One possible solution:
Mrs. Hudson's hand meets the minimum requirement of 8 points with:
a) 3 chows combining for 2 1-point fans (e.g. 2 Mixed Double Chows; Mixed Double Chow and Pure Double Chow; Pure Double Chow and Short Straight etc.);
b) 1 Melded Kong of a Terminal (i.e. 1 or 9), giving 2 points;
c) hand contains only 2 Suits (i.e. One Voided Suit) and No Honours, giving 2 points;
d) and she self-drew on a unique wait (i.e. Single Wait, Closed Wait, or Edge Wait), giving 2 points.
Additionally, Mrs. Hudson managed to get all 8 bonus tiles, so her hand is worth 16 points in total.
Mrs. Hudson’s Record
- Details
- Created on Friday, 20 May 2011 23:02
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
This week's Sherlock Holmes Mystery is about Mrs. Hudson's Record.
Sherlock Holmes and Dr. Watson were away from Baker Street 221 B for almost a month. They were involved into trapping some big criminals.
Meanwhile, Mrs. Hudson went to her first MCR tournament. One hand she won had tremendous success - not in terms points though in structure.
Dr. Watson bought the newspaper “Mahjong News” and began to read:
“Mrs. Hudson, MCR tournament debutant, has managed to collect a 16-point hand consisting only of 1-point fans! … Sherlock!” – he asked Mr. Holmes – “Is it really possible?”
“Watson, It’s rather elementary!” – answered Holmes and began to draw the hand.
Question: Please, re-construct Mrs. Hudson’s hand.
Comment: The problem is rather tough to solve. There are two conceptual solutions.
Comments (3)
1Saturday, 21 May 2011 23:23
Edwin Phua
2Sunday, 22 May 2011 13:49
Quentin
The other possible solution is a hand close to the former one:
Hand: (123, 456, 999)b 4d, melded Kan of 1d, 4d self drawn.
The number of points is the same, 1 Chow fan is replaced by a Terminal Pung.
Add 8 flowers and it is a Hand with 16 1-point fans !
Hand: (123, 456, 999)b 4d, melded Kan of 1d, 4d self drawn.
The number of points is the same, 1 Chow fan is replaced by a Terminal Pung.
Add 8 flowers and it is a Hand with 16 1-point fans !
3Sunday, 22 May 2011 20:08
Vitaly
Congratulations, both Edwin Phua and Quentin!
You have provided two available conceptual solutions.
Very quick and very correct :).
As an author I am a bit disappointed because it was considered this problem to be tough to crack down :(.
You have provided two available conceptual solutions.
Very quick and very correct :).
As an author I am a bit disappointed because it was considered this problem to be tough to crack down :(.
It is difficult to judge the difficulty of problems, but some did take me some time to solve.
I especially enjoyed the '32nd of December' and its fourth question.
Thanks to Vitaly for the problems, to Martin for hosting the "venue" and congrats to Sylvain and Scott for their success.
A: two kongs of the same suit (i.e. 8 one-suit tiles) and a kong of wind
So... winds are now suit tiles?
And they are in every suit?
Wow!
Looks like I've misunderstood the question and it was actually an easy one!
Was it because I was the only one to answer the question within the allotted time?
Just curious.
Thanks.
At first, it looks like each player had three pure melded kongs, two of them separated by two numbers (e.g. 1 and 4), and that their left-side neighbour is waiting for these two said kongs with a ryanmen (e.g. _23_).
But it turns out there are not enough tiles for that.
So, here's the trick:
Watson had: melded: 1111m 4444m 5555m, concealed: 23s EE.
Lestrade had: melded: 1111s 4444s 5555s, concealed: 23p SS.
Holmes had: melded: 1111p 4444p 5555p, concealed: 78m WW.
Mrs. Hudson had: melded: 6666m 9999m, concealed: RRRR(concealed kong) 23m NN, and erronously melded as flowers: 2223m.
It certainly "cut off all conceivable scenarios".