If only opposite corners are spanned, as in f03c, the shape propagates down the x axis forever. Thus we may assume the top two corners are spanned. If these are the only corners spanned, or if all 4 are spanned, we can assume the top row is contiguous, without gaps. Things get a little more complicated when three out of four corners are spanned. Leave the lower right corner open, and suppose some squares are missing from the top row. The first piece must rest against the origin in precisely this orientation. The bottom row proceeds uninterrupted, until it ends at the right, leaving the barrel of the gun to extend further to the right along the top row. A long barrel (length > 1) forces a second copy, a 180 degree rotation of the first, to fill in the slot at the right. This second piece creates isolated holes between itself and the floor. If the snub-nosed barrel has length one, and the gun is missing the second square of its top row, the second gun can fill in the gap with its shoulder rest, standing with its barrel pointing straight up. The gun that tiles the square at 0,2 must also be vertical, else it creates holes between itself and the first piece. If its barrel points up, then a reflected copy, shifted down and to the right, must fill the gap at 1,1. Yet both copies occupy the space 1,n. Thus the barrel points down and fills the gap at 1,1. This leaves a square at either 2,1 or 2,2 that cannot be filled without creating holes somewhere. Hence we may assume the top row is contiguous, as in fca0.
If neither lower corner is spanned, we say the gun has a leading "tab". Of course we orient the gun so that its barrel is at least as long as its tab. A blunt gun has no tab; the lower left corner is filled. Some squares are always missing from the bottom row of a blunt gun, else it has order two. A degenerate gun is blunt at both ends, having no barrel at all. The thickness of a blunt en is the number of contiguous squares in its bottom row, before the first gap. A thin blunt end has thickness 1, as in fcb0.
If not, and teach gun has a tab or barrel pointed down, the guns create a hole just above the floor that cannot be filled. So assume the left gun rests on its blunt end. To avoid creating holes, the right gun must place its barrel on the floor, immediately next to the left gun, and the guns must face each other, their teeth and gaps interlocking perfectly, as in fcd0. Such a gun has order 2, and is not considered here.
If the gun has a tab, we cannot fill 3 consecutive squares on the floor with vertical guns. The first two stand back-to-back, and the third cannot fit in. If the gun has a blunt end, we might fill 3 holes, or even 4, with back-to-back copies, but never 5.
Given a horizontal gap of width k, where k < n, and k >= 3 (or 5 if the gun is blunt), we cannot position a gun horizontally on its back along the k squares, nor can we fill the squares with vertical guns. We must back up and try another configuration.
Actually we must be a bit careful here, as shown by f870[k=3]. We must make sure the face of the gun doesn't contain a contiguous span of k squares. This is not a factor if the brackets extend up through 2 or more rows. Despite the plethora of given conditions, this is a useful dead-end criterion.
Having placed the first two pieces (call them a and b) against the x and y axies respectively, suppose 2,1 is covered. We must now tile 2,2. Yet the third copy (c) lines up with either a or b and creates holes. Thus a fails to tile 2,1.
Since c and b must not run parallel, c lies face-to-face with a. The right end of c extends 2 squares beyond a, leaving a thin gap just above the x axis that cannot be filled. Thus our gun cannot have two thick ends.
The gun has at least one thin end, thanks to {5}. Assume it has one thick end, and let a point its thick end at the origin. If a covers 2,1 then we argue as in {5}. So a does not cover 2,1, c interlocks with a, and c extends two squares to the right of a. This extension cannot be the thick end of c, else we could argue as in {5}. So both a and c have their thick ends pointed left. In other words, their ends are aligned. One is a vertical reflection of the other, shifted by 2 squares. Whenever this happens, an inductive argument forces a periodic pattern of alternating teeth and gaps of a uniform width, in this case the teeth have size 2. This persists until the thin end of c hangs over the thin end of a. Piece d must hook its thin end under the overhang created by c, so that d and a are horizontal reflections. Yet the first gap in d has length 2, and c only consumes one of these 2 squares. The other is consumed by e, which is a horizontal reflection of c. The thick end of e at the right forms an inaccessible slot with the x axis. Thus a degenerate gun must point its thin end at the origin.
Assume our degenerate gun has a thick end. By {5} and {6}, piece a has one thin end, pointed at the origin. Piece b tiles 0,2, and piece c tiles 1,1. Clearly the thin end of c points down. If the same is true of b, then b and c are aligned, and the shape has gaps of size 1, as in feaa, whence c has no thick blunt end. So the thick end of b points down, and b and c are interlocked in reverse. We say they are reverse aligned. Slicing off the thin end, the gaps at either end of our truncated gun correspond perfectly with the teeth at the other end. The simplest polyomino that can do this is f8c8, with n = 5. Note that n must be odd.
Place d at 2,n+1, where b and c meet. If d is vertical it traps another piece against the y axis, creating holes. So d runs horizontally, and we need to populate column 3 between d and a, running up the length of c. When n = 5 we can place two horizontal pieces to the right of c, filling the gap of height 4, but we always wind up with holes below d. This, combined with {3} for larger values of n, rules out a stack of horizontal guns between a and d. Assume piece e stands back-to-back against c, thus the height of the gap is exactly n. To ensure this, either the first gap in the face of a is wider than 2, or d is oriented face down with its thin end to the left, but not both.
Suppose the first gap in a, where c rests, has length k, for k > 4. Let piece q fill k,1, lying horizontally. Because b and c interlock in reverse, the thick end of q is exactly k squares wide. It cannot fit into the first gap of a along with c and e. Thus the thin end of q restss in this gap, which has width 5. That leaves the thick end of e, having width 5, dangling off to the right, one row above the x axis. Clearly this cannot be filled, hence k <= 4.
Suppose k = 4, and piece f fills the first gap of e by reverse alignment, similar to b and c. Now the thin end of f pokes into the underside of d. This gives d a thick end of size 2, yet it should have a thick end of size 4. So f and g protrude horizontally from the first gap in e, lying back-to-back. If the thin end of e sits in the gap of a, next to the thin end of c, then f is face-to-face with a, but not interlocked. This creates holes under f, so f and g are at the top of e, with one row of empty squares separating them from d. This only postpones the inevitable, for the square at 5,2 must still be tiled, and piece h cannot run vertically without colliding with f and g. Thus h rests on top of a and creates holes with a. Actually h might not trap any holes in a if n = 9. But it must still lie face down, so i can lie on top of it, back-to-back. This puts j in the gap of h, and traps a thin segment of length 9 against the x axis. Finally we conclude k = 2, which pushes e up one row into the first gap of d. Note that this gap corresponds to the first gap in a; both have width 2.
If f is interlocked with e, its thin end descends into a, so that one square separates the first 2 gaps of a. Conversely, if a single square terminates the first gap in a, an interlocked piece (f) must fill the second gap at 4,1. The corresponding second gap of d begins just northeast of f. If the gap in a continues beyond f, column 6 is vacant for n+1 squares. This cannot be tiled, so the second gap has length 2 as well. Section {4} forces another vertical piece g, standing back-to-back with f. This is the same situation we encountered 3 columns earlier. Piece h is interlocked with g iff the second gap in a is terminated by a single square. Continue through all the vertical interlocked pairs. This process must stop before the midpoint of a, because gaps of length 2 on the left correspond to teeth of length 2 on the right. For simplicity, assume the process stops at e. Both squares below e are filled by a, and d presents at least 2 squares to the right of the top of e. If a does not present another gap immediately to the right of e, we must fill 5,2 with a horizontal piece h. This is certain to trap holes in a, so let another gap begin at 5,1. Suppose h fills this gap running horizontally, extending beyond the end of a. If h and a are reversed, n,0 begins a 2×1 gap that cannot be filled. So assume they are aligned. The interlock of b and c tells us that the thick end of h, dangling off the end of a, meshes perfectly with the gaps and teeth at the left of a. The single gap in a, filled by the thin end of h, has its corresponding tooth at n,1. This makes n,0 inaccessible. There is a way around this; assume a drops off altogether, giving the shape f8c8. Let's pursue this particular shape.
Let piece f tile 5,0. If f is horizontal then let piece g tile 5,2, which must also be horizontal. This creates holes, so f is vertical, the thin end of e is embedded in d, and g snakes around the top of f. Let h protrude horizontally from the gap in f. If this gap is near the bottom, the back of h creates a thin gap with either g or the x axis. So h floats two rows above the x axis, and i tiles 7,0. Now h and i conspire to create holes, thus f8c8 doesn't tile anything. This was our last chance at a degenerate gun with a thick end.
By {7}, we may assume the degenerate gun has 2 thin ends. Since piece c must tile 1,1, b and c interlock. The first gap in a has length at least 2 to accommodate c. If c has a gap of length >= 2 at the bottom, it leaves a hole at 1,3. Thus the "last" gap of c has length 1, and c points down. If b and c are aligned, all gaps have size 1, contradicting the first gap in a. Thus the first gap of b points down, and c covers 1,3 and 1,4. In other words, the pieces are reverse aligned. The shape feb2 is the smallest example.
As in {7}, piece d, starting at 2,n+1, must be horizontal. If d is face down with its smaller gap pointed left, then e fills the gap at 3,n+1, and is interlocked with d. This restricts the column to the right of c to n-1, and invokes {4}. If the larger gap is pointed left, e stands vertically and nestles in this gap, just as c rests in the first gap of a. If d lies face up, e still stands back-to-back against c, but this time it extends down into the first gap of a. In either case, e has a gap of size one, next to one of its thin blunt ends, that must be filled by f, interlocked. Thus d is not face up, else the end of f crashes into its back. Instead, d lies face down, e nestles in the first gap of d, and f nestles in the second gap of a. From here the proof is similar to {7}. Interlocked pairs must march all the way down the length of a, yet this canot continue past the midpoint of a, since the teeth on the right side of a have width 2.
That's the last degenerate gun. Hereinafter, we may assume each gun has a barrel.
If a third piece is required to tile 1,1, it must do so using its blunt end. Hence it interlocks with either a or b. Reflecting the picture as necessary, we will assume c faces b.
Suppose a places its barrel at the origin. Now b and c are perfectly antiparallel. They either enclose holes, or the piece has order 2.
Suppose a places its blunt end (necessarily thin) at the origin. Now b must point up, else we have a hole at 1,2. This means b and c are aligned, giving a uniform gap length of 1. Yet the first gap in a has length 2.
Suppose b places its blunt end at the origin. The two pieces are aligned, and the shape has single gaps, as in ffa8. Since the piece doesn't have order 2, the barrel has length > 1. Thus a vertical piece fills in the slot between b and c, and creates a hole at 0,n.
Finally let the barrel of b touch the origin. This barrel has length 2 iff the blunt end is thin. In this case d, a reflected copy of b, tiles 0,n, and another piece e tiles 1,n+1. If e runs vertically, then d and e are aligned, giving a shape like fea8. As above, the slot between d and e is filled with a piece that creates inaccessible regions with the y axis. So e points into d, stretching just as far right as a. Now the column to the right of c creates an empty span that contradicts {4}. Thus the barrel has length > 2 and the blunt end is thick.
Suppose the barrel has length k, where k >= 5, and attempt to place a piece at 3,1. The blunt end has length k-1, and will not fit horizontally in the gap of length k-2 between the flat edge of c and the first tooth of a. Therefore {4} is in force.
Next assume k = 4 and slide d in position back-to-back with c. The blunt end has length 3, and e, a vertical reflection of b, fills the square at 0,n and runs up the y axis. If piece f at 2,n+1 is vertical it creates holes with e, so it runs horizontally across the barrels of c and d. Now the barrel of d defines a gap of length 4, which must sprout two more horizontal copies next to e. Now e cannot face down and interlock, because the remaining barrel of e will not accommodate the thick end of f. Nor can e lie face up without creating holes in f.
Finally we set k = 3, which places the barrel of d at 3,1. Place piece e next to d, at 4,2. Now e cannot place its back against a, nor can it align with a (due to its blunt end dangling off to the right), so e and d interlock, just as b and c interlock. Because the barrel of e extends up to row n+1, piece f at 0,n cannot be horizontal. Instead it runs up the y axis, a vertical reflection of b. Now f, c, d, and e define a 1×3 gap starting at 2,n+1. If g is vertical it creates holes with f. If g is horizontal it creates a hole at 4,n+1. This completes all the cases. One of the first two pieces tiles 1,1.
If the gun has a tab of length > 1, neither of the first two pieces can tile 1,1, which contradicts {9}. Thus the gun is blunt, or has a single square as leading tab.
Consider a gun with a thin blunt end and a long barrel (at least two in length). By {9}, the blunt end of b must tile 1,1. Let c tile 1,2. If c is vertical, it is aligned with b, giving a uniform gap size of 1. The barrel has length 2, else we have a hole at 2,1. The slot at 1,n is filled by the downward pointing barrel of piece d, turning 0,n+1 into a hole.
If c extends horizontally from 1,2, it is bound to create holes with a, unless a and c are aligned. This produces the same shape as above, gaps and teeth of size 1 and a barrel length of 2. Place piece d at 2,3. It cannot interlock with b, since its barrel would crash into c. Any other orientation leaves a hole at 1,4. In general this hole becomes inaccessible, unless b is f8a0. We will assume this shape and proceed.
After ruling out the vertical orientations, assume d points its barrel at b. We cannot place a piece at 1,4 without creating holes or slots by the y axis, So d points right, and e is aligned with d as it fills the hole at 1,4. Meantime f lies on its back at 5,0, its barrel sliding under c. Let g fill the gap at 6,1. Since d hangs over c, g points right and interlocks with f. This forces h to point right at 10,0. Let i fill the slot at 5,4 between d and e. If i is horizontal it creates an inaccessible region between itself and g. Thus i is vertical, and g spans a continuous gap of length 4 along row 3. Let j fill the gap in h at 11,1. If j stands up and closes the gap across g, {4} is invoked. Thus j points right and interlocks with h. As k fills the slot between h and j, it creates a hole with the x axis. This completes the proof. Henceforth all blunt guns are assumed to have thick ends or short barrels.
Let a place its thick blunt end at the origin, so that it covers 1,1, and let the barrel of b touch a at 0,2. Now c, based at 1,2, makes a perfect rectangle with b, or creates holes with a, or interlocks with a in reverse. Only the latter is reasonable, with a barrel of length k and a thick end of length k+1. The barrel of d slides its last square under the thick end of c at n,0, leaving a gap of length k-1. Assume k > 1. This gap cannot be filled horizontally by a blunt end of length k+1, so e stands up in the barrel of d. Using a similar argument, piece f, at 1,3, cannot interlock with b; it must lie back-to-back across c, bracketed by b and e. If f points left, k must equal 2, and we have a gap to fill at 2,4. This means g and b are aligned, all gaps and teeth have length 2, and the thick end of b has length 2, rather than 3. Therefore the blunt end of f embeds in b, and k >= 3. Now the barrel of f presents a gap of width k, or possibly k-1 if k = 3 and e faces left. Nothing interlocks with e, so e faces right and the gap is indeed k. This trips {3}. All this forces k = 1, giving the thick end a length of 2.
Recall that d still has its snub barrel under the thick end of c. Let piece e tile n+1,2, where c and d meet. If e and d are aligned, d cannot have a thick blunt end. To avoid holes between e and d, and to avoid {4} across the top of c, e must place its barrel into d at n+2,1 and face left, whence f can lie back-to-back atop c, its right end embedded in e. Note that this gap in e has a corresponding gap in b, one row higher. If f points right, 1,4 becomes inaccessible, hence f embeds its thick blunt end into e. If this end does not completely fill the gap in e, we must fill it with g. Now g cannot run horizontally across f or interlock with f, so g interlocks with e. This means the gap in e filled by f and g has length 4. The corresponding gap in b cannot be filled, so f sates the gap in e, which has length 2. This is not the last gap in e, since e contains a gap of size 1. Let g fill the corner where f and e meet. If g places its back to e it creates holes with the subsequent gaps in e. If g faces e they are aligned, but g is shifted up by 4. Now the top two teeth of g have size 2, with a gap of 1 in between, hence the second tooth crashes into the first tooth of e. So g and f interlock, and the blunt end of g fills the gap in b, just as the blunt end of f fills the gap in e. Now we have the same situation we had 3 rows below. Again, h must lie flat atop g with its blunt end completely filling a gap in e, which has a corresponding gap in b, which must be filled by i. This cannot continue past the midpoint of e, where gaps of length 2 give way to teeth of length 2. No longer can a gun, lying on its back, embed its blunt end in e, and we arrive at a contradiction.
Let the (necessarily thick) blunt end of b touch the blunt end of a at the origin. Suppose the blunt end has length k > 2. The piece c based at 2,2 creates holes with either a or b, or interlocks with b to make a perfect rectangle, or interlocks with a, and has a barrel of k-2. If k = 3, n,0 becomes inaccessible, so k >= 4. Piece d, based at 2,3, cannot run vertically, so it lies atop c. If d points right, piece e cannot fill 2,5 without creating holes somewhere, or making a perfect rectangle with d. So d points its barrel at b, and e tiles 2,4. This means e and b interlock, just like a and c. Let f stand back-to-back against e, tiling 3,4. Suppose f has its barrel down. If k > 4, then g must tile 4,4. Yet g and f cannot create a perfect rectangle, nor can g interlock with d, so k = 4. To fill 4,5, g and f must be reverse aligned, forcing a blunt end of length 1, not 4, or g and d are aligned, forcing a blunt end of length 2, not 4. Therefore f has its blunt end down, and k >= 5. If k = 5 then g cannot sit in the corner of f and d without creating holes somewhere, nor can it align with d, which gives d a thick end of 3. If k > 5, g cannot fill the gap between f and d, along the barrel of d. This completes the case k > 2.
The blunt end has length 2, c begins at 2,1, c cannot run vertically alongside b, c interlocks with a, c and a are aligned, and all gaps and teeth have length 2. If the barrel has length 2 the piece has order 2. If the barrel is longer than 2, c and a create a horizontal slot, which, when filled, creates inaccessible regions against the x axis. So the barrel has length 1, and d slides its barrel into n,0. Let e tile n+2,2. If e and d are aligned, 2n,0 becomes inaccessible, hence e is vertical. If e extends over the rightmost square of c, {4} is invoked. So e points its barrel at d and faces right (if based at n+2,2) or left (if based at n+3,1). In either case, f lies atop c, and points its barrel at b. The blunt end of f must not create a hole to its right, so e stands facing right. Let g fill the 2×2 gap at 1,4. If g and f interlock, let h fill n+1,5, the single gap between g, f, and e. Now the back of e forces h to stand upright on its barrel and face left, whereupon the top of g spans n-1, and invokes {4}. So g interlocks with b. Now h stands to the right of g, back-to-back, and i stands next to h, filling the first gap in f. However, h and i create holes, completing the proof.
Let the barrel of a touch the origin, while the blunt end of b rests on this barrel and covers 1,1. Let k be the length of the blunt end. With the longer barrel, {11} tells us k >= 2. Let d fill the square 1,k+1, the first gap in b. Now d traps a vertical gap, extending up to 2,k, which must be filled with a series of horizontal guns. Let c be the gun that lies atop a. Now c cannot lie face up without creating holes, so c and a interlock. If they are aligned, the piece has order 2. Hence we assume they are reverse aligned, and the barrel has length k+2.
Assume d protrudes horizontally from b. If d is face down, nothing can fill the gap at 2,k, so d lies back-to-back with c, and k = 2. If d embeds its barrel in b, this barrel spans a horizontal gap of 3 or 4, depending on the size of the gap in b. Let e nestle in the corner of b and d. If the gap in b has size > 1, e must form a rectangle with d, or align with b. The latter is possible, but only if all gaps and internal teeth have length 3. In this case the barrels of b and e create a vertical slot that, when filled, leaves inaccessible regions by the y axis. If the gap in b has size 1, e must interlock with b or d, or stand next to b. Only the first is plausible, yet e traps an L-shaped hole on the y axis that cannot be filled. Thus the blunt end of d embeds in b. A reflection of a, call it e, runs along the x axis starting at n,0, its blunt end under the protruding barrel of c. Let piece f fill the first gap of e at n+2,1. If f runs horizontally then the barrels of e and f form a slot which, when filled, creates holes along the x axis. If f, running vertically, faces left, it makes a gap at n+1,3, just above d. So f faces right, and the barrel of d defines a gap of length 4 that cannot be filled. From this we conclude that d runs vertically along b.
Now d and b are aligned and all gaps and teeth have length k. However, their barrels create a vertical slot that, when filled, creates inaccessible regions by the y axis. This completes the blunt-barrel case.
Combined with {12} and {13}, the blunt end of a gun is never used to tile 1,1.
We have determined that a gun cannot use its blunt end to tile 1,1, yet by {9}, one of the two guns pointed at the origin must tile this square. This can only be done by a gun lying on its back, such that its first tooth tiles 1,1. Thus every gun has a short tab or barrel. If there is no tab, as with a blunt gun, the barrel has length 1.
Let piece a, a blunt gun, point its short barrel at the origin, as it must by {15}. Suppose piece a also covers 2,1, and let c cover 2,2. Since c must not create holes with either b or a, it must interlock with a. Now c and a are aligned, all gaps have length 2, the blunt end of c extends 2 squares to the right of a, and n,0 is inaccessible. Thus a does not cover 2,1, and c must.
c runs horizontally, interlocked with a in reverse. Avoiding the shape of order 2, n is at least 6. The combination of c and a forms a parallelogram that might repeat forever down the x axis. Similarly, d could interlock with b, creating a parallelogram that might propagate up the y axis. Clearly this shape tiles the quadrant. We can't just tile and tile until we reach a dead end; we need a new approach.
Consider a corner of size n-1 square, like the origin. Point the barrels of a and b into the corner. Now c forms a parallelagram with a as above. If d forms a parallelagram with b, we have another corner. Otherwise d lies back-to-back atop c, with one end in the first gap of b. If this gap exceeds the end of d, we won't be able to fill it without trapping holes in d, so we have a perfect fit.
Suppose d points right, hence its blunt end, in b, is a 2×2 square. Let e fill the corner of b and d. If e interlocks with d, this is a reverse alignment 3 squares over from the alignment demonstrated by a and c. Using an inductive argument, all teeth have size 2, yet this cannot continue past the midpoint of d. If e points down and faces right, f is aligned with d. This alignment makes all teeth size 2, even the last tooth, which is actually size 1. Thus e faces left and interlocks with b, and f stands back-to-back with e. By comparing e and b, we find that the first gap in d is size 1, the second tooth is size 1, and the second gap is size 2. Thus f points down, and g, in the second gap, can only point right and align with d. Even if this alignment works, h makes a new corner with g, or it places its blunt end into f, and creates the same pattern we saw with b and d.
Next we assume d points into b, whence the first gap of b, like the first tooth, has size 1. This means we have a thin blunt end. Place the barrel of e into the first gap of d. If e faces right we have the same pattern we saw with b and d. So e faces left, but this traps holes against b. That leaves only a d e interlock, like the parallelagram formed by a and c. Place f in the gap between b and e, and even if the interlock works, e and f create a new corner.
Next place the blunt end of a at the corner. If c fills the first gap of a, and a and c are aligned, all teeth have the same size, which must be 1, and the shape has order 2. The pieces may be reversed aligned, hence the blunt end is thick. But remember, the piece also reverse aligns with itself at a different offset, as illustrated by the parallelagram at the (real) origin. Again, our inductive argument shows all teeth have a uniform size > 1, which is impossible.
With c standing up, the blunt end is thick, b slides in between c and the y axis, c and b interlock and align, and the teeth have size 1, even though the blunt end of a has size 2.
We have shown that corners propagate up and to the right forever. This finishes off the blunt guns. Hereinafter, all guns have tabs, and by {10}, those tabs have length 1.
Consider a gun formed by removing the lower corners from an n×2 solid rectangle. Thus the barrel and tab have lengths 1, and there are no gaps (e.g. ff7e). For n = 3 or 4, the order is 4, so assume n >= 5. Copies a and b meet at the origin as usual, making an artificial corner at 2,2. One wall of this corner is n-2 squares long, the other n-3. Piece c fits into this new corner, and faces the main diagonal. Piece d fills the gap between c and a or b. If c runs along the longer wall, and n = 5, d has the opportunity to face away from the main diagonal. But whether d faces toward or away from the diagonal, c and d create a new corner whose walls are at least 2 squares long. These corners propagate up and to the right forever.
Suppose the barrel and tab have length 1, and the teeth adjacent to these ends have lengths at least 2. Thanks to {17}, we can assume there is at least one gap in the face of the gun. The smallest example is fe6c. Once again a and b create an artificial corner at 2,2, that must be filled by c. Now c cannot place its back against either wall without creating holes, so let c interlock with a, and let d tile the square at 2,3. Now d cannot place its back to b without creating holes, and if d interlocks with b, d and c make a new corner, and the process begins again. Let d lie on its back immediately above c. Let e fill the gap between d and b. If e turns its back on b, it creates holes, so e and b interlock. Now we know the tooth next to the tab has length 3, and the tooth by the barrel has length 2. There are 2 or 3 squares above d, just to the right of e, that must be filled. Piece f cannot lie atop d, nor can it interlock with d, so it stands upright and fills the first square. Piece g must fill the next hole by interlocking with d. The result is the diagonal reflection of the configuration of e and d. These patterns repeat as corners proceed up to infinity.
Suppose the gun has a tab and short barrel, and the first and last teeth have length 1. With a at the origin and b standing upright at 0,1, piece c must fill 2,1. This places c and b front-to-back, and creates holes.
Let the barrel and tab have length 1, and let the tooth next to the tab have length k > 1. If a points its barrel at the origin, piece c must stand upright at 2,1, just as it did before. Therefore a points right, and b slides its barrel or tab into the tab of a.
Let c tile 2,2. To avoid holes, c must interlock with either a or b. Suppose the tab of b touches the tab of a, so that c and a interlock, setting k = 2. If d interlocks with b then c and d create a corner, just like the origin, and the process begins again. Yet d cannot place its back against either b or c without creating an inaccessible hole at 1,4.
Next let b point its barrel down into the tab of a. If b and c interlock, the square at 0,n+1 cannot be filled without creating a hole at 1,n+1. As above, c interlocks with a and k = 2. Now d lies atop c, filling the size 1 gap in b at 1,3. If another gap begins at 1,5, e fills this gap by placing its back against d, which creates holes. So b continues through 1,5, and e tiles 2,5. Suppose e is horizontal, whence it must interlock with d. If d and e are aligned, the thick teeth adjacent to their tabs mus collide. If they are reversed, they look like a and c. That shifts e one square to the right, where it cannot tile 2,5. Thus e is vertical. If e faces right, it can only avoid trapping holes in b if b has no more gaps. This gives a shape of fc68, which minimally meets our criteria. However, the piece that fills the gap above b faces left and creates a hole with the y axis. Thus e faces left and interlocks with b. If the barrel of d points left, e and d adopt the same configuration as c and b, and the process begins again. With the tab of d embedded in b, let f tile 3,5. If f is horizontal it creates holes with d, or interlocks with d, whereupon f and e make a perfect corner. If f stands back-to-back with e, 4,4 becomes inaccessible. That completes the proof. We need no longer consider guns with tabs and single-length barrels.
Let our gun have a long barrel, as required by {20}, and assume there are gaps in its face. Because of {9}, piece a must place its tab against the origin. Suppose the first gap in a occurs at 2,1, and suppose piece c fills this gap by standing vertically. To avoid creating holes, c and b are interlocked in reverse. Note that the barrel has length 2. This leaves a gap at 1,n that cannot be filled by a vertical gun. Let d, a vertical reflection of a, fill this gap. If a tiles 3,1, then the column next to c invokes {4}. Thus the first gap in a (and d) has size 2, and e stands back-to-back with c. If another gap opens at 5,1 and 5,n, piece f interlocks with e, just like b and c. But this cannot continue past the midpoint of a, where gaps of length 2 give way to teeth of length 2, so assume a tiles 5,1, as in ff4c. Verify that ff4c doesn't work; a is actually longer.
Let f tile 4,2 if the barrel of e points down, 4,n-1 otherwise. This can only be done by interlock with a or d. Let g lie back-to-back with f, and let h lie face-to-face with g. Verify that h sticks its tab into e, whence g and h are aligned. All teeth have width 1, and that doesn't look like a. This contradicts the assumption that c is vertical.
At this point a and c interlock, and b points its long barrel into a. If a and c are aligned, they form a slot at the right that, when filled, creates inaccessible regions with the x axis. So they are interlocked in reverse, and that drives the long barrel of c into b. Therefore the first gap in a occurs farther to the right.
Assume the first gap begins at 3,1, and let the tab of b touch the tab of a. If c tiles 2,2, and runs vertically, it must interlock with b in reverse. However, c protrudes one square at 2,n+1, and 0,n+1 becomes inaccessible. If a and c are aligned, all gaps and teeth have length 2. If the barrels have length 2, n,0 is inaccessible. If the barrels are longer, they define a slot that, when filled, creates inaccessible regions along the x axis. A reverse alignment would drive the barrel of c into b.
Now the barrel of b points down, and c tiles 1,2 by running horizontally. Now c interlocks with a, and n,0 becomes inaccessible. Thus the first gap occurs at 4,1, or farther to the right. The smallest example is ff74.
Verify that a and c must reverse align to form a parallelagram. If b and d do the same, you will notice the parallelagrams marching along the x and y axies. From here the proof is similar to that employed in {16}.
Start with a corner and place the tab of a at the origin. We know that a and c can reverse align, but that won't cover 1,2, so the tab of b fits into the tab of a. Now c and a are revers aligned, just as they were at the (real) origin. If d makes a parallelagram with b, we have a new corner, so let d lie flat atop c. Let e fill the corner of b and d. Verify that e and d cannot interlock. So e interlocks with b. This must be an alignment, hence all teeth and gaps have size 3, and the barrel has length 2. Place f at the corner of d and e. If d points left, f points up, g fills the corner of d and f by interlocking with d, and the pattern looks exactly like b and c. If f points down then g must lie on its back atop d, and that creates holes.
Hereinafter, we need not consider any guns with gaps in their faces. Henceforth a gun is completely characterized by two numeric parameters, its length n and the length of its barrel k, where k > 1 and k+1 < n.
Let a gun have a single square as its face and a barrel of length k >= 4. (Note that k = 1, 2, and 3 have orders 4, 10, and 92 respectively.) We shall show that 3 corner configurations inevitably lead to eachother, thereby propagating corners up and to the right forever.
0) A wall and floor of 4 squares each, like the origin.
1) Same as (0), but 0,0 is already populated, and the wall and floor are 5 and 6 squares long respectively.
2) Same as (1), but 1,0 is also populated.
Given (0), let a place its barrel at the origin. If a is face up, the span above the barrel of a trips {4}, so a lies face down. Place b at 0,1. If b is vertical, with its barrel down, it must face left, recreating (0). If b has its tab down then let c tile 1,1. If c points its barrel at b, the span above the barrel of c has length k-1, at least 3, and trips {4}. So c points right and creates (2).
If b lies atop a, it must point right, and c stands upright at 0,2. Now c and b produce either (1) or (2).
Suppose a has its tab at the origin. If b, standing at 0,1, faces left, it creates (1). So b faces right and embeds its tab in a (to avoid {4}), which creates (2).
Given (1), let a tile 1,0. If a faces left its back joins the floor to produce (0). If a faces right its tab is down, and b fills the gap at 2,0. This creates (1) or (2). If a lies face up it must point right, with b and c filling the gap of width 2. Together a and c create (2). If a is face down, and the back of a extends along row 0, it creates (0). Finally we assume the back of a runs along row 1. Place piece b at 0,2. If b is vertical, it joins a to produce (0), or it forces c to tile 1,2, thereby producing (2). If b is horizontal then c tiles 0,3 vertically, and c and b produce (1) or (2).
Given (2), let a tile 2,0. If a faces left, let b tile 3,0. If b is horizontal, it creates (0) with the back of a, or it forces c to stand with its back to a, whence b and c create (2). If b stands back-to-back with a, there is still enough floor to force c, tiling 4,0, to run horizontally. Thus c and b produce (1) or (2). If a faces right then 0,1 begins a 2×1 inaccessible gap. If a faces up it must point left, to avoid {4}. This forces b to lie face up on a, its barrel pointing right. Now c stands in the tab of b, and creates (1) or (2). Finally a is face down. If its back extends row 0, it creates (0). If its tab is at 1,1, b tiles 0,1 by standing upright, and b and a create (0) or (1).
This completely characterizes the guns with a single square on the second row.
Let a gun have a barrel of length 2 and a face of length > 2. (Note that f860 is known to have order 76.) Assume a 2×3 corner and let piece a tile the origin. If the barrel of a points left, and a lies face down, the back of a joins the wall to create another 2×3 corner. If a is face up, b and c fill the gap above the barrel of a. If c points up we have another corner. If c points down the resulting gap is filled by d. If d points right we have another corner. If d points left we have another gap that is filled by e. Now e can point up or down, like c, creating a corner or continuing the pattern. Eventually we find another corner.
If the tab of a touches the origin, let b stand at 0,1. If b faces right we have a situation that we dealt with in the previous paragraph. If b faces left we have another corner.
Next suppose a stands upright at 0,0, whence it is forced to face right. If a has its barrel down, b and c protrude horizontally from the gap in column 1, which is the same situation we dealt with earlier. If a has its tab down, b fills 1,0, and creates a corner with a, or produces a single gap that will eventually lead to a corner.
These 2×3 corners propagate up and to the right forever. That finishes the guns with barrel length 2.
Let the gun have a barrel of length at least j, where j >= 3. Thanks to {22}, we know the face is at least 2 in length. We will postpone fe70, barrel = face = 3. Consider the following configurations.
0) A j×j corner.
1) An n-1×n corner with the face of a gun on the floor.
2) Same as (1), but another face climbs the wall, starting at 0,1.
3) An n×(j-1) corner.
4) A wall of height n and a floor of width n-2,
with a shortened (by 1) face on the floor.
5) A wall of height n and a floor of width n-3,
with a shortened (by 2) face on the floor.
The face must be at least 3 for this case to be meaningful.
Starting with (0), let the barel of a touch the origin. {4} forces a to lie face down. This creates (3), which we will pursue at this time.
Let b stand at 0,1. If b faces left then a and b create another corner (0). If b faces right then a and the imaginary wall at b's back form (0). So b lies on its back. If b points left it trips {4}. This is because its barrel has length at least 3, and even if the face and barrel are the same length, the face of c cannot perfectly fill this gap, unless barrel = face = 3, which we are not dealing with here. So b points right, and c fills the gap at 0,2. If c faces left we have (1), and if c faces right , a joins with an imaginary wall behind c to form (0). That finishes off (3), but we're still expanding (0).
Let a lie on its back and point right, with b standing at 0,1. If b faces left we have (1), and if b faces right we have (2), unless b points its barrel into a and j = 3. Now c interlocks with a at 1,2, and d lies back-to-back with c. Yet this interlock is impossible, since barrel and face have different lengths when j = 3.
Starting with (1), let a tile the square on the floor to the right of the preexisting face. If a faces left, its back joins the floor to produce (3). If a faces right, an imaginary wall behind a joins the floor to produce (0). If a points left, its barrel sliding atop the preexisting face, let k be the number of squares between the barrel of a and the wall. If k = 0 we have (0). If k = 1, b stands at 0,1 and creates (1) with a. If k > 1 the gap cannot be tiled. Finally, the back of a might extend the preexisting face. If a lies face down we have (0), and if a lies face up the region along row 1 trips {4}.
Next consider (2), and base a just above the preexisting vertical face. If a lies face down it joins the wall to produce (3). If a lies face up, an imaginary floor under a joins the wall to produce (0). Suppose the back of a stands in column 0. If a faces right, column 1 trips {4}. If a faces left, we have (4). Finally let the barrel or tab of a slide down the face on the left wall, giving a gap of height k. Note that the face of b cannot fill this gap, so k > 2 trips {4}. If k = 2, b and c protrude horizontally, and c and a produce (1). If k = 1, b fills the gap, and joins a to produce (0) or (1). If k = 0 we have (5).
Finally consider (4) and (5), and place a just to the right of the abbreviated face on the floor. From here the reasoning is similar to (1). If a faces right we have (0), and if a faces left we have (3). If a lies face up it trips {4}. If a lies face down with its back along row 0 we have (0). If a lies across the shortened face on the floor, with a gap of size k, we have (0), (1), or an inaccessible gap.
In summary, there are no tiling guns with barrel length > 2, except perhaps fe70.
Take a row of seven squares and place a row of three squares above it, shifted by 1. This shape is unresolved. Nobody knows if this polyomino tiles a rectangle or not. Personally, I doubt it, but I cannot prove it.
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