A Cricket Puzzle
Jeremy Piets
puzzle. The puzzle was promptly provided by Nick Wedd (and I have
reproduced it at the end of this post).
At the time, I had a crack at the puzzle but gave up a couple of weeks
later having failed to solve it. Months passed, then last weekend during
a clean out I rediscovered the puzzle, which I had printed out. I sat
down again and had another go. This time I solved it.
What I want to do, though, is see if there is more than one solution -
or more than one way of arriving at the intended solution. So what I am
asking is:
(a) Has anyone solved this puzzle, or
(b) Would anyone like to have a go at solving it now
and compare their methods and answers?
Jeremy.
Here's the puzzle:
This problem was composed by Sir Arthur Eddington, Professor of
Astronomy at Cambridge, in 1938.
Extract from the score of a cricket match between Eastshire and
Westshire.
EASTSHIRE - SECOND INNINGS
A.A.Atkins 6
Bodkins 8
D.D.Dawkins 6
Hawkins 6
Jenkins (J.) 5
Larkins 4
Meakins 7
Hon.P.P.Perkins 11
Capt.S.S.Simkins 6
Tomkins 0
Wilkins 1
Extras 0
---
Total 60
---
BOWLING
Overs Maidens Runs Wickets
Pitchwell 12.1 2 14 8
Speedwell 6 0 15 1
Tosswell 7 5 31 1
The score was composed entirely of singles and fours. There were no
catches, no-balls or short runs.
Speedwell and Tosswell each had only one spell of bowling. Pitchwell
bowled the first over, Mr. Atkins taking the first ball. Speedwell was
the other opening bowler.
Whose wickets were taken by Speedwell and Tosswell?
Who was not out?
What was the score at the fall of each wicket?
Daniel Ashworth
Here's my worked solution:
(1) General observations:
- No extras, catches, run-outs or short-runs prevents batsman
changing ends except through scoring a single.
- Only the striker can have been dismissed since the wickets column
in the bowling figures totals ten and
the only methods of dismissal of the non-striker- Run Out and
Obstructing the Field- are not credited to the bowler.
- I've assumed 6 ball overs, though whether that is justifed given
the date at which the puzzle was created I'm unsure.
- Let the end which Atkins faced the first ball be called end A and
the other end B.
(2) Pitchwell bowled the 1st over. Speedwell bowled the 2nd. Since
Speedwell only has one spell he must also have bowled the 4th, 6th,
8th, 10th and 12th overs.
(3) Pitchwell bowled the final (26th) over of the innings (since he
has a partially completed over). Since Speedwell had finished his
spell by this point, Tosswell must have bowled the 25th over. Since
Tosswell only has one spell he must also have bowled the 23rd, 21st,
19th, 17th, 15th and 13th. Pitchwell must bowl all the other overs. So
the order in which the bowlers bowled and the ends to which they were
bowling are:
PSPSPSPSPSPSTPTPTPTPTPTPTP
ABABABABABABABABABABABABAB
(4) Tosswell bowled 7 overs, 5 of which were maidens. So he conceded
all his 31 runs in the other 2 overs. 31 runs can only be made up of 7
fours and 3 singles (any other combination eg. 6 fours and 7 singles
involves more than the maximum possible 12 scoring shots). Now
Tosswell can bowl to a maximum of five different batsmen (since there
can be two in during his first non-maiden over, two in during the
second plus one more since Tosswell gets a wicket). The only way that
these boundaries can be allocated to the batsmen is 2 to Bodkins, 2 to
Perkins and 1 apiece to three other batsmen.
(5) Since Bodkins' entire score is accounted for the fours he scores
off Tosswell, and Tosswell did not start bowling until the 13th over
of the innings, Bodkins cannot be out until at least the 13th over and
cannot have scored before that over.
(6) At least 19 runs must have been scored prior to Bodkins'
dismissal, since Speedwell has completed his spell before then
(costing 15) and Pitchwell had bowled 6 overs (costing at least 4
runs, since he has only 2 maidens). This requires Atkins, Dawkins and
Hawkins at least to be out by then.
(7) Atkins scored 6 runs. Since his partner hadn't rotated the strike,
he must have been dismissed at the end at which he started batting
(end A) and therefore was dismissed by Pitchwell. FOW (1) = 6
(8) Dawkins therefore started batting at end A and by the same
reasoning was dismissed by Pitchwell at that end. FOW (2) = 12. The
same goes for Hawkins. FOW (3) = 18. Jenkins also enters at end A.
(9) At most 25 runs could have been scored in the first 12 overs,
derived from the 15 conceded by Speedwell and the maximum of 10 that
Pitchwell could have conceded in his first 6 overs (any more would be
inconsistent with his number of maidens). So at most one more batsman
(Jenkins) could have been out in the first 12 overs. Since Speedwell
took a wicket that must have been the case. So Speedwell took Jenkins'
wicket. FOW (4) = 23. Jenkins must have been out at end B (since
Speedwell was bowling), so Larkins entered at end B.
(10) Now reviewing (4): Larkins, Meakins & Simkins must be the batsmen
who scored 1 boundary apiece, since they are the only batsmen who are
not out before the 13th over who scored enough runs to allow them to
have hit a boundary.
(11) Since Larkins and Bodkins can't have changed ends before one of
them is out since each scores only boundaries, the next wicket had to
be Bodkins out at end A to Tosswell. Bodkins will have scored his 8.
So FOW (5) = 31. Meakins entered at end A. Now referring to (4),
Meakins and Larkins each have to had scored a boundary in the rest of
the same over. So this requires the over be exactly: Bodkins-4,
Bodkins-4, Bodkins-out, Meakins-4, Meakins-1, Larkins-4. This
leaves Meakins on strike to Pitchwell at end B at the start of the
next over, with the score at 40 for 5.
(12) Referring again to (4) we know that Pitchwell has to have
dismissed Larkins and Meakins prior to Tosswell conceding
any more runs. This requires Meakins to have scored a single to get
Larkins onto strike. Larkins then gets out to Pitchwell. FOW (6) = 41.
(13) The next three scoring shots then must be Perkins-1, Meakins-1,
Perkins-1, since at this point all boundaries must have come off
Tosswell's bowling. Meakins is then dismissed by Pitchwell. FOW (7) =
44. Simkins enters at end B.
(14) The number of runs conceded by Pitchwell to this point was 12 (8
in his first 6 overs, 4 subsequently) so he had 2 runs further to
concede. Tosswell had to concede the remaining 14 runs in one over.
Since that 14 runs will require a single from each batsman the strike
cannot have been rotated prior to that happening (or else Perkins
won't have had a run left to score, allowing for his two boundaries).
So the 14 run over is the next scoring over.
(15) The next scoring shot must have been a single from Simkins since
he was on strike. Then Perkins was then dismissed by Pitchwell. FOW
(8) = 59.
(16) Tomkins entered on strike to Pitchwell and was dismissed for a
duck. FOW (9) = 59. Wilkins then scored the final run and Simkins was
dismissed. FOW (10) = 60. Wilkins was not out.
So the answers are:
Speedwell dismissed Jenkins. Tosswell dismissed Bodkins.
Wilkins was not out.
FOW: 6, 12, 18, 23, 31, 41, 44, 59, 59, 60.
I believe this answer is unique, assuming 6-ball overs. Let me know if
you have a different one.
Daniel
Jeremy Piets
Daniel Ashworth wrote:
That's what I got too. My solution pretty much followed those steps except I
didn't keep tabs of ends A and B. Instead I created a 26 row/11 column table
and noted each batsman's score each over and who was on strike at the start
of each over. Also I derived that Speedwell conceded 4 runs in each of 3
overs, and singles in the other 3 overs (based on the fact that only one
batsman could have scored in each of his overs and 4+4+4+1+1+1 = 15 is the
only combination that fits).
Jeremy.
--
The rabble await...
(Remove NOSPAM to reply)
Jeremy Piets
> Solution is spoiler space
>
Oops...kind of defeats the purpose really.
Sorry.