# Hamilternion Games

**Hamilternion Games (based on Hamilton's quaternion multiplication)**

**In praise of Hamilton and quaternions (RTE)**

Some basic familiarity with the 8-element non-Abelian group { ±1, ±i, ±j, ±k } is assumed.

The key equations are: i^{2} = j^{2} = k^{2} = ijk = -1

(or equivalently i^{2} = j^{2} = k^{2} = -1, ij = k, ji = -k, jk = i, kj = -i, ki = j, ik = -j).

See http://mathworld.wolfram.com/QuaternionGroup.html.

There are several online sources for computing quaternion products.

**Rough outline of basic Hamilternion card game**

Two players (Aodh & Bea, pronounced A and Bea, respectively) each get half of a shuffled deck. They each hold their cards so they can see their own card faces, and take turns “leading tricks” until one person (the winner, by definition) completely runs out of cards. The skill involved is in trying to set up situations (based on quaternion multiplication) where the other person has difficulty playing (namely getting rid of cards).

What’s a trick? First Aodh puts a random card C from his stash face up on the table and Bea does likewise with card D from her stash. A now gets to decide which order (CD or DC) the cards are played in, and then Bea has to play a 3^{rd} card which is dictated (in part) by the quaternion product CD or DC.

Aodh has tried to arrange it so that Bea is forced to play a card E which she (Bea) does not have. At the start of the game if Bea doesn’t have the required card, then Aodh must have it, so Aodh now plays as Bea is unable to do so, thereby Aodh gets rid of another card, and then Bea tries to play the new card F determined by the new quaternion product, namely CE (if DC had been played earlier) or ED (if CD had been played earlier).

If Aodh is clever, and has thought 2 moves ahead, then Bea will also be missing F, so Aodh has unloaded 3 cards and there are 4 cards are on the table, “a complete trick”, and those 4 cards are set aside face up, so that all 4 faces are visible. This new pile is the discarded pile.

Now it’s Bea’s turn to put a new card on the table face up, as does Aodh, but this time Bea gets to decide which order they are to be played in. She tried to arrange it so that she herself has the next two required cards, or failing that, at least one of them is in the discarded pile. Aodh can play cards from his own hand or from the discarded pile, the latter not decreasing his own card count.

Then it’s Aodh’s turn again. Etc.

Sometimes a trick ends before it’s four cards. For instance, if KC (King of Clubs) is followed by any other Club, the “product” explained below is KC again, which can’t be played at this stage, so those two cards are set aside face up in the discard pile.

Four (or three, or two) card tricks are completed one by one until… hopefully one person runs out of cards first.

The discarded pile is scattered face up and available to both players at all times.

__The quaternion product of two cards__

There are actually two components to this, suite and values. There are 4 suits, but 8 quaternions (in the 8-element quaternion group, we’re not considering the 4-dimensional real algebra). *The suit order for us is Clubs, Hearts, Spades and Diamonds, (CHaSeD), which we identify with 1, i, j, and k, respectively* (note the lower case letters here, not to be confused with the capital letters J and K representing Jack and King).

So the product CC (two Clubs) is C, and the product CH (a Club followed by a Heart) is H. The product DD of two Diamonds is -C, and a Heart followed by a Spade is HS is D, whereas a Heart followed by a Diamond HD is -S. Negatives we will come back to: they “get passed to the values”.

Values get multiplied (addition is another possibility) mod 13. So for example (3S)(7D) = 8H, compared to (7D)(3S) = -8H which we interpret as 5H.

This chart was kindly computed and shared by David Plaxco:

Tayley Cable for Hamilternion products

The image shows all possible combinations, where a product XY of two cards is found by looking up X in the column on the left and Y in the row across the top. Generally YX is not the same as XY (though both will have the same suit).

**The Hamilternion idea in action**

**Possible variations**

- Addition mod 13 (not multiplication) is used for value products.
- The person leading gets to decide if the follow up card uses addition or multiplication for the values (Don Chesley’s idea).
- The person leading has the other person select a (random?) card first then picks one based on that, also getting to decide which order they are considered in.
- The discarded cards are kept in separate face-up piles, and some kind of credit is assigned for having the larger one.
- The discarded cards are not used. This soon leads to dead ends, of course, and the winner could then be determined by considering who has the fewest cards left and/or by the sizes of separate discard piles.
- Instead of negative i j or k having the minus passed to the value as the value’s additive inverse mod 13, the multiplicative inverse mod 13 could be used (Kings which have value 0 would then need special treatment).
- This game can be modified to work with the dihedral group of order 8, where the CHaSeD suits represent some 4 of the 4 rotations and 4 flips.

**Comments**

"The idea of using the quaternion multiplication table as a foundation for rules of a card game seems a very fertile one, with many possible variations to explore!" (Don Chesley, MoMath, 10/10/2018)

"I played around with a possible computer implementation of it, and it seems like our computer overlords would be brilliant players (surprise, surprise)" (Harun Slijak, Trinity College Dublin, 11/10/2018)

Got comments or new variations of interest?

Or just a report of how it went playing this?

Please alert us ASAP!