I thought the maximum hand* was the same, but i realised my mistake, this is as i see the sollution
1 - 3 kongs
2 - 2 Concealed Kongs
3 - Going out on replacement tile
4 - Single Wait
5 - Last tile of wall
6 - 4 Concealed Pungs
7 - Big Three Dragons
8 - Seat Wind
9 - Round Wind
10- All Honors
11- Fully Concealed
12- 8 Flowers (1 is enough, since maximum of different fans is asked)
So a 12 different Fan solution
PS. The maximum scoring hand is instead of 3 kongs (not 4, that eliminates Single wait as a different fan ) you indeed score 4 kongs
Did you find Sherlock Holmes’s ‘Full House’?
- Details
- Created on Saturday, 19 May 2012 13:46
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Sherlock Holmes, Dr. Watson and Mrs. Hudson are sitting in the lounge at Baker Street 221 B.
Holmes asks Dr. Watson: “There are 81 fans in MCR rules. It is very interesting how many different fans one can place in a winning hand.”
“Watson”, Holmes continues, “I must leave to buy some tobacco. Let’s discuss your solution of the problem. I guess it should take no more than half a pipe.”
Question: Please, construct a hand containing as maximum of different fans.
Comments (12)
1Monday, 21 May 2012 15:51
Gertjan Davies
2Tuesday, 22 May 2012 05:28
Vitaly
Thank you, Gertjan, for your solution!
I would ask you to put your solution hand in tiles since it is very difficult to check.
Let's wait for other readers' replies.
I would ask you to put your solution hand in tiles since it is very difficult to check.
Let's wait for other readers' replies.
Wednesday, 23 May 2012 13:53
Vitaly
Do you mean, Harry, they are counted or not?
Answer: fan # 81 in Green Book is "Flower", so it is counted as separate fan.
Answer: fan # 81 in Green Book is "Flower", so it is counted as separate fan.
3Tuesday, 22 May 2012 14:33
Gertjan Davies
East round, player is east:
/RR/ /GG/ /DD/ EEE N and get a N as replacement tile from your third kong (/DD/ means concealed White Dragon kong) on the last tile of the wall, flowers are omitted in the example.
/RR/ /GG/ /DD/ EEE N and get a N as replacement tile from your third kong (/DD/ means concealed White Dragon kong) on the last tile of the wall, flowers are omitted in the example.
4Wednesday, 23 May 2012 06:48
Morten Andersen
Well how about: Replacing two of the dragon-kongs with kongs of nines and the pair of north with a pair in nines. That would remove the Big three Dragons and the all honors but would give you: All types, All terminals and Honors, Double Pung, Pung of terminals and pung of dragons resulting in three more fans total. Not sure this is the maximum yet but I think we must be close.
5Wednesday, 23 May 2012 09:53
Vitaly
I guess that there is some mess-up in "maximum hand" notions:
* first, having maximum points,
* second, having some other "maximum".
In case of our problem it was asked to provide "maximum different fans" hand but NOT any others.
Going slightly ahead, next Sherlock's problems was planned to highlight "maximum points" hand.
Gertjan, your hand fits very well to that "maximum points" hand but not, frankly, to "maximum fans" hand since it may be improved.
Morten, please, provide your hand in tiles so we could look at it and check.
And, thanks again, for your replies..
* first, having maximum points,
* second, having some other "maximum".
In case of our problem it was asked to provide "maximum different fans" hand but NOT any others.
Going slightly ahead, next Sherlock's problems was planned to highlight "maximum points" hand.
Gertjan, your hand fits very well to that "maximum points" hand but not, frankly, to "maximum fans" hand since it may be improved.
Morten, please, provide your hand in tiles so we could look at it and check.
And, thanks again, for your replies..
6Wednesday, 23 May 2012 12:35
Gertjan Davies
I don't mind being proved wrong, but All (Terminal) and Honors the one point Pungs (like non-seat or round wind and the terminal Pungs) are implicated and can't be counted.
Big three dragons is replaced by All types & Double pung & Pung of Dragon (so +2 fan solution wise)
Big three dragons is replaced by All types & Double pung & Pung of Dragon (so +2 fan solution wise)
7Wednesday, 23 May 2012 13:51
Vitaly
"Official" solution has 15 different fans and 23 fans in total (including Flowers).
8Wednesday, 23 May 2012 19:22
Morten Andersen
Bamboo: 9999 (concealed)
Circles: 9999 (concealed)
Characters: 99
Dragon: RRRR (concealed)
Wind: EEE (East is round and seat wind)(concealed).
Flowers: 8
Going out on 9 characters after making a kang and drawing the last tile from the wall.
1- 3 kangs
2- 4 concealed pungs
3- Terminals and honors
4- All types
5- Double Pung
6- Dragon Pung
7- Pung in Seat Wind
8- Pung in Round Wind
9- Pung in terminals (X2)
10-After a kang
11- Last tile Draw
12- Single Wait
13-Fully Concealed
14- 2 concealed kangs
15- Flower tile (*8)
Although this has probably not been sufficiently announced yet I don't actually think you can claim the two concealed kangs in the newest EMA rules though but up untill now I think that is the highest number of fan you can get.
Circles: 9999 (concealed)
Characters: 99
Dragon: RRRR (concealed)
Wind: EEE (East is round and seat wind)(concealed).
Flowers: 8
Going out on 9 characters after making a kang and drawing the last tile from the wall.
1- 3 kangs
2- 4 concealed pungs
3- Terminals and honors
4- All types
5- Double Pung
6- Dragon Pung
7- Pung in Seat Wind
8- Pung in Round Wind
9- Pung in terminals (X2)
10-After a kang
11- Last tile Draw
12- Single Wait
13-Fully Concealed
14- 2 concealed kangs
15- Flower tile (*8)
Although this has probably not been sufficiently announced yet I don't actually think you can claim the two concealed kangs in the newest EMA rules though but up untill now I think that is the highest number of fan you can get.
9Thursday, 24 May 2012 05:19
Vitaly
Morten, thank you for your 15-fan solution!
>> I don't actually think you can claim the two concealed kangs in the newest EMA rules ..
And what about EMA rules?
>> I don't actually think you can claim the two concealed kangs in the newest EMA rules ..
And what about EMA rules?
10Saturday, 26 May 2012 12:31
Quentin
I've got a much cheaper 15 different fan solution, accounting of the future EMA probable ruling about Kongs (2 concealed Kongs may only be counted is there are not more than two Kongs).
Exposed: EEEE (both seat and prevalent wind)
Concealed: RRRR
Standing: Dots 11113, Bams 99
Winning on Dots 2 self-picked after a kong, the three others being visible and no more tile remaining in the wall.
1. Pung of seat Wind (2)
2. Pung of prevalent Wind (2)
3. Pung of Dragon (2)
4. Pung of Dots 1 (1)
5. Two "melded" (i.e. non-concealed) Kongs (4)
6. Concealed Kong (2)
7. Two concealed Pungs (2)
8. Tile hog (2)
9. Outside hand (4)
10. No Characters (1)
11. Closed Wait (1)
12. Last Tile (4)
13. Lase Tile Draw (8)
14. On replacement Tile (8)
15. Flower*8 (1*8)
15 different fan whatever the Kong ruling, 22 fan and 51 points.
Exposed: EEEE (both seat and prevalent wind)
Concealed: RRRR
Standing: Dots 11113, Bams 99
Winning on Dots 2 self-picked after a kong, the three others being visible and no more tile remaining in the wall.
1. Pung of seat Wind (2)
2. Pung of prevalent Wind (2)
3. Pung of Dragon (2)
4. Pung of Dots 1 (1)
5. Two "melded" (i.e. non-concealed) Kongs (4)
6. Concealed Kong (2)
7. Two concealed Pungs (2)
8. Tile hog (2)
9. Outside hand (4)
10. No Characters (1)
11. Closed Wait (1)
12. Last Tile (4)
13. Lase Tile Draw (8)
14. On replacement Tile (8)
15. Flower*8 (1*8)
15 different fan whatever the Kong ruling, 22 fan and 51 points.
11Saturday, 26 May 2012 22:11
Quentin
Morten's hand only scores 14 different fan. You are not allowed to score a pung of terminals or 1-point pung of Honors if you score All Terminals and Honors (32).
12Sunday, 27 May 2012 11:31
Vitaly
One of possible 15-fan solutions is now available at this site.
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".