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Jim Randell 7:12 am on 18 June 2020 Permalink | Reply
Tags: by: Nick MacKinnon ( 60 )
Brainteaser 1819: Early bath
From The Sunday Times, 27th July 1997 [link]

There are 20 teams in one country’s premier league. They each play each other once in the first half of the season, and then they each play each other a second time in the rest of the season. Each team plays each Saturday of the season, earning three points for a win and one point for a draw. At the end of the season the bottom three teams are relegated and the top team wins the league championship.

Last season was the most boring ever. It was possible to determine the relegated teams well before the end of the season. In fact it would be impossible in any season to be able to determine the three relegated teams in fewer weeks.

Three further Saturdays after the relegations were determined the league championship was also determined when the league leaders were in a 0-0 draw and then found that they were unassailable. There were great celebrations that night.

At that time, how many points did the current top two teams have?

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

[teaser1819]
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John Crabtree and Jim Randell are discussing. Toggle Comments
- Jim Randell 7:12 am on 18 June 2020 Permalink | Reply
  
  After a tortuous chain of reasoning we arrive at the answer:
  
  In the first half of the season there are C(20, 2) = 190 matches, and then another 190 in the second half of the season. Each team plays 19 matches in the first half of the season, and another 19 matches in the second half of the season. The season lasts 38 weeks, with 10 matches played each week.
  
  At some point there are 3 teams doomed to relegation, as they cannot possibly catch up with any of the remaining 17 teams.
  
  Suppose in the first half of the tournament there are three teams that lose all their matches, except the matches they play amongst themselves, which are drawn.
  
  Each of the three doomed teams has only 2 points (from their draws with the other two doomed teams). If all the other 187 matches were won outright there are 187×3 = 561 points to distribute between the remaining 17 teams. This gives an average of 33 points per team. But, if one of the doomed teams were to win all their matches in the second half of the tournament they would end up with 2 + 3×19 = 59 points, so their relegation is not guaranteed by the end of the first half of the tournament.
  
  If we carry on for k weeks into the second half of the tournament, with each doomed team losing each of their matches, and each of the other matches being won, then how many weeks is it before enough points are accumulated, so that each of the 17 remaining teams are out of reach?
  
  There are (19 − k) weeks remaining and a doomed team could win all their remaining matches, so they would end with 2 + 3(19 − k) = 59 − 3k points.
  
  And the total number of points to be shared between the 17 remaining teams would be: 561 + 30k.
  
  This can happen when:
  
  (561 + 30k) / 17 > 59 − 3k
  561 + 30k > 1003 − 51k
  81k > 442
  k > 5.46…
  
  i.e. after the 6th week of the second half of the tournament, at the earliest.
  
  There would then be 13 weeks remaining, so a doomed team that had a sudden change of fortune could finish with 2 + 3×13 = 41 points. But by that stage in the tournament 561 + 30×6 = 741 points could have been scored by the remaining 17 teams, which is enough for each of them to have accumulated at 43 points each. So the doomed teams can indeed be doomed to relegation.
  
  Now after three more weeks, i.e. after the 9th week of the second half of the tournament, with 10 weeks remaining, the championship was determined. So the team with the most points must be more than 30 points ahead of the next highest team.
  
  If we go back to 6th week of the second half of the tournament, then 16 of the 17 non-doomed teams could have 42 points, and the other team could have 69 points. So they are 27 points ahead.
  
  If they win the following 2 weeks, and the week after that they draw (as we are told in the puzzle text), then after the 9th week they have 76 points, and in order to be unassailable the next highest team can have no more than 45.
  
  So now we need to make sure none of the other 16 teams can get more than 45 points by the end of week 9. An easy way is to suppose all the matches apart from the ones involving the future champions are drawn.
  
  Then at the end of the 7th week, the best any of the 16 challengers can do is 43 points. The future champions have 72 points, and there are 12 weeks remaining, so they are assailable.
  
  At the end of the 8th week, the best any of the challengers can do is 44 points. The future champions have 75 points, and there are 11 weeks remaining, so they are still assailable.
  
  At the end of the 9th week, the best any of the challengers can do is 45 points. The future champions have 76 points, and there are 10 weeks remaining, so they are now unassailable.
  
  So this is a viable scenario for the puzzle.
  
  Solution: The top team had 76 points, and the next highest team had 45 points.
  
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  - John Crabtree 5:14 pm on 19 June 2020 Permalink | Reply
    
    This teaser just works. After 6 weeks of the second half, 16 teams have 42 points and one has 69, ie a total of 741 for those teams and the maximum possible.
    
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Jim Randell 8:42 am on 16 June 2020 Permalink | Reply
Tags: by: Angela Newing ( 40 )

Teaser 2662: Number please

From The Sunday Times, 29th September 2013 [link] [link]

I finally got through to the operator of a large company and asked to be connected to an appropriate department. He tried eight different extension numbers before then finding the correct one. The eight numbers were 1933, 2829, 3133, 4630, 5089, 5705, 6358 and 6542. Curiously, each of these wrong numbers did have at least one correct digit of the correct extension in the correct position!

What was the correct extension number?

[teaser2662]

Jim Randell 8:42 am on 16 June 2020 Permalink | Reply

Here is a recursive program that starts with unassigned digits in all positions, and then at each step looks for a number that doesn’t already match and uses it to fill out one of the unassigned digits. It runs in 53ms.

Run: [ @repl.it ]

from enigma import (update, join, printf)

# find sequences that match ns in at least one position
# ds - candidate digits ('?' for unassigned)
# ns - remaining numbers to check
def solve(ds, ns):
  # are we done?
  if not ns:
    yield join(ds)
  else:
    # consider the digits of the next number
    ms = list(zip(ds, ns[0]))
    # if any digits already match move on to the next one
    if any(a == b for (a, b) in ms):
      yield from solve(ds, ns[1:])
    else:
      # set one of the unassigned digits
      for (i, (a, b)) in enumerate(ms):
        if a == '?':
          yield from solve(update(ds, [(i, b)]), ns[1:])

# numbers provided
numbers = "1933 2829 3133 4630 5089 5705 6358 6542".split()

# solve the puzzle, starting with 4 ?'s
for n in solve("????", numbers):
  printf("number = {n}")

Solution: The correct extension number is 6739.

A brute force search of all possible extension numbers is shorter, and only slightly slower:

from enigma import (subsets, join, printf)

# the numbers tried (each has at least one correct digit in the correct position)
numbers = "1933 2829 3133 4630 5089 5705 6358 6542".split()

# consider digits for the actual number
for ds in subsets("0123456789", size=4, select="M"):
  if all(any(a == b for (a, b) in zip(ds, n)) for n in numbers):
    printf("number = {n}", n=join(ds))

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GeoffR 10:29 am on 6 October 2020 Permalink | Reply


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

var 0..9: A; var 0..9: B; var 0..9: C; var 0..9: D;

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

% Each of these wrong numbers did have at least one correct
% digit of the correct extension in the correct position

% Extn Try 1 - 1933 
constraint sum ([A==1,B==9,C==3,D==3]) > 0;

% Extn Try 2 - 2829 
constraint sum ([A==2,B==8,C==2,D==9]) > 0;

% Extn Try 3 - 3133 
constraint sum ([A==3,B==1,C==3,D==3]) > 0;

% Extn Try 4 - 4630 
constraint sum ([A==4,B==6,C==3,D==0]) > 0;

% Extn Try 5 - 5089 
constraint sum ([A==5,B==0,C==8,D==9]) > 0;

% Extn Try 6 - 5705 
constraint sum ([A==5,B==7,C==0,D==5]) > 0;

% Extn Try 7 - 6358  
constraint sum ([A==6,B==3,C==5,D==8]) > 0;

% Extn Try 8 - 6542 
constraint sum ([A==6,B==5,C==4,D==2]) > 0;

solve satisfy;

output ["The correct extension number is " ++
show(A),show(B),show(C),show(D) ];

% The correct extension number is 6739
% % time elapsed: 0.03 s
% ----------
% ==========

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Jim Randell 12:18 pm on 14 June 2020 Permalink | Reply
Tags: by: Peter G Chamberlain ( 11 )

Teaser 2655: Sudoprime

From The Sunday Times, 11th August 2013 [link] [link]

The grid shows a cross-figure with two of its digits given. The eleven answers (five of them being “across” and the other six “down”) are all different prime numbers with no leading zeros. No digit appears more than once in any row, column or main diagonal.

What are the two largest of the eleven primes?

[teaser2655]

Jim Randell 12:19 pm on 14 June 2020 Permalink | Reply

See also: Enigma 1730, Enigma 1740, Enigma 1755.

I used the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle. The condition that there are no repeated digits in the rows, columns or diagonals can be expressed using the [[ --distinct ]] parameter, which lets us specify multiple groups of symbols, where none of the groups may contain repeated digits.

The following run file executes in 93ms.

Run: [ @replit ]

#! python -m enigma -rr

SubstitutedExpression

#
#  A # # # B
#  C D # E F
#  # G H I #
#  J K # L M
#  N # # # P
#

# the two values we are given
--assign="G,8"
--assign="N,7"

# the 11 answers are all different prime numbers
"is_prime(AC)"
"is_prime(BF)"
"is_prime(CD)"
"is_prime(EF)"
"is_prime(JK)"
"is_prime(LM)"
"is_prime(JN)"
"is_prime(MP)"
"all_different(AC, BF, CD, EF, JK, LM, JN, MP)"

"is_prime(DGK)"
"is_prime(EIL)"
"is_prime(GHI)"
"all_different(DGK, EIL, GHI)"

# no digit appears more than once in any row, column, or main diagonal
--distinct="AB,CDEF,GHI,JKLM,NP,ACJN,DGK,EIL,BFMP,ADHLP,NKHEB"

# answer is the two largest primes
--code="ans = lambda *vs: last(sorted(vs), 2, fn=tuple)"
--answer="ans(DGK, EIL, GHI)"

# suppress verbose output
--template=""

Solution: The two largest primes are: 821, 983.

The completed grid looks like this:

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GeoffR 3:21 pm on 14 June 2020 Permalink | Reply

% A Solution in MiniZinc - same grid as Jim
include "globals.mzn";

var 1..9:A; var 1..9:B; var 1..9:C; var 1..9:D; var 1..9:E; 
var 1..9:F; var 1..9:G; var 1..9:H; var 1..9:I; var 1..9:J; 
var 1..9:K; var 1..9:L; var 1..9:M; var 1..9:N; var 1..9:P; 

% Given values
constraint G == 8 /\ N == 7;

% Rows
var 11..97:CD = 10*C + D; 
var 11..97:EF = 10*E + F; 
var 11..97:JK = 10*J + K; 
var 11..97:LM = 10*L + M; 
var 101..997:GHI = 100*G + 10*H + I;
% Columns
var 11..97:AC = 10*A + C; 
var 11..97:JN = 10*J + N; 
var 101..997:DGK = 100*D + 10*G + K; 
var 101..997:EIL = 100*E + 10*I + L; 
var 11..97:BF = 10*B + F; 
var 11..97:MP = 10*M + P; 

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

% The eleven answers are all prime numbers
constraint is_prime(CD) /\ is_prime(EF) /\ is_prime(JK) /\ is_prime(LM)
/\ is_prime(GHI) /\ is_prime(AC) /\ is_prime(JN) /\ is_prime(DGK)
/\ is_prime(EIL) /\ is_prime(BF) /\ is_prime(MP);

% All the prime numbers are different
constraint all_different([CD,EF,JK,LM,GHI,AC,JN,DGK,EIL,BF,MP]);

% Top and bottom rows have different digits
constraint A != B /\ N != P;

% Other rows have different digits
constraint all_different([C,D,E,F]) /\ all_different([G,H,I])
/\ all_different([J,K,L,M]);

% All the columns have different digits
constraint all_different([A,C,J,N]) /\ all_different([D,G,K])
/\ all_different([E,I,L]) /\ all_different([B,F,M,P]);

% The diagonals have different digits
constraint all_different([A,D,H,L,P]) /\ all_different([B,E,H,K,N]);

solve satisfy;

output ["Three digit primes are : " ++ show([DGK,GHI,EIL])
++ "\nA, B, C, D, E, F = " ++ show([A,B,C,D,E,F])
++ "\nG, H, I, J, K, L = " ++ show([G,H,I,J,K,L])
++ "\nM, N, P = " ++ show([M,N,P]) ];

% Three digit primes are : [983, 821, 617]
% A, B, C, D, E, F = [6, 9, 1, 9, 6, 7]
% G, H, I, J, K, L = [8, 2, 1, 4, 3, 7]
% M, N, P = [1, 7, 3]
% ----------
% ==========
% Grid solution
% 6 # # # 9
% 1 9 # 6 7
% # 8 2 1 #
% 4 3 # 7 1
% 7 # # # 3

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Jim Randell 4:48 pm on 12 June 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 65 )
Teaser 3012: Number blind rage
From The Sunday Times, 14th June 2020 [link] [link]

After school, angry at getting “50 lines”, I kicked my satchel around. Impacts made my 11-digit calculator switch on. An 11-digit number was also entered and the display was damaged. Strangely, I found “dYSCALCULIA” displayed and saved this to memory (as shown).

After various tests I confirmed that all arithmetic operations were correct and the decimal point would appear correctly if needed. No segments were permanently “on”, two digits were undamaged, and for the other digits, overall, several segments were permanently “off”.

Retrieving “dYSCALCULIA”, I divided it by 9, then the result by 8, then that result by 7, then that result by 6. No decimal point appeared and the last result (at the right-hand side of the display) had three digits appearing as numerals.

What number was “dYSCALCULIA”?

[teaser3012]
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- Jim Randell 9:07 pm on 12 June 2020 Permalink | Reply
  I used the standard set of digits (as illustrated in Enigma 1701).
  
  This Python program runs in 90ms.
```
from enigma import (cproduct, join, div, printf)

# normal digits
normal = "0123456789"

# map digits to illuminated segments, arranged as:
#
#   0
# 1   2
#   3
# 4   5
#   6
#
# map digits to segments
f = lambda *ss: frozenset(ss)
seg = {
  # normal digits
  '0': f(0, 1, 2, 4, 5, 6),
  '1': f(2, 5),
  '2': f(0, 2, 3, 4, 6),
  '3': f(0, 2, 3, 5, 6),
  '4': f(1, 2, 3, 5),
  '5': f(0, 1, 3, 5, 6),
  '6': f(0, 1, 3, 4, 5, 6), # or could be (1, 3, 4, 5, 6)
  '7': f(0, 2, 5), # or could be (0, 1, 2, 5)
  '8': f(0, 1, 2, 3, 4, 5, 6),
  '9': f(0, 1, 2, 3, 5, 6), # or could be (0, 1, 2, 3, 5)
  # malformed digits
  'a': f(0, 1, 2, 3, 4, 5),
  'c': f(0, 1, 4, 6),
  'd': f(2, 3, 4, 5, 6),
  'l': f(1, 4, 6),
  'u': f(1, 2, 4, 5, 6),
}
# map segments to normal digits
norm = dict((seg[k], k) for k in normal)

# the display
display = "d45calcul1a"

# compute possible replacement (superset) digits for the symbols
r = dict((k, list(d for d in normal if seg[d].issuperset(seg[k]))) for k in set(display))

# choose possible replacement digits for the symbols
for ds in cproduct(r[x] for x in display):
  if ds[0] == '0': continue
  # two of the digits are unaffected
  if sum(x == y and x in normal for (x, y) in zip(display, ds)) < 2: continue
  # make the number
  n = int(join(ds))
  # and the result
  s = div(n, 9 * 8 * 7 * 6)
  if s is None: continue
  # remove broken segments from the result
  # x = original display, y = original digit, z = result digit
  rs = list(
    norm.get(seg[z].difference(seg[y].difference(seg[x])), '?')
      for (x, y, z) in zip(display[::-1], ds[::-1], str(s)[::-1])
  )
  # there should be 3 normal digits
  if len(rs) - rs.count('?') != 3: continue
  # output solution
  printf("{n} -> {s} -> {rs}", rs=join(rs[::-1]))
```
  Solution: dYSCALCULIA = 84588800688.
  
  The digits displaying 4 and 5 must be the undamaged ones.
  
  So the segments that are definitely broken are as shown below:
  
  There are 22 segments that are definitely broken, and a further 3 that we cannot determine if they are broken or not.
  
  The result of dividing the number by 9×8×7×6 = 3024 is 27972487, which would look something like this:
  
  (Only showing the segments that we definitely know to be broken, there are two segments shown lit that may be broken).
  
  The 2nd digit (7) displays as a 7. The 7th digit (8) displays as a 1. The 8th digit (7) displays as a 7.
  
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Jim Randell 11:18 am on 11 June 2020 Permalink | Reply
Tags: by: Ann Kindersley
Brain-Teaser 522: [Palindromic phone numbers]
From The Sunday Times, 13th June 1971 [link]

The other day I noticed some strange coincidences about the telephone numbers of my three friends.

Each number reads the same from left to right and vice versa. The first and third digit formed the owner’s age, as did the sum of all the digits of his number and, furthermore, his number was divisible by his age.

What were my friends’ ages?

This puzzle was originally published with no title.

[teaser522]
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GeoffR and Jim Randell are discussing. Toggle Comments
- Jim Randell 11:20 am on 11 June 2020 Permalink | Reply
  Assuming the ages cannot have a leading zero (which means the phone numbers can’t either) gives us a unique solution to the puzzle.
  
  This Python program uses the [[ SubstitutedExpression() ]] solver from the enigma.py library to try 3-, 4- and 5-digit numbers satisfying the required conditions. It runs in 54ms.
```
from enigma import (sprintf as f, join, SubstitutedExpression, subsets, all_different, printf)

# alphametic forms for possible palindromic phone numbers
numbers = [ 'ABA', 'ABBA', 'ABCBA' ]

# collect possible (number, age) solutions
ss = list()

# consider possible phone numbers
for n in numbers:
  # age is 1st and 3rd digit
  a = n[0] + n[2]
  
  exprs = [
    # sum of digits is the same as age
    f("{s} = {a}", s=join(n, sep=" + ")),
    # the number itself is divisible by the age
    f("{n} % {a} = 0"),
  ]

  # solve the alphametic expressions
  p = SubstitutedExpression(exprs, distinct="", answer=f("({n}, {a})"))
  ss.extend(p.answers(verbose=0))

# choose three solutions with different ages
for (A, B, C) in subsets(ss, size=3):
  if not all_different(A[1], B[1], C[1]): continue
  printf("A={A} B={B} C={C}")
```
  Solution: The ages are: 21, 23, 27.
  
  The corresponding phone numbers are: 28182, 28382, 28782.
  
  These are the only three solutions for 5-digit phone numbers, and there are no solutions for 3- or 4-digit phone numbers.
  
  LikeLike
- GeoffR 4:48 pm on 16 July 2021 Permalink | Reply
```
% A Solution in MiniZinc 
% assumes 5-digit tel. no. is ABCBA
include "globals.mzn";

var 1..9:A; var 0..9:B; var 0..9:C; 

var 10..100: age1;

var 10000..99999: ABCBA = 10001*A + 1010*B + 100*C; 

constraint age1 == 10*A + C /\ age1 == 2*A + 2*B + C
/\ ABCBA mod age1 == 0;

solve satisfy;

output ["Age = " ++ show(age1) ++ ", Tel. No. = " ++ show(ABCBA) ];

% Age = 21, Tel. No. = 28182
% ----------
% Age = 23, Tel. No. = 28382
% ----------
% Age = 27, Tel. No. = 28782
% ----------
% ==========
```
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Jim Randell 9:33 am on 9 June 2020 Permalink | Reply
Tags: by: Ernest J Luery, flawed ( 56 )
Brain-Teaser 521: [Names and occupations]
From The Sunday Times, 6th June 1971 [link]

There were six. Messrs. Butcher, Carpenter, Draper, Farmer, Grocer and Miller, who shared the fire-watching on Friday nights — three this week, three the next. By occupation they were (not necessarily respectively) a butcher, a carpenter, a draper, a farmer, a grocer and a miller.

Incidents were few and far between, until that Saturday morning when they found the “log” book signed by the Acting Deputy Assistant something or other, as follows: “All three present and fast asleep”.

Something had to be done about it: they decided to watch, to play whist, and to keep awake in the future. It was arranged thus: four did duty this week, next week two stood down and two others came in, and so on. Each did two turns in three weeks.

On the first Friday the carpenter, the draper, the farmer and the miller watched. Next week Mr Carpenter, Mr Draper, Mr Farmer and Mr Grocer played. On the third occasion Mr Butcher played against Mr Grocer, Mr Farmer against the butcher, and the miller against the draper. Each night the four cut for partners and kept them till morning.

If Mr Carpenter’s occupation is the same as the name of the one whose occupation is the same as the name of the one whose occupation is the same as the one whose occupation is the same as the name of the miller:

What is Mr Miller’s occupation?

As presented this puzzle has no solutions. In the comments I give a revised version that does have a unique answer.

This puzzle was originally published with no title.

[teaser521]
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- Jim Randell 9:33 am on 9 June 2020 Permalink | Reply
  I thought the tricky bit in this puzzle would be detangling the long obfuscated condition at the end (and decide if there was a “the name of” missing from it). But it turns out we can show that there are no solutions before we get that far, so the puzzle is flawed.
  
  In week 3 we have the following pairs (Mr Butcher + Mr Grocer), (Mr Farmer + the butcher), (the miller + the draper).
  
  But there are only four people (i.e. two pairs), so one of the pairs is repeated. The middle one doesn’t overlap with either of the outer two, so the outer two must refer to the same pair. i.e. (Mr Butcher + Mr Grocer) = (the miller + the draper).
  
  In week 2 we had Mr Carpenter, Mr Draper, Mr Farmer and Mr Grocer. And Mr Farmer and Mr Grocer stayed on for week 3, which means they can’t have done week 1.
  
  The jobs missing from week 1 are the butcher and the grocer, so: (Mr Farmer + Mr Grocer) = (the butcher + the grocer).
  
  But Mr Grocer cannot be one of (the miller + the draper) and also one of (the butcher + the grocer).
  
  So the described situation is not possible.
  
  However if we change the names in the second week to match the jobs from the first week (which I think makes for a more pleasing puzzle), and also insert the missing “the name of” into the final obfuscated condition we get the following revised puzzle:
  
  …
  
  On the first Friday the carpenter, the draper, the farmer and the miller watched. Next week Mr Carpenter, Mr Draper, Mr Farmer and Mr Miller played. On the third occasion Mr Butcher played against Mr Grocer, Mr Farmer against the butcher, and the miller against the draper. Each night the four cut for partners and kept them till morning.
  
  If Mr Carpenter’s occupation is the same as the name of the one whose occupation is the same as the name of the one whose occupation is the same as the name of the one whose occupation is the same as the name of the miller:
  
  What is Mr Miller’s occupation?
  
  Then we get a puzzle that does have solutions (although not the same as the published solution).
  
  The following Python program runs in 54ms.
  
  Run: [ @repl.it ]
```
from enigma import (irange, subsets, multiset, map2str, printf)

# labels for the names and jobs
labels = (B, C, D, F, G, M) = irange(0, 5)

# check the schedule for the 3 weeks
def check(w1, w2, w3):
  m = multiset()
  for w in (w1, w2, w3):
    w = set(w)
    # 4 different people each week
    if len(w) != 4: return False
    m.update_from_seq(w)
  # each person does 2 duties over the 3 weeks
  return all(v == 2 for v in m.values())

# map: job -> name
for n in subsets(labels, size=len, select="P"):

  # Week 1 = carpenter, draper, farmer, miller
  w1 = [n[C], n[D], n[F], n[M]]
  # Week 2 = Carpenter, Draper, Farmer, Miller
  w2 = [C, D, F, M] # NOT: [C, D, F, G]
  # Week 3 = Butcher + Grocer, Farmer + butcher, miller + draper
  # so: Butcher + Grocer = miller + draper
  if not(set([B, G]) == set([n[M], n[D]])): continue
  w3 = [B, G, F, n[B]]

  if not check(w1, w2, w3): continue

  # "Mr Carpenter's occupation is the same as the name of the one
  # whose occupation is the same as the name of the one whose
  # occupation is the same as the [name of] one whose occupation is
  # the same as the name of the miller"
  if not(n[n[n[n[n[M]]]]] == C): continue

  # output the map
  names = ("Butcher", "Carpenter", "Draper", "Farmer", "Grocer", "Miller")
  jobs = list(x.lower() for x in names)
  printf("{s}", s=map2str(((names[n], j) for (n, j) in zip(n, jobs)), sep="; ", enc=""))
```
  Solution: Mr Miller is the carpenter.
  
  There are three ways to assign the jobs to the names:
  
  Butcher=draper; Carpenter=butcher; Draper=farmer; Farmer=grocer; Grocer=miller; Miller=carpenter
  Butcher=miller; Carpenter=butcher; Draper=farmer; Farmer=grocer; Grocer=draper; Miller=carpenter
  Butcher=miller; Carpenter=farmer; Draper=butcher; Farmer=grocer; Grocer=draper; Miller=carpenter
  
  LikeLike
Jim Randell 5:14 pm on 5 June 2020 Permalink | Reply
Tags: by: Danny Roth ( 89 )
Teaser 3011: Optical illusion
From The Sunday Times, 7th June 2020 [link] [link]

George and Martha are studying optics. If you place an object a specific distance from a lens, an image will appear at a distance from that lens according the following formula:

The reciprocal of the object distance plus the reciprocal of the image distance is equal to the reciprocal of the focal length of the lens.

The object distance was a two-digit whole number of cm (ab). The image distance and the focal length of the lens were also two-digit whole numbers (cd and ef respectively). The six digits were all different and non-zero. Also, the object distance and the focal length were of the same parity and b was an exact multiple of d. Martha pointed out that the sum of those three two-digit numbers was a prime number.

What was that prime number?

[teaser3011]
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- Jim Randell 5:27 pm on 5 June 2020 Permalink | Reply
  I used uppercase letters for the alphametic symbols, as that is more usual.
  
  We can use the [[ SubstitutedExpression() ]] solver from the enigma.py library, to solve the relevant alphametic expressions. The following run file executes in 104ms.
  
  Run: [ @replit ]
```
#! python -m enigma -rr

SubstitutedExpression

--digits="1-9"

"fraction(1, AB, 1, CD) == (1, EF)"
"B % 2 == F % 2"
"div(B, D) > 1"
"is_prime(AB + CD + EF)"

--answer="AB + CD + EF"
```
  Solution: The sum of the numbers is 211.
  
  The object distance is 78 cm, the image distance is 91 cm and the focal length is 42 cm.
  
  (1 / 78) + (1 / 91) = (1 / 42)
  
  The optical constraint and multiple constraint (lines 12 and 14) are sufficient to arrive at a unique answer to the puzzle.
  
  However, if we remove the constraint that the symbols stand for different digits there is a further solution:
  
  (1 / 28) + (1 / 21) = (1 / 12)
  
  In this case the numbers sum to 61.
  
  LikeLike
- GeoffR 7:33 pm on 5 June 2020 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";

var 1..9: a; var 1..9: b; var 1..9: c; 
var 1..9: d; var 1..9: e; var 1..9: f;

constraint all_different ([a, b, c, d, e, f]);

var 10..99: ab = 10*a + b;  % object distance
var 10..99: cd = 10*c + d;  % image distance
var 10..99: ef = 10*e + f;  % focal length

predicate is_prime(var int: x) = 
x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x)))))
 ((i < x) -> (x mod i > 0));
 
% optical formula is  1/ab + 1/cd = 1/ef, so... 
constraint (cd * ef) + (ab * ef) == (ab * cd);

% parity is same for numbers ab and ef
constraint ab mod 2 == ef mod 2;

% b was an exact multiple of d
constraint b div d > 1 /\ b mod d == 0;

% sum of three two-digit numbers was a prime number
constraint is_prime(ab + cd + ef);

solve satisfy;

output ["Prime number is " ++ show(ab + cd + ef) ++
"\nab = " ++ show(ab) ++ ", cd = " ++ show(cd) ++ ", ef = " ++ show(ef)];
```
  LikeLike
Jim Randell 12:06 pm on 4 June 2020 Permalink | Reply
Tags: by: Graham Smithers ( 87 )
Teaser 2860: Cricketing greats
From The Sunday Times, 16th July 2017 [link] [link]

On each of the next seven evenings a different media pundit will advocate the merits of two cricketers. The pundits are Agnew, Blofeld, Dagnall, Mann, Mitchell, Norcross and Smith. The fourteen cricketers to be discussed are Ali, Anderson, Ball, Ballance, Broad, Carberry, Compton, Hales, Kerrigan, Patel, Stokes, Tredwell, Trescothick and Woakes. Each evening, looking at the names of the pundit and the two cricketers, then for any two out of the three names there are just two letters of the alphabet that occur (once or more) in both.

(a) Which cricketers will Dagnall advocate?
(b) Which cricketers will Norcross advocate?

[teaser2860]
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- Jim Randell 12:07 pm on 4 June 2020 Permalink | Reply
  See: Teaser 2959.
  
  I incorporated the [[ gangs() ]] function into the [[ grouping ]] code that is available in the enigma.py library, and we can use it to solve this puzzle.
  
  The following Python code runs in 53ms.
  
  Run: [ @repl.it ]
```
from enigma import (grouping, diff)

# gang leaders (in solving order)
pundits = ( 'Dagnall', 'Norcross', 'Agnew', 'Blofeld', 'Mann', 'Mitchell', 'Smith' )

# gang followers
cricketers = (
  'Ali', 'Anderson', 'Ball', 'Ballance', 'Broad', 'Carberry', 'Compton',
  'Hales', 'Kerrigan', 'Patel', 'Stokes', 'Tredwell', 'Trescothick', 'Woakes'
)

# selection function
fn = grouping.share_letters(2)

# find possible 2-gangs for Dagnall and Norcross
for (g1, g2) in grouping.gangs(2, pundits[0:2], cricketers, fn):
  # check the remaining gangs can be made
  for gs in grouping.gangs(2, pundits[2:], diff(cricketers, g1 + g2), fn):
    grouping.output_gangs(pundits, [g1, g2] + gs)
    # we only need one example
    break
```
  Solution: (a) Dagnall advocates Ball and Broad; (b) Norcross advocates Ballance and Woakes.
  
  The program checks that the remaining gangs can be formed, and it turns out that there is only one way to do this:
  
  Dagnall: Ball, Broad
  Norcross: Ballance, Woakes
  Agnew: Carberry, Tredwell
  Blofeld: Patel, Trescothick
  Mann: Anderson, Compton
  Mitchell: Ali, Kerrigan
  Smith: Hales, Stokes
  
  So we could just use the following even shorter program to find all possible groupings:
```
from enigma import grouping

# gang leaders
pundits = ( 'Dagnall', 'Norcross', 'Agnew', 'Blofeld', 'Mann', 'Mitchell', 'Smith' )

# gang followers
cricketers = (
  'Ali', 'Anderson', 'Ball', 'Ballance', 'Broad', 'Carberry', 'Compton',
  'Hales', 'Kerrigan', 'Patel', 'Stokes', 'Tredwell', 'Trescothick', 'Woakes'
)

# find 2-gangs
for gs in grouping.gangs(2, pundits, cricketers, grouping.share_letters(2)):
  grouping.output_gangs(pundits, gs)
```
  LikeLike
Jim Randell 8:11 am on 2 June 2020 Permalink | Reply
Tags: by: John Owen ( 34 ), corrected ( 10 ), flawed ( 56 )
Teaser 2566: A fulfilling strategy
From The Sunday Times, 27th November 2011 [link] [link]

I drove down a road with a number of petrol stations whose locations I saw on my map. I decided to check the price at the first station then fill up when I found one offering a lower price (or, failing that, the last one).

When I got home I noticed that I could arrange the prices (in pence per litre) into an ascending sequence of consecutive whole numbers of pence, plus 0.9p (i.e. 130.9p, 131.9p, 132.2p, etc). I also worked out the average price that I would expect to pay using this strategy, if I were to encounter this set of prices in an unknown order, and I was surprised to find that this value turned out to be a whole number of pence per litre.

How many petrol stations were there?

When the puzzle was originally published in The Sunday Times it had been edited into a form that meant there were no solutions. Here I have rephrased the middle paragraph so that it works as the setter originally intended.

[teaser2566]
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- Jim Randell 8:12 am on 2 June 2020 Permalink | Reply
  (See also: Teaser 2988).
  
  Here is a program that calculates the result directly. It runs in 50ms.
  
  Run: [ @replit ]
```
from enigma import (subsets, irange, inf, factorial, div, printf)

# strategy, take the first amount lower than the first n items
def strategy(s, n):
  t = min(s[:n])
  for x in s[n:]:
    if x < t: break
  return x

# play the game with k items
def play(k):
  # collect total amount
  t = 0
  # consider possible orderings of the items
  ps = list(1299 + 10 * p for p in irange(1, k))
  for s in subsets(ps, size=k, select="P"):
    t += strategy(s, 1)
  return t

# consider numbers of items
for k in irange(2, inf):
  t = play(k)
  # calculate the average expected price (in 10th pence)
  d = factorial(k)
  p = div(t, d)
  if p is not None and p % 10 == 0:
    printf("k={k} -> p={p}")
    break
```
  Solution: There were 5 petrol stations.
  
  And the expected average fuel price was 132.0 pence.
  
  It doesn’t actually matter what the prices actually are, all the petrol stations could adjust their prices by the same whole number of pence, and the expected average price would be adjusted by the same amount.
  
  Analytically:
  
  As determined in Teaser 2988 we can use the formula:
  
  S(n, k) = ((2n + 1)k − n(n + 1)) (k + 1) / 2(n + 1)k
  
  To determine the mean expected value of choosing from k boxes with values from 1..k, using the strategy of picking the next box that is better than any of the first n boxes (or the last box if there are no boxes better than the first).
  
  In this case n = 1:
  
  S(1, k) = (3k − 2) (k + 1) / 4k
  
  If we award ourselves a prize of value k for selecting the cheapest fuel, and a prize of value 1 for selecting the most expensive fuel, then we have the same situation as in Teaser 2988, and if p is expected average prize value, then the corresponding expected average fuel price is (130.9 + k − p).
  
  And k is an integer, so in order to make the expected average fuel price a whole number of pence we are looking for expected average prize value that is an integer plus 0.9.
  
  i.e. when the value of 10 S(1, k) is an integer with a residue of 9 modulo 10.
  
  10 S(1, k) = (15k + 5) / 2 − (5 / k)
  
  When k is odd, the first part gives us a whole number, so we would also need the (5 / k) part to give a whole number. i.e. k = 1, 5.
  
  When k is even, the first part gives us an odd number of halves, so we also need (5 / k) to give an odd number of halves. i.e. k = 2, 10.
  
  And we are only interested in values of k > 1, so there are just 3 values to check:
  
  k = 5: 10 S(1, k) = 39
  k = 2: 10 S(1, k) = 15
  k = 10: 10 S(1, k) = 77
  
  The only value for k that gives a residue of 9 modulo 10 is k = 5.
  
  And in this case our expected average prize is p = 3.9 so our expected average fuel price is 132.0 pence.
  
  LikeLike
Jim Randell 9:44 pm on 29 May 2020 Permalink | Reply
Tags: by: John Owen ( 34 )
Teaser 3010: Putting game
From The Sunday Times, 31st May 2020 [link] [link]

A putting game has a board with eight rectangular holes, like the example (not to scale), but with the holes in a different order.

If you hit your ball (diameter 4cm) through a hole without touching the sides, you score the number of points above that hole. The sum of score and width (in cm) for each hole is 15; there are 2cm gaps between holes.

I know that if I aim at a point on the board, then the ball’s centre will arrive at the board within 12cm of my point of aim, and is equally likely to arrive at any point in that range. Therefore, I aim at the one point that maximises my long-term average score. This point is the centre of a hole and my average score is a whole number.

(a) Which hole do I aim at?

(b) Which two holes are either side of it?

[teaser3010]
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Frits, Jim Randell, and Robert Brown are discussing. Toggle Comments
- Jim Randell 9:23 am on 30 May 2020 Permalink | Reply
  The 4cm diameter ball is not allowed to touch the sides of the hole, so we can consider the outside 2cm of each hole to be “out of bounds”. Which is the same as if the ball was a point, the gaps between holes are extended 2cm in each direction (so they become 6cm wide), and each hole is reduced by a corresponding 2cm on either edge.
  
  To be sure I satisfy all the conditions this Python program constructs all possible orderings for the holes, and then for each ordering looks at possible target positions. It looks for orderings that have a unique maximum targeting point that is at the centre of a hole, and that yields an integer average score per shot. Once an ordering passes these tests we record the hole that the target is in, along with the two holes on either side. This gives a unique target + left/right value (although there are many orderings that contain the three holes appropriately positioned).
  
  It runs in 1.58s, but I think this could be improved with some additional analysis.
```
from enigma import (irange, subsets, multiset, ordered, printf)

# available holes (points)
holes = [ 1, 2, 3, 4, 5, 6, 7, 8 ]

# layout for a sequence of holes
# return a list of: (region size (in mm), points scored) pairs
def layout(ps):
  for (i, p) in enumerate(ps):
    if i > 0: yield (60, 0) # gap between holes
    yield (110 - 10 * p, p) # hole size

# generate intervals ((a, b), p) from a layout
# where a, b are absolute distances, p is number of points scored
def intervals(ss):
  a = b = 0
  for (d, p) in ss:
    if b == 0:
      (a, b) = (0, d)
    else:
      (a, b) = (b, b + d)
    yield ((a, b), p)

# analyse the layout (working in 5mm steps)
# return list of (d, (v, r)) values, where:
# d = absolute target distance
# v + r/240 = max average score
def analyse(ss):
  # determine total length
  t = sum(d for (d, _) in ss)
  rs = list()
  # consider target positions (in 5mm steps)
  for x in irange(120, t - 120, step=5):
    # consider each zone
    r = 0
    for ((a, b), p) in intervals(ss):
      # overlap?
      if b < x - 120: continue
      if a > x + 120: break
      d = min(x + 120, b) - max(x - 120, a)
      r += p * d
    rs.append((x, r))
  # find maximum average value
  m = max(r for (_, r) in rs)
  return list((x, divmod(r, 240)) for (x, r) in rs if r == m)

# collect results
m = multiset()

# choose an order for the holes
for ps in subsets(holes, size=len, select="P"):
  # but ignore the order in the diagram
  if ps == (6, 3, 8, 7, 2, 5, 1, 4): continue
  # construct the sequence of (regions, points)
  ss = list(layout(ps))
  # analyse it
  rs = analyse(ss)
  # there should only be one max position
  if len(rs) != 1: continue
  (x, (v, r)) = rs[0]
  # and the average score should be an integer
  if r != 0: continue
  # find the interval x is in
  for ((a, b), p) in intervals(ss):
    # and check it is centred
    if p > 0 and a < x < b and b - x == x - a:
      # find the holes on either side
      i = ps.index(p)
      z = ps[i - 1:i + 2]
      printf("[{ps} -> target = {x} mm, avg pts = {v}; segment = {z}]")
      m.add((p, ordered(ps[i - 1], ps[i + 1])))

# output solution
for ((x, (l, r)), n) in m.most_common():
  printf("centre = {x}, left/right = {l}/{r} [{n} solutions]")
```
  Solution: (a) You should aim at the centre of the 5 point hole. (b) The 6 point and 8 point holes are either side of the 5 point hole.
  
  Here is a diagram of the arrangement:
  
  The “out of bounds” areas are indicated in red, and we score zero points if the centre of the ball lands in these.
  
  In the 24cm section centred on the 5 point hole we see that there is a 3cm zone where we can score 6 points, a 6cm zone where we can score 5 points, and a 3cm zone where we can score 8 points.
  
  The average expected score is thus: (6×3 + 5×6 + 8×3) / 24 = 72 / 24 = 3.
  
  LikeLike
  - Jim Randell 6:28 pm on 30 May 2020 Permalink | Reply
    Here is the program adapted to just consider the centre + left/right values. It finds there is only one segment that must appear in any solution (or its reverse), and this gives the required answer. It runs in 54ms.
```
from enigma import (irange, subsets, peek, printf)

# available holes (points)
holes = [ 1, 2, 3, 4, 5, 6, 7, 8 ]

# layout for a sequence of holes (in mm)
# return a list of: (region size, points scored) pairs
def layout(ps):
  for (i, p) in enumerate(ps):
    if i > 0: yield (60, 0) # gap between holes
    yield (110 - 10 * p, p) # hole

# generate intervals ((a, b), p) from a layout
# where a, b are absolute distances, p is the number of points scored
def intervals(ss):
  a = b = 0
  for (d, p) in ss:
    if b == 0:
      (a, b) = (0, d)
    else:
      (a, b) = (b, b + d)
    yield ((a, b), p)

# analyse the layout (working in 5mm steps)
# return list of (d, (v, r)) values, where:
# d = absolute target distance
# v + r/240 = max average score
def analyse(ss):
  # determine total length
  t = sum(d for (d, _) in ss)
  rs = list()
  # consider target positions (in 5mm steps)
  for x in irange(120, t - 120, step=5):
    # consider each zone
    r = 0
    for ((a, b), p) in intervals(ss):
      # overlap?
      if b < x - 120: continue
      if a > x + 120: break
      d = min(x + 120, b) - max(x - 120, a)
      r += p * d
    rs.append((x, r))
  # find maximum average value
  m = max(r for (_, r) in rs)
  return list((x, divmod(r, 240)) for (x, r) in rs if r == m)

# find triples with integer scores max average scores at the centre of the centre hole

# choose the centre hole and left/right holes
for (X, L, R) in subsets(holes, size=3, select="P"):
  if not (L < R): continue
  # create the layout
  ss = list(layout((L, X, R)))
  # analyse it
  rs = analyse(ss)
  # there should be only one max position
  if len(rs) != 1: continue
  (x, (v, r)) = rs[0]
  # and the max average score should be an integer
  if r != 0: continue
  # and it should be centred on X
  ((a, b), _) = peek(intervals(ss), 2)
  if a < x < b and b - x == x - a:
    printf("segment = ({L}, {X}, {R}) -> target = {x} mm, avg pts = {v}")
```
    I suppose really I should demonstrate that we can construct an ordering containing the required segment that satisfies all the requirements, but as my first program finds lots I don’t think it is necessary.
    
    LikeLike
- Robert Brown 10:02 am on 30 May 2020 Permalink | Reply
  
  If the middle & adjacent slots are too small, the ±12 cm range goes into the next slot, which is undefined. This reduces the solution space and quickly leads to what appears to be a unique result. But if you reduce the smaller adjacent slot by 1 and increase the other one, it still works. This would be valid if the smaller slot was the first one on the board, so a possible duplicate result?
  
  LikeLike
  - Jim Randell 1:11 pm on 30 May 2020 Permalink | Reply
    
    @Robert: I don’t think it is possible for the target area to extend beyond the centre hole and the holes on either side. The smallest holes are 7cm and 8cm and the distance from the centre of one to the edge of the other is always greater than 12cm, so I think we only need to consider centre + left/right configurations. (Which I’m looking at to give a more efficient program).
    
    LikeLike
- Robert Brown 10:52 am on 30 May 2020 Permalink | Reply
  
  Further to above. The ‘not to scale’ drawing masks the fact that low value slots are quite wide. I assume there are 8 values as per the drawing, with corresponding slot widths.
  So there is a solution where the mid value is low, and the adjacent values can take one of two different pairs (they have the same sum), and where the total length measured from the centre of the mid slot is > 12 in each case. Definitely two sets of results, doesn’t matter where they are on the board. Maybe I’m missing something?
  
  LikeLike
  - Jim Randell 6:55 pm on 1 June 2020 Permalink | Reply
    
    I’ve updated the diagram with a scale drawing. Including a ball.
    
    LikeLike
- Robert Brown 12:17 pm on 30 May 2020 Permalink | Reply
  
  This is not a good place to have a conversation. You have my email address, but I don’t have yours!
  Both of the above sets of results work, but each post has typos which I cannot correct. An in depth exploration reveals 5 different solutions, all with 3 different L M R values between 1 & 8, and with average values of 3, 4 or 5. In each case the minimum distance from the centre of the middle value (where he aims for) is at least 15.5 cm before going into the next (unknown) slot, plenty of room for the 2cm radius ball to have it’s centre up to 12 cm from that centre. So no need for any of the slots to be the first one on the board.
  
  LikeLike
  - Jim Randell 6:55 pm on 1 June 2020 Permalink | Reply
    
    @Robert: It’s a limitation of WordPress.com I’m afraid, but if there are any corrections you would like to make, just post a followup comment, and I will use it to correct errors in the original.
    
    LikeLike
- Frits 1:14 pm on 15 August 2025 Permalink | Reply
  Based on Jim’s diagram of the arrangement.
```
# number of holes, sum score+width, gap, ball diameter, aim deviation  
N, T, G, B, Z = 8, 15, 2, 4, 12

cm = lambda pts: T - pts - 2 * G

# length of deviation range
dev_cm = 2 * Z
rng = set(range(1, N + 1))
# look for a solution L, M, R
for M in rng:
  mid_cm = cm(M)
  mid_total = mid_cm * M
  # remaining cm to use for left and right holes
  rem_cm = dev_cm - mid_cm - 2 * G - 2 * B
  rem_rng = sorted(rng - {M})
  for R in rem_rng:
    right_cm = cm(R)
    # aim as far right as possible in order to maximise the score
    right = min(rem_cm, right_cm)
    # but aim must also be at the centre of a hole
    if 2 * right != rem_cm: continue
    mid_right_total = mid_total + right * R
    _, r = divmod(mid_right_total, dev_cm)
    left_total_needed  = dev_cm - r
    
    while True:
      L, r = divmod(left_total_needed, right)
      if L >= R: break
      if not r and L != M:
        print(f"answer: (a) {M}, (b) {L} and {R}")
      left_total_needed += dev_cm
```
  LikeLike
Jim Randell 12:53 pm on 28 May 2020 Permalink | Reply
Tags: by: Graham Smithers ( 87 )
Teaser 2863: Little time
From The Sunday Times, 6th August 2017 [link] [link]

Do you have a little time to try this Teaser? I have taken a four-figure number and a six-figure number and I have consistently replaced digits by letters to give the words LITTLE and TIME.

If you take the digits of TIME and write down all the four-figure numbers which use exactly those digits in some order and then add up all those numbers, then your total will be LITTLE.

What number is TIME?

[teaser2863]
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- Jim Randell 12:54 pm on 28 May 2020 Permalink | Reply
  The puzzle text doesn’t include the usual alphametic caveat that different letters stand for different digits, only that the digits are replaced consistently by letters (i.e. the same letter is replaced by the same digit). And if there are zero digits involved the 24 rearrangements of the 4 digits of TIME will include some sequences which don’t give 4 valid digit numbers.
  
  So this Python program considers all 4 digit values for TIME, and totals only those rearrangements corresponding to 4 digit numbers. It runs in 135ms.
  
  Run: [ @repl.it ]
```
from enigma import (irange, nsplit, nconcat, subsets, ordered, cached, printf)

# update a map m consistently with (k, v) pairs in ps
def update(m, ps):
  m = m.copy()
  for (k, v) in ps:
    x = m.get(k)
    if x is None:
      m[k] = v
    elif x != v:
      return None
  return m

# sum the rearrangements that give 4 digit numbers
@cached
def total(ds):
  t = 0
  for s in subsets(ds, size=len, select="mP"):
    if s[0] != 0:
      t += nconcat(s)
  return t

# consider 4 digit numbers for TIME
for TIME in irange(1000, 9999):
  # this maps the letters T, I, M, E to digits
  ds = nsplit(TIME, 4)

  # sum the rearrangements of TIME that give 4 digit numbers
  t = total(ordered(*ds))

  # the total is a 6 digit number
  ts = nsplit(t)
  if len(ts) != 6: continue

  # map letters to digits
  m = update(dict(zip("TIME", ds)), zip("LITTLE", ts))
  if m is None: continue

  # output solution
  printf("TIME={TIME} LITTLE={t}")
```
  Solution: TIME = 3578.
  
  And so LITTLE = 153318.
  
  It turns out that using a standard alphametic interpretation, and not worrying about whether the rearrangements always give 4 digit numbers will also give this answer. So:
  
  If we write down the sum of the 24 permutations of TIME, each letter appears in each column 6 times, the result of the sum is thus:
  
  6666 × (T + I + M + E) = LITTLE
  
  Which we can solve as an alphametic expression using the [[ SubstitutedExpression() ]] solver from the enigma.py library.
  
  The following run file gives the same solution in 92ms.
```
#! python -m enigma -rr

SubstitutedExpression

"6666 * (T + I + M + E) = LITTLE"

--answer="(TIME, LITTLE)"
```
  LikeLike
- GeoffR 8:47 am on 29 May 2020 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";

var 1..9: T; var 1..9: I; var 1..9: M;
var 1..9: E; var 1..9: L; 

var 1000..9999: TIME = 1000*T + 100*I + 10*M + E;

var 100000..999999: LITTLE = 100000*L + 10000*I
+ 1000*T + 100*T + 10*L + E;

% All letters T, I, M, E appear 6 times in 24 permutations
constraint  6666 * (T + I + M + E) == LITTLE;

solve satisfy;

output ["TIME = " ++ show(TIME) 
++ "\nLITTLE = " ++ show(LITTLE)];

% TIME = 3578
% LITTLE = 153318
% ----------
% ==========
```
  Easy to check the digits in TIME give the digits in LITTLE
```
little, cnt = 0, 0

# check value of LITTLE using TIME'S digits of (3,5,7,8)
from itertools import permutations
for p in permutations((3,5,7,8)):
    t, i, m, e = p
    time = 1000*t + 100*i + 10*m + e
    # Add current <time> total
    little += time
    # increase permutation count
    cnt += 1

print(f"LITTLE = {little} in {cnt} permutations")
# LITTLE = 153318 in 24 permutations
```
  LikeLike
Jim Randell 12:10 pm on 26 May 2020 Permalink | Reply
Tags: by: J Cameron ( 2 )
Brain-Teaser 520: [Picnic allocations]
From The Sunday Times, 30th May 1971 [link]

“What’s inside it?” asked the Mole wriggling with curiosity.

“There’s cold chicken inside it”, replied the Rat briefly: “cold-tongue-cold-ham-cold-beef-pickled-gherkins-salad-french-rolls-cress-sandwiches-potted-meat-ginger-beer-lemonade-soda-water…”

“Oh, stop”, cried the Mole in ecstasies. “This is too much for one picnic. We can have another tomorrow on what’s left”.

“Do you really think so?” inquired the Rat seriously. “Let’s see. There’s only salad-pickled-gherkins-french-rolls-and-soda-water enough for two days: so if we have ham today we’ll have beef tomorrow; if we have potted meat today we’ll have cress sandwiches tomorrow; and if we have tongue today we’ll have lemonade tomorrow”.

“If we save the cress sandwiches for tomorrow we’ll have the beef today; if we keep the potted meat for tomorrow we’ll have the ginger beer today; and if we keep the lemonade for tomorrow we’ll have the ham today”. The Mole was entering into the spirit of the thing.

“In any event we’ll have the lemonade and ginger beer on different days, and likewise the beef and the chicken”, Rat shrieked excitedly.

“And if we have the chicken and cress sandwiches together, we’ll have the potted meat the day after we have the tongue”. The Mole rolled on his back at the prospect. “And we’ll eat every scrap”.

Which of the eight items did they save for the second day?

This puzzle was originally published with no title.

[teaser520]
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Jim Randell, John Crabtree, and Jim Randell are discussing. Toggle Comments
- Jim Randell 12:10 pm on 26 May 2020 Permalink | Reply
  The following Python program checks all possible assignments of days to the 8 items in question, and then checks to find which assignments satisfy all the conditions.
  
  It runs in 71ms (on my new laptop). (Internal runtime is 192µs).
```
from enigma import (subsets, implies, printf)

# choose days for the 8 items in question
for (CC, CT, CH, CB, CS, PM, GB, LE) in subsets((1, 2), size=8, select="M"):

  # check the statements:

  # R: "if we have ham today we'll have beef tomorrow"
  if not implies(CH == 1, CB == 2): continue

  # R: "if we have potted meat today we'll have cress sandwiches tomorrow"
  if not implies(PM == 1, CS == 2): continue

  # R: "if we have tongue today we'll have lemonade tomorrow"
  if not implies(CT == 1, LE == 2): continue

  # M: "if we save the cress sandwiches for tomorrow we'll have the beef today"
  if not implies(CS == 2, CB == 1): continue

  # M: "if we keep the potted meat for tomorrow we'll have the ginger beer today"
  if not implies(PM == 2, GB == 1): continue

  # M: "if we keep the lemonade for tomorrow we'll have the ham today"
  if not implies(LE == 2, CH == 1): continue

  # R: "we'll have the lemonade and ginger beer on different days ..."
  if not (LE != GB): continue

  # R: "... and likewise the beef and the chicken"
  if not (CB != CC): continue

  # M: "if we have the chicken and cress sandwiches together, we'll
  # have the potted meat the day after we have the tongue"
  if not implies(CC == CS, PM == CT + 1): continue

  # output solution
  printf("CC={CC} CT={CT} CH={CH} CB={CB} CS={CS} PM={PM} GB={GB} LE={LE}")
```
  Solution: The cold beef, potted meat and lemonade was saved for the second day.
  
  LikeLike
  - John Crabtree 8:10 pm on 25 June 2020 Permalink | Reply
    
    There are 9 statements for 8 unknowns. If Ch is true for chicken today, and ~Ch is true for chicken tomorrow,then the statements may be written as boolean equations:
    1. H.~B + ~H = 1
    2. P.~Cr + ~P = 1
    3. T.~L + ~T = 1
    4. ~Cr.B + Cr = 1
    5. ~P.G + P = 1
    6. ~L.H + L = 1
    7. L XOR G
    8. B XOR Ch
    9. [(Ch XNOR Cr).~P.T] + (Ch XOR Cr) = 1
    
    Combining equations 5, 7, 6, 1, 4, 2 and 8 gives
    1 = (~P.G + P).( L XOR G).(~L.H + L).(H.~B + ~H).(~Cr.B + Cr).(P.~Cr + ~P).(B XOR Ch)
    = ~P.G.( L XOR G).(~L.H + L).(H.~B + ~H).(~Cr.B + Cr).(P.~Cr + ~P).(B XOR Ch) +
    ….P.(P.~Cr + ~P).(~Cr.B + Cr).(H.~B + ~H).(~L.H + L).( L XOR G).(B XOR Ch)
    = ~P.G.~L.H.~B.Cr.Ch + P.~Cr.B.~H.L.~G.~Ch
    Combing this with equation 9 gives ~P.G.~L.H.~B.Cr.Ch.T = 1
    Equation 3 is satisfied, but is not required to solve the teaser by this method
    
    The items to be saved for tomorrow are given by ~B,~L and ~P, ie cold Beef, Lemonade and Potted meat.
    
    Starting with any assumption other than T or ~T and working through the equations 1-2, and 4-8 in some order and finally 9 gives either a consistent solution,or a contradiction (reductio ad absurdum). If one starts by assuming T, then equation 3 is used before the others. If ~T is assumed, then equation 9 gives Ch.~Cr + ~Ch.Cr = 1, and both conditions have to be checked to show that ~T is not true
    
    LikeLike
  - Jim Randell 3:12 pm on 15 November 2025 Permalink | Reply
    Or using the [[ SubstitutedExpression ]] solver from the enigma.py library.
    
    The code generated from the following run file has an internal runtime of 75µs.
```
#! python3 -m enigma -rr

SubstitutedExpression

--distinct=""
--digits="1,2"

--macro="@CC = {A}"  # cold chicken
--macro="@CT = {B}"  # cold tongue
--macro="@CH = {C}"  # cold ham
--macro="@CB = {D}"  # cold beef
--macro="@CS = {E}"  # cress sandwiches
--macro="@PM = {F}"  # potted meat
--macro="@GB = {G}"  # ginger beer
--macro="@LE = {H}"  # lemonade

# R: "if we have ham today we'll have beef tomorrow"
"implies(@CH == 1, @CB == 2)"

# R: "if we have potted meat today we'll have cress sandwiches tomorrow"
"implies(@PM == 1, @CS == 2)"

# R: "if we have tongue today we'll have lemonade tomorrow"
"implies(@CT == 1, @LE == 2)"

# M: "if we save the cress sandwiches for tomorrow we'll have the beef today"
"implies(@CS == 2, @CB == 1)"

# M: "if we keep the potted meat for tomorrow we'll have the ginger beer today"
"implies(@PM == 2, @GB == 1)"

# M: "if we keep the lemonade for tomorrow we'll have the ham today"
"implies(@LE == 2, @CH == 1)"

# R: "we'll have the lemonade and ginger beer on different days ..."
"@LE != @GB"

# R: "... and likewise the beef and the chicken"
"@CB != @CC"

# M: "if we have the chicken and cress sandwiches together, we'll
# have the potted meat the day after we have the tongue"
"implies(@CC == @CS, @PM == @CT + 1)"

# output day for each dish
--template="CC=@CC CT=@CT CH=@CH CB=@CB CS=@CS PM=@PM GB=@GB LE=@LE"
--solution=""
--verbose="1-H"
```
    LikeLike

Jim Randell 5:08 pm on 22 May 2020 Permalink | Reply
Tags: by: Victor Bryant ( 148 )

Teaser 3009: Head count

From The Sunday Times, 24th May 2020 [link] [link]

My grandson and I play a simple coin game. In the first round we toss seven coins and I predict how many “heads” there will be whilst my grandson predicts the number of “tails”. After the tossing I score a point for each head plus a bonus of ten if my prediction was correct — my grandson scores likewise for the tails. We then repeat this with six coins, then five, and so on down to a single coin. After each round we keep cumulative totals of our scores. In one game, for over half of the pairs of cumulative scores, my grandson’s total was like mine but with the digits in reverse order. In fact he was ahead throughout and at one stage his cumulative total was double mine — and he had an even bigger numerical lead than that on just one earlier occasion and one later occasion.

List the number of heads in each of the seven rounds.

[teaser3009]

Jim Randell 11:03 pm on 22 May 2020 Permalink | Reply

Originally I missed the significance of the word “numerical”, and got no solutions. But careful interpretation of the puzzle text does lead to a single solution.

I wrote a recursive function that keeps track of all the conditions as it goes along.

The following Python 3 program runs in 320ms.

Run: [ @repl.it ]

from itertools import product
from enigma import irange, nsplit, join, cached, printf

# does A mirror B?
mirror = cached(lambda A, B: nsplit(A) == nsplit(B)[::-1])

# play the game starting with a round of k coins,
# where A plays heads, B plays tails, and B is always ahead
# ms = number of points B is ahead of A
# s = index in ms when B = 2A
# rv = count number of scores that are reverse of each other
def play(k, tA=0, tB=0, ms=[], s=None, rv=0, ss=[]):
  # are we done?
  if k == 0:
    if s is not None and rv > 3:
      # there must be exactly 1 round after s where B is further ahead
      if sum(x > ms[s] for x in ms[s + 1:]) == 1:
        yield ss
  else:
    # consider the number of heads, and predictions for heads and tails
    for (n, h, t) in product(irange(0, k), (0, 1), (0, 1)):
      # calculate the points for each player
      a = n + h * 10
      b = k - n + t * 10
      # and the total points
      A = tA + a
      B = tB + b
      m = B - A
      # B is always ahead
      if not(m > 0): continue
      # look for cases where B = 2A
      s_ = s
      if B == 2 * A:
        # it only happens once
        if s is not None: continue
        # there must be exactly 1 previous round where B is further ahead
        if not(sum(x > m for x in ms) == 1): continue
        s_ = len(ms)
      # are A, B mirrored scores?
      rv_ = rv + mirror(A, B)
      # in the final 4 rounds we must have encountered some mirrored scores
      if k < 5 and rv_ < 5 - k: continue
      # play the remaining rounds
      yield from play(k - 1, A, B, ms + [m], s_, rv_, ss + [(n, h, t, A, B)])

# play the game, starting with 7 coins
for ss in play(7):
  # output the rounds
  (pA, pB) = (0, 0)
  s = list(i for (i, (n, h, t, A, B)) in enumerate(ss) if B == 2 * A)[0]
  for (i, (n, h, t, A, B)) in enumerate(ss):
    k = 7 - i
    fs = [ f"lead = {B - A}" ]
    if i == s: fs.append("double")
    if mirror(A, B): fs.append("mirror")
    printf(
      "{k} coins: heads={n}; predictions = ({h}, {t}); points = ({a}, {b}); totals = ({A},  {B}); {fs}",
      a=A - pA, b=B - pB, fs=join(fs, sep="; "),
    )
    (pA, pB) = (A, B)
  printf()

Solution: The number of heads in each of the rounds was: 0, 2, 4, 4, 3, 1, 1.

The full description of the rounds is:

k   h  t  |  H   T  |  a   b  |  A   B
7:  0  7  |  -  10  |  0  17  |  0  17  [lead >16]
6:  2  4  | 10   -  | 12   4  | 12  21  [mirror]
5:  4  1  |  -  10  |  4  11  | 16  32  [double, lead =16]
4:  4  0  |  -   -  |  4   0  | 20  32
3:  3  0  |  -   -  |  3   0  | 23  32  [mirror]
2:  1  1  | 10  10  | 11  11  | 34  43  [mirror]
1:  1  0  |  -  10  |  1  10  | 35  53  [mirror, lead >16]

(k = number of coins; h, t = number of heads/tails; H, T = bonus points; a, b = points in round; A, B = cumulative totals)

However, keeping the cumulative totals always using 2 digits gives us three further solutions where the totals 03 and 30 are used as mirrored pairs:

k   h  t  |  H   T  |  a   b  |  A   B
7:  1  6  |  -  10  | 01  16  | 01  16
6:  2  4  |  -  10  | 02  14  | 03  30  [mirror, lead >16]
5:  3  2  | 10   -  | 13  02  | 16  32  [double, lead =16]
4:  4  0  |  -   -  | 04  00  | 20  32
3:  3  0  |  -   -  | 03  00  | 23  32  [mirror]
2:  1  1  | 10  10  | 11  11  | 34  43  [mirror]
1:  1  0  |  -  10  | 01  10  | 35  53  [mirror, lead >16]


k   h  t  |  H   T  |  a   b  |  A   B
7:  2  5  |  -  10  | 02  15  | 02  15
6:  1  5  |  -  10  | 01  15  | 03  30  [mirror, lead >16]
5:  3  2  | 10   -  | 13  02  | 16  32  [double, lead =16]
4:  4  0  |  -   -  | 04  00  | 20  32
3:  3  0  |  -   -  | 03  00  | 23  32  [mirror]
2:  1  1  | 10  10  | 11  11  | 34  43  [mirror]
1:  1  0  |  -  10  | 01  10  | 35  53  [mirror, lead >16]


k   h  t  |  H   T  |  a   b  |  A   B
7:  3  4  |  -  10  | 03  14  | 03  14
6:  0  6  |  -  10  | 00  16  | 03  30  [mirror, lead >16]
5:  3  2  | 10   -  | 13  02  | 16  32  [double, lead =16]
4:  4  0  |  -   -  | 04  00  | 20  32
3:  3  0  |  -   -  | 03  00  | 23  32  [mirror]
2:  1  1  | 10  10  | 11  11  | 34  43  [mirror]
1:  1  0  |  -  10  | 01  10  | 35  53  [mirror, lead >16]

The program will find all 4 solutions if we replace the calls to [[ nsplit(x) ]] with [[ nsplit(x, 2) ]] in the function [[ mirror() ]] (line 5).

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Robert Brown 11:57 am on 24 May 2020 Permalink | Reply

Turns out there’s a simple manual solution. After each section (7,6,5 etc) there’s a total sum for all heads & tails. Last digit is different in each case. Adding bonuses makes no difference.
Need to find 4 sections that end in reversible numbers, so just look for reversible pairs where the sum of the pair has the same last digit. Only works for 4 sections, QED.

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Jim Randell 12:59 pm on 21 May 2020 Permalink | Reply
Tags: by: Victor Bryant ( 148 )
Teaser 2864: Sequence of squares
From The Sunday Times, 13th August 2017 [link] [link]

I started with a rectangle of paper. With one straight cut I divided it into a square and a rectangle. I put the square to one side and started again with the remaining rectangle. With one straight cut I divided it into a square and a rectangle. I put the square (which was smaller than the previous one) to one side and started again with the remaining rectangle. I kept repeating this process (discarding a smaller square each time) until eventually the remaining rectangle was itself a square and it had sides of length one centimetre. So overall I had divided the original piece of paper into squares. The average area of the squares was a two-figure number of square centimetres.

What were the dimensions of the original rectangle?

[teaser2864]
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- Jim Randell 1:00 pm on 21 May 2020 Permalink | Reply
  If we start with the 1×1 cm square and replace the removed squares, we find the sequence of sizes of squares is:
  
  1, 1, 2, 3, 5, 8, 13, 21, …
  
  i.e. a Fibonacci sequence. [ @Wikipedia ]
  
  So we can start with the two 1×1 cm squares and build up the original rectangle square by square, until find one where the mean area of the squares is a two digit integer as required.
  
  This Python program runs in 82ms.
```
from enigma import printf

# initial a x b rectangle, n = number of squares, A = total Area
(a, b, n, A) = (1, 1, 2, 2)
while True:
  # calculate the mean area of the squares
  (m, r) = divmod(A, n)
  if m > 99: break
  # construct the rectangle
  (a, b) = (b, a + b)
  if r == 0 and m > 9:
    # output solution, when m is a 2 digit integer
    printf("[n={n}] rectangle = {a} x {b}, mean area = {m}")
  # move on to the next square
  A += b * b
  n += 1
```
  Solution: The original rectangle was 13 cm × 21 cm.
  
  So in total 6 cuts were made, producing 7 squares.
  
  The total area of the 7 squares is 273 sq cm, so the mean area is 39 sq cm.
  
  Manually:
  
  If:
  
  F(n) = nth Fibonacci number
  S(n) = sum of squares of F(n) = F(n) F(n + 1)
  M(n) = mean of squares of F(n) = F(n) F(n + 1) / n
  
  Then we can calculate:
```
 n  F(n)  S(n)  M(n)
 1    1     1    1
 2    1     2    1
 3    2     6    2
 4    3    15    3.75  
 5    5    40    8
 6    8   104   17.3...
 7   13   273   39
 8   21   714   89.25
 9   34  1870  207.7...
10   55  ...
```
  And the answer is apparent.
  
  If we draw a quarter circle through each square we can make a Fibonacci Spiral:
  
  LikeLiked by 1 person
Jim Randell 5:13 pm on 15 May 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 65 )
Teaser 3008: Three-fruit fractions
From The Sunday Times, 17th May 2020 [link] [link]

The owner of the old curiosity shop repaired an antique mechanical fruit machine having three wheels of identical size and format. Afterwards each wheel was independently fair, just as when new. Each wheel’s rim had several equal-sized zones, each making a two-figure whole number of degrees angle around the rim. Each wheel had just one zone showing a cherry, with other fruits displayed having each a different single-figure number (other than one) of zone repetitions.

Inside the machine were printed all the fair chances (as fractions) of getting three of the same fruit symbol in one go. Each of these fractions had a top number equal to 1 and, of their bottom numbers, more than one was odd.

What was the bottom number of the chance for three cherries?

[teaser3008]
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- Jim Randell 7:26 am on 16 May 2020 Permalink | Reply
  I assumed the wheels are identical duplicates of each other, which gives a unique solution.
  
  This Python 3 program runs in 49ms.
```
from enigma import (Rational, irange, divisors_pairs, seq2str, printf)

Q = Rational()

# decompose <t> into numbers between 2 and 9
def decompose(t, m=2, M=9, s=[]):
  if not (t < m or t > M):
    yield s + [t]
  for x in irange(m, min(t - m, M)):
    yield from decompose(t - x, x + 1, M, s + [x])

# each wheel is divided into n equal sized divisions
# each spanning d degrees
for (n, d) in divisors_pairs(360, every=1):
  if d < 10: break
  if d > 99: continue

  # probability of 3 cherries
  p = Q(1, n)**3

  # consider the number of other fruits (all different and between 2 and 9)
  for ns in decompose(n - 1):
    # calculate the probabilities of getting 3 of each of the other fruits
    ps = list(Q(x, n)**3 for x in ns)
    # probabilities are all 1/N
    if not all(x.numerator == 1 for x in ps): continue
    # and more than one N is odd
    if sum(x.denominator % 2 == 1 for x in ps) < 2: continue

    # output solution
    printf("{n} divisions, {d} degrees -> 1 + {ns}; p={p}, ps={ps}", ns=seq2str(ns), ps=seq2str(ps))
```
  Solution: There is a 1 / 5832 chance of getting 3 cherries.
  
  Each wheel is divided into 18 sectors. Each sector subtends 20°.
  
  There are 4 fruits, each is allocated 1, 2, 6, 9 sectors on each wheel.
  
  The probabilities for each of the 4 fruits are: 1/5832 (= 1/18³), 1/729 (= 1/9³), 1/27 (= 1/3³), 1/8 (= 1/2³).
  
  LikeLike
Jim Randell 10:42 am on 12 May 2020 Permalink | Reply
Tags: by: RBJT Allenby ( 4 )
Teaser 2865: Seventh heaven?
From The Sunday Times, 20th August 2017 [link] [link]

I have a modern painting by the surrealist artist Doolali. It is called “Seventh Heaven” and it consists of a triangle with green sides and a red spot on each of its sides. The red spots are one seventh of the way along each side as you pass clockwise around the triangle. Then each of the red spots is joined by a straight blue line to the opposite corner of the triangle. These three blue lines create a new triangle within the original one and the new triangle has area 100 sq cm.

What is the area of the green triangle?

[teaser2865]
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- Jim Randell 10:43 am on 12 May 2020 Permalink | Reply
  We have encountered similar problems to this before (see: Enigma 1313, Enigma 320, Enigma 1076).
  
  The generalisation of this puzzle is known as Routh’s Theorem [ @wikipedia ], which states that the ratio of the area of the central triangle to the original triangle, where the sides are divided in the ratios 1:x, 1:y, 1:z is given by the formula:
  
  R = (xyz − 1)² / ((xz + x + 1)(xy + y + 1)(yz + z + 1))
  
  I made a note giving the areas of the various subdivisions here [ rouths-theorem.pdf ].
  
  For this puzzle we have: x = y = z = 6, which gives the ratio as:
  
  R = (x − 1)² / (x² + x + 1) = 25/43
  
  The smaller triangle has an area of 100, so the larger triangle has an area of: 100 × 43/25 = 172.
  
  Solution: The area of the green triangle is 172 sq cm.
  
  Or using a program:
```
from enigma import (rational, printf)

# ratio ABC/XYZ in Routh's theorem
def routh(x, y, z):
  a = x * y * z - 1
  b = (x * y + y + 1) * (y * z + z + 1) * (z * x + x + 1)
  return rational(a * a, b)

# compute the ratio of the area of the inner and outer triangles
R = routh(6, 6, 6)

# output the solution, given the inner triangle has area 100
printf("outer triangle = {x} [R={R}]", x=100 / R)
```
  LikeLike
Jim Randell 3:59 pm on 7 May 2020 Permalink | Reply
Tags: by: Howard Williams ( 45 )
Teaser 3007: Paving stones
From The Sunday Times, 10th May 2020 [link] [link]

James has decided to lay square block paving stones on his rectangular patio. He has calculated that starting from the outside and working towards the middle that he can lay a recurring concentric pattern of four bands of red stones, then three bands of grey stones, followed by a single yellow stone band. By repeating this pattern and working towards the centre he is able to finish in the middle with a single line of yellow stones to complete the patio.

He requires 402 stones to complete the first outermost band. He also calculates that he will require exactly 5 times the number of red stones as he does yellow stones.

How many red stones does he require?

[teaser3007]
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- Jim Randell 4:33 pm on 7 May 2020 Permalink | Reply
  Assuming an x by y rectangle, with x > y.
  
  If there are k repeats of the (red, grey yellow) bands, then y = 16k − 1.
  
  And the first band of red stones requires 2x + 2y − 4 = 402 stones, so x + y = 203.
  
  Here is a constructive program that counts the number of stones in each band. It runs in 78ms.
```
from enigma import irange, inf, printf

# calculate the number of stones with <k> repeats
# and the initial band being <x> by <y>
def stones(k, y, x):
  d = dict(R=0, G=0, Y=0)
  for i in ("RRRRGGGY" * k):
    if y < 2:
      d[i] += x * y
      break
    d[i] += 2 * (x + y - 2)
    x -= 2
    y -= 2
  return (d['R'], d['G'], d['Y'])

# consider the number of repeats of the bands
for k in irange(1, inf):
  y = 16 * k - 1
  x = 203 - y
  if not(x > y): break
 
  # calculate the number of stones needed
  (R, G, Y) = stones(k, y, x)
 
  # output solution
  if 5 * Y == R:
    printf("k={k}, y={y} x={x}; R={R} G={G} Y={Y}")
```
  Solution: James requires 5520 red stones.
  
  The patio consists of 6 red/grey/yellow layers, finishing with a single row of 14 yellow stones:
  
  The number of stones required is: red = 5520, grey = 3636, yellow = 1104; total = 10260.
  
  The outer band measures 108×95 stones.
  
  LikeLike
- Jim Randell 9:33 am on 10 May 2020 Permalink | Reply
  Each band requires 8 fewer stones than the previous band, so we can just start with the outermost band (402 stones) and work our way in. When we get to the number blocks for the yellow band we can adjust the number to account for a single line to see if we have reached the middle of the design, and if not carry on. We don’t need to do any analysis to work out the dimensions of the patio, or consider the number of repeats.
  
  This gives us a simpler, and slightly shorter, program.
  
  Run: [ @repl.it ]
```
from enigma import irange, chunk, printf

# number of blocks in outer band
n = 402

# record Red, Grey, Yellow blocks required
R = G = Y = 0

# work inwards in chunks of 8 bands
# each band has 8 less blocks than the previous band
for s in chunk(irange(n, 1, step=-8), 8):
  if len(s) < 8: break
  (r1, r2, r3, r4, g1, g2, g3, y) = s
  R += r1 + r2 + r3 + r4
  G += g1 + g2 + g3
  # the middle is a single row
  m = 1 + y // 2
  if 5 * (Y + m) == R:
    printf("R={R} G={G} Y={Y}", Y=Y + m)
  # otherwise, carry on
  Y += y
```
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Jim Randell 10:54 am on 7 May 2020 Permalink | Reply
Tags: by: Danny Roth ( 89 ), flawed ( 56 )
Teaser 2869: Cubic savings
From The Sunday Times, 17th September 2017 [link] [link]

In 2009 George and Martha had a four-figure number of pounds in a special savings account (interest being paid into a separate current account). At the end of the year they decided to give some of it away, the gift being shared equally among their seven grandchildren, with each grandchild getting a whole number of pounds. At the end of the following year they did a similar thing with a different-sized gift, but again each grandchild received an equal whole number of pounds. They have repeated this procedure at the end of every year since.

The surprising thing is that, at all times, the number of pounds in the savings account has been a perfect cube.

What is the largest single gift received by any grandchild?

[teaser2869]
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GeoffR and Jim Randell are discussing. Toggle Comments
- Jim Randell 10:54 am on 7 May 2020 Permalink | Reply
  This puzzle is marked as flawed, as there are two possible solutions.
  
  I assumed the number of grandchildren remained constant (at 7) during the eight years in question (2009 – 2016).
  
  The amounts in the savings account are perfect cubes, that differ by multiples of 7, so we can collect cubes by their residue modulo 7, and consider the sets for each residue to look for 9 amounts that satisfy the remaining conditions.
  
  This Python program runs in 76ms.
  
  Run: [ @replit ]
```
from enigma import (group, powers, mod, subsets, tuples, printf)

# group the cubes (up to 4 digits) in descending order, by their residue mod 7
cubes = group(powers(21, 1, 3, step=-1), by=mod(7))

# choose values for the eight gifts made (2009 - 2016)
for s in cubes.values():
  for vs in subsets(s, size=9):
    if s[0] < 1000: continue
    # and calculate the sequence of gifts
    gs = tuple((a - b) // 7 for (a, b) in tuples(vs, 2))
    if len(set(gs)) != len(gs): continue # gift amounts are all different
    # output solutions
    m = max(gs)
    printf("{vs} -> {gs}, max = {m}")
```
  Solution: There are two solutions. The largest amount is received by a grandchild is £ 292, or £ 388.
  
  If we start with 5832 (= 18³) in the savings account, and then give presents of (248, 103, 292, 86, 31, 64, 8, 1) to each grandchild then the amounts remaining in the savings account are:
  
  5832 (= 18³)
  4096 (= 16³)
  3375 (= 15³)
  1331 (= 11³)
  729 (= 9³)
  512 (= 8³)
  64 (= 4³)
  8 (= 2³)
  1 (= 1³)
  
  However, starting with 8000 (= 20³) and giving (163, 278, 388, 67, 104, 112, 13, 14), leaves amounts of:
  
  8000 (= 20³)
  6859 (= 19³)
  4913 (= 17³)
  2197 (= 13³)
  1728 (= 12³)
  1000 (= 10³)
  216 (= 6³)
  125 (= 5³)
  27 (= 3³)
  
  One way to rescue the puzzle is to exclude 1 as an amount given to the grandchildren (or remaining in the account).
  
  Another way is to require that none of the amounts given to any grandchild is itself a perfect cube.
  
  Either of these restrictions eliminate the first solution, leaving a unique answer of £ 388.
  
  LikeLike
- GeoffR 8:28 am on 10 May 2020 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";

% Yearly account balances
var 1000..9999: S1; var 1..9999: S2; var 1..9999: S3;
var 1..9999: S4; var 1..9999: S5; var 1..9999: S6;
var 1..9999: S7; var 1..9999: S8; var 1..9999: S9;

% yearly gifts 2009 - 2016
var 1..1400: G1; var 1..999: G2; var 1..999: G3;
var 1..999: G4; var 1..999: G5; var 1..999: G6;
var 1..999: G7; var 1..999: G8; 

% All yearly gifts are different values
constraint all_different ([G1, G2, G3, G4, G5, G6, G7, G8]);

% Set of cubes up to a maximum of four digits
set of int: cube = {n * n * n | n in 1..21};

% S1 is largest yearly account balance
constraint increasing([S9,S8,S7,S6,S5,S4,S3,S2,S1]);

% All yearly account balances are cubes
constraint S1 in cube /\  S2 in cube /\  S3 in cube 
/\ S4 in cube /\  S5 in cube /\  S6 in cube
/\ S7 in cube /\  S8 in cube /\ S9 in cube;

% Yearly amounts to distribute amongst seven grandchildren
constraint (S1 - S2) mod 7 == 0 /\ (S2 - S3) mod 7 == 0 /\ 
(S3 - S4) mod 7 == 0 /\ (S4 - S5) mod 7 == 0
/\ (S5 - S6) mod 7 == 0 /\ (S6 - S7) mod 7 == 0 
/\ (S7 - S8) mod 7 == 0 /\ (S8 - S9) mod 7 == 0;

% Yearly total gifts to each of seven grandchildren
constraint G1 == (S1 - S2) div 7 /\ G2 == (S2 - S3) div 7
/\ G3 == (S3 - S4) div 7 /\ G4 == (S4 - S5) div 7
/\ G5 == (S5 - S6) div 7 /\ G6 == (S6 - S7) div 7
/\ G7 == (S7 - S8) div 7 /\ G8 == (S8 - S9) div 7;

% Maximum yearly gift
var 10..1000: maxv;
maxv = max([G1, G2, G3, G4, G5, G6, G7, G8]);
 
solve satisfy;

output ["Yearly account balances: " ++ 
show([S1, S2, S3, S4, S5, S6, S7, S8, S9])] ++ 
["\nYearly gifts(£): " ++ show ([G1, G2, G3, G4, G5, G6, G7, G8])]
++ ["\nMax gift = " ++ show(maxv)] ;

% Yearly account balances: [5832, 4096, 3375, 1331, 729, 512, 64, 8, 1]
% Yearly gifts(£): [248, 103, 292, 86, 31, 64, 8, 1]
% Max gift = 292
% ----------
% Yearly account balances: [8000, 6859, 4913, 2197, 1728, 1000, 216, 125, 27]
% Yearly gifts(£): [163, 278, 388, 67, 104, 112, 13, 14]
% Max gift = 388
% ----------
% ==========
```
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Jim Randell 9:07 am on 5 May 2020 Permalink | Reply
Tags: by: Andrew Skidmore ( 89 )

Teaser 2870: Lucky dip

From The Sunday Times, 24th September 2017 [link] [link]

My bank PIN consists of four different digits in decreasing order. I used this PIN to help me choose my six lottery numbers. I wrote down all the two-figure numbers that used two different digits from the PIN. Just six of those numbers were in the range from 10 to 49 and so they were my lottery choices. In fact the sum of the six is a perfect square. If you knew that square it would now be possible to work out my PIN.

What is my PIN?

[teaser2870]

Jim Randell 9:08 am on 5 May 2020 Permalink | Reply

This Python program runs in 87ms.

Run: [ @repl.it ]

from enigma import (subsets, irange, is_square, filter_unique, unpack, nconcat, printf)

# record (<PIN digits>, <2 digit numbers>, <square root of the sum of 2 digit numbers>)
rs = list()

# choose the 4 digits (in decreasing order)
for ds in subsets(irange(9, 0, step=-1), size=4, select="C"):
  # collect 2-digt numbers between 10 and 49
  ns = tuple(n for n in map(nconcat, subsets(ds, size=2, select="P")) if 9 < n < 50)
  # there are exactly 6 of them
  if len(ns) != 6: continue
  # and their sum is a perfect square
  s = is_square(sum(ns))
  if s is None: continue
  printf("[ds = {ds}, ns = {ns}, s = {s}]")
  rs.append((ds, ns, s))

# find PIN digits unique by s
(rs, _) = filter_unique(rs, unpack(lambda ds, ns, s: s), unpack(lambda ds, ns, s: ds))

# output solutions
for (ds, ns, s) in rs:
  printf("PIN = {pin}, numbers = {ns}, sum = {s}^2", pin=nconcat(ds), ns=sorted(ns))

Solution: The PIN is 5420.

There are 5 candidate PINs:

5420, numbers = (45, 42, 40, 25, 24, 20), sum = 14²
7320, numbers = (37, 32, 30, 27, 23, 20), sum = 13²
7410, numbers = (47, 41, 40, 17, 14, 10), sum = 13²
8621, numbers = (28, 26, 21, 18, 16, 12), sum = 11²
9521, numbers = (29, 25, 21, 19, 15, 12), sum = 11²

Only the first of these has a unique sum.

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GeoffR 5:04 pm on 5 May 2020 Permalink | Reply

from itertools import permutations

from collections import defaultdict
D = defaultdict(list)

def is_sq(n):
  c = int(n**(1/2) + 0.5)
  return (c**2 == n) 

for P1 in permutations (range(10),4):
  a, b, c, d = P1
  # choose four different digits in decreasing order
  if a > b > c > d:
    t = 1000*a + 100*b + 10*c + d   # PIN
    # find numbers in range 10...49 from 4 digits
    for P2 in permutations((a, b, c, d), 2):
      x, y = P2
      n = 10 * x + y
      if 10 <= n <= 49:
        D[t] += [n]

for k,v in D.items():
  # find 6 lottery numbers
  if len(v) == 6:
    # check sum of 6 numbers is a square
    t2 = v[0] + v[1] + v[2] + v[3] + v[4] + v[5]
    if is_sq(t2):
      print(k, v, t2)

# 5420 [45, 42, 40, 25, 24, 20] 196 << PIN = 5420
# 7320 [37, 32, 30, 27, 23, 20] 169
# 7410 [47, 41, 40, 17, 14, 10] 169
# 8621 [28, 26, 21, 18, 16, 12] 121
# 9521 [29, 25, 21, 19, 15, 12] 121

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Jim Randell 10:47 am on 3 May 2020 Permalink | Reply
Tags: by: Sir James Joint ( 6 )
Brain-Teaser 519: Badberg clock
From The Sunday Times, 23rd May 1971 [link]

At the village of Badberg, hidden away in the Alps, there is a town clock of a certain antiquity. The maker was a wealthy but inexpert amateur. After the ornate instrument had been installed it was found that the great hands stood still during the equal intervals between each stroke of the hour on the massive bell. Naturally, the clock was always slow.

The placid villagers became accustomed to the erratic behaviour of their timepiece; only after the death of its donor did his nephew dare to tackle the problem. Finding it impossible to alter the striking mechanism, he ingeniously modified the movement to run at a higher constant speed so that the hands showed the correct time at least when the clock struck certain hours.

Unfortunately, the hands still stop for the same period between successive strokes of the bell, but the villages can now see and hear the correct time every six hours.

At what hours does the clock make its first stroke correctly?

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

[teaser519]
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- Jim Randell 10:47 am on 3 May 2020 Permalink | Reply
  The clock only pauses between the strokes, so at 1 o’clock it doesn’t pause. At 2 o’clock it pauses for 1 interval. At 3 o’clock, 2 intervals. And so on until 12 o’clock when it pauses for 11 intervals. So the total time lost in a 12 hour period is 66 intervals.
  
  And the speed of the clock is adjusted so that the time lost in these 66 intervals is made up during the 12 hours. So we need to find adjacent 6 hour periods each containing 33 intervals. And there are only 6 pairs of periods to consider.
  
  Run: [ @repl.it ]
```
from enigma import irange, tuples, printf

# "clock" arithmetic
clock = lambda n, m=12: (n % m) or m

# interval count for each hour
ps = list(irange(0, 11))

# find a consecutive sequence of 6 intervals that is half the total
t = sum(ps)
for (i, s) in enumerate(tuples(ps, 6)):
  if sum(s) * 2 == t:
    (h1, h2) = (i + 1, i + 7)
    printf("{h1}, {h2}", h1=clock(h1), h2=clock(h2))
```
  Solution: The clock is correct at 4 o’clock and 10 o’clock.
  
  In striking 4, 5, 6, 7, 8, 9 o’clock the clock pauses for 3 + 4 + 5 + 6 + 7 + 8 = 33 intervals.
  
  In striking 10, 11, 12, 1, 2, 3 o’clock the clock pauses for 9 + 10 + 11 + 0 + 1 + 2 = 33 intervals.
  
  LikeLike

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