answer (and I assume the puzzle is meant to have a unique answer):
  
  (1) Although the letters used in Roman numerals (I, V, X, L, C, D, M) are all distinguishable whatever the orientation, the puzzle requires that when letters are read they are in a “sensible” orientation. So, I and X can be read correctly upside down. However this is not sufficient to solve the puzzle, we also require that X can be read correctly in any orientation (which may not be the case if a serif font is used, as is often the case with Roman numerals).
  
  (2) The mechanism to read the numbers is that the front face is read, the cube is then rotated through 90° about a major axis to change the front face, which is then read, and the same move is repeated until the four faces in a band around the cube have been read in order (and a final move would bring the cube back to the starting position).
  
  (3) The numbers read should be valid Roman numerals in the “usual” form. (I used the [[ int2roman() ]] and [[ is_roman() ]] functions in the enigma.py library to ensure this). I didn’t allow “IIII” to represent 4 (although if you do it doesn’t change anything).
  
  (4) Each of the three numbers read is different. (Possibly this is meant to be indicated by the use of “another”).
  
  We are not told what direction the cube is rotated, so this Python program considers rotations in each direction for the 3 numbers. Although note that for some numbers (e.g. VIII (= 8), where the second and fourth letters are the same), the direction of the rotation doesn’t matter.
  
  I used the [[ Cube() ]] class (originally from Teaser 2835) to keep track of the rotation of the cube (and orientation of the faces).
  
  This Python program considers possible values for the second and third numbers (which cover bands of the cube that include all faces, so leave the cube fully specified), and then reads the first number and looks for cases where this gives a valid Roman numeral.
  
  It runs in 78ms. (Internal runtime is 11ms).
  
  Run: [ @replit ]
```
from enigma import (irange, int2roman, is_roman, join, seq2str, printf)
from cube import (Cube, U, D, L, R, F, B, face_label as f2l)

# valid orientations for roman digits
valid = dict((k, {0}) for k in "IVXLCDM")
valid['I'] = {0, 2}  # I can be upside down
valid['X'] = {0, 1, 2, 3}  # X can be in any orientation

# collect 4-letter roman numerals less than 100
n2r = dict()
for n in irange(1, 99):
  r = int2roman(n)
  if len(r) == 4:
    n2r[n] = r

# read values from the front of cube <c>, and apply each of the specified moves <ms>
# while match the values in <vs> (can be '?')
# return (<updated cube>, <values read>)
def read(c, ms, vs):
  rs = list()
  while True:
    # read the front face
    (v, vs0) = (c.faces[F], vs[0])
    # blank face?
    if v is None:
      v = vs0
      if v != '?':
        # fill out the new value
        c = c.update(faces=[(F, v)], orientations=[(F, 0)])
    else:
      # check it is in a valid orientation and a valid value
      if (c.orientations[F] not in valid[v]) or (vs0 != '?' and vs0 != v):
        return (None, None)
    rs.append(v)
    # any more moves?
    if not ms: break
    c = c.rotate(ms[0:1])
    ms = ms[1:]
    vs = vs[1:]
  return (c, join(rs))

# start with an empty cube
c0 = Cube(faces=[None] * 6)

# choose a 4-letter roman for the second number
for (n2, r2) in n2r.items():
  # choose the direction of rotation
  for d2 in [U, D]:
    # read the second number
    (c1, _) = read(c0.rotate([L]), [d2] * 3, r2)
    if c1 is None: continue
    # reset to original position
    c1 = c1.rotate([d2, R])

    # choose a 4-letter roman for the third number
    for (n3, r3) in n2r.items():
      # the first and second numbers sum to less than 100
      if n2 + n3 > 99: continue
      # choose a direction of rotation
      for d3 in [L, R]:
        # read the third number
        (c2, _) = read(c1, [d3] * 3, r3)
        if c2 is None: continue
        # all faces should now be filled out
        assert None not in c2.faces
        # reset to original position
        c2 = c2.rotate([d3])

        # choose a direction for the first number
        for d1 in [U, D]:
          # read the first number
          (_, r1) = read(c2, [d1] * 3, '????')
          if r1 is None: continue
          # is it a valid roman?
          n1 = is_roman(r1)
          if not n1: continue
          # check numbers are all different
          if len({n1, n2, n3}) != 3: continue

          # output solution
          printf("{r1} ({n1}) [{d1}]; {r2} ({n2}) [{d2}]; {r3} ({n3}) [{d3}]; {cube}",
                 d1=f2l[d1], d2=f2l[d2], d3=f2l[d3], cube=seq2str(c2.faces))
```
  Solution: The third Roman numeral seen was: LXII (= 62).
  
  The first two numbers were: LXIX (= 69) and XXIX (= 29), and the second and third sum to 91.
  
  Note that both of the first two numbers can be read using either direction of rotation (which is why my code finds 4 different ways to the answer), but the final number only works in one direction.
  
  The cube has faces (U, D, L, R, F, B) = (X, I, X, X, L, I).
  
  If we allow repeated numbers there is also a solution with the cube (U, D, L, R, F, B) = (X, I, X, X, X, I), which gives the numbers XXIX (= 29), XXIX (= 29), XXII (= 22).
  
  LikeLike
- Lise Andreasen 12:07 am on 24 July 2024 Permalink | Reply
  
  Especially (1) the orientation of X is interesting.
  
  LikeLike

Jim Randell 12:59 pm on 21 July 2024 Permalink | Reply
Tags: by: B W M Young ( 10 )

Brain-Teaser 420: [The Island of Squares]

From The Sunday Times, 25th May 1969 [link]

Here is the Island of Squares with its 22 provinces, and its capital in the middle of the central province.

The Geographer Royal has been instructed to colour the provinces red, blue and yellow, in such a way that no two provinces with the same colouring have a common boundary-line.

There is one further complication. Four main roads are planned each of which will go straight from the capital to one corner of the island crossing the corners of provinces on the way; one of these roads must never be in a blue province, and another must at no corner move from a yellow to a blue province or vice versa.

What colours Geographer Royal will the use for provinces 1 and 2?

This puzzle was originally published with no title.

[teaser420]

Jim Randell 1:00 pm on 21 July 2024 Permalink | Reply

This Python program colours the provinces so that no two adjacent regions share a colour, and then checks to see if there are two roads that meet the requirements.

It runs in 68ms. (Internal runtime is 4.5ms).

Run: [ @replit ]

from enigma import (peek, update, join, union, map2str, printf)

# adjacency matrix for provinces
adj = {
  1: {2, 3, 4, 5},
  2: {1, 6, 7},
  3: {1, 4, 14},
  4: {1, 3, 5, 8, 15},
  5: {1, 4, 6, 9, 10},
  6: {2, 5, 7, 10, 13},
  7: {2, 6, 13, 18, 21, 22},
  8: {4, 9, 11, 16},
  9: {5, 8, 10, 11},
  10: {5, 6, 9, 12},
  11: {8, 9, 12, 17},
  12: {10, 11, 13, 17},
  13: {6, 7, 12, 18},
  14: {3, 15, 19},
  15: {4, 14, 16, 20},
  16: {8, 15, 17, 20},
  17: {11, 12, 16, 18, 21},
  18: {7, 13, 17, 21},
  19: {14, 20, 22},
  20: {15, 16, 19, 21, 22},
  21: {7, 17, 18, 20, 22},
  22: {7, 19, 20, 21},
}

# colour regions <rs> with colours <cs>
def colour(rs, cs, m=dict()):
  # are we done?
  if not rs:
    yield m
  else:
    # choose an uncoloured region
    r = peek(rs)
    # choose a colour
    xs = cs.difference(m[x] for x in adj[r] if x in m)
    for x in xs:
      yield from colour(rs.difference([r]), cs, update(m, [(r, x)]))

# regions for each road
roads = dict(
  NW=[11, 16, 20, 19],
  NE=[11, 17, 21, 7],
  SE=[11, 10, 6, 2],
  SW=[11, 8, 4, 1],
)

# check roads
def check(m):
  # map roads to colours
  r = dict((k, join(m[x] for x in v)) for (k, v) in roads.items())
  # find roads the do not pass through a blue province
  R1 = set(k for (k, v) in r.items() if not ('B' in v))
  if not R1: return
  # find roads that do not pass directly from blue to yellow (or vice versa)
  R2 = set(k for (k, v) in r.items() if not ('BY' in v or 'YB' in v))
  if not R2: return
  # check we can find one of each
  if len(union([R1, R2])) < 2: return
  return r

# colour the regions
for m in colour(set(adj.keys()), set("RBY")):
  # and check the roads
  r = check(m)
  if r:
    # output solution
    printf("{m}", m=map2str(m))
    printf("-> {r}", r=map2str(r))
    printf()

Solution: Province 1 is red. Province 2 is yellow.

The complete map is shown below:

There is one province that can be either red or blue (On the extreme left of the diagram. I numbered it 14, and gave it both colours).

The NE road passes through no blue, and the SW road does not pass directly from a blue to a yellow (or vice versa).

We can see that there is no point trying to colour the central province blue (as that would make blue appear in every road), but I think the code is fast enough without this.

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