This page is devoted to *Star, the latest game in the series (Mudcrack) Y, Poly-Y and Star, games by Ea Ea (that's me, previously known as Craige Schensted). *Star is what those other games wanted to be.
Kadon is planning to produce a beautiful *Star set this year.
Mark Thompson is planning to host an Internet *Star tournament. If you would like to play *Star on the Web, go to the online page.
The present page is a trial run for the booklet to go with the Kadon *Star set. Any suggestions are very welcome.
Rules for *Star
*Star is an abstract strategy connection board game for two players, the end product of the evolution from Hex to Y to Poly-Y to Star to *Star. It is played on the board shown here.
The players take turns, each player on their turn filling in one cell (not previously colored) with their color. Of course, if the players continued to play until the board was entirely filled in, then the game would end at that point. But almost certainly the game will end with the players agreeing on what the score is long before the board is completely filled in.
The 50 cells on the perimeter of the board (pericells) each contain one "peri". The 5 peries in the 5 corner pericells each have one "quark" associated with them. A player has two goals: to capture peries and quarks, and to connect "stars" together.
If a player has a connected region in their color which "owns" two or more peries, then that region is a "star". Both players can use the special star-shaped "bridge" at the center of the board to make a connection, but neither player may play in it. A star "owns" all of the peries which it contains and all of the peries which it encloses which are not owned by another star. In addition the player who has 3 or more of the 5 quarks gets an additional peri.
Each pericell on the perimeter of the board contains a peri, indicated on this figure by a little star.
Here Blue has two groups. One of these groups is a star owning 3 peries, containing two and enclosing a third. The other blue group is not a star. It is a mere "spark" containing one peri. A group cannot "own" just one single peri. The peri which this blue group contains is owned by the red star which encloses it. The Red group is a star owning 5 peries and a quark (the player who owns the peri in a corner cell also owns the quark in that cell) --- containing 4 peries (and a quark) and enclosing the peri which Blue's spark contains.
There are several very different looking ways of scoring *Star which lead to exactly the same results as to who wins and by what margin.
I will discuss the latter two in more detail now.
- In one each star is given a score depending on the number of peries it owns, the number of "arms" it has and whether or not it uses the bridge. This method of scoring can be considered to be a generalization of the scoring for Star used in the 1983 article in Games magazine in which each star has a definite value independent of how many stars either player has. It is the method which I gave as the basic method on this page until AUG22,01, but it is unnecessarily complicated so now I am opting for ...
- a method in which the score for a star is just the number of peries it owns, but a player is given a "reward" which is twice the difference in the number of stars the two players have (negative for one player and positive for the other).
- In the third principal method a player's score is the difference between his score and the other player's score in either of the other two methods.
You get one point for each peri you own (including the peri you might get for owning three or more quarks). In addition you get a "reward" for connecting your stars together equal to twice the difference between the number of stars the other player has and the number of stars you have.
Note that a "reward" is not always something you desire, as in "He got the reward due him for all of his evil deeds." Also note that the difference between Poly-Y, Star and *Star is that in Poly-y there is no reward for connecting stars together, in Star the reward is equal to the difference in the number of stars which the two players have, while in *Star the reward is equal to twice that difference.
Your "reward" is positive if the other player has more stars, and negative if you have more stars. In short your score is
the number of peries you own
plus (or minus) your reward.
The sum of the two player's scores is equal to the number of peries on the board (including one "quark peri") --- 51 on the "tournament" board, 31 on the "junior" board (the board heavily outlined with 7 cells along each edge in the middle of the tournament board). So the winning scores are 26 and 16 on the two boards.
There is an alternative way of counting which the editor of Games magazine, R. Wayne Schmittberger, likes to use. His approach is to subtract his coplayer's score from his score. If you use this approach your score is
the number of peries you own
minus the number of peries your coplayer owns
plus (or minus) your REWARD.
Remember to count a "quark peri" for the player with more quarks.Your REWARD is your reward minus your coplayer's reward (which is the negative of your reward). So your REWARD for connecting stars together is four times the difference between the number of stars your coplayer has and the number of stars you have, positive if the other player has more stars, and negative if you have more stars. If you choose to use this method of keeping score then you win if your score is greater than zero, lose if it is less than zero.
In this example Blue has one star which uses the 5-pointed bridge in the center to connect two parts. This star contains and thus owns 7 peries and two quarks. Red has three stars. The large one using the bridge owns 8 peries (containing 5 and enclosing 3) and two quarks.
Another Red star owns 3 peries (containing two and enclosing the third) and a quark. Red's third star contains and thus owns 2 peries.
All told Red owns 14 peries (including one "quark peri" because Red has 3 quarks). But Blue has been much more effective at joining stars together --- Blue has only one star whereas Red has three. Thus Red gets a "reward" of minus four. Red's score is 14-4=10. Blue gets a reward of plus four, so Blue's score is 7+4=11. Blue wins. As always on this small "beginner's board", the sum of the two player's scores is 21.
Note that the Red star which owns only two peries does not benefit Red at all. The negative "reward" Red gets for having another star exactly cancels out the two peries. The score would be exactly the same if Blue occupied one or both of these cells, or if neither player did. This is a positive feature which sets *Star apart from Star. Such a "trivial" easy-to-make star is worthless in *Star whereas it is worth one point in Star.
In the next example both players have 3 stars. Blue has a 2-peri star. But, unlike Red's 2-peri star in the previous example, this 2-peri star is not worthless because it also owns a quark --- which in this case is crucial because it gives Blue a third quark and hence one more point for the quark peri.
Red also has a 2-peri star here which is not worthless because it separates two of Blue's stars. If this Red star were colored Blue then Red's loss of two peries would not be compensated by an increase in Red's reward as it was in the previous example because here, unlike the previous example, both players would have one less star. There is a Red "group" (of one cell) which contains just one peri. This is just a spark, not a star --- and the peri which it contains is owned by the Blue star which encloses it, which thus owns 5 peries. Red owns 10 peries. Blue owns 11 peries (including the quark peri). They have the same number of stars, so the rewards are zero and Blue wins, 11 to 10.
In the above two examples any of the unoccupied cells could be filled in by either player with absolutely no effect on the score, so obviously the game is over. However the players usually stop when both players can see what the score will be if both players play correctly.
In the first example the game might have stopped at the point shown in this figure. Here if Red plays on any cell marked A, B, C or D then Blue can defend by playing in the other cell with the same label. And if Blue plays in any of the cells marked E, F, G or H then Red can play in the other cell with that label. The score will be the same as in the first example. These are examples of the "two-way stretch", one of the most elementary tactics.