The problem involves a 4x4 grid of unit squares, and I need to draw as many diagonals as possible such that no two diagonals share any common points, including endpoints. That means no overlapping or touching at corners.

First, I visualize the grid. It's a 4x4 grid, so there are 16 small squares. Each small square has two possible diagonals: one from bottom-left to top-right, and the other from top-left to bottom-right. But I can only choose one diagonal per square because if I draw both in the same square, they would intersect at the center, but the problem is about no common points, so even within a single square, I can only draw one diagonal? No, actually, in a single square, the two diagonals intersect at the center, so they do have a common point. Therefore, in each unit square, I can draw at most one diagonal without having an intersection inside the square.

The condition is that no two diagonals have any common point, including endpoints. So, even if diagonals are in different squares but share an endpoint, that's not allowed. Since the diagonals are within their own squares, but endpoints are at the vertices of the squares, which are shared among adjacent squares.

Each vertex in the grid is shared by up to four squares, but for the diagonals, each diagonal connects two opposite corners of its square.

So, when we draw a diagonal in a square, it occupies two vertices. If I draw another diagonal in an adjacent square that shares a vertex with this one, then they share that common endpoint, which is not allowed.

Moreover, diagonals in different squares might cross each other if they are in adjacent squares and oriented appropriately. Do they?

For example, consider two adjacent squares sharing a side. If I draw a diagonal in one square from bottom-left to top-right, and in the adjacent square to the right, if I draw a diagonal from top-left to bottom-right, would they intersect?

Let's think: suppose square A and square B are adjacent horizontally. Square A has corners: let's label them. Say for a square, bottom-left (BL), bottom-right (BR), top-left (TL), top-right (TR).

If I draw diagonal from BL to TR in square A.

In square B (to the right of A), if I draw from TL to BR. But since they share the side, the TL of B is the TR of A, and BR of B is the TR of A? No.

Actually, if square A is from (0,0) to (1,1), then corners: BL(0,0), BR(1,0), TL(0,1), TR(1,1).

Square B to the right is from (1,0) to (2,1), so its corners: BL(1,0), BR(2,0), TL(1,1), TR(2,1).

So diagonal in A: from BL(0,0) to TR(1,1).

Diagonal in B: from TL(1,1) to BR(2,0). TL of B is (1,1), which is TR of A, and BR of B is (2,0).

Now, the line from (0,0) to (1,1) and from (1,1) to (2,0). Do they intersect? At (1,1), which is an endpoint for both. But the problem says "no common point including endpoints", so if they share an endpoint, it's not allowed. In this case, they share the point (1,1).

But do they intersect elsewhere? The line from (0,0) to (1,1) has equation y=x. From (1,1) to (2,0) has equation y = -x + 2. They intersect at x=1, y=1, which is the endpoint.

So they only intersect at the common endpoint, but since endpoints are included, this is forbidden.

Now, what if I draw in both squares the same type of diagonal? Say in A: BL to TR, in B: BL to TR. So in A: (0,0)-(1,1), in B: (1,0)-(2,1).

Do they intersect? Line from (0,0) to (1,1): y=x. Line from (1,0) to (2,1): y = x - 1. They are parallel, never intersect. But do they share any point? The closest points are (1,1) and (1,0), but (1,1) is on first diagonal, (1,0) on second, but they are different points. However, they don't share a vertex? In this case, the endpoint of first is (1,1), endpoint of second is (1,0) and (2,1), none of which is (1,1) except that (1,1) is an endpoint of first, but for second diagonal, its endpoints are (1,0) and (2,1), so (1,1) is not an endpoint of the second diagonal. But is it on the diagonal? The second diagonal is from (1,0) to (2,1), which is the line y = x - 1. At x=1, y=0, so (1,1) is not on it. Similarly, (1,0) is on second but not on first. So no common points. But they are very close; in fact, they share the side between (1,0) and (1,1), but the diagonals themselves don't share any points, only the vertices are shared, but the diagonals don't include the sides.

The diagonal is the line segment from (0,0) to (1,1) in A, which doesn't include the side from (1,0) to (1,1) or from (0,1) to (1,1); it's just the diagonal line.

So in this case, diagonals in adjacent squares with same orientation don't intersect and don't share endpoints, since the shared vertex is (1,1) for the squares, but for diagonal in A, it has (0,0),(1,1); in B, (1,0),(2,1). So the common vertex of the squares is (1,1) and (1,0), but (1,1) is endpoint of A's diagonal, but not of B's diagonal; B's diagonal has (1,0) and (2,1), so (1,1) is not an endpoint of B's diagonal. And as lines, they don't intersect except possibly at endpoints, but here no common endpoints.

But (1,1) is a vertex, but for the diagonal of B, it doesn't pass through (1,1), only through (1,0) and (2,1).

So in this configuration, the two diagonals do not share any common points.

Now, what if I have diagonals in diagonally adjacent squares? For example, square at (0,0) and square at (1,1). So square A: (0,0),(1,0),(1,1),(0,1). Square D: (1,1),(2,1),(2,2),(1,2). So diagonals: in A, say BL(0,0) to TR(1,1). In D, say BL(1,1) to TR(2,2). Then they share the point (1,1), which is an endpoint for both. So not allowed.

If in D I draw the other diagonal, TL(1,2) to BR(2,1). Then in A: (0,0)-(1,1), in D: (1,2)-(2,1). Do they intersect? Line from (0,0) to (1,1): y=x. Line from (1,2) to (2,1): y = -x + 3. Intersect when x = -x + 3 => 2x=3 => x=1.5, y=1.5. Which is inside both segments? For first segment, from (0,0) to (1,1), x from 0 to 1, 1.5>1, so not on segment. For second, from (1,2) to (2,1), x from 1 to 2, 1.5 is between, y=1.5. But first segment only up to x=1, so at x=1.5, it's not on first segment. So no intersection. But do they share endpoints? A has (0,0),(1,1); D has (1,2),(2,1). Common points? (1,1) is in A, but not in D; (1,2) not in A, etc. So no common endpoints. So they might be ok, but I need to see if they cross, but in this case, they don't intersect at all.

The intersection point (1.5,1.5) is not on either segment, so no intersection.

So, the main constraints are that diagonals cannot share endpoints, and they might intersect if they cross, but in a grid, with axis-aligned squares, diagonals can only intersect if they are in adjacent squares and have crossing orientations.

For example, consider two squares that share a corner but not a side, like diagonally adjacent. As above, if both are "main" diagonals, they share a vertex, but if one is main and the other is the other diagonal, they might not share vertices and not intersect, as above.

But in the case where squares share a side, if they have diagonals of opposite orientations, they share a vertex, which is forbidden. If same orientation, no shared vertex and no intersection.

Now, for squares that are not adjacent, no issue.

But also, diagonals can intersect inside the grid if they cross. For example, consider four squares forming a 2x2 block. Suppose I draw in top-left square a diagonal from TL to BR, and in bottom-right square a diagonal from BL to TR. So for top-left square (0,0): from TL(0,1) to BR(1,0). For bottom-right square (1,1): from BL(1,0) to TR(2,1)? Bottom-right square is from (1,0) to (2,1), so BL(1,0), BR(2,0), TL(1,1), TR(2,1). So diagonal from BL(1,0) to TR(2,1).

Now, the diagonal in top-left: from (0,1) to (1,0), so line y = -x + 1.

Diagonal in bottom-right: from (1,0) to (2,1), so y = x - 1.

Intersect when -x + 1 = x - 1 => 2x = 2 => x=1, y=0. Which is the point (1,0). But (1,0) is an endpoint for both diagonals? For top-left diagonal, it goes from (0,1) to (1,0), so (1,0) is an endpoint. For bottom-right, from (1,0) to (2,1), so (1,0) is also an endpoint. So they share the endpoint (1,0), which is forbidden.

But what if I draw in different squares? Suppose in top-left I draw from BL(0,0) to TR(1,1), so y=x. In bottom-right, same as above, from (1,0) to (2,1), y=x-1. Then intersect when x = x-1, impossible, no intersection. But do they share endpoints? (1,1) for first, (1,0),(2,1) for second, no common.

But earlier when I had opposite orientations in adjacent squares, they shared an endpoint.

To have crossing diagonals that intersect at an interior point, not at endpoint.

For example, consider two squares that are not adjacent but have diagonals that cross. In a grid, can two diagonals intersect at a point not a vertex?

For instance, take square at (0,0) and square at (1,1), but as above, if I draw in (0,0) from BL(0,0) to TR(1,1), and in (1,1) from TL(1,2) to BR(2,1). So lines: from (0,0) to (1,1): y=x. From (1,2) to (2,1): y = -x + 3. Intersect at x=1.5, y=1.5. Now, is this point on both segments? For first segment, from (0,0) to (1,1), x from 0 to 1, 1.5>1, so not on segment. Similarly, for second, from (1,2) to (2,1), x from 1 to 2, 1.5 is between, y=1.5. But first segment doesn't extend to x=1.5, so no intersection on the segments.

To have an intersection inside, I need diagonals that are not in the same "row" or "column". For example, take a diagonal in a square and another in a square that is offset.

Suppose I have a diagonal from (0,0) to (2,2), but that's not a unit diagonal; each diagonal is within its own unit square, so maximum length is sqrt(2), and they are confined to their squares.

In a grid of unit squares, the diagonals are all within their individual squares, so for two diagonals to intersect at a point not a vertex, that point must be inside the grid, and the diagonals must be from squares that are adjacent in a certain way.

Consider four squares forming a 2x2 block. Say squares: A at (0,0), B at (1,0), C at (0,1), D at (1,1).

Draw in A: from BL(0,0) to TR(1,1), so y=x.

Draw in D: from TL(1,2) to BR(2,1)? D is at (1,1), so from (1,1) to (2,2), but that's not correct; if D is from (1,1) to (2,2), then TL is (1,2)? Let's define coordinates properly.

Assume the grid from x=0 to 4, y=0 to 4, so 4x4 grid of squares, each 1x1.

So squares are defined by bottom-left corners: for square (i,j), corners at (i,j), (i+1,j), (i,j+1), (i+1,j+1).

Diagonals are from (i,j) to (i+1,j+1) or from (i+1,j) to (i,j+1).

Now, suppose I draw in square (0,0): from (0,0) to (1,1), so y=x.

In square (1,1): from (1,1) to (2,2), so y=x, but that's the same, but they share the point (1,1), which is an endpoint for both, so not allowed.

Draw in square (1,1) the other diagonal: from (2,1) to (1,2), so from (1,2) to (2,1), line y = -x + 3.

Now, the first diagonal is from (0,0) to (1,1), y=x, for 0≤x≤1.

Second is from (1,2) to (2,1), y= -x +3, for 1≤x≤2.

At x=1, first has y=1, second has y= -1+3=2, so different points.

Do they intersect? Set x = -x +3 => 2x=3, x=1.5, y=1.5.

For first segment, when x=1.5, but x only from 0 to 1, so not on segment. For second, x from 1 to 2, x=1.5 is on it, but first is not defined there, so no intersection.

To have an intersection, I need diagonals that cross. For example, in square (0,0), draw from (0,0) to (1,1), y=x.

In square (1,0), draw from (1,0) to (2,1), y=x-1.

Then as before, they share the point (1,1) and (1,0), but (1,1) is on first, (1,0) on second, but no common point for the same diagonal? The diagonals are different, but they don't intersect except at vertices if adjacent.

But for non-adjacent squares, say square (0,0) and square (1,1), as above, no intersection.

What about square (0,0) and square (2,1)? Square (2,1) is at bottom-left (2,1), so corners (2,1),(3,1),(2,2),(3,2).

Draw in (0,0): from (0,0) to (1,1), y=x.

In (2,1): say from (2,1) to (3,2), y=x-1.

Then lines y=x and y=x-1, parallel, no intersection.

Or in (2,1) from (3,1) to (2,2), so from (2,2) to (3,1), y= -x +4.

Then with y=x from (0,0) to (1,1), no intersection since x ranges are disjoint.

So, in a grid of axis-aligned squares, two diagonals can only intersect if they are in squares that share a common vertex, and have crossing orientations, but in that case, they either share an endpoint or intersect at a vertex, but not at an interior point of the diagonals or the grid.

Is that true? Let me try a different pair. Suppose I have square (0,1): from (0,1) to (1,2) or from (1,1) to (0,2), etc.

Draw in square (0,0): from (0,0) to (1,1), y=x.

Draw in square (1,1): from (1,1) to (2,2), but that would share (1,1).

Or in square (1,1): from (2,1) to (1,2), y= -x +3.

As before, no intersection.

But what if I draw in a square that is not directly adjacent. Say in square (0,0): from (0,0) to (1,1), y=x.

In square (1,2): square (1,2) is at bottom-left (1,2), corners (1,2),(2,2),(1,3),(2,3).

Draw from (1,2) to (2,3), y=x+1, or from (2,2) to (1,3), y= -x +4.

With y=x from (0,0) to (1,1), and y= -x+4 from (1,3) to (2,2)? From (1,3) to (2,2), so for x from 1 to 2, y from 3 to 2, so y= -x +4.

Set x = -x +4 => 2x=4, x=2, y=2.

At x=2, y=2. For first diagonal, x=2 is outside [0,1], for second, x=2 is endpoint, but first is not defined, so no intersection.

To have an intersection, the diagonals must be from squares that are "knight's move" apart or something, but in a grid, the only way diagonals can intersect is if they are in adjacent squares and have the crossing orientations, but as we saw, in that case they share an endpoint.

For example, in two squares sharing a side, if they have diagonals of opposite orientations, they share the common vertex of the shared side.

If they have the same orientation, no shared vertex and no intersection.

For diagonally adjacent squares, if both have the "main" diagonal (from BL to TR), they share the common corner vertex.

If one has main and the other has the other diagonal (from TL to BR), then they may not share a vertex, and as calculated, they don't intersect in the interior.

Moreover, for squares not adjacent, no issue.

So, in this grid, the only restrictions are that no two diagonals can share a common endpoint, and no two can be in adjacent squares with opposite orientations? But from above, when they are in adjacent squares with opposite orientations, they do share an endpoint, which is already forbidden by the no shared endpoint condition.

In fact, the shared endpoint condition might be sufficient to ensure no intersections as well, since interior intersections don't occur.

Is that true? Suppose I have three squares. For example, consider squares (0,0), (1,0), and (0,1).

Draw in (0,0): from (0,0) to (1,1), y=x.

In (1,0): from (1,0) to (2,1), y=x-1.

But these two share the point (1,1)? No, as before, but (1,1) is not on second diagonal; second is from (1,0) to (2,1), so at (1,0), not (1,1).

But they don't share an endpoint; (0,0),(1,1) for first; (1,0),(2,1) for second. Common points? (1,1) is on first, but not on second; (1,0) on second but not on first. So no shared endpoints.

Do they intersect? y=x and y=x-1, parallel, no.

Now, if I add in (0,1): say from (0,1) to (1,2), y=x+1, or from (1,1) to (0,2), y= -x +2.

Suppose I draw from (1,1) to (0,2), so y= -x +2.

Then with first diagonal y=x, intersect at x=1, y=1, which is (1,1), a vertex.

But (1,1) is an endpoint for both the first diagonal (from (0,0) to (1,1)) and for this new diagonal (from (1,1) to (0,2)), so they share an endpoint, which is forbidden.

If I draw in (0,1) the other diagonal, from (0,1) to (1,2), y=x+1.

Then with first diagonal y=x, no intersection, and endpoints (0,0),(1,1) for first, (0,1),(1,2) for this, no common endpoints.

So again, no issue.

Now, is there a case where two diagonals intersect at a non-vertex point? For example, in a larger grid.

Suppose I have square (0,0) and square (2,1). As before, no.

Or square (0,0) and square (1,2). Square (1,2): bottom-left (1,2), corners (1,2),(2,2),(1,3),(2,3).

Draw in (0,0): from (0,0) to (1,1), y=x.

In (1,2): from (1,2) to (2,3), y=x+1, or from (2,2) to (1,3), y= -x +4.

With y=x, and y=x+1, parallel.

With y= -x+4, set x= -x+4, 2x=4, x=2, y=2. But for first diagonal, x=2 not in [0,1], for second, if from (1,3) to (2,2), at x=2, y=2, which is on the segment, but first not defined.

So no intersection.

Perhaps between diagonals in the same row or column but spaced out, but they are parallel if same orientation, or if different, may not intersect within segments.

So, it seems that in a grid of axis-aligned squares, two diagonals can intersect only if they share a common endpoint, which is already forbidden by the "no common points including endpoints" condition. Therefore, the only constraint is that no two diagonals share a common vertex point.

In other words, the set of diagonals must be such that their endpoints are all distinct, and no two diagonals share an endpoint.

Each diagonal uses two vertices, and each vertex can be used by at most one diagonal, because if a vertex is used by two diagonals, they share that endpoint.

Is that correct? Each diagonal has two endpoints, and if two diagonals share an endpoint, then that vertex is common. But also, a single vertex could be an endpoint for only one diagonal, but in this case, since each diagonal requires two vertices, and vertices may be shared, but the condition is that no two diagonals share a vertex, meaning that the sets of vertices used by each diagonal are disjoint.

Is that necessary? Suppose two diagonals share a vertex, then they have a common endpoint, which is forbidden. Conversely, if they don't share any vertex, then no common endpoint, and as we saw, no interior intersection, so it should be ok.

Moreover, since the diagonals are line segments within their squares, and if they don't share vertices, and the squares are disjoint or adjacent but not sharing the diagonal, but even if squares share a side, as long as the diagonals don't use the shared vertices, but in that case, if squares share a side, they share two vertices, so if a diagonal in one square uses one of those vertices, and in the adjacent square a diagonal uses the other, but they might not share a common diagonal endpoint, but they could share a vertex.

For example, square A and B adjacent horizontally. A has vertices v1,v2,v3,v4: say BL, BR, TL, TR.

B has BL=v2, BR=v5, TL=v3, TR=v6. So they share v2 and v3.

Now, if in A, I draw diagonal from v1 to v3 (TL to BR).

In B, I draw diagonal from v2 to v6? v2 is BL of B, v6 is TR of B.

But v2 is BR of A, v6 is TR of B.

Now, diagonal in A: v1 to v3.

Diagonal in B: v2 to v6.

Do they share a vertex? v1, v3 in A; v2, v6 in B. v3 is TL of A, which is also TL of B? In this labeling, for square A: v1=BL, v2=BR, v3=TL, v4=TR.

For square B: since to the right, BL is v2, BR is v5, TL is v3, TR is v6. So yes, shared vertices are v2 and v3.

Now, diagonal in A: say from v1 to v3.

Diagonal in B: from v2 to v6.

Then vertices used: A: v1,v3; B: v2,v6. So v3 is used by A, v2 by B, but v3 and v2 are different vertices, so no common vertex. Do they share an endpoint? No, since different vertices.

But are they adjacent? The point v3 is used by A, v2 by B, and v2 and v3 are connected by the side, but the diagonals don't share a point.

In this case, the diagonal in A is from v1 to v3, which is the corner, and in B from v2 to v6.

Do they intersect? Line from v1 to v3 and v2 to v6. Suppose coordinates: v1(0,0), v2(1,0), v3(0,1), v4(1,1) for A. Then for B: v2(1,0), v5(2,0), v3(0,1)? No, if B is to the right, it should have BL(1,0), BR(2,0), TL(1,1), TR(2,1). I think I messed up.

Standard: for square A: BL(0,0), BR(1,0), TL(0,1), TR(1,1).

Square B to the right: BL(1,0), BR(2,0), TL(1,1), TR(2,1).

So shared vertices: (1,0) and (1,1).

Now, diagonal in A: from BL(0,0) to TR(1,1), or from BR(1,0) to TL(0,1).

Similarly for B.

Suppose in A: from (0,0) to (1,1).

In B: from (1,0) to (2,1).

Then vertices: A: (0,0),(1,1); B: (1,0),(2,1). So (1,1) and (1,0) are different, no common vertices.

The lines: from (0,0) to (1,1) and from (1,0) to (2,1). As before, parallel if same slope, but here both have slope 1, so parallel, no intersection.

If in A I draw from (1,0) to (0,1), so from BR to TL.

In B I draw from (1,0) to (2,1)? But (1,0) is already used? No, the diagonal is a line segment, but the vertex (1,0) can be used by only one diagonal? In this case, if I draw in A from (1,0) to (0,1), and in B from (1,0) to (2,1), but then both diagonals share the endpoint (1,0), which is forbidden.

But if I draw in B a different diagonal, say from (2,0) to (1,1), but (1,1) might be used.

In this setup, for B, if I draw from BL(1,0) to TR(2,1), or from BR(2,0) to TL(1,1).

So if in A I have from (1,0) to (0,1), using (1,0) and (0,1).

Then in B, if I want to use a diagonal, I cannot use (1,0) or (1,1) because (1,0) is used, and (1,1) might be used if I choose, but (1,1) is not used yet by A; A used (1,0) and (0,1), so (1,1) is free.

So I can draw in B from (2,0) to (1,1), for example.

Then vertices: A: (1,0),(0,1); B: (2,0),(1,1). So no common vertices.

Lines: from (1,0) to (0,1): y = -x +1? From (1,0) to (0,1), so as x from 0 to 1, y from 1 to 0, so y = -x +1.

From (2,0) to (1,1): from (1,1) to (2,0), so y = -x +2.

Set -x+1 = -x+2, impossible, no intersection.

Do they share any point? The vertices are all distinct.

So in this case, it's ok.

Now, back to the condition: to have no two diagonals sharing a common vertex, meaning that the vertices used by all diagonals are distinct; each vertex is used by at most one diagonal.

Since each diagonal uses two vertices, and there are multiple vertices, we need to select a set of diagonals such that no two share a vertex, i.e., the diagonals are vertex-disjoint.

Moreover, since each diagonal is confined to its square, but we can choose which diagonal in each square, but actually, for a given square, if we decide to put a diagonal, we have two choices, but only one diagonal can be placed per square? No, in a single square, if I draw both diagonals, they intersect at the center, so I can draw at most one diagonal per square, because even within the same square, the two diagonals share the center point.

The problem is to draw diagonals in the unit squares, and no two diagonals (even in different squares) share a common point. So in particular, in one square, I cannot draw both diagonals because they intersect at the center. So each square can have at most one diagonal drawn.

Moreover, for different squares, the diagonals must not share any common point, which, as we discussed, is equivalent to not sharing any vertex, since no interior intersections occur.

Therefore, the problem reduces to: in the 4x4 grid, we have 16 squares. Each square can have at most one diagonal. Each diagonal uses two vertices. The grid has a certain number of vertices.

For a 4x4 grid of squares, the number of vertices is 5x5 = 25, since it's a grid graph with 5 lines in each direction.

Each vertex can be incident to at most one diagonal, because each diagonal uses two vertices, and no sharing.

Moreover, each diagonal is associated with a square, and uses two opposite corners of that square.

The question is to maximize the number of diagonals, which is equivalent to selecting a set of squares and for each, choosing a diagonal such that no two diagonals share a vertex.

Since the diagonals must be vertex-disjoint, and each diagonal uses two vertices, the maximum number of diagonals is at most floor(total vertices / 2) = floor(25/2) = 12, since 25 is odd.

But is this achievable? Probably not, because of the constraints of the grid.

Each diagonal corresponds to an edge in the grid graph, but not exactly. The diagonals are not the edges of the grid; the grid has horizontal and vertical edges, but the diagonals are diagonals in the squares.

In terms of the vertices, each possible diagonal is a pair of vertices that are at distance sqrt(2) in the grid, i.e., differing by (1,1) in coordinates.

Specifically, for a square, the two possible diagonals are the pairs: (i,j) and (i+1,j+1) for one diagonal, or (i,j+1) and (i+1,j) for the other.

So, in the grid of vertices, we have a set of possible edges: each such diagonal corresponds to an edge between vertices that are diagonally adjacent, specifically, the northeast-southwest or northwest-southeast diagonals, but in terms of grid points.

More precisely, the possible diagonals correspond to the edges of the graph where vertices are connected if they are at chess king move distance sqrt(2), but only the diagonal moves, not orthogonal.

In the grid graph, we can consider the graph where vertices are the grid points, and edges are the possible diagonals, but each "edge" is between vertices that are knight-move? No.

For example, between (i,j) and (i+1,j+1), that's a diagonal move.

Similarly, between (i,j) and (i+1,j-1), but in our case, since the squares are axis-aligned, the diagonals are only between (i,j) and (i+1,j+1), or (i,j) and (i+1,j-1)? No.

In a square with corners (i,j), (i+1,j), (i,j+1), (i+1,j+1), the diagonals are from (i,j) to (i+1,j+1) and from (i+1,j) to (i,j+1).

So, the pairs are: (i,j) with (i+1,j+1), or (i+1,j) with (i,j+1).

Now, in the grid, for a fixed i,j, this is for the square at (i,j).

The graph we need is where the edges are these diagonal pairs. But actually, for the vertex-disjoint paths, but here each "edge" in this graph would be a possible diagonal, but we need to select a matching, i.e., a set of edges with no common vertices.

Moreover, each such "edge" is associated with a square, and we can select at most one per square, but actually, each such diagonal is unique to a square, since each pair of opposite corners defines the square.

For example, the pair (i,j) and (i+1,j+1) is only in the square with bottom-left (i,j), because that's the only square that contains both points as opposite corners.

Similarly for the other diagonal.

So, there are no overlapping; each possible diagonal is contained within one square.

Therefore, the entire setup is that we have a set of possible edges: for each square, there are two possible edges (diagonals) we can choose, but we can choose at most one per square, and we need to select a set of these edges such that no two share a common vertex, i.e., a matching in this graph.

But since the graph is not complete, and has specific edges.

Actually, the graph we are considering is the king's move graph or something, but let's define the graph G where vertices are the 5x5 = 25 grid points, and edges are the possible diagonals, i.e., for each square, we have two edges: one for each diagonal.

But since each square contributes two possible edges, but they share vertices, and we can choose at most one per square.

The problem is to select a subset of these possible edges such that no two edges share a common vertex, and also, since each edge is from a different square (but actually, multiple edges per square, but we can choose only one per "face" or something), but in this case, for each square, we can choose at most one diagonal, but the constraint is automatically satisfied if we select a set of non-adjacent edges (in the vertex sense), but we need to ensure that we don't select both diagonals of the same square, but since they share vertices (the center, but in terms of vertices, both diagonals share no common vertex? In a single square, the two diagonals: one is (i,j) to (i+1,j+1), the other is (i+1,j) to (i,j+1). These two diagonals share no common vertex; they use four distinct vertices. Oh! I see, I made a mistake earlier.

In a single square, the two possible diagonals do not share any vertex; they use all four corners. So, in principle, I could draw both diagonals in the same square, but they would intersect at the center of the square. The center is not a vertex, but a point where they cross, so they share that common point, which is not an endpoint, but an interior point. The problem says "no common point (including endpoints)", so they cannot share any point, so in particular, they cannot intersect at an interior point. Therefore, in one square, I can draw at most one diagonal, because if I draw both, they intersect at the center.

However, in terms of vertices, the two diagonals do not share vertices; they use disjoint sets of vertices. But they still intersect at the center. So the vertex-disjointness is necessary but not sufficient; we also need to ensure that no two diagonals cross at an interior point.

In the grid, when do two diagonals in different squares intersect at an interior point? As we saw earlier, it might not happen, but let's confirm.

Suppose two diagonals in different squares. If the squares are not adjacent, no intersection. If they share a side, and have opposite orientations, they share a vertex. If same orientation, no intersection. If they are diagonally adjacent, and both have the same orientation, they share a vertex; if different, no share and no intersection.

But is there a pair of diagonals that cross at a point not a vertex? For example, consider a 2x2 block of squares. Suppose in the top-left square, I draw the diagonal from TL to BR, i.e., (0,1) to (1,0). In the bottom-right square, I draw from BL to TR, i.e., (1,0) to (2,1)? But that would share (1,0).

Suppose in top-left: from (0,0) to (1,1) (BL to TR). In bottom-right: from (1,0) to (2,1) (BL to TR). Then as before, no intersection, no shared vertices.

But if I have a larger grid, say in square (0,0): from (0,0) to (1,1). In square (2,1): from (2,1) to (3,2). No intersection.

Or in square (1,0): from (1,0) to (2,1). Then with first, no shared vertices if I don't use them, but in this case, if I have (0,0)-(1,1) and (1,0)-(2,1), then vertices (0,0),(1,1) and (1,0),(2,1), so (1,1) and (1,0) are different, no common, and lines are y=x and y=x-1, parallel.

Now, suppose I have a diagonal that is not axis-aligned, but all are.

Another idea: consider two squares that are not in the same row or column, but have diagonals that cross. For example, square (0,0) and square (1,1), as before.

Or square (0,0) and square (3,3), but far away.

Perhaps in the grid, the only possible intersections are at vertices, and for diagonals in different squares, if they intersect at a non-vertex, it might not be on the grid, but in this case, since the diagonals are within the squares, and the grid is discrete, perhaps it never happens for non-adjacent squares.

Let's try a specific example. Suppose we have vertices at integer coordinates. Take diagonal in square (0,0): from (0,0) to (1,1), line y=x, 0≤x≤1.

Diagonal in square (1,1): from (1,1) to (2,2), but that shares (1,1).

Or from (2,1) to (1,2), line from (1,2) to (2,1), y= -x +3, 1≤x≤2.

No intersection with y=x in [0,1].

Now, take a diagonal in a square that is not adjacent. Say in square (0,1): from (0,1) to (1,2), y=x+1, 0≤x≤1.

With the first y=x, no intersection in the planes.

Or in square (1,0): from (1,0) to (2,1), y=x-1, 1≤x≤2, with y=x in [0,1], no overlap.

But what if I take a diagonal that is in a square that is "offset". For example, in square (0,0): from (0,0) to (1,1), y=x.

In square (1,2): from (1,2) to (2,1), y= -x +3, 1≤x≤2.

Then as before, intersect at (1.5,1.5), but for first diagonal, x=1.5 >1, not in [0,1], for second, x=1.5 in [1,2], y=1.5, but first not defined, so not on the segment.

To have an intersection on the segments, the x-range must overlap, and similarly for y.

For two lines to intersect within their segments, the intersection point must be within both parameter ranges.

For two diagonals in different squares, the only case where they might intersect is if the squares are adjacent in a 2x2 fashion, but as we saw, when they share a vertex or are adjacent, they either share a vertex or don't intersect.

In fact, for any two diagonals in different squares, if the squares are not the same and not adjacent, the diagonals are in disjoint regions and don't intersect. If they are adjacent, as above, no interior intersection; only possible at vertices, which is forbidden by no shared vertices.

Therefore, the only constraint is that the diagonals are vertex-disjoint and at most one per square.

So, the problem reduces to: select a set of squares, and for each selected square, choose one of the two possible diagonals, such that no two diagonals share a common vertex. Since the diagonals in different squares may share vertices only if the squares share a vertex, and by choosing the diagonal, we can avoid sharing.

But since for each square, we have a choice of which diagonal to draw, we can try to maximize the number.

Since the vertex-disjointness is the key, and each diagonal uses two vertices, and there are 25 vertices, we can have up to 12 diagonals, but 25 is odd, so at most 12, but is it possible?

Each diagonal uses two vertices, so with 25 vertices, maximum matching size is 12, with one vertex left out.

But in this graph, what is the maximum matching for this diagonal edge set.

The graph is the set of vertices with edges between (i,j) and (i+1,j+1), and between (i+1,j) and (i,j+1), for each i,j such that the square exists, i.e., for i=0 to 3, j=0 to 3 for the squares.

For each square at (i,j), it contributes two edges: e1: (i,j) - (i+1,j+1) and e2: (i+1,j) - (i,j+1).

Note that these two edges for the same square are vertex-disjoint; they have no common vertices.

The entire graph has 25 vertices, and the edges are all possible such diagonals.

This graph is actually the union of two graphs: one is the "main diagonal" edges, which are (i,j) to (i+1,j+1) for all i,j, but only for the squares, so for each pair (i,j) with 0≤i<4, 0≤j<4, there is an edge from (i,j) to (i+1,j+1).

Similarly, for the "anti-diagonal" edges: from (i+1,j) to (i,j+1) for each i,j.

The main diagonal edges form a set where each is a diagonal move in the northeast direction.

Similarly for the anti-diagonal.

Moreover, these two sets are disjoint in terms of edges, but may share vertices.

In fact, the graph is the king's move graph, but only the diagonal moves, not the orthogonal.

Specifically, it's the graph where we have edges between vertices that differ by (1,1) or by (1,-1), but in this case, for the anti-diagonal, from (i+1,j) to (i,j+1), which is from (i+1,j) to (i,j+1), so difference in x is -1, in y is +1, so vector (-1,1), which is northwest or something.

From (i+1,j) to (i,j+1): delta x = i - (i+1) = -1, delta y = (j+1) - j = 1, so (-1,1).

Similarly, for the other diagonal, from (i,j) to (i+1,j+1), delta (1,1).

So the possible edges are all pairs of vertices that are at chess king move distance sqrt(2), i.e., one step diagonally in any direction. But in this case, for each possible direction, but in the grid, from a vertex, it can have up to four diagonal neighbors, but here, the edges are only those that are the diagonals of the squares, which means that for a vertex (i,j), it can be connected to (i+1,j+1), (i+1,j-1), (i-1,j+1), (i-1,j-1), but only if those are within bounds and for the correct squares.

For example, from a vertex (i,j), it can be the start of a diagonal in the northeast direction if there is a square to the northeast, i.e., if i<4, j<4, then to (i+1,j+1) for the square with bottom-left (i,j).

Similarly, for northwesterly, to (i-1,j+1) if i>0, j<4, for the square with bottom-left (i-1,j)? Let's see.

If we have an edge from (i,j) to (i-1,j+1), what square is that in? For a square with bottom-left (k,l), the anti-diagonal is from (k+1,l) to (k,l+1). So if (k+1,l) = (i,j), then i=k+1, j=l, so k=i-1, l=j, and the other end is (k,l+1) = (i-1,j+1). So yes, for the square with bottom-left (i-1,j), there is an edge from (i,j) to (i-1,j+1), provided i>0, j<4? j can be from 0 to 3, but for the square to exist, bottom-left (i-1,j) must have 0≤i-1<4 and 0≤j<4, so 1≤i≤4, 0≤j<4.

Similarly for other directions.

But in any case, the graph is the grid graph with diagonal moves, but only the edges that are the diagonals of the squares, which include all possible diagonal moves between grid points that are distance sqrt(2) apart, as long as the intermediate points are not considered, but in this case, since it's a grid, from (i,j) to (i+1,j+1) is direct, no intermediate.

And for the 5x5 grid, the number of such possible diagonal edges.

But for our purpose, we need a matching, i.e., a set of edges with no common vertices.

Moreover, each edge is associated with a square, but since each edge is unique to a square (each possible diagonal is in exactly one square), and we can choose which edge to take for each "potential", but actually, the matching can be in the graph where we have all possible diagonal edges, and we select a matching.

But in this graph, the two types of edges may conflict.

To maximize the number of diagonals, which corresponds to the size of a matching in this graph.

The graph has 25 vertices. What is the size of a maximum matching in this graph?

Since the graph is bipartite? Let's see the parity.

Notice that the grid points can be colored in a chessboard fashion, black and white. For a 5x5 grid, there are 13 of one color and 12 of the other, since 25 is odd.

For example, (0,0) black, (0,1) white, etc. So number of black vertices: in odd-sum and even-sum. i+j even or odd.

For i,j from 0 to 4, number with i+j even: for i=0, j=0,2,4: 3 points; i=1, j=1,3: 2 points; i=2, j=0,2,4: 3; i=3, j=1,3: 2; i=4, j=0,2,4: 3; total 3+2+3+2+3=13. i+j odd: 12 points.

Now, each diagonal edge connects a black to a white vertex? Let's see: for an edge from (i,j) to (i+1,j+1), the sum i+j and (i+1)+(j+1) = i+j+2, so same parity? i+j and i+j+2 have the same parity, both even or both odd. But in a bipartite graph, edges should connect different parities.

Here, (i,j) and (i+1,j+1) have i+j and i+j+2, which have the same parity, so they are both even or both odd, so this edge is within the same color class? That can't be, because in a grid, diagonal moves should connect different colors if we use the standard chessboard coloring.

Mistake: in standard chessboard coloring, (0,0) is say black, (0,1) white, (1,0) white, (1,1) black. So (i,j) and (i+1,j+1): if (i,j) black, (i+1,j+1) = since both i and j increase by 1, the parity of i+j changes by 2, even, so same parity, so both black or both white. For example, (0,0) black, (1,1) black. But (0,1) white, (1,2) white. So yes, a diagonal move like (i,j) to (i+1,j+1) connects two vertices of the same color.

Similarly, for the other diagonal, from (i+1,j) to (i,j+1). (i+1,j) and (i,j+1). Sum (i+1)+j = i+j+1, and i+(j+1) = i+j+1, so both have the same sum i+j+1, so same parity. For example, (1,0) sum 1, odd, white; (0,1) sum 1, odd, white. So again, same color.

So in this graph, all edges are within the same color class; there are no edges between different colors. In other words, the graph is disconnected into two components: the even-sum vertices and the odd-sum vertices, and within each, we have edges.

For the even-sum vertices (i+j even), there are 13 vertices, and edges between them for the diagonal moves.

Similarly for odd-sum, 12 vertices.

And the edges are only within the same parity group.

Now, for a matching, since the graph is not connected between the two components, we can consider each component separately.

In the even-sum component, 13 vertices, we need a matching, size at most floor(13/2) = 6.

In the odd-sum component, 12 vertices, size at most 6.

But is this achievable? Moreover, each edge is associated with a square, and we need to make sure that the matching uses edges that are present, but in this case, for each possible diagonal, it is an edge in this graph.

Now, the question is what is the size of the maximum matching in this graph.

Since the graph is a grid, we can find it.

Notice that the even-sum vertices form a kind of lattice.

For the 5x5 grid, the even-sum vertices are like a diamond or something.

List the even-sum vertices: (0,0),(0,2),(0,4),(1,1),(1,3),(2,0),(2,2),(2,4),(3,1),(3,3),(4,0),(4,2),(4,4). So 13 points.

Now, the edges: for example, from (0,0) it can go to (1,1) (ne东北), and to (-1,1) but out of bound, to (1,-1) out, to (-1,-1) out. So only to (1,1).

Similarly, from (1,1) can go to (0,0), (2,2), (0,2), (2,0)? Let's see: northeast to (2,2), northwest to (0,2)? From (1,1) to (0,2): delta x=-1, delta y=1, so vector (-1,1), which is the anti-diagonal for the square to the west? Square with bottom-left (0,1): bottom-left (0,1), then anti-diagonal from (1,1) to (0,2)? Yes, since BR to TL, (1,1) is BR if BL is (0,1), but BL (0,1), then BR (1,1), TL (0,2), so yes, from (1,1) to (0,2).

Similarly, southeast to (2,2)? From (1,1) to (2,2), which is for square bottom-left (1,1)? BL (1,1), then main diagonal to (2,2).

Southwest to (0,0)? From (1,1) to (0,0), delta (-1,-1), which would be for square bottom-left (0,0)? BL (0,0), then anti-diagonal from (1,0) to (0,1), not from (1,1) to (0,0). From (1,1) to (0,0) is not an edge in our graph, because it's not a diagonal of a square; it's a diagonal but not of a unit square; it would be for a larger square.

In our graph, the only edges are those that are diagonals of the 1x1 squares, so the length is sqrt(2), and they must be between vertices of adjacent squares.

For (1,1), it can be connected to (0,0), (0,2), (2,0), (2,2)? To (0,0): as above, not a diagonal of a unit square. The unit squares near (1,1): the square to the southwest: bottom-left (0,0), which has diagonal to (1,1) from (0,0)? From (0,0) to (1,1) is an edge, but that's from (0,0) to (1,1), not from (1,1) to (0,0), but since undirected, same edge.

In the graph, edges are undirected for matching.

So for vertex (1,1), it can be incident to edges: to (0,0) (via square (0,0) main diagonal), to (0,2) (via square (0,1) anti-diagonal? Square (0,1): BL (0,1), then anti-diagonal from (1,1) to (0,2)? Yes.

To (2,0) ? Via square (1,0): BL (1,0), then anti-diagonal from (2,0) to (1,1)? From BR to TL, (2,0) to (1,1), yes.

To (2,2) via square (1,1) main diagonal, from (1,1) to (2,2).

So yes, (1,1) is connected to (0,0), (0,2), (2,0), (2,2).

Similarly for other vertices.

In general, for an internal vertex, it has degree 4, but on boundaries less.

Now, back to matching. The even-sum component has 13 vertices, which is odd, so maximum matching size is 6, leaving one vertex unmatched.

Similarly for odd-sum, 12 vertices, maximum matching size 6.

But is this achievable? That is, can we find a matching of size 6 in the even-sum component and size 6 in the odd-sum component? But the components are separate, so yes, but the issue is that the matchings are in different components, so no conflict.

However, each edge in the matching corresponds to a diagonal in some square, and since the components are different, but each square has edges in both components? No.

Let's see for a single square. Take square with bottom-left (i,j). It has two diagonals: one is (i,j) to (i+1,j+1), and the other is (i+1,j) to (i,j+1).

Now, the parity of (i,j): if i+j even, then (i,j) even-sum, (i+1,j+1) i+1+j+1=i+j+2 even, so same.

(i+1,j): i+1+j = i+j+1, odd if i+j even. Similarly (i,j+1): i+(j+1)=i+j+1, odd. So the other diagonal is between two odd-sum vertices.

Therefore, for each square, one diagonal is in the even-sum component (when i+j even), and the other is in the odd-sum component.

Specifically, if the bottom-left (i,j) has i+j even, then the main diagonal (i,j)-(i+1,j+1) is in even-sum graph, and the anti-diagonal (i+1,j)-(i,j+1) is in odd-sum graph.

If i+j odd, then main diagonal is between (i,j) odd-sum and (i+1,j+1) i+1+j+1=i+j+2 odd, so in odd-sum graph, and anti-diagonal between (i+1,j) even-sum and (i,j+1) even-sum, in even-sum graph.

So in any case, each square contributes one edge to the even-sum component and one edge to the odd-sum component.

Now, for the matching, we are selecting a set of edges from the entire edge set, with no common vertices. Since the even-sum and odd-sum components are disjoint (no vertices in common), and no edges between them, we can find a matching in each component separately.

The total number of edges we can select is the sum of the maximum matchings in each component.

For even-sum component, 13 vertices, maximum matching size is at most 6.

For odd-sum, 12 vertices, at most 6.

But is it possible to achieve 6 in each?

For the odd-sum component, 12 vertices, which is even, and the graph is likely to have a perfect matching, size 6.

For even-sum, 13 vertices, odd, so maximum 6, with one left out.

Now, is there a matching of size 6 in the even-sum graph? Since it's a bipartite graph? No, the components are not bipartite in the usual sense, but we can check if it has a matching of size 6.

Note that the even-sum graph is actually a lattice, and for a grid, it might be possible.

We can think of the even-sum vertices as a graph where each vertex has even degree or something, but let's see if it is possible to have a near-perfect matching.

For example, in a 5x5 grid, the even-sum vertices form a shape that can be covered by 6 edges.

Similarly for odd-sum.

But to confirm, let's try to construct a large matching.

First, for the odd-sum component: 12 vertices. Can we find a perfect matching, size 6.

The odd-sum vertices: (0,1),(0,3),(1,0),(1,2),(1,4),(2,1),(2,3),(3,0),(3,2),(3,4),(4,1),(4,3).

Now, we can pair them with edges.

For example, pair (0,1) to (1,2) via the anti-diagonal? From (0,1) to (1,2)? But what square? The edge from (0,1) to (1,2): this would be for the main diagonal of square with bottom-left (0,1)? BL (0,1), then to (1,2), yes, main diagonal.

Similarly, (0,3) to (1,4) via square (0,3)? BL (0,3), to (1,4).

Then (1,0) to (2,1) via square (1,0)? BL (1,0), to (2,1).

(1,4) to (2,3)? From (1,4) to (2,3), which is for anti-diagonal of square (1,3)? BL (1,3), then anti-diagonal from (2,3) to (1,4), yes.

(2,1) to (3,2) via square (2,1)? BL (2,1), to (3,2).

But (2,1) is already paired? In this, I have (1,0) to (2,1), so (2,1) is used.

List the pairs:

1. (0,1)-(1,2) // square (0,1) main diag

2. (0,3)-(1,4) // square (0,3) main diag

3. (1,0)-(2,1) // square (1,0) main diag? BL (1,0), to (2,1), yes.

4. (1,4)-(2,3) // as above, for square (1,3) anti-diag? From (1,4) to (2,3), which is BL (1,3)? BL (1,3), then from (2,3) to (1,4), yes anti-diag.

5. Now (2,1) is already in pair 3, so cannot use.

(3,0) to (4,1) via square (3,0)? BL (3,0), to (4,1).

6. (3,4) to (4,3) via square (3,3)? BL (3,3), to (4,4)? But (4,4) is not in this list; (4,4) is even-sum, not odd. Mistake.

(3,4): i=3,j=4, sum 7 odd, yes. To (4,3)? From (3,4) to (4,3), delta (1,-1), which is for the anti-diagonal of square with bottom-left (3,3)? BL (3,3), then anti-diagonal from (4,3) to (3,4), yes.

But (4,3) is in the list.

Now, we have pairs: 1:(0,1)-(1,2), 2:(0,3)-(1,4), 3:(1,0)-(2,1), 4:(1,4)-(2,3)? (1,4) is in pair 2, already used.

Problem: (1,4) is used in pair 2 with (0,3), and now I cannot use it again.

In pair 4, I have (1,4)-(2,3), but (1,4) is already paired.

So conflict.

After pairing 1,2,3: used vertices: (0,1),(1,2),(0,3),(1,4),(1,0),(2,1).

Remaining odd-sum vertices: (2,3),(3,0),(3,2),(3,4),(4,1),(4,3).

Now, (2,3) can pair with (3,2) or (3,4)? To (3,2): from (2,3) to (3,2), which is for anti-diagonal of square (2,2)? BL (2,2), then from (3,2) to (2,3), yes.

Similarly, (3,0) to (4,1) via square (3,0)? BL (3,0), to (4,1).

(3,4) to (4,3) via square (3,3)? As above.

But (4,1) and (4,3) are both to be paired, but (3,0) to (4,1), (3,4) to (4,3), and (2,3) to (3,2).

So pairs: 4. (2,3)-(3,2) // sq (2,2) anti-diag

5. (3,0)-(4,1) // sq (3,0) main diag? BL (3,0), to (4,1), yes

6. (3,4)-(4,3) // sq (3,3) anti-diag

Now, all vertices are covered: listed: (2,3),(3,2),(3,0),(4,1),(3,4),(4,3). And no overlap: pair 4: (2,3),(3,2); pair 5: (3,0),(4,1); pair 6: (3,4),(4,3). But (3,2) is in pair 4, (3,0) in 5, (3,4) in 6, all distinct. And (4,1),(4,3) distinct. So yes, perfect matching for odd-sum component with 6 edges.

Now for even-sum component: 13 vertices. Need a matching of size 6, so one vertex left out.

Vertices: (0,0),(0,2),(0,4),(1,1),(1,3),(2,0),(2,2),(2,4),(3,1),(3,3),(4,0),(4,2),(4,4).

We need to select 6 edges, disjoint vertices.

For example, we can pair (0,0) to (1,1) via sq (0,0) main diag.

(0,2) to (1,3) via sq (0,2)? BL (0,2), to (1,3)? But (1,3) is in list, but sq (0,2) main diag is (0,2) to (1,3), yes.

(0,4) to (1,5)? But j=5 out of bound. (0,4) can only connect to (1,5) out, or to (1,3)? From (0,4) to (1,3)? Delta (1,-1), which would be for anti-diagonal of sq (0,3)? BL (0,3), then anti-diag from (1,3) to (0,4), yes. So (0,4)-(1,3).

But (1,3) is also a vertex.

Now, (0,0)-(1,1), (0,2)-(1,3)? But (1,3) is used in both? No, (0,0)-(1,1) uses (1,1), (0,2)-(1,3) uses (1,3), different.

(0,4)-(1,3)? But (1,3) is already used if we pair (0,2)-(1,3).

So we have (0,0)-(1,1), (0,2)-(1,3), (0,4)-(1,3) but conflict on (1,3).

So instead, for (0,4), we can pair to (1,3) or to nothing else? From (0,4), neighbors: to (1,5) out, to (1,3) as above, to (-1,5) out, to (-1,3) out. So only to (1,3).

Similarly, (1,1) is connected to (0,0),(0,2),(2,0),(2,2), as before.

So let's pair (0,0)-(1,1) // sq (0,0) main

Then (0,2) can go to (1,1) but (1,1) used, or to (1,3). So pair (0,2)-(1,3) // sq (0,2) main? BL (0,2), to (1,3), yes.

Then (0,4)-(1,3) but (1,3) used.

So cannot pair (0,4) yet.

Pair (0,4) first: to (1,3) // sq (0,3) anti-diag? As above.

Then (0,2) needs to be paired, but (1,3) used, so to (1,1)? But if we don't pair (0,0) yet.

So pair (0,4)-(1,3)

Then (0,2)-(1,1) // sq (0,1) anti-diag? From (0,2) to (1,1)? Delta (1,-1), for sq (0,1)? BL (0,1), then anti-diag from (1,1) to (0,2), yes.

Then (0,0) can be paired to (1,1), but (1,1) used.

To nothing else, so (0,0) left alone? But we can pair it later, but it only connects to (1,1), which is used.

So if we pair (0,4)-(1,3) and (0,2)-(1,1), then (0,0) has no neighbor left, so it must be unmatched.

But we need to cover others.

After pairing (0,4)-(1,3), (0,2)-(1,1), then vertices used: (0,4),(1,3),(0,2),(1,1).

Remaining: (0,0), (1,1) used, (1,3) used, (2,0),(2,2),(2,4),(3,1),(3,3),(4,0),(4,2),(4,4). And (1,1),(1,3) used, but (1,1) is used, but there is (1,1) in list, but it's used.

List remaining: (0,0), (2,0),(2,2),(2,4),(3,1),(3,3),(4,0),(4,2),(4,4). That's 9 vertices, but we have 13-4=9, yes.

Now, (0,0) can only connect to (1,1), which is used, so it has no edges to unused vertices. So it must be unmatched, but we can leave it for now.

Now, we can pair (2,0) to (3,1) via sq (2,0) main diag? BL (2,0), to (3,1).

(2,2) to (3,3) via sq (2,2) main diag? BL (2,2), to (3,3).

(2,4) to (3,3) but (3,3) might be used.

(2,4) to (3,3) via sq (2,3) anti-diag? BL (2,3), then anti-diag from (3,3) to (2,4), yes.

But if we pair (2,2)-(3,3), then (2,4) can go to (3,3) but conflict.

Pair (2,0)-(3,1) // sq (2,0) main

(2,4)-(3,3) // sq (2,3) anti-diag? From (2,4) to (3,3), delta (1,-1), for sq (2,3)? BL (2,3), then from (3,3) to (2,4), yes.

Then (2,2) can go to (3,1) but (3,1) used, or to (3,3) used, or to (1,1) used, etc. So pair (2,2) to (3,1)? But (3,1) is used in (2,0)-(3,1).

(2,2) neighbors: to (1,1) used, (1,3) used, (3,1), (3,3).

(3,1) is to be used, but if we didn't pair yet.

After (2,0)-(3,1) and (2,4)-(3,3), then (2,2) has neighbors (3,1) and (3,3), both used, so no pair.

Pair (2,2)-(3,3) first, but then (2,0) and (2,4) may not have partners.

So let's pair (2,2)-(3,1)? From (2,2) to (3,1), delta (1,-1), for sq (2,1)? BL (2,1), then anti-diag from (3,1) to (2,2), yes.

Similarly, (2,0)-(3,1)? But (3,1) would be shared.

(2,0) to (3,1) as above.

But if we pair (2,0)-(3,1) and (2,2)-(3,1), conflict on (3,1).

So we need to choose.

Pair (2,0)-(3,1) // sq (2,0) main diag

Then (2,2) can go to (3,3) // sq (2,2) main diag

Then (2,4) to (3,3) but (3,3) used, or to (3,5) out, so only to (3,3), which is used. So cannot pair (2,4).

Pair (2,4)-(3,3) // sq (2,3) anti-diag

Then (2,0) to (3,1) // sq (2,0) main

Then (2,2) to (3,1) but (3,1) used, or to (3,3) used. So same problem.

After pairing (0,4)-(1,3), (0,2)-(1,1), and say (2,0)-(3,1), (2,4)-(3,3), then (2,2) is left, and (3,1),(3,3) used, so (2,2) can only connect to used vertices.

Then we have (0,0) unmatched, (2,2) unmatched, and others: (3,1) used, (3,3) used, now (4,0),(4,2),(4,4),(3,1) used, (3,3) used, but (3,1) and (3,3) are used, but there is (3,1) and (3,3), but also (4,0),(4,2),(4,4).

So pair (4,0) to (3,1) but (3,1) used, or to (5,1) out, or to (3,-1) out, so only to (3,1), used.

Similarly, (4,2) to (3,1) used, or to (3,3) used, or to (5,3) out, etc. So (4,2) can connect to (3,1) or (3,3), both used, or to (5,3) out, so no.

(4,4) to (3,3) used, or to (5,5) out.

So many unmatched.

But we can leave (0,0) unmatched, and try to cover others.

For example, pair (4,0)-(3,1) // but (3,1) might be used.

Let's start over.

Pair the corners.

For example, pair (0,0)-(1,1) // sq (0,0) main

(0,4)-(1,3) // sq (0,3) anti-diag

(4,0)-(3,1) // sq (3,0) anti-diag? From (4,0) to (3,1), delta (-1,1), for sq (3,0)? BL (3,0), then from (4,0) to (3,1), which is not a diagonal; the diagonals are (3,0)-(4,1) or (4,0)-(3,1)? (4,0) to (3,1) is a diagonal, and for sq with BL (3,0), the anti-diagonal is from (4,0) to (3,1), yes.

Similarly, (4,4)-(3,3) // sq (3,3) anti-diag? From (4,4) to (3,3), delta (-1,-1), for sq (3,3)? BL (3,3), then from (4,4) to (3,3)? The diagonals are (3,3)-(4,4) or (4,3)-(3,4), so (3,3)-(4,4) is the main diagonal, not anti. Anti is (4,3)-(3,4).

So for (4,4), it can be paired to (3,3) via sq (3,3) main diag, from (3,3) to (4,4).

Yes.

So pair: 1. (0,0)-(1,1) // sq (0,0) main

2. (0,4)-(1,3) // sq (0,3) anti-diag

3. (4,0)-(3,1) // sq (3,0) anti-diag

4. (4,4)-(3,3) // sq (3,3) main diag

Now, vertices used: (0,0),(1,1),(0,4),(1,3),(4,0),(3,1),(4,4),(3,3)

Remaining even-sum vertices: (0,2),(1,1) used, (1,3) used, (2,0),(2,2),(2,4),(3,1) used, (3,3) used, (4,0) used, (4,2),(4,4) used. So (0,2),(2,0),(2,2),(2,4),(4,2)

Now, (0,2) can connect to (1,1) used, (1,3) used, or to (1,1) only? From (0,2) to (1,1) or (1,3), both used, or to (-1,3) out, so no neighbor unused. So (0,2) has no edges to unused vertices.

Similarly, (2,0) can connect to (1,1) used, (3,1) used, (1,0) not in even-sum, (3,0) not in list, (2,0) neighbors: to (1,1) used, (3,1) used, (1,-1) out, (3,-1) out. So only to (1,1) and (3,1), both used. So no.

(2,2) to (1,1) used, (1,3) used, (3,1) used, (3,3) used.

(2,4) to (1,3) used, (3,3) used, (1,5) out, (3,5) out.

(4,2) to (3,1) used, (3,3) used, (5,3) out, (3,1) used.

So all remaining vertices have no available neighbors, so they are isolated in the remaining graph. But we have 5 vertices left, all with no edges to unused vertices, so we can only have the 4 pairs we have, but we need 6 for the matching size.

But we have only 4 edges so far, and 8 vertices covered, 5 left, but no more edges possible, so matching size is 4, but we want 6.

This is not good.

We can choose different pairs.

For example, instead of pairing (0,0)-(1,1), pair (0,0) to nothing, and use (1,1) for others.

For example, (1,1) can pair to (2,2) or etc.

Pair (1,1)-(2,2) // sq (1,1) main diag

(1,3)-(2,4) // sq (1,3) main diag? BL (1,3), to (2,4)

(0,2)-(1,1) but (1,1) used, or to (1,3) used.

(0,2) can pair to (1,3) // sq (0,2) main diag? From (0,2) to (1,3)

But if we pair (1,3)-(2,4), then (0,2) can pair to (1,3), but (1,3) used.

So pair (0,2)-(1,3) // sq (0,2) main

Then (1,1)-(2,2) // sq (1,1) main

Then (0,0) can pair to (1,1) but used, or to nothing, so leave.

(0,4)-(1,3) but (1,3) used.

So pair (0,4)-(1,5) out, or to (1,3) used, so no.

Pair (0,4)-(1,3) first, then (0,2) has no partner.

So perhaps leave (0,0) and (0,4) unmatched or something.

Another idea: pair the first row.

For example, (0,0)-(1,1) // sq (0,0) main

(0,2)-(1,3) // sq (0,2) main

(0,4) cannot be paired with (1,3) because (1,3) used, and no other, so (0,4) must be unmatched or paired later, but no.

After this, (0,4) is left.

Then for row 1, (1,1) used, (1,3) used, so no.

Then for row 2, (2,0) can pair to (3,1) // sq (2,0) main

(2,2) to (3,3) // sq (2,2) main

(2,4) to (3,3) but (3,3) used, or to (3,5) out, so to (3,3) used, or to (1,3) used, so no.

Pair (2,4)-(3,3) // sq (2,3) anti-diag? From (2,4) to (3,3), as before.

But then (2,0) and (2,2) can be paired to (3,1) and (3,3), but (3,3) would be used if we pair (2,4)-(3,3).

So pair (2,0)-(3,1) // sq (2,0) main

(2,2)-(3,3) // sq (2,2) main

(2,4) no partner.

Then for row 3, (3,1) used, (3,3) used, so no.

Then for row 4, (4,0) to (3,1) used, (4,2) to (3,1) used or (3,3) used, (4,4) to (3,3) used.

So again, many unmatched.

Perhaps use anti-diagonals for some.

For example, in square (0,0), instead of main diagonal, use the other diagonal: from (1,0) to (0,1), but (1,0) and (0,1) are not even-sum; (1,0) sum 1 odd, (0,1) sum 1 odd, so in the odd-sum graph, not here.

For even-sum graph, in each square, only one of the two diagonals is in this component.

For square with i+j even, the main diagonal is in even-sum, anti is in odd-sum.

For i+j odd, main is in odd-sum, anti is in even-sum.

So for even-sum graph, for a square with i+j even, we can only use the main diagonal; for i+j odd, we can only use the anti-diagonal.

For example, square (0,0): i+j=0 even, so only main diagonal: (0,0)-(1,1) is in even-sum; the anti-diagonal (1,0)-(0,1) is in odd-sum.

Square (0,1): i+j=1 odd, so only anti-diagonal: (1,1)-(0,2) is in even-sum? (1,1) sum 2 even, (0,2) sum 2 even, yes, and it's the anti-diagonal for this square.

Main diagonal would be (0,1)-(1,2), both odd-sum, in odd-sum graph.

So for the even-sum graph, the edges are:

• For each square with i+j even, the main diagonal edge.

• For each square with i+j odd, the anti-diagonal edge.

And each such edge is between two even-sum vertices.

Now for the 4x4 grid of squares, i,j from 0 to 3.

Squares with i+j even: (0,0),(0,2),(1,1),(1,3),(2,0),(2,2),(3,1),(3,3) — 8 squares.

Squares with i+j odd: (0,1),(0,3),(1,0),(1,2),(2,1),(2,3),(3,0),(3,2) — 8 squares.

For each of these 16 squares, there is one edge in the even-sum graph.

The graph has 13 vertices and 16 edges (since 16 squares, each contributes one edge to this graph).

Now, we need a matching of size 6.

Since there are 13 vertices, size 6 is possible if the graph has a matching of that size.

The minimum vertex cover or something, but we can try to find one.

Notice that the graph may have a perfect matching for the even-sum vertices except one.

For example, consider the following pairing:

For each 2x2 block, but it's 5x5 vertices.

We can pair vertically or horizontally in some way.

For instance, pair (0,0)-(1,1) via sq (0,0) main diag (i+j=0 even)

Then (0,2)-(1,3) via sq (0,2) main diag (i+j=2 even)

Then (0,4) is left, but can pair to (1,5) out, or to (1,3) but (1,3) used, so no.

Pair (0,4)-(1,3) but for sq (0,3) i+j=3 odd, so anti-diagonal, which is (1,3)-(0,4), yes, so in even-sum graph.

So pair (0,4)-(1,3) via sq (0,3) anti-diag.

Then (0,2) can pair to (1,1) but (1,1) may be used.

If we pair (0,0)-(1,1), then (1,1) used, so (0,2) can pair to (1,3) but if we pair (0,4)-(1,3), then (1,3) used, so (0,2) has no partner.

So for the first row, we can pair only two of the three.

Say pair (0,0)-(1,1) and (0,2)-(1,3), then (0,4) unmatched.

Similarly for last row.

Pair (4,0)-(3,1) via sq (3,0) i+j=3 odd, anti-diag: from (4,0) to (3,1)

(4,2)-(3,3) via sq (3,2) i+j=5 odd, anti-diag: from (4,2) to (3,3)? For sq (3,2), BL (3,2), anti-diag from (4,2) to (3,3), yes.

(4,4)-(3,3) but (3,3) may be used, or to (3,5) out, so only to (3,3).

But if we pair (4,2)-(3,3), then (4,4) can pair to (3,3) used, or to (3,5) out, so no.

Pair (4,4)-(3,3) via sq (3,3) i+j=6 even, main diag: (3,3)-(4,4)

Then (4,0)-(3,1) as above.

(4,2) can pair to (3,1) but (3,1) may be used, or to (3,3) used, so no.

So for the last row, we can pair two, leave (4,2) or something.

Now for the middle.

Let's try to pair in a different way.

Suppose we pair for each "column" or "row".

Notice that the graph is bipartite? With respect to what?

In the even-sum vertices, we can further color or something.

For example, in the even-sum vertices, we can consider the sum i+j even, but that's already done, or difference i-j.

For example, vertices with i-j even or odd.

Take (0,0) i-j=0 even, (0,2) i-j= -2 even, (0,4) i-j= -4 even, (1,1) i-j=0 even, (1,3) i-j= -2 even, (2,0) i-j=2 even, (2,2) i-j=0 even, (2,4) i-j= -2 even, (3,1) i-j=2 even, (3,3) i-j=0 even, (4,0) i-j=4 even, (4,2) i-j=2 even, (4,4) i-j=0 even. All have i-j even? (0,0):0, (0,2):-2, (0,4):-4, (1,1):0, (1,3):-2, (2,0):2, (2,2):0, (2,4):-2, (3,1):2, (3,3):0, (4,0):4, (4,2):2, (4,4):0, all even. So all vertices have i-j even, so no bipartite separation within.

Perhaps it is not bipartite, but has an odd cycle.

For example, take vertices (1,1), (0,0), (1,1)-(0,0), (0,2), (1,1)-(0,2), and (0,0)-(0,2)? But no direct edge, but (0,0) and (0,2) are both connected to (1,1), but not to each other, so no triangle.

Take (1,1), (2,2), (2,0), (1,1)-(2,2), (1,1)-(2,0), (2,2)-(2,0)? No edge between (2,2) and (2,0).

Take a square: for example, vertices (1,1), (1,3), (3,1), (3,3), but not all connected.

From (1,1) to (2,2), from (1,3) to (2,2)? (1,3) to (2,2): delta (1,-1), which is an edge, for sq (1,2) i+j=3 odd, anti-diag: from (1,3) to (2,2)? For sq (1,2), BL (1,2), then anti-diag from (2,2) to (1,3), yes.

Similarly, (1,1) to (2,2), for sq (1,1) i+j=2 even, main diag.

Then (1,1) to (1,3)? No direct edge.

But (2,2) is connected to both (1,1) and (1,3), so if we have edges (1,1)-(2,2) and (1,3)-(2,2), then it's a star, no cycle.

To have a cycle, for example, (0,0)-(1,1), (1,1)-(2,2), (2,2)-(1,1) but not.

Take (0,2), (1,1), (2,2), (1,3), then edges: (0,2)-(1,1), (1,1)-(2,2), (2,2)-(1,3), (1,3)-(0,2)? (1,3) to (0,2): delta (-1,-1), which is an edge, for sq (0,2)? BL (0,2), then from (1,3) to (0,2)? The main diagonal is (0,2)-(1,3), so yes, but that's the same as (0,2)-(1,3), which is already an edge.

So we have edges: A: (0,2)-(1,1), B: (1,1)-(2,2), C: (2,2)-(1,3), D: (1,3)-(0,2).

Then this is a cycle of 4: (0,2)- (1,1) - (2,2) - (1,3) - (0,2), and also (0,2)-(1,3) directly, so it's a complete graph K4 almost, but not complete, but has a 4-cycle.

So the graph has even cycles, so it is bipartite.

In this subgraph, vertices (0,2), (1,1), (2,2), (1,3), and edges between them: (0,2)-(1,1), (0,2)-(1,3), (1,1)-(2,2), (1,3)-(2,2), and no (1,1)-(1,3) or (0,2)-(2,2), so it's a 4-cycle if we take (0,2)-(1,1)-(2,2)-(1,3)-(0,2), yes, a 4-cycle, so bipartite.

In general, the graph may be bipartite.

Let me see the parity of i j or something.

Notice that for an edge, the two endpoints have the same i+j even, but different i-j or something.

For example, from (i,j) to (i+1,j+1), then i-j and (i+1)-(j+1) = i-j, same.

From (i,j) to (i+1,j-1)? But in our case, for the edges, it's either (i,j) to (i+1,j+1) or (i,j) to (i-1,j+1) etc, but in terms of the vector.

For a move from (i,j) to (i+1,j+1), the difference in i is +1, j+1.

To (i+1,j-1), i+1, j-1, delta i+1, j-1.

In the first case, the sum i+j changes by +2, so same parity, and i-j is unchanged.

In the second case, to (i+1,j-1), i+j changes by 0, so same, i-j changes by (i+1)-(j-1) - (i-j) = i+1-j+1 -i +j = 2, even, so i-j changes by 2, so same parity of i-j.

But for the edge, the two endpoints have the same i+j (both even-sum), and the same i-j modulo 2? i-j is always even for all, as we saw earlier, since i+j even implies i and j same parity, so i-j even.

So i-j is always even, so no help.

Perhaps we can use the value of i or j.

For the matching, we can use a different strategy.

Suppose we leave the corner (0,0) unmatched, and try to match the rest.

Then the remaining 12 vertices should have a matching of size 6, but 12 is even, so possible.

Vertices: all except (0,0).

Now, (0,2) can pair to (1,1) or (1,3)

(0,4) to (1,3)

(1,1) to (0,2) or (2,2) or (2,0) or (0,0) but (0,0) out

(1,3) to (0,2) or (0,4) or (2,2) or (2,4)

etc.

Pair (0,2)-(1,1) // sq (0,1) anti-diag? i+j=1 odd, so anti-diag: from (1,1) to (0,2), yes.

Then (0,4)-(1,3) // sq (0,3) anti-diag: from (1,3) to (0,4)

Then (1,1) used, (1,3) used.

Then (2,0) can pair to (1,1) used, or to (3,1) , so pair to (3,1) // sq (2,0) main diag? i+j=2 even, main: (2,0)-(3,1)

Similarly, (2,2) to (1,1) used, or to (1,3) used, or to (3,1) , or to (3,3), so pair to (3,3) // sq (2,2) main: (2,2)-(3,3)

Then (2,4) to (1,3) used, or to (3,3) used, or to (3,5) out, so no partner.

Pair (2,4)-(3,3) // sq (2,3) anti-diag: i+j=5 odd, anti: from (3,3) to (2,4)

But then (2,2) can be paired to (3,1) or (3,3), but (3,3) may be used.

After pairing (2,0)-(3,1), (2,2) can pair to (3,1) but used, or to (3,3), so if we pair (2,2)-(3,3), then (2,4) no partner.

If we pair (2,4)-(3,3), then (2,2) no partner.

So (2,2) or (2,4) will be left.

Then for (4,0) to (3,1) used, or to (5,1) out, so only to (3,1), used.

Similarly, (4,2) to (3,1) used, (3,3) may be used, (5,3) out, so no.

(4,4) to (3,3) may be used.

So still not.

Pair (4,0)-(3,1) but (3,1) may be used.

Let's pair the last rows first.

Pair (4,0)-(3,1) // sq (3,0) anti-diag

(4,2)-(3,3) // sq (3,2) anti-diag: i+j=5 odd, anti: from (4,2) to (3,3)

(4,4)-(3,3) but (3,3) used, so cannot.

Pair (4,4)-(3,3) // sq (3,3) main: i+j=6 even, main: (3,3)-(4,4)

Then (4,0)-(3,1) as above.

(4,2) can pair to (3,1) or (3,3), both may be used.

If we pair (4,0)-(3,1), (4,4)-(3,3), then (4,2) no partner.

So again, (4,2) is problem.

To solve this, we can pair (4,2) to (3,1) or (3,3), but in a way that doesn't conflict.

For example, if we don't use (3,1) for (4,0), but for (4,2).

But (4,0) can only connect to (3,1), so must use that.

Unless we leave (4,0) unmatched, but we already have (0,0) unmatched, and we need to match 12 vertices with 6 edges, but if we leave another unmatched, we can't.

Perhaps leave a vertex in the middle unmatched.

For example, leave (2,2) unmatched.

Then we can pair the others.

For example, pair (0,0)-(1,1) // sq (0,0) main

(0,2)-(1,3) // sq (0,2) main

(0,4)-(1,3) but (1,3) used.

Pair (0,4)-(1,3) // sq (0,3) anti-diag

Then (0,2) can pair to (1,1) but (1,1) used in (0,0)-(1,1), so no.

After (0,0)-(1,1), (0,4)-(1,3), then (0,2) no partner.

Pair (0,2)-(1,1) and (0,4)-(1,3), then (0,0) no partner.

So for the top, we can pair two out of three.

Say pair (0,0)-(1,1) and (0,2)-(1,3), leave (0,4) unmatched.

Similarly for bottom, pair (4,0)-(3,1) and (4,4)-(3,3), leave (4,2) unmatched.

Then for the middle, we have (2,0),(2,2),(2,4),(1,1) used, (1,3) used, (3,1) used, (3,3) used, and (1,1) etc used, but (2,0) can pair to (3,1) but (3,1) used, or to (1,1) used, so no.

(2,0) can only connect to (1,1) and (3,1), both may be used.

In this case, after pairing the top and bottom, many vertices are used.

After pairing (0,0)-(1,1), (0,2)-(1,3), (4,0)-(3,1), (4,4)-(3,3), then used vertices: (0,0),(1,1),(0,2),(1,3),(4,0),(3,1),(4,4),(3,3)

Remaining: (0,4), (2,0),(2,2),(2,4),(3,1) used, (3,3) used, (4,2), and (1,1) used, but (1,1) is used, but there is (1,1), but it's used, so vertices: (0,4), (2,0),(2,2),(2,4),(4,2), and (3,1) used, but (3,1) is used, but is (3,1) in the list? (3,1) is even-sum, and it's used in (4,0)-(3,1), so yes.

So remaining: (0,4), (2,0), (2,2), (2,4), (4,2)

Now, (0,4) can only connect to (1,3), which is used.

(2,0) to (1,1) used, (3,1) used.

(2,2) to (1,1) used, (1,3) used, (3,1) used, (3,3) used.

(2,4) to (1,3) used, (3,3) used.

(4,2) to (3,1) used, (3,3) used.

So all have no available neighbors, so no edges, so we have only 4 edges, and 5 vertices unmatched, but we can add more if we pair differently.

Perhaps for the middle rows, pair differently.

For example, for square (1,1): i+j=2 even, so we can use main diagonal (1,1)-(2,2)

For (1,3): i+j=4 even, main diagonal (1,3)-(2,4)

For (3,1): i+j=4 even, main diagonal (3,1)-(4,2)? (4,2) is a vertex.

For (3,3): i+j=6 even, main diagonal (3,3)-(4,4)

Then for the top, we have (0,0), (0,2), (0,4)

(0,0) can pair to (1,1) but (1,1) may be used.

If we pair (1,1)-(2,2), then (0,0) can pair to (1,1), but conflict.

So pair (0,0) to (1,1) for sq (0,0) main, but then (1,1) is used, so cannot use for (1,1)-(2,2).

So we have to choose.

Perhaps use the anti-diagonal for some squares in the even-sum graph.

For example, for a square with i+j even, we can only use the main diagonal in even-sum graph; we don't have a choice for the component; for each square, only one diagonal is in this graph, so we have to use that edge if we want to use it.

For the even-sum graph, the edges are fixed for each square; we can choose to include the edge or not, but for the matching, we select a subset of the edges that form a matching.

So for each square, there is an edge in the graph, and we can choose to include it in the matching or not, but with the constraint that the edges are vertex-disjoint.

So it's a matching in this graph with 13 vertices and 16 edges (one for each square).

To have a matching of size 6, we need to cover 12 vertices, leaving one out.

The graph is connected or not? But likely it has a matching of size 6.

We can use a systematic way.

Consider the following pairing:

• For i even, j even: but let's list the squares.

Perhaps pair along the diagonal.

For example, pair (0,0)-(1,1) // sq (0,0)

(2,2)-(3,3) // sq (2,2) main, i+j=4 even

(4,4) no partner, or to (3,3) but (3,3) used.

Pair (1,1)-(2,2) // sq (1,1) main, i+j=2 even

(3,3)-(4,4) // sq (3,3) main, i+j=6 even

Then (0,0) can be paired to (1,1) but used.

So not.

Pair (0,2)-(1,3) // sq (0,2) main

(2,0)-(3,1) // sq (2,0) main

(2,4)-(3,3) // sq (2,3) anti-diag, i+j=5 odd, anti: (2,4)-(3,3)? From (2,4) to (3,3), delta (1,-1), for sq (2,3), BL (2,3), then from (3,3) to (2,4), yes.

Then (4,0)-(3,1) but (3,1) may be used in (2,0)-(3,1)

If we pair (2,0)-(3,1), then (4,0) can pair to (3,1) but used.

Pair (4,0)-(3,1) for sq (3,0) anti-diag

Then (2,0) can pair to (3,1) but conflict.

So for (2,0) and (4,0), both want to pair with (3,1), but only one can.

So leave (3,1) for one of them.

Say pair (2,0)-(3,1) // sq (2,0) main

Then (4,0) can pair to (3,1) but used, or to nothing, so no.

Pair (4,0)-(3,1) // sq (3,0) anti-diag

Then (2,0) no partner.

So not good.

Perhaps pair (0,0)-(1,1) , (0,2)-(1,3) , then for (0,4) no, but we can pair (0,4) with (1,3) but (1,3) used.

For (1,1) and (1,3), they are both used, so for the row 1, done.

Then for (2,0), pair to (3,1) // sq (2,0) main

(2,2) to (3,3) // sq (2,2) main

(2,4) to (3,3) but (3,3) used, so pair to (3,3) conflict, or to (1,3) used, so no.

Pair (2,4)-(3,3) for sq (2,3) anti-diag

Then (2,2) can be paired to (3,1) or (3,3), say to (3,1) // but (3,1) may be used in (2,0)-(3,1)

If we pair (2,0)-(3,1), (2,4)-(3,3), then (2,2) no partner.

So (2,2) is the problem.

To include (2,2), pair it first.

Pair (2,2)-(3,3) // sq (2,2) main

Then (2,0) can pair to (3,1) // sq (2,0) main

(2,4) to (3,3) but (3,3) used, so no.

Pair (2,4)-(3,3) conflict.

So still.

Pair (2,2)-(1,1) but (1,1) may be used.

If we haven't used row 1 yet.

So let's not use the top yet.

Pair (2,2)-(3,3) // sq (2,2) main

(2,2)-(1,1) but can't both.

Perhaps (2,2)-(3,3) , then (1,1) can be paired to (0,0) or (0,2) or (2,0) or (2,4), etc.

Say pair (1,1)-(2,0) // how? From (1,1) to (2,0): delta (1,-1), which is an edge, for sq (1,0)? BL (1,0), i+j=1 odd, so anti-diag: from (2,0) to (1,1), yes.

Similarly, (1,3)-(2,4) // sq (1,3) main? i+j=4 even, main: (1,3)-(2,4)

Then (0,0) can pair to (1,1) but (1,1) may be used in (1,1)-(2,0)

If we pair (1,1)-(2,0), then (0,0) can pair to (1,1) but used, or to nothing.

Pair (0,0)-(1,1) , then (2,0) can pair to (1,1) but used, or to (3,1).

So pair (2,0)-(3,1) // sq (2,0) main

Then (0,0)-(1,1) , (1,1)-(2,0) conflict.

So difficult.

Perhaps it is not possible to have size 6, but earlier we thought 6 is the maximum.

For the odd-sum component, we had a perfect matching with 6 edges.

For even-sum, perhaps we can have 6.

Let's try to list a matching.

Suppose we pair:

1. (0,0)-(1,1) // sq (0,0) main

2. (0,2)-(1,3) // sq (0,2) main

3. (0,4)-(1,3) but (1,3) used.

Instead, for (0,4), pair to (1,3) but conflict.

For sq (0,3) i+j=3 odd, anti-diag: (1,3)-(0,4), so we can pair (0,4)-(1,3) , but then (0,2) cannot use (1,3).

So for the top, we can have only two: say (0,0)-(1,1) and (0,2)-(1,3), leave (0,4) unmatched, or (0,0)-(1,1) and (0,4)-(1,3), leave (0,2) unmatched, or (0,2)-(1,3) and (0,4)-(1,3) conflict.

So only two for the top row vertices.

Similarly for the bottom.

But for the bottom, (4,0)-(3,1) // sq (3,0) anti-diag

(4,2)-(3,3) // sq (3,2) anti-diag: from (4,2) to (3,3)

(4,4)-(3,3) but (3,3) used, so leave (4,4) or pair to (3,3) conflict.

So pair (4,0)-(3,1), (4,2)-(3,3), leave (4,4) unmatched.

Then for the middle, we have (2,0),(2,2),(2,4),(3,1) used, (3,3) used, and (1,1) used, (1,3) used.

So (2,0) can pair to (3,1) used, or to (1,1) used, so no.

(2,2) to (1,1) used, (1,3) used, (3,1) used, (3,3) used, so no.

Same for (2,4).

So no.

After pairing the top and bottom, we have used (1,1),(1,3),(3,1),(3,3), and (0,0),(0,2),(4,0),(4,2), and unmatched (0,4),(2,0),(2,2),(2,4),(4,4)

Then we can try to pair among them, but (0,4) has no neighbor, etc.

Perhaps pair some of the middle first.

For example, pair (1,1)-(2,2) // sq (1,1) main

(1,3)-(2,4) // sq (1,3) main

(3,1)-(4,2) // sq (3,1) anti-diag? i+j=4 even, so for even-sum, since i+j even, we can only use main diagonal: (3,1)-(4,2)? From (3,1) to (4,2), delta (1,1), for sq (3,1) main diag, yes.

Similarly, (3,3)-(4,4) // sq (3,3) main: (3,3)-(4,4)

Then now used (1,1),(2,2),(1,3),(2,4),(3,1),(4,2),(3,3),(4,4)

Remaining: (0,0),(0,2),(0,4),(2,0),(2,2) used, (4,0), and (2,0) etc.

Vertices: (0,0),(0,2),(0,4),(2,0),(4,0) and (2,2) used, but (2,0) is there.

(0,0) can pair to (1,1) but (1,1) used.

(0,2) to (1,1) used, (1,3) used.

(0,4) to (1,3) used.

(2,0) to (1,1) used, (3,1) used.

(4,0) to (3,1) used.

So all have no partners, so only 4 edges, but we have 8 vertices covered, 5 left, but no edges.

Not good.

With this, we have 4 edges, but we need 6.

Perhaps add more.

For example, before pairing the middle, pair some on the top.

But (0,0) can only pair to (1,1), which is used in (1,1)-(2,2).

So conflict.

So perhaps for the even-sum graph, the maximum matching size is 5 or 6.

Let's calculate the minimum vertex cover or use a formula.

Notice that the graph may have a matching of size 6.

After some search, I recall that in a grid, for such a graph, the maximum matching can be achieved by taking a "staircase" or something.

For a 5x5 grid, the even-sum vertices can be matched as follows:

• Pair (0,0)-(1,1)
• (0,2)-(1,3)
• (0,4) cannot, so pair (0,4) to (1,3) but conflict.

Pair (0,4)-(1,3) for sq (0,3) anti-diag

• (0,2)-(1,1) for sq (0,1) anti-diag
• then (0,0) no partner.

Then for (2,0)-(3,1) for sq (2,0) main

• (2,2)-(3,3) for sq (2,2) main
• (2,4)-(3,3) but (3,3) used, so pair (2,4)-(3,3) conflict, or to (1,3) used, so no.

Then for (4,0)-(3,1) but (3,1) used in (2,0)-(3,1), so conflict.

Pair (4,0)-(3,1) for sq (3,0) anti-diag

• then (2,0) no partner.

So not.

Perhaps pair (1,1)-(2,2) for sq (1,1) main

• (1,3)-(2,4) for sq (1,3) main
• (3,1)-(4,2) for sq (3,1) main (i+j=4 even, main: (3,1)-(4,2))
• (3,3)-(4,4) for sq (3,3) main: (3,3)-(4,4)
• then for (0,0) to (1,1) but (1,1) used
• (0,2) to (1,1) used, (1,3) used
• (0,4) to (1,3) used
• (2,0) to (1,1) used, (3,1) used
• (2,2) to (1,1) used, (1,3) used, (3,1) used, (3,3) used
• (4,0) to (3,1) used
• (4,2) to (3,1) used, (3,3) used
• so only 4 edges.

But we can add (0,0)-(1,1) but conflict.

So perhaps include (0,0)-(1,1) and not use (1,1)-(2,2), but then (2,2) may not be paired.

Sacrifice (2,2) or something.

Suppose we do not pair (1,1)-(2,2), but pair (0,0)-(1,1) , and (2,2)-(3,3) , then (1,1) is used, so (1,3) can be paired to (2,2) but (2,2) may be used in (2,2)-(3,3), so conflict.

Pair (0,0)-(1,1) , (2,2)-(3,3) , (0,2)-(1,3) , (0,4)-(1,3) conflict.

Pair (0,0)-(1,1) , (2,2)-(3,3) , (0,4)-(1,3) , then (0,2) no, and (1,3) used, so (1,1) used, so for (1,3) , it is used, but (2,4) can be paired to (1,3) used or to (3,3) used, so no.

Then for (3,1)-(4,2) , (3,3) used in (2,2)-(3,3), so (3,1) can be paired to (4,2) , but (4,2) is there.

Pair (3,1)-(4,2) for sq (3,1) main

• (3,3)-(4,4) for sq (3,3) main
• then (4,0) to (3,1) used, so no.

So still only 4 or 5.

Perhaps it is not possible, and the maximum is 5 for the even-sum graph.

For example, pair:

1. (0,0)-(1,1) // sq 0,0

2. (0,2)-(1,3) // sq 0,2

3. (0,4)-(1,3) but conflict.

Instead, for (0,4), pair to (1,3) but (1,3) may be used.

After 1 and 2, (1,1) and (1,3) used, so (0,4) no.

So for top, only two.

Then pair 3. (2,0)-(3,1) // sq 2,0

4. (2,2)-(3,3) // sq 2,2

5. (2,4)-(3,3) but (3,3) used, so cannot.

Pair 5. (2,4)-(3,3) for sq 2,3 anti-diag, but (3,3) would be used in 4, so conflict.

Pair 5. (2,4) to nothing or to (1,3) used.

So only 4.

Pair 3. (1,1)-(2,2) // sq 1,1

4. (1,3)-(2,4) // sq 1,3

5. (0,0)-(1,1) but (1,1) used.

Not.

Let's list a matching that works for 5 edges.

For example:

• (0,0)-(1,1) // sq 0,0

• (0,2)-(1,3) // sq 0,2

• (2,0)-(3,1) // sq 2,0

• (2,2)-(3,3) // sq 2,2

• (4,2)-(3,1) but (3,1) used in 3.

• (4,0)-(3,1) used.

• (4,4)-(3,3) used in 4.

• (0,4)-(1,3) used in 2.

• (2,4)-(3,3) used.

• (4,2) can pair to (3,1) used or (3,3) used.

• (4,0) to (3,1) used.

• (4,4) to (3,3) used.

• (0,4) to (1,3) used.

So no fifth edge.

Pair (3,1)-(4,2) for sq 3,1 main (i+j=4 even)

Then with 1,2,3,4 as above, but 3 is (2,0)-(3,1), so (3,1) used, so cannot pair (3,1)-(4,2).

Pair (3,1)-(4,2) instead of (2,0)-(3,1), but then (2,0) no partner.

So perhaps:

1. (0,0)-(1,1) // sq 0,0

2. (0,2)-(1,3) // sq 0,2

3. (3,1)-(4,2) // sq 3,1 main

4. (3,3)-(4,4) // sq 3,3 main

5. (2,2)-(3,3) but (3,3) used in 4.

6. (2,0) to (3,1) but (3,1) used in 3.

7. (1,1)-(2,2) but (1,1) used in 1.

8. (1,3)-(2,2) or (2,4)

Say 5. (1,3)-(2,2) // how? From (1,3) to (2,2): delta (1,-1), for sq 1,2? BL (1,2), i+j=3 odd, so anti-diag: from (2,2) to (1,3), yes.

Then (2,2) is used, but in 5, we have (1,3)-(2,2), so (2,2) used.

Then (2,0) can be paired to (3,1) but (3,1) used in 3.

(2,4) to (3,3) used in 4.

(0,4) to (1,3) used in 2 or 5.

(4,0) to (3,1) used.

So edges: 1,2,3,4,5, so 5 edges, covering 10 vertices, leaving (0,4),(2,0),(2,4),(4,0) unmatched, but that's 4 vertices, but we have 13-10=3 vertices unmatched? 5 edges cover 10 vertices, 13-10=3 vertices left: (0,4),(2,0),(2,4),(4,0) are 4, but (4,0) is in, but in this matching, which vertices are covered?

1. (0,0),(1,1)

2. (0,2),(1,3)

3. (3,1),(4,2)

4. (3,3),(4,4)

5. (1,3),(2,2) – (1,3) is in 2 and 5? Conflict! (1,3) is used in 2 and in 5.

Mistake.

In 2, we have (0,2)-(1,3), so (1,3) used, then in 5, (1,3)-(2,2), but (1,3) already used.

So cannot.

In 5, if we pair (1,3)-(2,2), but (1,3) is already used in 2.

So not.

After 1 and 2, (1,3) is used, so for 5, we can pair (2,2)-(1,1) but (1,1) used in 1.

Or (2,2)-(3,3) , etc.

So perhaps:

1. (0,0)-(1,1)

2. (0,2)-(1,3)

3. (3,3)-(4,4) // sq 3,3

4. (2,2)-(3,3) but (3,3) used in 3.

5. (2,0)-(3,1) // sq 2,0

6. (4,0)-(3,1) but (3,1) may be used in 4.

If we pair 4. (2,0)-(3,1) , then 5. (4,0)-(3,1) conflict.

5. (4,2)-(3,1) used.

So no.

Perhaps for the even-sum graph, the maximum matching size is 5, not 6.

For example, pair:

• (0,0)-(1,1) // sq 0,0

• (0,4)-(1,3) // sq 0,3 anti-diag (i+j=3 odd)

• (2,0)-(3,1) // sq 2,0 main (i+j=2 even)

• (2,4)-(3,3) // sq 2,3 anti-diag (i+j=5 odd)

• (4,0)-(3,1) but (3,1) used in 3.

• (4,4)-(3,3) but (3,3) used in 4.

• (0,2)-(1,1) used or (1,3) used.

• (2,2) to (1,1) used, (1,3) used, (3,1) used, (3,3) used.

• (4,2) to (3,1) used, (3,3) used.

So only 4 edges.

With (0,0)-(1,1), (0,4)-(1,3), (2,0)-(3,1), (2,4)-(3,3), then we can add (4,2)-(3,1) but (3,1) used, or (4,0)-(3,1) used, or (4,4)-(3,3) used, or (2,2)-(3,1) used, etc.

So no.

Add (1,1)-(2,2) but (1,1) used.

So perhaps 4 is the best for now, but we can do better.

Pair (0,2)-(1,1) // sq 0,1 anti-diag (i+j=1 odd)

Then (0,0) can pair to (1,1) but conflict.

Not.

Let's use (4,2)-(3,1) for sq 3,1 main? i+j=4 even, main: (3,1)-(4,2), so yes, but (3,1) and (4,2).

Then with previous.

Perhaps it is possible with 5.

For example:

1. (0,0)-(1,1) // sq 0,0

2. (0,2)-(1,3) // sq 0,2

3. (0,4)-(1,3) conflict.

4. (2,2)-(3,3) // sq 2,2

5. (2,0)-(3,1) // sq 2,0

6. (4,0)-(3,1) conflict.

7. (4,4)-(3,3) conflict with 3.

8. (4,2)-(3,1) but (3,1) may be used in 4.

If we don't pair 4, then pair 4. (3,1)-(4,2) // sq 3,1

5. (3,3)-(4,4) // sq 3,3

Then vertices covered: 1: (0,0),(1,1)
2: (0,2),(1,3)
4: (3,1),(4,2)
5: (3,3),(4,4)
3 not paired, but (2,2) not paired, (2,0) not, etc.

So covered: (0,0),(1,1),(0,2),(1,3),(3,1),(4,2),(3,3),(4,4) — 8 vertices.

Remaining: (0,4),(2,0),(2,2),(2,4),(4,0)

Now, can we pair some of them? For example, (0,4) to (1,3) but (1,3) used in 2.

(2,0) to (1,1) used or (3,1) used.

(2,2) to (1,1) used, (1,3) used, (3,1) used, (3,3) used.

(2,4) to (1,3) used, (3,3) used.

(4,0) to (3,1) used.

So no, and (0,4) is there, so no edge for it.

But we can add a pair like (2,2)-(3,3) but (3,3) used in 5.

Or (0,4)-(1,3) used.

So still only 4 edges in this matching, but we have 5 edges listed: 1,2,4,5, and no 3, so 4 edges.

1,2,4,5 are 4 edges.

To get a fifth, we can pair (2,2)-(3,1) but (3,1) used in 4.

Or (1,1)-(2,2) but (1,1) used in 1.

So not.

After 1 and 2, we have (1,1) and (1,3) used, so for (2,2), it can pair to (1,1) used, (1,3) used, (3,1), (3,3).

So pair (2,2)-(3,3) // sq 2,2 main

Then (2,0) to (3,1) // sq 2,0 main

Then (2,4) to (3,3) but (3,3) used in (2,2)-(3,3), so conflict.

Pair (2,4)-(3,3) for sq 2,3 anti-diag, then (2,2) can be paired to (3,1) or (3,3), say to (3,1) // but (3,1) may be used in (2,0)-(3,1)

If we pair (2,0)-(3,1), then (2,2) to (3,1) conflict.

So pair (2,0)-(3,1), (2,4)-(3,3), then (2,2) no partner.

Then for the bottom, (4,0)-(3,1) used, (4,2)-(3,1) used or (3,3) used, (4,4)-(3,3) used.

So edges: 1,2, and 3. (2,0)-(3,1), 4. (2,4)-(3,3), and 5. (2,2) not paired, but we can pair (4,2)-(3,1) but used, or etc.

So only 4 edges.

To include (2,2), pair (2,2)-(3,3) , then (2,0) to (3,1), (2,4) to (3,3) used, so no.

So perhaps the maximum for even-sum is 5, and for odd-sum 6, but since the components are separate, total 11, but 5+6=11, and 25-11*2=3 vertices unmatched, which is odd, so possible.

But in the odd-sum, we have a perfect matching with 6, so for even-sum, if we can get 5, total 11.

Is 11 possible? But earlier we thought 12 is upper bound, but 11<12, so possible.

For even-sum, let's find a matching of size 5.

For example:

• (0,0)-(1,1) // sq 0,0

• (0,2)-(1,3) // sq 0,2

• (0,4)-(1,3) conflict.

• (0,4) not paired.

• (2,0)-(3,1) // sq 2,0

• (2,2)-(3,3) // sq 2,2

• (2,4)-(3,3) conflict.

• (4,0)-(3,1) conflict with (2,0)-(3,1)

• (4,2)-(3,1) used or (3,3) used

• (4,4)-(3,3) used

So only 4.

Include (4,4)-(3,3) but (3,3) may be used.

Suppose we pair (3,3)-(4,4) // sq 3,3

Then (3,1)-(4,2) // sq 3,1

Then (2,2)-(3,3) but (3,3) used.

(2,2)-(3,1) but (3,1) may be used in (3,1)-(4,2)

If we pair (3,1)-(4,2), then (2,2) can be paired to (3,1) used or to (3,3) used or to (1,1) or (1,3).

Say pair (2,2)-(1,1) // sq 1,1 main? From (1,1) to (2,2), for sq 1,1, i+j=2 even, main diag.

Then (1,1) may be used or not.

Assume we pair:

1. (3,3)-(4,4) // sq 3,3

2. (3,1)-(4,2) // sq 3,1

3. (2,2)-(1,1) // sq 1,1

4. (2,0)-(3,1) but (3,1) used in 2.

5. (2,0)-(1,1) but (1,1) used in 3.

6. (0,0)-(1,1) used.

So for top, pair 4. (0,0)-(1,1) conflict.

Pair 4. (0,2)-(1,3) // sq 0,2

5. (0,4)-(1,3) conflict.

So perhaps:

1. (0,0)-(1,1) // sq 0,0

2. (0,2)-(1,3) // sq 0,2

3. (3,3)-(4,4) // sq 3,3

4. (3,1)-(4,2) // s q 3,1

5. (2,2)-(3,3) but (3,3) used in 3.

6. (1,1)-(2,2) but (1,1) used in 1.

7. (2,0)-(3,1) but (3,1) used in 4.

8. (2,4)-(3,3) used.

9. (4,0)-(3,1) used.

So no fifth edge.

With this, we have 4 edges, covering 8 vertices, 5 left.

But we can add (1,3)-(2,2) for sq 1,2? i+j=3 odd, so for even-sum, anti-diag: from (2,2) to (1,3), yes.

So pair 5. (1,3)-(2,2) // sq 1,2 anti-diag

Then vertices: 1: (0,0),(1,1)
2: (0,2),(1,3)
5: (1,3),(2,2) – (1,3) is in 2 and 5? Conflict.

In 2, we have (0,2)-(1,3), so (1,3) used, then in 5, (1,3)-(2,2), but (1,3) already used.

So not.

If we haven't used (1,3) in 2, but in this setup, it is.

So perhaps for the top, only use one pair.

For example, pair only (0,0)-(1,1) for top.

Then for (0,2), pair to (1,3) // sq 0,1 anti-diag: from (1,3) to (0,2)? For sq 0,1, BL (0,1), anti-diag from (1,1) to (0,2), not to (0,2) from (1,3).

From (0,2) to (1,3) is for sq 0,2 main, as before.

So pair (0,2)-(1,3) for sq 0,2.

Then (0,4)-(1,3) conflict.

So still.

I am considering that for the 4x4 grid, the maximum number of diagonals with no common points is 8, not 12.

Perhaps 8 is the answer.

For example, in each 2x2 block, we can place one diagonal, but not more.

Or in a checkerboard pattern.

Suppose we place diagonals in all squares of one orientation, but they may share vertices.

For example, if we place the main diagonal in all squares, then in a row, adjacent squares share a vertex, and if they have the same orientation, they share a vertex, but in this case, for adjacent squares horizontally, if both have the main diagonal, they share the vertex at the common side, but in the diagonal, for left square, it uses the top-right vertex, for right square, it uses the bottom-left vertex, which are different, so no common endpoint, and as we saw, no intersection, so it might be ok.

For example, in a row, for squares (i,j) for j=0 to 3, place the main diagonal: from (i,j) to (i+1,j+1).

Then for fixed i, the diagonals are: for each j, (i,j)-(i+1,j+1).

Then for adjacent squares, say j and j+1, the diagonal in j: (i,j)-(i+1,j+1), in j+1: (i,j+1)-(i+1,j+2).

Do they share a common vertex? (i,j+1) is in the second, (i+1,j+1) is in the first, so different vertices. And no common endpoint. Do they intersect? As before, no.

Similarly for rows.

But for the same i, the diagonals are parallel and separated, so no intersection.

Now for the grid, if we do this for all i,j, then in each 2x2 block, the diagonals may conflict at vertices.

For example, in square (0,0): (0,0)-(1,1)
Square (0,1): (0,1)-(1,2)
Square (1,0): (1,0)-(2,1)
Square (1,1): (1,1)-(2,2)

Then vertices like (1,1) is used in (0,0) diagonal and in (1,1) diagonal, so shared endpoint, which is not allowed.

So cannot do that.

We can do every other square or something.

Perhaps a matching in the grid.

Since each diagonal uses a vertex, and vertices are shared, but in the vertex-disjoint condition, each vertex can be used by at most one diagonal, so it's like covering vertices with edges, but with the constraint that each edge is from a different square, but in this case, since each edge is unique to a square, it's just a matching in the grid graph with the diagonal edges.

And as we have, with 25 vertices, maximum matching size 12, but may not be achievable.

For the 4x4 grid of squares, the number of vertices is 5x5=25.

The graph has edges as described.

The maximum matching size can be calculated.

In such a grid, the maximum number of vertex-disjoint diagonals is 8 for a 4x4 grid.

For example, place diagonals in a way that no two share a vertex.

For instance, in each 2x2 block, place one diagonal, but 4x4 has 4 such blocks, so 4, but we can do better.

Place in all squares but with a pattern that avoids shared vertices.

Suppose we place the main diagonal in square (0,0): (0,0)-(1,1)
Then in (0,1): cannot use because (0,1) and (1,1) might be used, but (0,1) is not used yet, (1,1) is used, so if we place in (0,1) the anti-diagonal: (1,1)-(0,2), but (1,1) is already used, so not allowed.

Place in (0,1) the main diagonal: (0,1)-(1,2), then (0,1) is new, (1,2) new, and (1,1) is used, but not in this diagonal, so ok, as long as no common vertices.

So (0,0): (0,0)-(1,1)
(0,1): (0,1)-(1,2)
Then vertices (0,0),(1,1) for first, (0,1),(1,2) for second, all distinct so far.

Then for (0,2): if place main: (0,2)-(1,3), vertices (0,2),(1,3) new.
(0,3): (0,3)-(1,4), but j=4 may be out if grid is to 4, but in 5x5, y=0 to 4, so (1,4) is ok.

Then for row 1, square (1,0): can place, say anti-diagonal: (2,0)-(1,1), but (1,1) is used in first diagonal, so not allowed.
Place main: (1,0)-(2,1), vertices (1,0),(2,1) new.
Similarly, (1,1): if place anti-diagonal: (2,1)-(1,2), but (1,2) is used in (0,1) diagonal, and (2,1) may be used in (1,0) if we place it.
If we place (1,0): (1,0)-(2,1), then (2,1) used.
Then for (1,1): if place anti-diagonal: (2,1)-(1,2), but (2,1) used, (1,2) used, so not.
Place main: (1,1)-(2,2), then (1,1) used in first, so not.

So for (1,1), cannot place any diagonal because both possible vertices (1,1) and (1,3) etc are used or may be used.

(1,1) is used, and for main diagonal, it uses (1,1) and (2,2), so (1,1) is used, so cannot.

So in this way, for the first two columns, we have 4 diagonals: (0,0), (0,1), (0,2), (0,3) all with main diagonal, using vertices (0,0),(1,1), (0,1),(1,2), (0,2),(1,3), (0,3),(1,4)

Then for row 1, we can try to place in (1,0): if we place anti-diagonal: (2,0)-(1,1), but (1,1) used, so not.
Place main: (1,0)-(2,1), then (1,0),(2,1) new.
Then for (1,1): cannot, as above.
For (1,2): place anti-diagonal: (2,2)-(1,3), but (1,3) may be used in (0,2) diagonal? In (0,2): (0,2)-(1,3), so (1,3) used.
So (2,2)-(1,3) uses (1,3) used, not allowed.
Place main: (1,2)-(2,3), then (1,2),(2,3) new, and (1,2) not used yet? In the list, (1,2) is not used by any diagonal so far. So can place.
Similarly for (1,3): place anti-diagonal: (2,3)-(1,4), but (1,4) may be used in (0,3): (0,3)-(1,4), so (1,4) used, so not.
Place main: (1,3)-(2,4), then (1,3) used in (0,2) diagonal, so not allowed.

So for row 1, we can place (1,0): (1,0)-(2,1) main, and (1,2): (1,2)-(2,3) main, but not (1,1) or (1,3).

Then for row 2, and so on.

So far, diagonals in row 0: all 4 squares.
Row 1: squares (1,0) and (1,2), so 2 squares.
Total 6 diagonals.

Vertices used: from row 0: (0,0),(1,1), (0,1),(1,2), (0,2),(1,3), (0,3),(1,4)
From row 1: (1,0),(2,1), (1,2),(2,3) – but (1,2) is used in (0,1) diagonal? In (0,1): (0,1)-(1,2), so (1,2) is used, but in (1,2) diagonal, it uses (1,2) and (2,3), so (1,2) is already used, so conflict.

I forgot that (1,2) is already used in the diagonal of (0,1).

So when I try to place in (1,2) the main diagonal, it uses (1,2), which is already used, so not allowed.

So cannot place in (1,2).

For (1,0): (1,0)-(2,1), ok, (1,0) and (2,1) new.
For (1,1): cannot.
For (1,2): if place, uses (1,2) which is used, or if anti-diagonal: (2,2)-(1,3), but (1,3) is used in (0,2) diagonal, so not.
For (1,3): similarly, (1,3) used, so cannot.
So only one in row 1: (1,0).

Then for row 2, etc.

So with 4 in row 0, 1 in row 1, then perhaps more in lower rows.

But it might be less than 8.

Perhaps use a different pattern.

Suppose we place the anti-diagonal in all squares where i+j even, and main where i+j odd, or something.

But each square can have only one, and we need to choose which one to use.

Perhaps the maximum is 8 for 4x4.

For a n x n grid of squares, the maximum number of vertex-disjoint diagonals is n* (n+1)/2 or something, but for n=4, 45/2=10, but 102=20 < 25, so possible.

But in our case, with the no intersection, but since no intersection occurs, only vertex-disjointness matters, so it should be the size of a maximum matching in this graph.

For a 5x5 grid of vertices, with edges between vertices at chess king move distance sqrt(2), i.e., the diagonal adjacency, then the graph is the same as the king's move graph but only diagonal moves.

In this graph, the maximum matching size can be found.

For a grid, it is known that for an m x n grid, the maximum number of vertex-disjoint diagonals or something.

Perhaps for this problem, the answer is 8.

Let's assume that and box it.

Perhaps 9 or 10.

Another idea: in each 2x2 block of squares, we can place two diagonals that don't share vertices.

For example, in squares (0,0), (0,1), (1,0), (1,1).

Place in (0,0): main diagonal (0,0)-(1,1)
In (1,1): anti-diagonal (2,1)-(1,0)? For sq (1,1), anti-diagonal from (2,1) to (1,0)? But (1,0) is a vertex.

From (1,0) to (2,1) for sq (1,0) main, but for (1,1) anti: from (2,1) to (1,0), yes.

Then vertices: (0,0),(1,1) for first, (2,1),(1,0) for second, all distinct.

Then in this 2x2, we have two diagonals, no shared vertices.

Similarly, we can do for other 2x2 blocks.

For 4x4 grid, we can have four such 2x2 blocks, each with two diagonals, so 8 diagonals.

And no two share a vertex, since the blocks are adjacent but the vertices are different.

For example, between block (0,0) to (1,1) and block (2,0) to (3,1), they share the vertices (2,0),(2,1),(3,0),(3,1), but in the first block, we use (0,0),(1,1),(1,0),(2,1) for the two diagonals? In above, for first block: sq (0,0): (0,0)-(1,1), sq (1,1): (2,1)-(1,0) -- so vertices (0,0),(1,1),(1,0),(2,1)

For block (2,0) to (3,1): say sq (2,0): main (2,0)-(3,1), sq (3,1): anti (4,1)-(3,0) or something.

Vertices for second block: (2,0),(3,1) for first diagonal, and for the other, say in sq (3,1): anti-diagonal from (4,1) to (3,0), so (4,1),(3,0)

So vertices: (2,0),(3,1),(4,1),(3,0)

Now, compare to first block: (0,0),(1,1),(1,0),(2,1)

Common vertices? (2,1) in first, (2,0) in second, different. (1,0) in first, (3,0) in second, different. So no common vertices.

Similarly for other blocks.

So for the 4x4 grid, we can divide into four 2x2 blocks: blocks 1: i=0-1,j=0-1; block 2: i=2-3,j=0-1; block 3: i=0-1,j=2-3; block 4: i=2-3,j=2-3.

In each block, place two diagonals as above: for each 2x2 block, in the bottom-left square place the main diagonal, in the top-right square place the anti-diagonal.

For block 1: sq (0,0) main: (0,0)-(1,1)
sq (1,1) anti: (2,1)-(1,0) // as above

For block 2: sq (2,0) main: (2,0)-(3,1)
sq (3,1) anti: (4,1)-(3,0) // since for sq (3,1), BL (3,1), anti-diagonal from (4,1) to (3,0)

For block 3: sq (0,2) main: (0,2)-(1,3)
sq (1,3) anti: (2,3)-(1,2) // for sq (1,3), BL (1,3), anti-diagonal from (2,3) to (1,2)

For block 4: sq (2,2) main: (2,2)-(3,3)
sq (3,3) anti: (4,3)-(3,2) // for sq (3,3), BL (3,3), anti-diagonal from (4,3) to (3,2)

Now, list all diagonals and vertices:

1. (0,0)-(1,1) // sq (0,0)

2. (2,1)-(1,0) // sq (1,1) anti, but (1,0) and (2,1)

3. (2,0)-(3,1) // sq (2,0)

4. (4,1)-(3,0) // sq (3,1) anti

5. (0,2)-(1,3) // sq (0,2)

6. (2,3)-(1,2) // sq (1,3) anti

7. (2,2)-(3,3) // sq (2,2)

8. (4,3)-(3,2) // sq (3,3) anti

Now, list all vertices used: from 1: (0,0),(1,1)

2: (2,1),(1,0)

3: (2,0),(3,1)

4: (4,1),(3,0)

5: (0,2),(1,3)

6: (2,3),(1,2)

7: (2,2),(3,3)

8: (4,3),(3,2)

Now, are there any common vertices? Look at (1,1) in 1, is it in others? 2 has (2,1),(1,0), not (1,1). 3 has (2,0),(3,1), not. 4 has (4,1),(3,0), not. 5 has (0,2),(1,3), not. 6 has (2,3),(1,2), not. 7 has (2,2),(3,3), not. 8 has (4,3),(3,2), not. So (1,1) only in 1.

(1,0) in 2, not in others.

(2,1) in 2, not in others? 3 has (2,0),(3,1), not (2,1). 7 has (2,2), not. so ok.

(2,0) in 3, not in others.

(3,1) in 3, not in others.

(4,1) in 4, not.

(3,0) in 4, not.

(0,2) in 5, not.

(1,3) in 5, not.

(2,3) in 6, not.

(1,2) in 6, not.

(2,2) in 7, not.

(3,3) in 7, not.

(4,3) in 8, not.

(3,2) in 8, not.

So all 16 vertices used are distinct, no common vertices.

And we have 8 diagonals.

Now, is there any intersection? As we discussed earlier, no, because no two diagonals share a vertex, and in this grid, no two cross at an interior point.

So this works.

Can we have more? 9 diagonals would require 18 vertices, but there are only 25, 25-18=7 left, but since each diagonal requires 2 vertices, and 9*2=18<25, so possible in terms of vertices, but likely not because of the grid structure.

For example, the vertex (1,1) is used, but in this configuration, it's used, and other vertices are used.

Suppose we try to add another diagonal, say in square (0,1). But square (0,1) is not used. Its vertices are (0,1),(1,1),(0,2),(1,2). But (1,1) is used in diagonal 1, (0,2) is used in diagonal 5, (1,2) is used in diagonal 6, (0,1) may be free? Is (0,1) used? In the list, not used yet. But if we place a diagonal in (0,1), say main: (0,1)-(1,2), but (1,2) is used in diagonal 6 (since diagonal 6 is (2,3)-(1,2), so (1,2) used). Anti-diagonal: (1,1)-(0,2), but (1,1) used in 1, (0,2) used in 5. So both diagonals use vertices that are already used, so cannot add any more in this block.

Similarly for other squares. So probably 8 is the maximum.

For a 4x4 grid, with 8 diagonals, no two share a vertex or cross.

So the answer should be 8.

In the problem, it's a 4x4 table, so 16 squares, and we have 8 diagonals.

And in the initial example, with inner size 2x2 and outer 4, they used 3 diagonals, but for larger, we can have more.

So I think 8 is achievable and likely maximum.

So the answer is 8.