Hanabi set-packing puzzle

Background reading: my previous post on computerized Hanabi hint strategies.

Suppose our computer player is looking at the public information known about a given player’s hand. (Ignore the actual contents of the player’s hand; we can’t use it because it’s not public knowledge. Also ignore any information we know about other players’ hands.) For each of the cards in the hand, our knowledge about that card’s identity can be summarized as a table of possibilities (this is CardPossibilityTable in InfoBot’s source code). In fact, let’s not even consider the value of the card, but merely its color.

For example, here’s one possible table representing the public information about the colors of the target player’s hand:

Red White Yellow Green Blue
Newest 1 1 1 1 .
. 1 1 . .
. . 1 1 .
Oldest 1 . . . .

This table reports that the player’s oldest card is definitely red; their next card is either yellow or green; their next card is either white or yellow; and their newest card is definitely not blue.

Now consider all the color hints we could give this player. There are only (at most) five such hints: “red”, “white”, “yellow”, “green”, and “blue”. Some of these hints might be known-impossible to give (for example, “blue” is definitely not a valid hint in the example above). Some hints might be known-possible to give (for example, “red” is definitely a valid hint in the example above). In many cases, we won’t know for sure whether a given hint (e.g. “white”) is possible or not, just from the public information.

For any given hand, we can partition the five color hints into one or more sets, such that each set in the partition definitely contains at least one valid hint. For example, in the above example, a trivially correct partition is

(red, white, yellow, green, blue)


but a better (higher-cardinality) partition is

(red, blue), (white, yellow, green)


Notice that it doesn’t matter into which subset we place the known-invalid hint “blue”. Also notice that the target player doesn’t necessarily have any white cards at all; nor any yellow cards at all; nor any green cards at all; but we know that he must have at least one white or yellow or green card, and so the set (white, yellow, green) must contain at least one valid hint, even though we can’t say for sure which of the three hints is the valid one.

Notice also that this partition is maximum-cardinality. It is possible that the target player’s hand is composed entirely of red and yellow cards, such that there are only two valid color hints. And we found a partition into two sets. Can’t do better than that!

The puzzle is: to find an algorithm that can compute a maximum-cardinality partition for any input table of possibilities.

Solution

This problem can be reduced to the problem of selecting a maximum-cardinality set of rows from the table, such that the columnwise sum of all the selected rows does not exceed (1, 1, 1, 1, 1).

Red White Yellow Green Blue
Oldest 1 1 1 1 .
. 1 1 . .
. . 1 1 .
Newest 1 . . . .
Sum of bolded rows 1 . 1 1 .

Another way to express the “columnwise sum” constraint is to say that the rows we pick must all be disjoint.

This is known as a set packing problem, and the general case (for arbitrarily many “colors” and arbitrarily many “cards in hand”) is known to be NP-complete.

Once we find our maximal disjoint set (which can be done by a brute-force search of all 24 possible subsets of 4 rows), we turn each row into a set of colors —

(red), (yellow, green)


— and then produce a partition of the full color space by placing the untouched colors (blue and white, in our example) into the partition at arbitrary spots.

(red, blue), (yellow, green, white)


Q.E.D.!

Posted 2018-04-09