Shanten: Difference between revisions

Shanten (view source)

Revision as of 16:25, 6 September 2024

147 bytes added , 6 September

→‎Accurate correction (perfect shanten): make the final sentence a bit easier for humans to understand

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 Calculating shanten with the above formula will give a proper result, where -1 indicates a complete hand, while any positive integer is the distance from tenpai.
-However, when the result is 0, there is a chance that there are no existing tiles that could lead to creating a winning hand. If a hand is 0-shanten, but has all 4 copies of its own winning tiles, as in the below example, then it usually does not count as "tenpai" for the purposes of gameplay. (This only accounts for tiles in the hand, not tiles discarded or in other players' hands.)
+However, when the result is 0, there is a chance that there are no existing tiles that could lead to creating a winning hand. If a hand is 0-shanten, but has all 4 copies of its own winning tiles, as in the below example, then it usually does not count as "tenpai" for gameplay purposes. (This only accounts for tiles in the hand, not tiles discarded or in other players' hands.)
-:{{#mjt:3456p111m}} {{#mjt:3'336'66p}}
+:{{#mjt:3456p111m}} {{#mjt:3'336'66p}} Waiting for: {{#mjt:36p}}
-If ''accurateShanten'' = 0, for all possibilities of tile: keep lowest ''accurateShanten''(''hand'' + ''tile'') value (-1 or 0) and add 1 (for 0 or 1). The hand above, when any other tile is added to it, is never complete. A proper test would never make a 5th copy of 3p or 6p.
+The above hand has no possible tiles that it could draw to form a winning shape; there doesn't exist a 5th 3-pin or 6-pin.
+Therefore, when ''accurateShanten'' = 0, then make the following correction:
+* Count the shanten the hand has after drawing every possible remaining tile. (A proper test would never make a 5th copy of a tile.)
+* Take the lowest shanten from the above calculation, then add 1 to it.
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