World

Game

Of

Sprouts

Association

Ever since a former math teacher passed me a book filled with mathematical fun, I became obsessed with a game that was not at all the purpose of his giving me that book. You’ve guessed what game I’m talking about... The book contained the original Sprouts (now referred to as Princeton Sprouts) and Brussels Sprouts, which is a solved game as I’ve read some days before. Well, and another version which, lacking a name, I’ve been calling "Black and White Sprouts", for the possible moves which I’ll explain further down. This version is normal Sprouts, but with the exception that the spot you add to a newly drawn line is possible rather than mandatory. You can leave it out if you like, of course only immediately after drawing the new line. Several people have told me that this seems to make Sprouts only more complicated, but as I’ll show that those rules do change Sprouts, but make the game not a bit less intuitive to play and, above all, fun. And, to say it right away, I’ve solved it.

Basic theory: The key to this version are the liberties each spot and therefore region contains. As well, keep in mind that in one turn you are able to take away either one or two liberties - by drawing a line and adding or leaving out the spot. In a normal game (I’ll refer only to normal games) you therefore will lose if you leave your adversary 2 or 3 liberties, for he will in either case leave you with only one, which is an exhausted position. That means that if you leave your opponent with 4 liberties, you will win: He will return either 3 or two liberties to you, which you’ll finish off. 4 liberties are a so-called "Killing Point" (KP), as you have seen. Another Killing Point are 7 liberties, because for every move (only open moves so far) your opponent makes, you will be able to reach 4 liberties and be on the winning track again. The formula for KPs is therefore 3n-2, where n is an integer above zero. 1, 4, 7, 10, 13 etc. are KPs. This means that the first move in a game starting with n spots (3n liberties) must be a line without a new spot, subtracting 2 liberties to form 3n-2 again.

Biospheres: I’m happy you’re still with me! "Are there KPs for different regions or even biospheres?" you might ask, and yes, there are. For a simple explanation, let me remind you that when you pass a KP to your opponent, you will make the last move. It would then be your opponent’s turn, but he cannot move anymore unless there are two biospheres with both KPs in them. Then, he has to move again in the remaining vital biosphere and you’ll still win. If one biosphere has a life force of 7 and another one has got one of 4, the two have KPs in them and you’ll win. Now, what if one biosphere has got 2 liberties left, and another one does also? You’ll win, too, since your adversary has got no choice what to do. Following the logic, 5/2 is also a Killing Point: If the opponent erases the sphere containing the two liberties, you’ll extract one liberty from the large field, leaving him again with 4 - KP. If he does anything in the large field, you’ll take care that the game gets back to 2/2 - KP, which makes 5/2 a Killing Point for all possible actions of the opponent lead to another KP or defeat. The same works with 3/6, by the way. You see that it’s the difference of a multiple of 3 that decides the game. Any biosphere’s life force can be expressed by x+3n. In the case of life force 7, life force=1+(3*2). In the case of 6, life force=0+(3*2), and so on. The term expressed with "x" is the "base" of the sphere. If there are two spheres that fit the formula and share the same base, you have got a KP: 4/4, 4/7, 7/13, 2/2, 2/5, 8/5 or 6/12 are just examples. Divide by 3 and if both spheres yield the same undividable fraction, it’s all fine and dandy. One more thing: There can be any number of biospheres, it can be always determined if we’re dealing with a KP or not. Just check if there are pairs of spheres that share the same base. The "classic" Killing Points may of course stand alone without forming the pair. So 5/8/7 is a KP and so on.

Spheres and Pivots: Let’s start the very final part of this paper with a hypothesis: If a sphere contains one or more pivots, is therefore somehow (indirectly) connected to the outside, it cannot be a KP. Since I know that "Just trust me" may not completely convince you, remember that a sealed biosphere can very well be a KP with the right base. Now, a sphere with a single outer pivot can have three different bases (1 to 3, 4 would have the base of 1 again). The pivot’s single liberty is not included in the life force of the biosphere, so if a pivot is on a line separating two spheres with equal bases, just eliminate it by drawing a line from the pivot to some point inside the sphere, but do not forget to paint a new spot on the line to keep the bases at balance. Here’s your KP. Take spheres a and b, for example, and their bases. All is fine if (base a - base b)=0. If (base a - base b)=1 or =-1, then do not draw the extra spot on the new line and choose into which sphere you’ll draw the line. You’ll subtract one liberty there, bringing the bases back to balance. In a sphere with 3 pivots, this time really trust me or try it out yourself, no sphere with 1 or 2 pivots on it can ever be KP, it’s suicide.

Coming to an end, I hope you’ve seen why I call this variant "Black and White", because a move either enables your opponent to pass you a Killing Point or yourself. One can only win if the other makes mistakes or if one moves first: Unfortunately, it struck me that this Sprouts version has got a winning strategy: It is clear that Left has an advantage because he may make his first move subtracting 2 liberties, leaving Right with 3n-2 ("4+ 1()2 I"). Now, it could be possible that Right saves the day by creating an open sphere ("1()2 [3,4] II"). We have got a sphere with two pivots now, which can be eliminated by connecting themselves, taking their third liberty and yielding two separate biospheres with 3n/3m liberties in them - KP. A last chance would be a closed move "1(5)2 [3,4]", which means again separating a new region (2 outer pivots) and creating a new spot on it (now 3 outer pivots). If you try possible answers by Left on paper, you’ll see that such an enclosure is suicide for Left can counter with a move that leads to a KP again.

Apart from this method of empirical proof it is possible to formulate the solvability of Black and White Sprouts theoretically: Since the existence of Killing Points, there are either white (good) or black (bad) moves, which will lead to reaching a KP or to enabling the opponent to do so. Consequently, there is no ambivalent combination possible that wouldn’t determine already the game’s outcome. Such a combination, if it existed, couldn’t lead to a decision (for it would be either a KP itself or suicide), it could only lead to a new ambivalent combination and therefore ultimately to a draw, which in Sprouts is not possible. And since every combination is therefore black or white, so must be the initial game setup, already determining the winner. Saving this version by alternating the game setup like introducing a few spots with more or less liberties won’t change anything.

Since this game is solved, I might now switch to Princeton Sprouts. I hope, though, that my insights may be of worth to some interested in this version of the game, and maybe aid in research for other Sprouts versions. I believe that the Black and White Theory of the inexistence of ambivalent points (given a draw is impossible) and therefore the original (and definitive) advantage for one player can be proven in Princeton Sprouts as well, if not for any game.