Two miniatures, two solutions
- Details
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Two Sherlock Holmes miniatures by Vitaly Novikov published on the same date. They were not too difficult (that's why they are called 'miniatures' for in the first place). Here are the answers.
Miniature #1: Negative Fan
Question: Please, explain what has happened when a fan was added to a hand and hand value diminished.
Answer. The main fan was undoubtedly “Chicken Hand”. Though a player missed to add another fan (costing less than 8 pts.) which was added by the other player. Hence, total hand value decreased (“Chicken Hand” is no longer a valid fan).
What might be that “missed” fan? Here is some list:
• wait (any of three),
• Last Tile,
• Tile Hog
…And we assume that a player missed a fan unintentionally.
Miniature #2: Mrs. Hudson’s New Fan
Question: Please, describe Mrs. Hudson’s starting hand and explain what is this new fan about.
Answer. Sherlock Holmes said: “Since you have looked at your tiles up and down and the hand is knitted, I assume that your starting hand is…"
“Wonderful! And the new fan is Knitted Reversible Tiles!” – exclaimed Dr. Watson.
I used the following definition of "knitted": whenever tile in hand occurs no more than ONE time.
In your comment you have used term "isolated" which has much stronger constraint: not only tile occurs in hand only once but also there are no any two-tile "joints" (for instance, Bam 2-3).
Summarizing, let's consider published solution as a certain way to solve mystery -- not only one possible :)?.
My definition of ‘knitted’ implies "suit tiles belonging to separate Knitted sequences (for example, 1-4-7 of Bamboos, 2-5-8 of Characters, and 3-6-9 of Dots)", which are the core of the three knitted fans as written in the official rule book.
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".