NxN Go

Similar to the last problem from the appendix in that we’re taking an actual game and extending it.

The problem: NxN Go.  This is problem GP10 in the appendix.

The description: Given an integer N, and a position (consisting of black piece locations, white piece locations, and black piece locations) on an NxN go board, and the name of the current player’s turn, does white have a forced win on the game?

Example: Instead of a true example, what I think is most relevant here is a little discussion of the rules and basic strategy of Go.  In Go, players take turns playing stones on a grid, with the goal of surrounding spaces on the board with pieces of your color:

(all pictures are taken from the tutorial at https://www.pandanet.co.jp/English/learning_go/learning_go_2.html):

In this case, white has surrounded 9 spaces (stones are placed at intersections, including the edges of the board).

If a stone of the other color can be surrounded on all 4 sides by a stone of your color, the stone is captured, scoring you a point at the end of the game:

In this case, the black play at “1” captures the white piece (and likely gains black an additional point for having surrounded territory at the end of the game)

More than one piece can be surrounded at a time, but not all configurations of stones can be surrounded.  Most notably, a configuration with two “eyes” (or empty holes)  cannot be captured, for example:

Here, the white play at “1” creates two eyes.  There is no way for black to surround all of the white pieces since they would have to play in both empty holes, and as soon as black plays in one of the holes, the black piece is surrounded and immediately captured.

So, the main goal of Go (and the proof) revolves around creating “safe” structures that contain two eyes and using them to capture as much territory as possible.

Reduction: The paper by Lichtenstein and Sipser reduce from Planar Geography.  The idea is to have a set of safe territory for White, and another threatened set of stones large enough that if it can be made safe, White will win, but if it can be captured, Black will win.  Here’s the picture from the paper:

Black pieces surround white all the way around, so the only escape would be for white to extend the “pipe” on the left around to the safe white spaces (or some other group of two eyes, which will keep the large group alive).  Then we encode the vertices and edges in the graph as sets of stones, where each “choice” of going through an edge is reflected by a choice of where stones are placed.

There are many types of subgraphs and corresponding board positions in the paper, here’s just one of them:

(Graph position)

(Board position)

Here’s the general structure of the arguments that show how the play “has” to go through this vertex.  Suppose we are coming from the top, and white wants to go left.

  • If White doesn’t play at 1, 2, or 3 first, Black will play at 2.  White will now have to play at 1 to keep the middle vertical strip (which connects back to the big threatened set of pieces) alive.  But then Black plays at 3 and takes them all anyway.
  • Even if White plays at 3, Black wins by playing at 1, then White moves to 2, then Black plays at 5
  • If White does play at 1 or 2, (let’s say 1, because that will take us left), Black has to respond at the other point (so, 2 for us).  If they don’t, White plays at 2, and 3 black stones below the 1 and 2 are captured, and White will be able to connect to the two eye group below.
  • After Black plays at 2, white needs to go to 3 to build a line of white stones coming in from the top, and going out to the left.  Black plays at 4 to stop white from connecting to the group of 2 eyes on the left.

As a result, the “edge” going through this vertex and coming out to the left has been chosen.

Difficulty: 7. I think this is easier to see than the NxN checkers reduction, but still takes a lot of cases and structures to realize.

Planar Geography

In the process of reading up on the next problem in the appendix, I saw this nice reduction for a generalization of Geography that we used last week and will use next week.  So I thought it would make a nice addition to post it here.

The problem: Planar Geography.  This problem is not in the appendix (though the reference to this problem and the paper with this reduction appears in the notes to the “Generalized Geography” problem).

The description: Given an instance of the Generalized Geography problem, but where the graph G is planar, does Player 1 have a forced win?

Example: The example we had in the Generalized Geography problem is actually a planar graph:

Reduction: The paper by Lichtenstein and Sipser which has the reduction for next week’s NxN Go problem has this in the ramp-up.

What we’re doing to do is re-build the graph that was created in the Generalized Geography reduction.  That graph may not be planar. Any time the graph has two edges that cross:

We replace it with the following construction

(figures are on pages 395 and 396 of the paper).

The important things to realize are these:

1) If we use this on the graph generated from the original Geography Construction, it will never be the case that we will use both crossing edges in one solution.  This is an important point that isn’t really addressed in the paper.  The Lichtenstein and Sipser paper has a slightly different reduction for Generalized Geography where it’s easier to see.

2) Suppose it’s player 1’s turn, and they enter the intersection going up.  Then it will be their turn again once they hit the middle point.  The point of the extra loop structures (instead of just adding a vertex at the intersection point) is to create a situation where if player 1 decides to turn right, player 2 can turn down, and make player 1 lose.  This works in both directions.

Difficulty: I was hoping that this would be a good difficulty 4 or so problem in the middle of all of these really hard ones, but the whole “you’ll never use this intersection twice” problem makes the construction hard to see, and it’s a tough thing to think about why it matters, even if you told the student the rule.  Maybe this is a 5 if you give them that information.

NxN Checkers

Back from my trip with a simple problem to explain, but a hard reduction to do.

The problem: NxN checkers.  This is problem GP10 in the appendix.

The description: Given a position on an NxN checkerboard, does black have a forced win?  It turns out the reduction will also work if we restrict the board to only having kings on the board (and so no “un-kinged” pieces)

Example: The “NxN” requirement is there since on a standard 8×8 checkerboard, there is a finite set of moves, and so theoretically you could solve the problem in O(1) time (for a really large constant factor, of course).  The starting configuration adds extra rows and columns of pieces to the board, still leaving two blank rows in between the two pieces.

So, let’s do an example on a 4×4 board.  The starting configuration is this:

* *
O O

(the dashes are empty spaces, * is Black, O is White)

Here is a configuration of pieces that will lead to a black win:

*
O
*

If it’s Black’s turn, they should move the piece in the second row up to either location on the first row (recall that all pieces are kings).  Then White’s only move is to go to the space Black just vacated, where it will be jumped, giving Black the win.

Reduction:

The paper by Fraenkel, Garey, Johnson, Schaefer, and Yesha contains a pretty detailed description of the reduction, which contains lots of complicated structures.  I’ll just give the general idea here.

The reduction is going to be from Geography, which is still NP-Complete even if the graph is bipartite and planar.  They create several structures to help them build their instance of the checkers game.

The first is what the call a phalanx– an open rectangle of (say) White kings that surround the (say) Black pieces.  The idea is that since there is no way for the Black pieces to jump anything in the rectangle, then White can “shrink” the phalanx towards Black, running them out of room to maneuver.  Here is a picture of a small phalanx on a 6×6 board:

O O O
O O O
O X O
O O
O O
O O O

..notice that whatever Black does, they will be captured on their next turn.  This remains true no matter how many Black pieces are trapped inside the phalanx, and no matter how much open space is inside the phalanx (White can use their moves to shrink it over time).

The key to the reduction is to build a set of interlocking “potential” phalanxes- situations where a Black king may be able to escape the phalanx.  If it can, Black can jump White’s pieces and win, but if it can’t, the phalanx will engulf Black and they will lose.  The geography instance is placed in the center of these potential phalanxes in such a way that a Black king can “escape” the Geography instance if and only if Black can win the geography game.  The reason why the Geography graph had to be planar was so that we could directly represent the vertices in the graph as positions on the checkerboard.  The reason why the Geography graph had to be bipartite was so that edges going from the first vertex set to the second could be all Black pieces, but the edges going from the second set to the first could be all White pieces.

The game starts with black at the “start vertex” for the geography problem, and jumping a line of White checkers:

When a vertex has more than one possible exit, that leads to more than one possible set of checkers to jump for the other player:

(This is part of figure 10 from the paper.  Here, after White jumps down the chain of Black pieces, Black can choose the chain of White pieces to jump through.)

The construction takes advantage of the rule in checkers (which I was not aware of until I was in my twenties!) that if a player can make a jump, they must make a jump.  So as long as players can jump checkers along these chains (alternately, as long as they can follow edges in the geography graph), they will.  As soon as a player cannot make a jump they will be able to deal with the Black king that can either escape the phalanx structure (and win for Black) or trap it (and win for White).

This is the general idea of the reduction, there are a lot of details that I am glossing over.

Difficulty: 8.  This is a bit hard to see and very hard to come up with, and it’s very easy to get lost in the weeds of the details.  I do like the way that the “removal” of edges from the Geography problem is modeled by the actual removal of pieces from the checkerboard, though.

Annihilation

I’m going to be out of town for a few weeks, so the next few posts might be delayed from my already slower schedule.

The problem: Annihilation.  This is problem GP9 in the appendix.

The description: G&J’s description is a little obscure, so we’ll go with the one in the paper by Fraenkel and Yesha that has the reduction.

Given a directed graph G=(V, E), and r subsets of E, E1 through Er.  The subsets may not be disjoint, but each edge of E is in at least one subset.

We’re also given r different types of tokens, placed on vertices of the graph.  Each token type corresponds to one of the r subsets of E.  A player moves by taking a token (of type i) and choosing an edge (u,v) in set Ei, where u is the current position of the token.  The token is moved to vertex v in the graph.  If 2 tokens ever meet on the same vertex, both are “annihilated” and removed from the game.  A player loses if they cannot make a move.  Does player 1 have a forced win?  (Though note that the Fraenkel and Yesha paper actually proves whether player 2 has a forced win)

Example: Here is a simple example that will hopefully be useful in the reduction that follows:

In this graph, the red edges are in E1 and the blue edges are in E2.  The red vertices currently hold a type 1 token, and the blue vertex holds a type 2 token.

If player 1 makes any move except moving the token on g to the vertex h, player 2 will be able to move a red and a blue token onto the same space, annihilating both.  Then there will be just 2 moves left in the game (the 2 remaining red pieces moving along their edges), the first moved by player 1, the second moved by player 2, then player 1 will have no move and will lose.

If player 1 moves from g->h, then there are 4 more moves in the game (the h->i edge, and the three red edges).  Thus, player 1 will get the last move and will win.

Reduction: Fraenkel and Yesha use Minimum Cover.  I’ll note again here that he reduction will show that the Cover instance is true if and only if player two has a forced win.

So we’re given a collection of sets Si where i goes from 1 to m, and an integer K.  We’re going to build a directed acyclic bipartite graph R= (V, E):

  • The graph has vertices xi and yi for i from 1 to K.
  • The graph has one vertex for each set Si and one vertex for each element ei in the union of the sets.
  • The graph also has two “special” vertices a and b.
  • “Type 1” edges go from all xi to its corresponding yi, and from each yi to all Si vertices.
  • “Type 2” edges go from a to all ei vertices, from all ei vertices to all set vertices that contain that element, between all e and x vertices, and from all x and S vertices to b.
  • Type 1 pieces start on all x vertices.  There is 1 type 2 piece, and it’s on a.

Here is the example used in the paper for the covering problem {{e1,e2}, {e2,e3}, {e3,e4}, {e1,e3,e5}, {e6}} and K=4:

An arrow going to a circled group of vertices represents a group of edges going to all vertices in the group.

Notice that without annihilation, the path a type 1 piece takes is from some x vertex to its y vertex, and from there to some S vertex (2 moves), and the path of the type 2 piece is 3 moves (either a->some e-> some x ->b or a->some e->some S ->b), so there is an odd number of moves, and thus player 1 wins if no annihilations happen.  Player 2 wins if a type 1 and type 2 piece collide someplace (on an x or S vertex).

This is because if 2 pieces of different types collide, we remove an odd number of moves from the game:

  • If they collide on an x vertex, we remove the 2 moves the type 1 piece can make, and the move the type 2 piece could make from x->b
  • If they collide on an S vertex, we remove the one move the type 2 piece makes from S->b

On player 1’s move, they will have to move all pieces off of the x vertices before moving the type 2 token off of the a vertex.  Otherwise, after player 1 moves a->e, player 2 can move e->x (to some x that hasn’t left its starting space yet).

So,  player 1 starts by moving some xi->yi.  Player 2 will move the piece from yi to the next set in the cover.  Recall that there are k different x and y vertices.  So what will happen is that the k vertices that comprise the cover will have tokens on them.

Once all of the type 1 pieces are some S vertex, player 1 will have to move the type 2 piece from a to some e vertex.  If there is a cover, no matter what e vertex player 1 chooses, player 2 will be able to move the token to an S vertex that contains that e element.  If there is no cover, player 1 will be able to choose an e vertex that has no e->S (or e->x) move that causes an annihilation, and player 1 will win.

Difficulty: 7.  This is a very cool reduction, and you can see from the picture how it works.  It’s fun to see how all of the edges and sets work out.

Alternating Maximum Weighted Matching

This is a “private communication” problem I haven’t been able to solve or find an actual reduction for.  I think I have the start of a reduction, though, but I haven’t put in the time to work out the details.  Hopefully, I’m on the right track.

The problem: Alternating Maximum Weighted Matching.  This is problem GP8 in the appendix.

The description: Given a weighted graph G=(V,E), with positive weights, and a positive bound B, each player chooses an edge from E.  No two chosen edges can share an endpoint.  Player 1 wins if the weights of the edges chosen ever exceed B.  Does player 1 have a forced win?

Example: Here’s a graph:

If B=11, and it’s player 1’s turn, then they can’t remove the highest-cost edge (d,e) because every edge is incident on d or e, so player 2 would have no moves, and the game would end with cost 10.

If player 1 removes one of the 6-cost edges (let’s say (a,d)) we’re left with:

..and so player 2 will have to take one of the remaining 6-cost edges, bringing the total cost of edges removed to 12.

Reduction (sort of): So, like I said at the top, the reference in G&J is to a “private communication” by Dobkin and Ladner.  I couldn’t find the actual result published anywhere.  I actually emailed both Dobkin and Ladner to ask if they remembered what the reduction was, but their response (reasonably) was “It’s been 40 years, I have no idea”.

But, I thought about it for a little while, at least, and came up with what I thought are the beginnings of an idea.  I didn’t have the time (or, perhaps, the ability) to get all of the details right, but this feels like a start, at least:

We’re going to reduce from One in Three 3SAT.  Each variable will be represented as a pair of vertices xi and xia, connected by a weight 1 edge.  The negation of the variable (~xi and ~xia) will also be vertices and also connected by a weight 1 edge.  The vertices xi and ~xi will be connected by a “large” edge of weight 10.

From each clause, we build up a component of a graph that looks like this:

The weights of 10 and 1 might not be right, think of them as “large” and “small” weights.  Each of the xi variables corresponds to the actual variables in the formula, we only include the variables that correspond to the clause we’re looking at.

(One thing I did wrong was to assume that player 1 loses if the edge cost goes over B.  So in what follows, player 1 is trying to keep the score low.  We can fix this by making it player 2’s turn and swapping the roles)

The idea is that player 1 will choose either the x1-x1a edge or the ~x1-~x1a edge to “fix” the value of the first variable.  If the variable shows up in the clause (for example, they chose the x1-x1a edge in the diagram above), this will eliminate the edges (x1a,c1), (x1a, x2a), and (x1a, x3a) from being able to be chosen.

Then player 2 will want to choose an “expensive” edge.  He’ll choose the edge (x2a,x3a)

Then we’ll move on to the next variable.  It’s again player 1’s turn to decide on a setting of x2.

The idea is that each clause will have it’s “ci” vertex connect to that central “home” vertex by an expensive edge, so if after all of the variables have been given their values, there still is an edge to the home vertex, it will be chosen, and that amount will be the amount that sends the total cost over the bound.  (So right now, I’m thinking of the bound being something like 11*N +1 for a problem with N variables, at least until the extra things below get added).

What still needs to be done is the detail work (and probably extra edges and vertices) to ensure that players have to choose the edges in the order I specify (i.e., not doing so loses a player the game immediately).  It’s entirely possible that doing so will make this whole construction wrong.  But I like the idea behind it, at least

Difficulty: N/A.  I don’t want to call this a 10, even though it stumped me.  I think if my idea is right, it’s not that hard.

Alternating Hitting Set

I think this might be the last of the problems we do from the Schaefer paper we’ve used so much recently.

The problem: Alternating Hitting Set.  This is problem GP7 in the appendix.

The description: Given some universe set B, and a collection C of subsets of B.  Players take turns choosing an element from B.  Once enough elements are chosen to make all sets in C have at least one element chosen, the player who made that move loses.  Does player 1 have a forced win?

Example: Let B= {1,2,3}, and C = {{1,2,3}, {2,3}, {1,3}, {1,2}}.  Player 1 can win by choosing 1, which hits all sets in C except {2,3}.  Since player 2 has to pick either a 2 or 3, they will hit that set and lose.

Reduction: Schaefer remarks that this is just a special case of Variable Partition Truth Assignment. with sets instead of variables.  Here’s how that goes:

Suppose we have an instance of Variable Partition Truth Assignment, so a CNF formula with all positive variables.  We just create a set B with one element per variable, and the clauses become the sets in C.  Then picking an element of B is the same as setting a variable to true.  Hitting a set in C is the same as making the clause true.

Difficulty: 3.  Maybe this problem is close enough to Variable Partition Truth Assignment that it didn’t need its own article.  On the other hand, it’s nice to see an easy reduction for once.

 

Sift

Finally, we’re ready to go back to the problem in the appendix.

The problem: Sift.  This is problem GP6 in the appendix.

The description: Suppose we have some set X, with collections of subsets of X A and B.  The sets A and B have no elements in common.  Players take turns choosing elements from X until every subset in A (in which case player 2 wins) or every subset in B has a chosen element (in which case player 1 wins).

We do need to have a rule for what happens if an element is chosen that makes both players lose (because the element chosen intersects with the last subsets in both A and B).  In this case, the player who made the move loses.

Example: Suppose X was {1,2,3,4,5,6}.  A = {{1,2}, {3,4}, {5,6}} and B = {{1,2,3}, {4,5,6}}.  Then A could force a win by choosing element 3.  This intersects the second set in A, and the second set in B.  So the current “live” subsets are:

A = {{1,2}, {5,6}}

B = {{1,2,3}}

Player B needs to pick 5 or 6, or they immediately lose.  But once that happens, player A can choose 3, which means the set of elements chosen {3,4,5} (or {3,4,6}) intersects with every set in B, but not every set in A, so A wins.

Reduction: The Shaefer paper reduces from Avoidance Truth Assignment, which means we start with a CNF formula with no negated literals.  Let’s assume the formula has m clauses and n variables.  For each clause k define the set Sk to be the indices of the variables in clause k.   (So if clause 1 was: x∨ x4 ∨ x7, then S1 = {2,4,7}).  Let “Set 1” = all of the Sk sets of even length, and “Set 2” = all of the Sk sets of odd length.  If n is even, give player 1 “Set 1” and player 2 “Set 2”.  If n is odd, reverse the allocation.

Notice that “Set 1” is the set of all clauses with an even number of distinct variables in them, and “Set 2” is the set of all clauses with an odd number of distinct variables in them.  Recall from our discussion of the Avoidance Truth Assignment problem that the player whose turn it was could win that game if we had an even number of unplayed variables, and an odd number of unplayed variables in every clause  (or an odd number of unplayed variables, with an even number in every clause).  What our “Set 1” and “Set 2” constructions are doing is listing out those clauses and assigning them to each player.  Once one set runs out, then we have the winning case for one of the 2 players.  So this construction of Sift is equivalent to the original Avoidance problem.

Difficulty: 8.  I wish there was a good proof of this odd/even claim he makes.  I don’t 100% buy his intuitive argument.

Avoidance Truth Assignment

This is the second of the two proofs in Schaefer’s paper that will get us the Sift reduction.

The problem: Avoidance Truth Assignment.  (My name for it).  This problem does not appear in the G&J appendix.  It is named “Gavoid (POS CNF)” in Schaefer’s paper.

The description: Given a CNF formula with no negated literals (and not necessarily 3 literals per clause), players take turns choosing a variable to make true.  A player loses if after their play the formula becomes true, even if all unchosen variables are set to false.  Does player 1 have a forced win?

Example: Suppose the formula was (x1 ∨ x2) ∧ (x1 ∨ x3) ∧ (x2 ∨ x3 ∨ x4).  Then player 1 could force a win by choosing x2, which makes clauses 1 and 3 true.  The player who chooses x3 will lose, so player 2 should pick one of x2 or x4.  But then player 1 chooses the other, leaving player 2 forced to pick x3, making the second clause (and the whole formula) true.

Reduction: Schaefer reduces from Pre-Partitioned Truth Assignment.  So we start with a logical formula given as a set of clauses, and the variables are partitioned into two sets, one for each player.  We’ll call player 1’s variables x1 .. xn and player 2’s variables y1 through yn.   We’ll assume (or modify the formula to make it happen) that each clause contains at least one x variable, and that every variable occurs both positively and negatively as a literal someplace in the formula.

Our new formula will have:

  • 4 copies of each variable (xi turns into xi, ~xi, xi‘, and ~xi
  • One clause (xi ∨ ~xi) for each x variable
  • One clause (yi ∨ ~yi ∨ yi‘) for each y variable
  • For each clause in the original formula: if it contains an unnegated variable xi (or yi), then our new clause will contain xi and xi‘.  (Or yi and yi‘).  If it contains a negated literal, we’ll use the 2 “negated” variables instead.  This clause will probably have more than 3 elements in it, but that’s ok.

Shaefer notes that if we hit a point in the game where we have:

  1. An even number of unplayed variables, and
  2. An odd number of unplayed distinct variables remaining in every clause.

or:

  1. An odd number of unplayed variables, and
  2. An even number of unplayed distinct variables remaining in every clause

..then the player who’s turn it is can win by picking some unsatisfied clause and picking any variable that does not satisfy it.  So the strategy for the first player is to create the first situation (by satisfying all clauses with an even number of variables), and the strategy for the second player is to create the second situation (by satisfying all clauses with an odd number of variables).  Since the first kind of clause we made (with the x variables) has an even number of variables, player 1 will want to satisfy those clauses as fast as possible.  Similarly, since the second kind of clause we made (with the y variables) has an odd number of variables, palyer 2 will want to satisfy those clauses as quickly as possible.  If this happens, we are exactly imitating the Pre-Partitioned Truth Assignment game.

The remainder of the reduction is a detailed proof explaining the above argument in detail (showing, for example, what happens when players do not play this way).

Difficulty: 8  Maybe they’re getting a little easier for me because I’ve been doing so much of them.  Or maybe because I’m still skipping all of the low-level details.

Pre-Partitioned Truth Assignment

The next game in the appendix is “Sift”, but the Schaefer paper we’ve been using has that reduction depend on two other problems, so we’ll do those first.   Schaefer calls this problem “G%free(CNF)”, I made up the name we’re using.

The problem: Pre-Partitioned Truth Assignment.  This problem is not in the appendix.

The description: Given a CNF formula A, where the variables in A are partitioned into two equal size sets, V1 and V2.   On player 1’s turn player 1 chooses a variable in V1 and chooses to make it true or false,  On player 2’s turn, player 2 chooses a variable in V2 and chooses to make it true or false.  Player 1 wins if the variable assignments made by both players make A true.  Does A have a forced win?

Example: Here’s the equation from last time: {(x1 ∨ x2 ∨ x3), (~x1 ∨~x2∨ ~x3), (~x4, x1, ~x2)}  Lets say that V1 = {x1, x2} and V2 = {x3. x4}.  Then player 1 can force a win by setting x2 to false on their first turn, making the second and third clauses true.  Since player 2 can only set x3 or x4 on their turn, they can’t stop player 1 from setting x1 to true on their next turn, making all clauses true.

Here’s another example where player 1 loses: {(x1  ∨ x3 ∨ x4), (~x1 ∨ x3 ∨ x4), (x2 ∨ ~x3 ∨ ~x4).  If we again have  V1 = {x1, x2} and V2 = {x3. x4}, then player 2 wins by setting both x3 and x4 to false because no matter what player 1 does, they have to set x1 somehow, which will make one of the first two clauses false.

Reduction: (This problem is Theorem 3.8 in Schaefer’s paper).  Just like in most of the reductions in the Schaefer paper, we start with an instance of QBF where we have variables x1 through xn (n is assumed to be even), and the formula alternates with “There exists” (starting with x1, on all of the odd-numbered variables up through xn-1) and “For all” (starting with x2, on all of the even-numbered variables up through xn) quantifiers on the variables, leading into a CNF formula A0.  We need to build up an instance of Pre-Partitioned Truth Assignment by creating a new formula A’ and the variable sets V1 and V2.  For each variable in the QBF formula xi, our formula has 3 variables: xi, yi, and zi.  V1 holds xi, yi, and zi+1 when i is odd, and V2 holds xi, yi, and zi-1 if  i is even.

The formula is pretty complicated but is built to force zi to be set by one player to the same value of either xi or yi by the other player (and so must be set after the corresponding xi or yi is chosen).  The optimal move will be to play zi immediately after the corresponding xi or yi, forcing the choice back on the other player.  So that player will play the other of xi or yi (rather than open up some other variable’s x or y, only to be responded with by the corresponding z).   Then it will be the other player’s turn to choose the next xi+1 or yi+1.

The values of xi correspond to the literals in the original QBF formula, so assuming players play “correctly”, player 1 takes the role of the “there exists” player in the QBF formula, and player 2 takes the role of the “for all” player in the formula.  And if there is a way to set the x variables in the new formula to make it satisfiable, then there is a way to make the original QBF formula satisfiable.

Difficulty: 9.  For this problem, even Schaefer admits the proof is hard to follow, so he spends a page explaining what the goal of the construction is before diving into the proof.  I do like the reappearance of the “triples” of moves where a meaningful decision by a player is followed by 2 forced moves, making the next meaningful decision be placed on the other player.

Variable Partition Truth Assignment

We’re going to dive into a couple of logic games for the next couple of posts before we come out with I think of as more “game-like” games.

The problem: Variable Partition Truth Assignment.  This is problem GP5 in the appendix.

The description: Given a set U of variables, and a set C of clauses over U, players play a game where they alternate choosing variables from U.  The variables Player 1 chooses will be set true, and the variables Player 2 chooses will be set to false.  Player 1 wins if all clauses in C are satisfied.  Can player 1 force a win?

Example: Here is a small example: The clauses will be: {(x1 ∨ x2 ∨ x3), (~x1 ∨~x2∨ ~x3), (~x4, x1, ~x2)}

Player 1 can force a win by choosing x1, which will be set to true.  This satisfies both clauses 1 and 3.  Player 2 does not want to choose x2 or x3 (since any variable player 2 chooses will be set to false, making clause 2 true), so they’ll choose x4.  Then player 1 picks either of x2 or x3 (clause 2 is still not satisfied, because the variable must be set to true by player 1), then player 2 must set the other one to false, satisfying clause 2.

Reduction: The paper by Schaefer that we’ve been going through calls this “GPOS (POS CNF)” and focuses on the subcase where we’re given a formula where all of the variables are positive.  If even that case is NP-Complete, then the generalized case where we allow negated literals is also NP-Complete.  The reduction in the paper is from Sequential Truth Assignment, specifically the case where the problem has 3 variables per clause.  So we’re given an input that is a set of 2n variables, which we can view as alternating between ∃ and ∀, and a set of m clauses with at most 3 literals per clause.  We’re going to build a new formula A’ out of many pieces.  The variables are all positive, but some of the variable names will correspond to negated literals.  We’ll have 2n “x” variables, 2n “~x” variables, and 2n “u” variables.  He then builds up a pretty crazy formula.  The point of the formula is to force a 6-move sequence between the players:

  • On move 6k+1 (so the first move of “round” k), player 1 chooses either x2n-2k or ~x2n-2k to become true.
  • The next move, player 2 chooses the other of that pair to become false.
  • The next move, player 1 chooses u2n-2k to become true
  • On moves 6k+4 through 6k+6, we repeat the same sequence, but this time, player 2 can choose whether to make x2n-2k-1 or ~x2n-2k-1 false.  Then we end with player 2 choosing u2n-2k-1 to become false.

Notice how each set of 3 moves corresponds to one truth assignment of one variable in the Sequential Truth Assignment game (which alternates between player 1 and player 2 choosing the assignment).  Also, notice how player 1 choosing a variable xi corresponds to setting it true in the Sequential Truth Assignment game, and player 1 choosing a ~xi variable to be set to true corresponds to making the corresponding Sequential Truth Assignment variable false. (The opposite assignments occur when player 2 chooses)

The hard part of the proof (and of the formula) involves all of the various ways the other player can force a win if the sequence above is not followed.

Difficulty: 9.  The formula in the paper is very hard to follow, which is why I didn’t go into it here- I think the important part is how it all works out.  Since I couldn’t think of a way to explain it without going on for pages, I figured just explaining the idea was a better strategy.