I must admit that there is NO solution for answer 2 since the above proposed "official" one contradicts Rules at the extent all guys who discussed the issue in mystery statement beleive.
Certain Rules clarifications though are needed to be made..
88 Points from Scratch - the solution
- Details
- Created on Monday, 06 June 2011 16:12
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Here follow the solution to Holmes's questions about 88 points hands cannot be made from the start without the help of other players.
Remember the mystery? Dr. Watson was reading an article in “Mahjong News” about the MCR London championship.
“At the last championship something incredible happened. Professor Moriarty has managed to win with a hand containing 88 points fan, just from the start, with no help of the wall or opponents.”
“You see, not all 7 existing 88 pts. fans hands you can declare just from the very start with no help from the wall or opponents”, said Holmes.
“Surely, you cannot declare “Four Kongs” without replacing four tiles from the wall.”
“There is one more fan”, replied Holmes.
Question 1 (for experts): Which 88-pts. fan (beside “Four Kongs”) East cannot declare without help from the wall?
Question 2 : How can East win then with this 88-pts. fan not giving a move to opponents?
…the answers!
Answer 1. When tiles are dealt from the wall East receives 14 tiles as-a-bundle having no opportunity to separate them. For the fan #4 “Nine Gates” a player in order to show 9-sided wait must have only 13 tiles and the 14th should fit the hand. Since East has 14 tiles no way “Nine Gates” may be declared straight.
Answer 2. To win with “Nine Gates” 14 starting tiles should contain a Flower. After Flower replacement 13+1 tile can declare “Nine Gates”.
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".