foobar

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

2H for both #1 and #2. Difficult to answer #3 without context, but my guess is that bidding 2H would be a deliberate systemic deviation as stated?

Aren't almost all existing responses to 2D, already tailor made for the stated design goal of repurposing it as GF or 5+ spades with 12+? I suppose you can be in a little bit of a pickle with a min hand with short clubs over 2S/2N, but that's a specific case, and the chances are that you'll have some other rebid. IMO, it seems best to leave 2D "as is" and see if the new semantics offer improvements over 2H/2S.

Couple of films for starters (warning: neither of them are for the squeamish): Raw (French, 2017): A vegetarian goes to college and accidentally tastes meat; enough said, but highly recommend by this vegan 😁 Wake In Fright (Oz,1971): A cult film set in the Oz outback

Think that for 2♦ you a better off folding the two tail voids into 3♠ and making 3N as 4261. This flows naturally into 4♣ asking QPs for the more common shape, and then 3N is TP. Over 2♣, you can fold the voids into 3♥. which leaves you similarly placed with 3♠ for 4261 and now 4♣ makes the same QP ask. You can then use 3N for the 7411 or choose to show two ranges for 4261.

Think you can improve the mnemonics by modeling it after the IMP 2♣...2♦....3♣ and moving things down by two steps, i.e.: 2♦....2♥: 2S: 4♣ OR SS ♦ with high (♠) shortness OR SS ♦ bal with 2=2=7=2 .....3♣: 4♣, then 3♥+ as below .....3♦: SS ♦ with short ♠ .....3♥: 2=2=7=2 .....3♠ / 3N: Vacant...can probably put something here 2N: SS ♦ with mid (♥) shortness OR SS ♦ bal with no singleton (implies 2-card ♠ tolerance) .....3♦: SS ♦ with short hearts .....3♥: 2=2=6=3 .....3♠: 2=3=6=2 .....3N: 3=2=6=2 3♣: 4♥, then 3♥+ as below 3♦: Single suited with ♦, short ♣ (6331, 7(32)1, 3370) ......3♠: 3136 ......3N: (32)71 ......4C: 3=3=7=0 3♥: 4♠, high (♥) short (add 4=1=7=1) 3♠: 4♠ low (♣) short, 4=2=6=1 3N: 4♠, low short 4=3=6=0 4C: 4♠=2=7=0 It does resolve some the voids above 3N, but they are relatively rare, and if relayer is still interested, we are likely looking at slam.

Couple of comments: 1) One of the really nice things about the IMP 2♣ - 2♦ relay is the sheer mnemonic elegance. Basically, one simply bids 4-card suits up the line, and it's more or less symmetric after that, with the extra values and 4♠ being the icing on the cake. 2) This structure seems symmetric to 2♦, which may ease the above concern a little bit, but it seems like there are still some oddities like clubbing the low shortness first with the "other suits" (instead of the standard scheme of high to low)

As awm noted, it may be impossible to fit in everything without removing at least some hand type (probably one 6-4)? Regardless, the proof of the pudding is in the bidding :D, so if you want to try it out sometime...

straube and I were bidding a few 1♥ hands with awm's structure and really liked it. Couple of questions for awm: 1) Are there some weak hands (say in 6-7 HCP range) with < 3♥ that will pass 1♥ rather than bid 1♠? 2) Over 1♥ - 1♠, when holding 3 (4)♠ with longer 6♥, is the preferred rebid 1N or some # of ♥? 3) If 4th hand bids over 1♥ - 1♠, are Xs by both hands generally for takeout?

Toss up between ♦ or ♠. Might prefer the former given that the extra length makes leading 8♦ a little safer.

I was thinking along similar lines...basically, for doubletons and singletons use inverse-classic DCB (scanning for A/K only, but stopping if held). Obviously this can introduce some ambiguity in case of Qx, KQ, AQ, but perhaps it can be offset by the relay captain's holding and the early K-parity?

4D....4H (relay; even K parity...could be S(AKQ)...K(D|C) among others) 4S....4N (relay; nothing in hearts; even spade...S(AKQ) is out the window. S(KQ)...D(A)...C(K) or the above? Can relay, but won't be resolved: 5C....5S (relay; odds in ♦+[♣)

4C...4D (relay; even K parity....at this point we can have two black kings + rounded queens) 4H...4N (relay; nothing in S, even D, so both black kings go out of the window. Must hold A(D+Q) perforce and by induction A(H) (since KQ of clubs is impossible) To be frank, I don't know if I can actually make this deduction at the table, but it seems promising.

Agree with all three above. Also think that awm's idea of always using K=2 makes logical sense if you consider that it reduces the number of ambiguous combinations, especially in conjunction with early K-parity (but with no exceptions for the stiff). One tweak to consider for the short suit scanning is to skip with nothing in both singletons and doubletons and then revert to normal scanning order for 4+ card suits. On a side note, when scanning the strong hand, my conjecture is that A-parity first might make more sense, but it's just speculation at this point.

Using my cocktail: 4C - 4D [relay; even K-parity) 4♥ - 4♠[relay; K ♦ perforce) 4N - 5♣ (relay; K ♣ perforce] 5♦ - 5♥ (relay; no Q♥, so Q♠)

Can you really place the cards? Inferring ♠KQ seems to follow, but how can one disambiguate between K♦/K♣ at this point?

Here's a Byzantine method with the following rules: 1) K-parity first (with K=2 QPs always) 2) Singletons and doubleton suits are scanned first, with ties broken in rank order. Skip with nothing or stop with A/K 3) 4+ cards suits are scanned in length order with ties broken in rank order. Stop with even and continue with odd 4♦ - 4♥ (even K-parity] 4♠ - 5♦ (nothing in ♥, ♠, even ♥) AQ of ♥, Q♣ perforce

Here's a Byzantine method with the following rules: 1) K-parity first (with K=2 QPs always) 2) Singletons and doubleton suits are scanned first, with ties broken in rank order. Skip with nothing or stop with A/K 3) 4+ cards suits are scanned in length order with ties broken in rank order. Stop with even and continue with odd 4♣ - 5♦ (relay; odd K-parity, nothing in ♥, nothing in ♣, odd ♠, even ♦) Perforce A♠, KQ of ♦

Here's a Byzantine method with the following rules: 1) K-parity first (with K=2 QPs always) 2) Singletons and doubleton suits are scanned first, with ties broken in rank order. Skip with nothing or stop with A/K 3) 4+ cards suits are scanned in length order with ties broken in rank order. Stop with even and continue with odd 3♠: 7 QPs 4♣ - 4♥ (relay; odd K-parity, K♠) -> At this point, slam is dubious 4♠ - 5♦ (relay; odd hearts (Q♥, even ♣) Since we must have the C(A) to make up the remaining 4QPs, C(Q) follows.