Mrs. Hudson’s Problem #2 - the solution
- Details
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Surely, this was a difficult question in our series of Sherlock Holmes Mahjong Mysteries. Yet, some readers solved it quite quick!
This was the situation:
Mrs. Hudson said that she would not collect even a Pair, not meld, not collect Flowers. “I hope that I would not discard a tile to somebody’s mahjong otherwise that would offend others”, she said.
Then, she drew the last tile from the wall. Then she asked Mr. Wan Tao, what tile she could safely discard in order to not give mahjong?
Mr. Tao looked looked at the hands iof all four players and said: “Mrs. Hudson, you can discard a safe tile due to the fact that all of your 14 single tiles would suit to somebody’s mahjong. Even more, despite there is no even a pair in your hand you have mahjong, so, please, declare it!”
So, the question was: reconstruct all four hands.
And the expert question was: provide a solution under the condition of a dead hand for Lestrade.
Answer. Mrs. Hudson’s hand (“Greater Honors and Knitted Tiles”):
Surely one of gentlemen’s hands (Mr. Holmes’s one) should be “Thirteen Orphans” wait (13 single Terminals or Honors) calling for the following 7 Honors and 2 Terminals out of Mrs. Hudson’s hand:
Now, two other hands have to wait for , , , and .
Two other hand are very simple “Pure-suiter-waits” with waiting tiles 
(plus a Pair) in “Craks” hand and 
(plus a Pair) in “Dots” hand.
Answer (for experts). Let’s start with the same Mrs. Hudson’s and Mr. Holmes’ hands.
Now, to cope with wait on sparse tiles in two suits like , , , , we need to utilize structure with co-called “chameleons” (see additional section below on “chameleons”). Dr. Watson’s hand is (hand is concealed):
Strictly speaking 8 pts. for mahjong the above hand is gaining after fan “Last Tile Claim”.
Chameleons
By definition, “chameleon” (term was first used in 2005 at my site http://www.mahjong-co.narod.ru/waits_analysis.html) is a structure with number of one-suited tiles = 3*X+2 (i.e., 2, 5, 8, 11) which form a legal regular mahjong structure in both occasions:
1. NO tiles added (“Chameleon sits straight”) – “X” groups by 3 tiles (normally, Chows) and a Pair,
2. 1 tile added (“Chameleon puts out its tongue”) – “X+1” groups by 3 tiles (normally, Chows) and a Pung.
In a case Dr. Watson’s hand,
may be split as:
1. + + -- no tiles added, or
2. + + when is added,
3. + + when is added,
4. + + when is added.
And may be split as:
1. + -- no tiles added, or
2. + when is added.
3. + when is added.
Dr. Watson’s hand consists of two chameleons each of them can “sit straight” (no tile added) or “put out its tongue” (a tile is added to form legal structure).
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".