How to explain the yaku

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How to explain the yaku

Post by Referee » Tue Jul 27, 2010 3:11 pm

Though it's a minor point, if I were to explain the yaku I'd explain Ryanpeikou as 1 han (plus the two Iipeikou, total 3 han), same way that Shousangen doesn't exclude the Yakuhai, and Hon routou doesn't exclude the Toitoi (or Chiitoitsu).

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Re: Let's Start Again

Post by Rosti » Tue Jul 27, 2010 5:41 pm

Referee wrote:Though it's a minor point, if I were to explain the yaku I'd explain Ryanpeikou as 1 han (plus the two Iipeikou, total 3 han)
Personally I'd explain it as 1+1+1, indicating accumulative han.

Same for Honroutou, which I'd class as 2+2+2 closed, or 1+2+2 open (for Chantaiyao, Toitoi and Honroutou respectively).

Though I think aside from this method it's better to state it as one han rather than three, just to stop things being counted twice.
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Re: Let's Start Again

Post by b4k4ni04 » Tue Jul 27, 2010 6:53 pm

Honroutou and Chanta do not stack. As one allows runs/sequences, while the other is entirely sets.

As it is named, Ryanpeikou is quite literally 2 Iipeikou (ii, ryan, san, suu, uu, ryuu, chii, paa, chuu)
So I see it a little silly to count the individual Iipeikou when the name of the yaku has already done that for you.
Also, it is not like Honroutou or Shousangen that are worth 2-han, but you eventually get 2 more han either 1 a piece from two dragon yaku, or 2-han for Toitoi. (oh, aha you get this ^^;.)

When counting yaku with Ryanpeikou, you don't count it as Iipeikou 1-han, Iipeikou 1-han, Ryanpeikou 1-han. It is Ryanpeikou 3-han. It is convenient to learn to count it as Two Iipeikou with an extra han, but overall you're creating a redundancy. Know that you count the highest number of occuring "pei kou" and score it at that. This is why I group Iipeikou, Chitoitsu, and Ryanpeikou in the same yaku "group".

With Iipeikou you have three pairs, those three pairs are in sequence, and you hand is closed; this gives you 1-han. For Chitoitsu, you have seven pairs and a closed hand for 2-han. If it just so happens that six of your seven pairs in your closed hand create two three-pair sequences, you have Ryanpeikou and 3-han. (And if you have seven pairs all in sequence in the same suit you've got a helluva hand, if not Daisharin. ;).) So, of these three, should you qualify, count the single yaku that grants you the most han.

tl;dr: If you have Ryanpeikou, you do not have Iipeikou. As Ryanpeikou is literally 2 Iipeikou. You need not count Iipeikou should you have Ryanpeikou. 3-han.

Okay I'm rambling by now, I hope this was useful :/.
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Re: Let's Start Again

Post by Referee » Tue Jul 27, 2010 9:56 pm

I know what ryanpeikou is. I was just saying that it makes more sense the other way around.

Ryanpeikou is two iipeikou, and then the iipeikou don't count. Shousangen is two dragon sets and a dragon pair, but the dragon pairs still count. See the problem?

I'd rather explain consistency: Count all the yaku you can see.
So Ryanpeikou (1 han) + Iipeikou (1 han) + Iipeikou (1 han) = 3 han (as it should be). It's consistent with Shousangen (2 han) + Yakuhai (1 han) + Yakuhai (1 han) = 4 han; Honroutou (2 han) + Chitoitsu (2 han) = 4 han; and Honroutou (2 han) + Toitoi (2 han) = 4 han.

I wonder whether I'm making myself clear or not.

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Re: Let's Start Again

Post by Rosti » Tue Jul 27, 2010 10:36 pm

b4k4ni04 wrote:Honroutou and Chanta do not stack. As one allows runs/sequences, while the other is entirely sets.
Of course they stack. How could you possibly get Honroutou (all terminals/honours) without getting Chantaiyao (each meld contains a terminal/honour tile) as a prerequisite?

The fact that one allows runs and the other doesn't is why they stack, rather than them being the exact same yaku. You can't get Honroutou without Chanta (or Toitoi), while you can get Chanta without Honroutou. That's why they're separate yaku, and it doesn't stop them stacking for Honroutou. It doesn't matter about the specifics, so long as having one implies you must also have the other (the reverse doesn't matter) then they stack.

b4k4ni04 wrote:It is convenient to learn to count it as Two Iipeikou with an extra han, but overall you're creating a redundancy.
~b4k4.
No you're not. You're solving potential confusion by making it clear that the three han for Ryanpeikou includes a han for each Iipeikou. Ryanpeikou is a bad example, because it's fairly obvious that it has to contain Iipeikou twice, as it's practically the definition of the yaku. For things like the aforementioned Honroutou though, it's not necessarily so obvious, because there are certain implied yaku (toitoi) where it might not be completely intuitive that it is a prerequisite and things get counted twice, because it's viewed as six plus two for toi toi as well, rather than six including the two for toi toi.
With big hands that have several yaku and things start to get complicated, it's far better, especially with beginners, if the number of han are shown in a way that's easier to trace back and see where every single han awarded comes from. Displaying a breakdown has pretty much no negative aspects, and has the positive of removing a lot of potential confusion, and I don't really see why you'd object to it so much aside from a "this is how I'm used to doing things, so this is how I think they should be done" perspective. It's far more logical, intelligent and clear when things are displayed more simplistically.

And standard mahjong games, when displaying the calculated score, do not do it your way either. Tenhou for example would display:

Code: Select all

対々           2
全帯么         2
混老頭対々和    2
and not

Code: Select all

混老頭対々和    6
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Re: Let's Start Again

Post by Archon_Wing » Wed Jul 28, 2010 12:11 am

Honroutou is actually a total of 4
http://reachmahjong.com/en/2007/11/in-g ... nd-honors/
; I'm quite sure Tenhou shows it as

混老頭 2
対々和 2

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Re: Let's Start Again

Post by b4k4ni04 » Wed Jul 28, 2010 12:32 am

Referee:
Yeah I got ya, but it looks like you have Suupeikou the way you've counted it ;P. *shrugs* That's my main issue with counting it like that. It's okay to learn it that way (1-han for each Iipeikou, and a bonus 1-han for having a total Ryanpeikou) But in the long run I see it better to recognize it as Ryanpeikou, e.g. "You do not have 1 and 1 apples, you have 2 apples."
Heh, they're dragons, they're just that awesome ;P. But yeah, it does create an inconsistency. Live with it, I'm not coming up with any cute analogies to explain it away.


Rosti:
*shrugs* kay, mebbe i'm wrong on Chanta + Honrou, I havent played enough recently to remember -3-;;. ... Thank you. Archon~~ ^^;.

So, uh. Are you sure that's how I score yaku? All in a bunch like that. :/...? I'm in favor of Honrou+Toitoi being shown as 2+2 or Shousangen shown as 2+1+1. <<;. It's like we've got different versions of the same textbook, we're on the same page number and seeing different things, when we really should be on the same page :/. I should clarify and say, that I'm fine with using it to teach beginners, but be sure to someday get them to Ryanpeikou 3-han.
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Re: Let's Start Again

Post by Archon_Wing » Wed Jul 28, 2010 1:00 am

I think life would be easier if one just sees iipekou defined as "exactly one pair of two identical sequences" and ryanpeikou as "exactly two pairs of two identical sequences". In fact that's what the names of those patterns are saying!

Thus they can never occur together then and they are seperate yaku, just like how 3 concealed triplets and 4 concealed triplets are different.

And yea, the apple analogy is best here. If you have 2 apples, you don't have 1 apple, and it's a waste of time to say you have 1 apple and another apple.

Fortunately, there aren't really too many implied patterns, so I think it's best to head for the simplest path for the few exceptions, namely the yakuhai which do stack.

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Re: Let's Start Again

Post by Rosti » Wed Jul 28, 2010 12:18 pm

Archon_Wing wrote:Honroutou is actually a total of 4
http://reachmahjong.com/en/2007/11/in-g ... nd-honors/
; I'm quite sure Tenhou shows it as

混老頭 2
対々和 2
It's six, because it's impossible to get Honroutou with chi in the hand, which means it's also going to be seven pairs or toitoi as well, both of which add another two. Counting yaku that stack (ie. yaku that you automatically must also have to get Honroutou) it is 2 (Chanta) + 2 (Toitoi/Seven Pairs) + 2 (Honroutou) = 6

Which demonstrates my original point that 2+2+2 is a good way of displaying it, because if you think about it and track where each number comes from then you're not going to get it wrong ;)

We've gotten way off topic though :P
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Re: Let's Start Again

Post by Archon_Wing » Wed Jul 28, 2010 4:40 pm

Rosti wrote:
Archon_Wing wrote:Honroutou is actually a total of 4
http://reachmahjong.com/en/2007/11/in-g ... nd-honors/
; I'm quite sure Tenhou shows it as

混老頭 2
対々和 2
It's six, because it's impossible to get Honroutou with chi in the hand, which means it's also going to be seven pairs or toitoi as well, both of which add another two. Counting yaku that stack (ie. yaku that you automatically must also have to get Honroutou) it is 2 (Chanta) + 2 (Toitoi/Seven Pairs) + 2 (Honroutou) = 6

Which demonstrates my original point that 2+2+2 is a good way of displaying it, because if you think about it and track where each number comes from then you're not going to get it wrong ;)

We've gotten way off topic though :P
It is not 6 no matter how you calculate it. Please click on the link in the post you quoted. And this is exactly why you shouldn't be counting it like that.

I'll just quote Garthe from the link here:
This hand is actually only worth 2 points but because it will also always contain at least one other 2 point hand (All pairs or All triples) it is essentially worth 4 points.
Last edited by Archon_Wing on Wed Jul 28, 2010 4:44 pm, edited 3 times in total.

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Re: How to explain the yaku

Post by Referee » Wed Jul 28, 2010 7:24 pm

That is true, Honroutou is 4 han (at least). Garthe's quote is one way to explain, and since consistency is good, that's why I'm calling for "Ryanpeikou is worth 1 point, but because it will also contain two Iipeikou, it is essentially worth 3 points."

PS: Good job with the thread split.

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Re: How to explain the yaku

Post by burke » Thu Jul 29, 2010 5:52 am

Regarding Honroutou and Chanta, this seems to be one of those situations where both sides are somewhat right. According to this page (by Wei-Hwa Huang) which details the Tokyo University scoring (http://www.ofb.net/~whuang/ugcs/gp/mahj ... #dirtyterm):
3.1.4.3 (5) DIRTY TERMINALS (honroutou [混老頭])
Also called "honroo" for short, or "honroutoitoi," as the hand is also a [Toitoi] hand.
All tiles are ends. Technically, only two multipliers -- add two for the [Toitoi] and one for the [Chanta] (but no bonus, even if the hand is closed) to get a total of 5.
If the hand is a [Chiitoitsu], it's only four multipliers -- two from [Honroutu], two for the [Chiitoitsu]. Sometimes such a hand is called "honroochiitoi."
Some variants play this as only four multipliers total instead of five.
I've highlighted the relevant parts and changed the confusing English terminology to the more familiar Japanese. Interestingly these rules only award 1 han for the Chanta in this case.

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Re: How to explain the yaku

Post by Rosti » Thu Jul 29, 2010 11:32 am

Hmm, so after some research, according to Tenhou Honroutou doesn't include Chanta, in which case, fair enough, I was wrong.

http://arcturus.su/rondb/entry.pl?id=d9 ... aee8bc50eb

Still, I find it a bit odd because it means Honroutou isn't really a disctinct yaku, or at least it offers pretty much no extra bonus (except for an extra han when open) over just Chanta and ToiToi, which sems a bit poor, given how difficult it is to get a hand that's entirely terminals and honours compared to having terminals and honours in each meld. If the only added bonus to the pure terminals and honours hand over one in each meld is two extra han for a toi-toi that you'd have gotten anyway, then that seems like a pretty undervalued hand. I'd have expected it to be far worth more than the psuedo-limit (by which I mean mangan, not yakuman) that it is.

Still, none of this really affects my original point that yaku which stack should be displayed in some way that makes it clear how they stack. If anything it just strengthens the argument because there's clearly a lot of scope for misunderstanding and confusion, especially given that most of these yaku don't pop up particularly often :P
IMO, if it's being displayed in a table or something, it should be 2+2 or whatever. If it's not, then it should be "It's X han, but as it will also include Y it's effectively Z han."
Last edited by Rosti on Thu Jul 29, 2010 11:49 am, edited 1 time in total.
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Re: How to explain the yaku

Post by Senechal » Thu Jul 29, 2010 11:46 am

Chinitsu is worth 9 when closed, 7 when open, because you always have honitsu implied.
This is wrong. Honitsu = 1 suit + winds and dragons. Chinitsu = 1 suit, no winds and dragons. You can't have yes and no at the same time.

Honrou implies chanta.
This is also wrong for the exact same reason. Chanta implies and requires a sequence in it. Honrou by definition can't have a sequence.

Chanta is a hand where you have a lot of flexibility, like tanyao, to build hands with a staggering amount of usable tiles (only 456 are really blocked, you can always mount something with 123 sequences or 789 sequences). You can call like mad and get only 1 han, or you can build it closed for 2 han. You can get a chanta hand with yakuhai or not, and the usual jazz.

Honrou is a hand in which you are limited to calling only terminal tiles. You get 2 han for it whether it's open or closed. Because of the possibilities, you'll only be able to have a Toitoi or 7 Pair hand with it, but neither are directly implied. So doing this will be either mangan+ or 6400/9600.

---------------------

Wei-Hwa Huang's site compiled stuff relating to Toudai-shiki which is a style of mahjong in the way EMA nashi-ari is a style. Is this an authoritative list? It was made in 1998 and updated in 2001. It was made as an attempt to be a better reference than whatever else was on the English internet. And in many ways, it was at that time, doesn't mean it's error-proof. There's also talk of sextuple and octuple "superlimit" hands. This is the first and only place I have seen so far talking about hands worth yakuman and a half (and we're not counting dealer 1.5x points here). From what I've heard, there are a fair amount of differences with Toudai play compared to standard ari-ari mahjong. It's good to have a resource, but bad to assume it's correct. As a reference, there's one thing that has to be said about references: a solid reference would not have listed a zillion variations (well, you know, you can start with 25k, 26k, 27k or 30k) nor would sections 4 and 5 be incomplete. It would also not have a list of non-yaku hands in it, at the end or anywhere (saying ippatsu isn't counted is one thing but that list is another thing entirely).

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Re: How to explain the yaku

Post by Rosti » Thu Jul 29, 2010 11:56 am

Senechal, you should revise your post, given that I revised mine ;)

Though interesting that you brought up Chinitsu with that example, given that I've personally always viewed it as being three han, and then an included +2 or +3 because of Honitsu, depending on whether it's open or closed, to bring it up to five or six han in total. The results ultimately end up the same anyway (the Honroutou thing was that I legitimately believed it would five or six on it's own in total, same as Chinitsu, rather than four, because the reference sheet I have is wrong and lists it as such), so it doesn't matter hugely as it's just how you end up looking at things to get to the same end result.
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