Mrs. Hudson have a Greater Honors and Knitted Tiles:
1s 258p 369w ESWN CFP
Lestrade’s hand is dead.
The two other hands would be:
Thirteen Orphans on thirteen waits
and:
22234567p 33345w
Mrs. Hudson’s Problem #2
- Details
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
A new problem of Mrs. Hudson's. The last time, she had to win before the three gentlemen she played against (see ‘Mrs. Hudson’s Problem’), since her winning hand had the regular structure 3-3-3-3-2. Now she makes a vow not to collect even a pair in a hand.
“I would not collect even a Pair, not a meld, not collect Flowers. I hope that I would not discard a tile to somebody’s mahjong, otherwise that would offend others.”
In such a way a whole deal passed. Mrs. Hudson took the last tile from the wall. She called for Mr. Wan Tao.
“Mr. Tao, be so kind and help me. What tile could I safely discard in order to not give mahjong?”
Wan Tao looked carefully at all three gentlemen’s hands and then he sentenced:
“No way, 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!”
Question 1: Please, reconstruct all four hands. (There exist many solutions, please, provide one).
Question 2 (for experts): Please, provide a solution under the condition of one dead hand (i.e., Lestrade’s hand).
Comments (3)
1Tuesday, 10 April 2012 17:15
Sylvain MALBEC
2Tuesday, 10 April 2012 21:25
Morten Andersen
Well if she has a hand of 14 different tiles it must be greater or lesser honors and knitted tiles.
Lets say she has 2,5,8 in bamboo, 1,4,7 of dots, 9 characters and the 7 seven honors.
Another player has been lucky enough to wait for 13 orphans on 13 chances and would go out on all the honors, the 1 and the 9.
A third player has 34567888 of bamboo and 56777 of dots thus waiting on the remaining tiles.
The last player can then have a dead hand and Mrs. Hudson would still feed on all her tiles.
Lets say she has 2,5,8 in bamboo, 1,4,7 of dots, 9 characters and the 7 seven honors.
Another player has been lucky enough to wait for 13 orphans on 13 chances and would go out on all the honors, the 1 and the 9.
A third player has 34567888 of bamboo and 56777 of dots thus waiting on the remaining tiles.
The last player can then have a dead hand and Mrs. Hudson would still feed on all her tiles.
3Wednesday, 11 April 2012 09:57
Quentin Porcherot
So, obviously Mrs Hudson was knitting, and she fished Greater Honors and Knitted Tiles.
More precisely, her hand was, e.g., b147 c258 d9 + 7 honors.
Although Lestrade had a dead hand, the two other players could win on any of her 14 different tiles as:
one had each honor and terminal in his hand, missing the pair for 13 Orphans: a 13-sided wait on any terminal or honor.
the other had, e.g., b44456 c34567888, a 5-sided wait (chameleon) on b4, b7, c2, c5 and c8.
Very nice problem, although I found the solution in 3 minutes.
More precisely, her hand was, e.g., b147 c258 d9 + 7 honors.
Although Lestrade had a dead hand, the two other players could win on any of her 14 different tiles as:
one had each honor and terminal in his hand, missing the pair for 13 Orphans: a 13-sided wait on any terminal or honor.
the other had, e.g., b44456 c34567888, a 5-sided wait (chameleon) on b4, b7, c2, c5 and c8.
Very nice problem, although I found the solution in 3 minutes.
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".