Bob cannot guarantee a win. Ann's strategy will essentially be to prevent the formation of top-right corners above the x-axis, and to prevent the formation of bottom right corners below the x-axis. Ann's first move will be from $(0,0)$ to $(1,0)$, if any of Bob's subsequent moves are on the x-axis Ann will make an arbitrary move along the x-axis. Otherwise, pair up edges which form the top right corner of a unit square above the x-axis or form the bottom right corner of a unit square below the x-axis. If Bob orients some edge not on the x-axis, Ann will orient the edge paired with it such that either both edges point towards or both edges point away from the endpoint they share. This strategy prevents the formation of top right corners which are part of cycles above the x-axis and prevents the formation of bottom right corners which are part of cycles below the x-axis. But any cycle has at least one top right corner above at least one bottom right corner, which is impossible. Thus Bob cannot create a cycle as desired.

The answer is no; Ann can prevent him from forming a cycle. Suppose Ann makes the first move at some horizontal edge $e$, say. Then we partition all edges other than $e$ into $L$-pairs as shown in the following figure (with $e$ in green): [asy][asy] size(6cm); defaultpen(linewidth(1.1)); path p = (1,0)--(0,0)--(0,1); for (int i=-2; i<4; ++i) { for (int j=-2; j<4; ++j) { if ((i==0) && (j>=0)) { draw(shift(i,j+1)*rotate(-90)*p, lightred, Margins); } else { draw(shift(i,j)*p, lightblue, Margins); } } } draw( (0,0)--(0.9,0), heavygreen); [/asy][/asy] Then, every time Bob plays on an edge in a pair, Ann plays the on the other edge in the pair, orienting it so that either both arrowheads point towards the center, or neither of them do. We claim this strategy works. And indeed, no oriented cycle can ``use a marked corner'', and by construction this prevents the cycle from existing at all. Remark: If Bob goes first instead, then the simpler tiling of lower-left $L$'s works fine, without the need for the green/red perturbation above. Solving this case makes the above solution more motivated.

This reminds me of a game called "Maker-Breaker Percolation" (at least that's what I've heard it called). There are two players Maker and Breaker who color edges of the infinite coordinate lattice with Maker in red and Breaker in blue. Maker's goal is to create a contiguous path in red from the origin which reaches an arbitrarily high taxi-cab distance (or a fixed taxi cab distance in variations) from the origin. Breaker tries to prevent this. Maker goes first. The strategy for Maker to get a contiguous path of a fixed distance $D$ from the origin uses a pairing strategy almost exactly like the one given in the above posts using $L$-pairs. Maker works within the diamond centered on the origin with taxi cab distance/radius $D+1$ and matches Breaker's moves with the corresponding edge in an $L$-pair if it has not been taken. If it has (it can be proven to have been done by Maker in red) or Breaker makes a move outside of the diamond, then Maker moves arbitrarily within the diamond. Eventually, all lattice edges within the diamond will be colored and Maker will have a path of taxi cab distance $D$. [asy][asy] size(6cm); defaultpen(linewidth(1.1)); path p = (1,0)--(0,0)--(0,1); for (int i=-2; i<4; ++i) { for (int j=-2; j<4; ++j) { draw(shift(i,j+1)*p, lightred, Margins); } } draw((-2,4)--(0,4), black); draw((-2,3)--(-1,3), black); draw((-2,3)--(-2,5), black); draw((-1,4)--(-1,5), black); draw((-2,2)--(-2,-1), black); draw((-1,1)--(-1,-1), black); draw((0,0)--(0,-1), black); draw((-2,1)--(-1,1), black); draw((-2,0)--(0,0), black); draw((-2,-1)--(1,-1), black); draw((2,-1)--(4,-1), black); draw((2,-1)--(2,0), black); draw((3,-1)--(3,1), black); draw((3,0)--(4,0), black); draw((3,3)--(4,3), black); draw((2,4)--(4,4), black); draw((2,4)--(2,5), black); draw((3,3)--(3,5), black); dot((1,2)); path q = (4,2)--(1,5)--(-2,2)--(1,-1)--(4,2)--(1,5); draw(q, green, Margins); [/asy][/asy] There are other variations on this, like letting Breaker have two moves at a time. In that case, Breaker can win using a pairing strategy where we pair sets of three edges that are at $90^{\circ}$ and $180^{\circ}$ angles from each other (and Breaker will never have to move a very small distance from the origin -- I forget, but I think it's something like taxi cab distance $5$ or less). Since the threshold for winning seems to be between giving Breaker $1$ move (a "1:1" game) and $2$ moves (a "1:2" game), you can consider intermediate choices, like giving Breaker an extra move every other turn (a "1:1.5" game). In these cases though, I don't know who wins.

Proposed by Maxim Didin

Is that T3?

I assume that orientation means directed edges, correct me if I'm wrong. Then the problem of cycles can be destroyed by simply preventing the formation of any one corner (two perpendicular vectors), which by induction will not allow a full cycle. Denote Ann's first move as vector $v_1$, then after Bob attempts to make this a cycle by placing $v_2$ at the tail of $v_1$, Ann can place her next move $v_3$ at the tail of Bob, but only matching tails so the vectors cannot be added. This strategy would work for any orientation as placing head to head vectors or tail to tail vectors prevents a summation of vectors which avoids a cycle. Thus Bob cannot guarantee a win if Ann plays optimally.

S.Das93 wrote: ...Denote Ann's first move as vector $v_1$, then after Bob attempts to make this a cycle by placing $v_2$ at the tail of $v_1$, Ann can place her next move $v_3$ at the tail of Bob, but only matching tails so the vectors cannot be added. This strategy would work for any orientation as placing head to head vectors or tail to tail vectors prevents a summation of vectors which avoids a cycle. .. But, after Ann puts $v_3$ at the tail of Bob's vector $v_2,$ Bob can place $v_4$ opposite to $v_3$ so that $v_2,v_4$ are part of a cycle. That is, preventing Bob to turn left, he can turn right. It's quite possible that after some moves Bob manages to make double threat and Ann couldn't prevent avoiding two cycles at the same time! Your argument does not covers this possibility. Note that the destructor always wins. It doesn’t matter who makes the first move. If it is Bob, let WLOG he puts the vector $(0, 0), (1, 0)$. Then Ann places some vector on $x$ axe, and after that she plays with the same strategy as in #2.

oops does this work The answer is no, Bob cannot win. Ann's strategy is as follows: she first orients a vertical edge downwards and calls it the live edge and also designates some region of the grid as the red zone, which is initially empty. Procedure 1: If Bob orients an edge sharing the same "lower left" corner as the live edge—suppose this has coordinates $(a,b)$—as the live edge such that the two oriented edges form a path (i.e. don't both point towards or away from the corner), then Ann takes the upwards and rightwards rays from that corner and adds the quarter-plane bounded by them (including the rays themselves) into the red zone. She then picks some point $(x,y)$ such that $x+y>a+b=1000$ and $(x,y)$ does not lie in the red zone—the rest of the algorithm makes it clear that this is always possible and orients the edge above it downwards, making it the new live edge. Procedure 2: If Bob orients the same edge as described before, but does it in a way such that the two oriented edges either point towards or away from the "lower left" corner, then Ann does not modify the red zone but repeats the picking operation and makes the resulting edge the new live edge. Procedure 3: If Bob orients an edge which lies entirely inside the red zone and not along an edge, then Ann will take its "upper right" corner and find the other edge with the same "upper right" corner (which must be unoriented so far, else Bob's move wouldn't be legal) and orient it such that the two oriented edges either both point towards or away from the corner. Note that Bob always orients the first, third, etc. of these edges while Ann orients the second, fourth, etc. Procedure 4: If Bob orients an edge whose "lower left" corner lies completely outside the red zone and is different from that of the live edge, then Ann compares the sum of $x$ and $y$ coordinates of this new "lower left" corner and the "lower left" corner of the live edge, picks the one that is smaller, and then draws the other edge so that they either point towards or away from the corner. Procedure 5: If Bob orients an edge which lies on the boundary of the red zone, then Ann ignores it, takes the "lower left" corner of the live edge, and orients the other edge sharing this "lower left" corner such that they either point towards or away from the corner and removes the edge's status as live (so for the time being there are no live edges). Then Ann completely changes her strategy: now, if Bob orients an edge whose lower left corner lies outside the red zone, then Ann orients the other edge sharing its "lower left" corner such that they both point towards or away. If Bob orients an edge lying entirely in the red zone and not along an edge, Ann does the same thing she did in procedure 3. If Bob orients another edge which lies on the boundary, she then does the same thing she did in procedure 1, ignoring the recent addition and creating a new live edge far away. The proof that this works is the following: if a cycle exists, it must contain some L-shaped part which is actually inside the red zone ("usually", but not necessarily, this will be one of the corners—all of the L-shaped oriented parts which lie outside the red zone are not valid since they either both point towards or away from their corner), whose "lower left" corner must be the point on the cycle with minimal $x+y$ value. The cycle must also have some ㄱ-shaped part whose edges don't both point towards or away from the "upper right" corner, which also lies inside the red zone, but this is not possible: once a red zone is established in some location, these ㄱ-shaped parts can never be created, since procedure 3 exists. On the other hand, it is not possible one of these ㄱ-shaped parts to be created before a red zone is established somewhere containing it, since it turns out that when the red zone is expanded to include the points above/to the right of some "lower left" corner, that corner must actually have (nonstrictly) maximal $x+y$ value, since the live edge (if it exists, which it must if the red zone is to be expanded) always has maximal $x+y$ value among the edges not in the red zone and every point added to the red zone has higher $x+y$ value than the "lower left" corner of the live edge. Therefore we can never find both a good L-shaped part and a good ㄱ-shaped part to form a cycle from, done. $\blacksquare$ Remark: very intense u3 flashbacks rn

No, Bob cannot guarantee a win. Without loss of generality, suppose Ann begins by orienting a horizontal edge $e$. Call a hook a set of two edges sharing a vertex oriented with terminal angle $0$ and $\pi/2$ with respect to that vertex. Notice that Ann can partition the edges of the grid into hooks and reflections of hooks about $e$. If Bob orients an edge in a hook or anti-hook, Ann can always orient the remaining edge such that the two edges never form a path through the hook. On the other hand, we can check that any oriented cycle in the lattice must contain a path through either a hook or one of the anti-hooks, so Bob can never create an oriented cycle. Remark: Although the solution seems simple, it was extremely hard to find. Game theory problems are like this sometimes.