TEACH US

The text is given away by the first line of the poem, so this round is to help teach you the basic mechanic. Each hand/number pair encodes a letter; the main puzzle is to discover the encoding scheme.

First, notice that the number is always in the range 1-14. Since a Tichu hand consists of 14 cards, this suggests that the numbers correspond to indices within a hand. Indexing gives a card, and we can associate (non-special) cards with the numbers 1-13. The image hand direction will give another bit, together making for 26 possibilities.

Given the information presented, the most natural indexing scheme is order of play. However, it's not clear in which of the four hands we should index.

Knowing the clued phrase, some experimentation yields:

• When the hand image points up, we should index into the hand of the first player who went out (player 3).
• When the hand image points down, we should index into the hand of their partner (third to go out, player 1).
• On the top row, a hand pointing down means second half of alphabet.
• On the bottom row, a hand pointing up means second half of alphabet.

This fully specifies a way to extract a letter. For example, on the top row, we had a 5 and a downwards-facing hand. This means to look at the 5th card played by player 1, which was a 7. A downwards-facing hand on the top row means to add 13, yielding 20. The 20th letter of the alphabet is T. Continuing like that, one recovers the first line of the poem:

This round was mainly to make sure the solver could execute the mechanic. Doing so yields the clue phrases lucky and skilled.

In this round, we must for the first time fill in cards to make the round sensible.

• First, we can see by counting known scores that the protagonist team is players 2 and 4 (where 1 is the player with the mahjong). By counting the number of points the opponents score, we see that they must score exactly 5 additional points. The only possibility for this is for the unknown card in trick 3 to be a 5. Their remaining tricks must not earn them points.
• The first tractor in trick 2 cannot contain a pair of fives, and so must contain a pair of threes.
• Also, by counting the total cards played, we see that only 1 card in the deck is left unplayed.
• The third play in trick 2 must be either a bomb or tractor. The only possibility for bombs is a bomb of 4s. In this case, the earlier tractor in that trick is with 2s and 3s. Now there remain a single 2 and 3 left. But they are unplayable, since there is no 4 to make a straight, so this is impossible. Thus the play must be a tractor, and we see by process of elimination it must be 99TT.
• Now, the play of 6 cards, by process of elimination, cannot be consecutive pairs. The only possible straight is 234567.
• The triple in trick 8 now must be of 4s.
• What remains is straightforward.

This yields the following result:

1(A) A . . . 2233 5566 . . 99TT . . . 5 7 . 8 Q A . . . 88 . . JJ QQ . . . JP(J) . . . D(R) . . - d - 444 . - . 6789T . - 9TJQK . . - 234567 . . - KK AA . - . 2 K - . . - - 3

Then, extracting yields the clue phrases wise and good-willed.

• First, notice of the 5 unplayed cards, four of them are the 4 special cards. Dog is recognizable by the trick pattern, and the rest are recognizable by the choice in parentheses. We will assume no special cards, throughout.
• Notice that in the second trick, both hands must be full houses, since no straight (bomb) can end in a 4.
• In trick 5, the only possibilities are 445566, and 456789. 456789 is impossible since a fourth 4 and 9 prevents the 7-card straight from being possible. Thus it is 445566.
• Since all 4s are used, the straight must be 56789TJ.
• By process of elimination, the card in trick 6 is an A
• Now, by process of elimination, the full houses are 7s over 9s, and 8s over 4s.
• So of the 2 remaining aces, at least one is played. In fact, only there is only 1 valid spot, in the last trick. And the last unplayed card (besides the special cards) is also an ace.
• The last 3 cards are straightforward.

And so we have

222333 . . . 77799 88844 . . . 2 3 9 A . . - - QQQQ . . . TTT JJJ . . . 445566 . . . 6 9 . A - KKKK . - . 56789TJ . - . A

Then, extracting yields the clue phrases husband and wife.

• First, notice that player 2 plays a ten in trick 9. Thus they must play a ten in trick 1.
• Now, the last trick can have no more tens, and thus is a straight, 6789T.
• Trick 2 also can't have more tens, and thus is a straight, 567P(8)9T
• Trick 7 contains a bomb, either jacks or aces. Suppose it were aces. Then trick 9 has JQK, trick 1 has JK, and so trick 11 is impossible. Thus it is jacks.
• Trick 9 is thus Q K A, and so trick 1 has K A, and so trick 11 an A
• The remaining cards are 223344556677888. There can be at most two full houses from those cards, and so the second play in trick 3 must be a straight. Thus it contains 456. This means all 6s are used, so the first play cannot be a full house either. Thus it is a straight, 23456.
• So trick 6 must be 7788
• Lastly, the trick 3 straight must be 23456, a straight flush bomb!

And so we have

1(T) T K A . . . 567P(8)9T . . . 23456 . . 23456 . . . d - 44 99 . . QQ . . . 7788 . . . KK JJJJ . . . 2233 . . . T Q K A . . . D(L) . . . 5 Q A . - . 6789T

And extracting yields the clue phrases grand vows and fulfilled.