This problem was proposed by me, Santiago Rodriguez from Colombia. I hope that you enjoy it. For those of you who are interested in the story behind the problem, my inspiration was a beautiful board game called Takenoko. In that game a gardener grows some bamboo and a panda eats it following some rules similar to the ones on the proposed problem. One time I was playing the game and realized that the interaction between the panda and the gardener could make for a fun math Olympiad problem. After some time I got what became the C3. But of course I had to change the pandas and the bamboo to something more Norwegian to send the problem to IMO.

Redacted

For context, this problem appeared during the US IMO Training Camp with the following flavor text: MOP Test 1 wrote: Alice and Bob play a game on a $2022 \times 2022$ grid with a $0$ initially written in each cell. The players alternate turns, with Alice going first. On her turn, Alice chooses a cell and increment the numbers of all cells sharing at least one vertex with the chosen cell by $1$. On his turn, Bob chooses any four cells in the grid and decrements their numbers by $1$. Determine the largest $K$ such that Alice can ensure that at some point, there are $K$ cells in the grid whose numbers are at least $10^6$.

This is just a easier version of JMO 2023 P4 lol... also @2above, I think it still requires proof that Alice's strategy will attain the claimed answer.... The answer is $5 \cdot \left( \frac{2022}{3} \right)^2$. The lumberjack can mark squares like this, and make sure that no tree ever grows in these squares: [asy][asy] for(int x = 0; x < 9; x+=3) { for(int y = 0; y < 9; y+=3) { fill((x,y)--(x+2,y)--(x+2,y+2)--(x,y+2)--cycle, red); } } for(int x = 0; x <= 9; ++x) { draw((x,0)--(x,9)); } for(int y = 0; y <= 9; ++y) { draw((0,y)--(9,y)); } [/asy][/asy] (This is doable, since each turn, the gardner touches at most four of these squares.) Now we prove the gardner can actually obtain this many tall trees. In particular, the gardner just splits up the grid into 3x3 squares, and repeatedly goes around tending to each of the $\left( \frac{2022}{3} \right)^2$ squares. After $\left( \frac{2022}{3} \right)^2 \cdot n$ gardner-lumberjack turns, the gardner has touched each square $n$ times. For the sake of contradiction, assume that at this point, the lumberjack has managed to stop $4 \cdot \left( \frac{2022}{3} \right)^2 + 1$ trees from growing. Then, he must have touched each of those trees at least $n-(10^6 - 1)$ times. So, he has made a total of \[ \left( 4 \cdot \left( \frac{2022}{3} \right)^2 + 1 \right) \cdot \left( n-(10^6 - 1) \right)\]cuts. But for sufficiently large $n$, this is more than $4 \cdot \left( \frac{2022}{3} \right)^2 \cdot n$, which is the number of cuts the lumberjack has made at this point. This is a contradiction, hence the lumberjack can only prevent $4 \cdot \left( \frac{2022}{3} \right)^2$ trees from growing.

Complete_quadrilateral wrote:

This is not exactly true: what if Bob always tries to cut the trees in the top left square, even when Alice moves outside? The problem is not difficult, but I would recommend making sure your solution is correct before making loud claims

IAmTheHazard wrote: Complete_quadrilateral wrote:

This is not exactly true: what if Bob always tries to cut the trees in the top left square, even when Alice moves outside? The problem is not difficult, but I would recommend making sure your solution is correct before making loud claims Rip

Solved with OronSH, pikapika007. The answer is $\boxed{5\cdot 674^2 = 2271380}$. The strategy for the lumberjack is to partition the board into 3x3 grids (call these the good 3x3 grids), and for each good 3x3 grid, color the topmost row and leftmost cell in the middle row. There are four of them in each 3x3 grid (not necessarily good), so when the gardner selects (at most) a 3x3 square, the lumberjack can cut the four (or less) squares that the gardner added to. This means that these squares will never become majestic, and there are $4\cdot 674^2$ of them, so $5\cdot 674^2$ squares are remaining. Thus, the lumberjack can guarantee that at most $5\cdot 674^2 $ trees can be majestic. Let $N = 1817105000001$. Now we show that the gardner can guarantee that there are at least $5\cdot 674^2$ majestic trees. Partition the board similar to above. The strategy is to touch the center of each good 3x3 square $N$ times. This means that each square has had its height increased $N$ times. This takes $674^2 \cdot N$ turns for the gardner, so the lumberjack is allowed to cut $4\cdot 674^2 \cdot N$ trees (not necessarily distinct) because they get four per turn. Suppose that after this, there were only at most majestic $5\cdot 674^2 - 1$ trees, so at least $4\cdot 674^2 + 1$ trees that are not majestic. For a square to not be majestic, the lumberjack must have cut it at least $N - 10^6$ times. Therefore, the lumberjack must have made at least $(4\cdot 674^2 + 1)(N - 10^6)$ cuts. However, we have \begin{align*} (4\cdot 674^2 + 1)(N - 10^6) = (4\cdot 674^2 + 1)( 1817105000001 - 10^6) \\ = 4\cdot 674^2 \cdot N + 1817105000001 - 10^6 \cdot (4\cdot 674^2 + 1) \\ = 4\cdot 674^2 \cdot N + 1 > 4\cdot 674^2 \cdot N, \\ \end{align*}which is absurd since the lumberjack can make at most $4\cdot 674^2 \cdot N$ cuts. Therefore the gardner can guarantee at least $5\cdot 674^2$ majestic trees, so we are done.

We claim the answer $K=5\cdot 674^2.$ Plug $k=674^2.$ Gardener's strategy. Split the garden onto squares $3\times 3,$ and numerate them by $1,2,\ldots ,k.$ Let the gardener consequently picks squares number $1,2, \ldots, k,$ and increases heights of all trees in picked $i-$th square for $n_i=10^6\cdot 2^{k-i}\cdot \binom{9}{5}^{k-i+1}$ times. By PHP after all these moves being done (and after lumberjack's responding moves being done) there would exist a set $S_i$ of five trees in the $i-$th square, all of which have a height at least $H_i=n_i/ \binom{9}{5}.$ Next, after all $$N_i=\sum_{j=i+1}^{k} \left( 10^6\cdot 2^{k-j} \cdot \binom{9}{5}^{k-j+1} \right) \leq 10^6\cdot (2^{k-i}-1) \cdot \binom{9}{5}^{k-i}$$other moves (for squares $i+1,$ $i+2\ldots$) height of any tree from $S_i$ equals at least to $$H_i-N_i\geq 10^6 \cdot 2^{k-i} \cdot \binom{9}{5}^{k-i}\geq 10^6.$$Thus, we totally have at least $5$ majestic trees in each of $k$ squares $\Box$ Lumberjack's strategy. Consider the following coloring of the garden in grey (limited by a square $9\times 9$) [asy][asy] unitsize(15); for(int x=0; x < 8; x = x + 3){ for(int y=1; y < 8; y = y + 3){ fill((x,y)--(x+2,y)--(x+2,y+2)--(x,y+2)--(x,y)--cycle, grey); } } for(int i = 0; i < 10; ++i){ draw((i,0)--(i,9)); draw((0,i)--(9,i)); } for(int j=0; j<10; j=j+3){ draw((j,0)--(j,9),linewidth(1.5)); draw((0,j)--(9,j),linewidth(1.5)); } [/asy][/asy] If the gardener increases heights of trees surrounding the corner cell ($4$ trees), let lumberjack just decreases all their heights. Otherwise, observe that the gardener has increased heights of exactly four grey cells - let lumberjack picks these cells on the next move. Thus trees in grey cells can't have a height above $1,$ and the total number of majestic trees is at most $5k$ $\Box$

I'm so sorry. The answer is $5\cdot 674^2$. To prove this color every top-left $2\times 2$ square after tiling the grid with $3\times 3$ squares. Whatever the gardener chooses, there are four colored squares so the lumberjack can cut them down. Now, using the same $3\times 3$ squares, the gardener plants $10^{10^{1}}$ times in the first square. Clearly, at least five of those are majestic now. Before that, the gardener plants $10^{10^{2}}$ times in the second square. There will be at least five majestic plants in the second square and during planting in the first square, the lumberjack cannot remove majesticness of the second square. Before all of that, the gardener plants $10^{10^{3}}$ times in the third square. During planting in the first and second squares, the lumberjack cannot remove majesticness of the third square. Repeating this process until we get to planting $10^{10^{674^{2}}}$ times, we are done.

Clean. Answer is $\frac{5}{9} \cdot 2022^2$. For the upper bound, simply cut the intersection of the gardener's region with the squares whose coordinates are both $0, 1 \pmod 3$. For the lower bound, simply grow every center of every $3 \times 3$ region, then keep doing this overlapping over the previous growths each time. It is easy to show that doing this large enough times means that the cutter simply does not have enough total cuts to cut any more than $\frac{4}{9} \cdot 2022^2$ of these really tall trees, even if they can do all the cuts at the very end to their liking.

Label rows and columns in the usual fashion; color cells red if neither row nor column number is a multiple of $3$. Let $P=2022^2$. Evidently any move by the gardener intersects at most four red cells, hence the lumberjack can guarantee (by moving on said cells) that they all remain $0$. Hence the answer is at most $\frac{5}{9}P$. Now suppose the gardener uses $\frac{1}{9}P$ moves to increase the heights of all trees by $1$ unit. The lumberjack can decrease the heights of $\frac{4}{9}P$ trees (with multiplicity). If at most $\frac{5}{9}P-1$ trees are majestic by repetition of this algorithm $c$ times, then the lumberjack was able to keep some $\frac{4}{9}P+1$ trees non-majestic. But the sum of the heights of these trees increases by at least $1$ after each repetition, hence is unbounded, a contradiction.

"this is a very skull strat" - mathboy100 Solved with shreyasharma. The gardener can ensure $\boxed{K = \frac{5}{9} \cdot 2022^2}$ majestic trees. Gardener's Strategy Consider partitioning the $202 2 \times 2022$ grid into $674^2$ individual $3 \times 3$ squares. Also define $f(x) = \underbrace{1000^{1000^{1000^{\dots}}}}_{x \text{ times}}$. Now let the gardener proceed with the following algorithm, repeating $674^2$ times. We provide instructions for the $k$-th loop of the step. Let the gardener choose a $3 \times 3$ square he has not chosen yet. This is his $k$-th $3 \times 3$ subsquare. Note that the gardener is guarenteed to be able to raise $5$ or the $9$ trees in this $3 \times 3$ subsquare to some arbitrary height, as at any given moment, the hunter may lower at most $4$ squares. Thus the gardener should raise some $5$ trees in this block of $3 \times 3$ squares to a height of $f(1,000,000 - j + 1)$. Now we claim that this is sufficient. Indeed consider the $2$-nd loop of this step. The trees in the $1$-st subsquare have already been raised to $f(1,000,000)$. Now say the lumberjack chooses to lower trees in subsquare $1$, rather than attack the trees the gardener is currently raising in subsquare $2$. We claim that this has no effect. Indeed note that in the $f(1,000,000 - 1)$ moves the gardener takes to raise the trees in subsquare $2$, the lumberjack may reduce the height of $4$ of the $5$ trees in subsquare $1$ by at most $f(1,000,000 - 1)$. However due to the enormous gap in $f(1,000,000) > f(1,000,000 - 1)$ we are guarenteed the trees in subsquare $1$ still have a height greater than $10^6$. Indeed, even if the lumberjack attacks the squares in square $1$ while the gardener raises all other $476^2 - 1$ subsquares, he will be unable to sufficiently reduce the height of the trees in subsquare $1$ to make them non-majestic. A similar argument follows for all subsquares due to size reasons. More simply put, the gardeners strategy is to raise the $k$-th subsquare to an unimaginably large height, although sufficiently lower than the heights of the previous $k-1$ subsquares, so that over the course of a finite game, the lumberjack may never destroy trees the gardener has already made majestic, only limit trees that the gardener is in the process of making. $\square$ Lumberjack's Strategy Consider coloring green every $1 \times 1$ square that has either an $x$ or $y$ coordinate divisible by $3$. Consider marking every other square red. On the lumberjack's move, he will cut down any trees lying on red squares that the gardener has raised on his previous turn. Clearly this strategy suffices, as there are exactly $4$ red squares in any $3 \times 3$ square, and these $4$ squares will never grow beyond a height of $1$. Hence the lumberjack may restrict the gardener to $\frac{5}{9} \cdot 2022^2$ majestic trees. $\square$ Remark: As you may see, we went a bit insane and did a random size strategy disregarding any ``nice" coloring arguments that are objectively ``better".

The answer is $\boxed{5\cdot 674^2 = 2271380}$ My Solution is same as @above

I like this <3

The answer is $\boxed{2271380}$. The Gardener's Strategy: Divide the grid into $3\times 3$ sub-grids. We claim the gardener can get $5$ majestic trees in each sub-grid. Start in one sub-grid. Each turn is equivalent to increasing the height of $5$ trees by $1$ and keeping the others constant. So eventually by PhP there will be $5$ trees in that sub-grid with arbitrarily large height. Now by induction if we can get $5$ cells on each of $k$ $3\times 3$ grids of arbitrary height then we can add one more $3\times 3$ grid without worrying what it does to the others. The Lumberjack's Strategy: It is easy to come up with a coloring with exactly $4$ shaded squares in each $3\times 3$ sub-grid. If the lumberjack always cuts these squares down then they will remain at $0$.

$\boxed{\text{Claim : For } 3k \times 3k \text{ grids, we can have could always achieve } 5 \cdot k^2}$ We will proceed this by induction For the case $k =1$ is trivial We suppose that it is true for all $ k < m$. We wanna show that it is also true for $m$ We will call a very tall tree majestic such that it could not turn back to non-majestic before we finish our task \textit{Immortal} We will let the gardener start at the lower corner. $\boxed{\text{SubClaim : If in a} 3k \times 3k \text{ grids, we can have could always achieve } 5 \cdot k^2 \text{majestic tree, then we can make those} 5 \cdot k^2 \text{Immortal}}$ We will prove this SubClaim first, then the Claim. For $k$ =1, gardener repeat the same square $x$ times. So, the total height will be $5x$. With each tree could be at most $x$. Suppose it is possible for the lumberjack to fix the height of $6$ tree. But, we get that the remaining tree could have at most $3x$ height, resulting in the six tree to have at least one variable by $x$. Thus, we take $x$ to infinity, resulting in $5$ immortal trees. For induction case, suppose this is true for all $k < n$, we wanna show that it is true for $k=n$ By hypothesis, we get that for any $3n-3 \times 3n-3$ subgrids, there will always be $5 \cdot (n-1)^2$ So, there are $2n-1$ of $3 \times 3$ grids. From hypothesis, for each $3 \times 3$ grids, we could create $5$ immortal trees. So, $10n-5$ more immortal tree. At last, there would be $5n^2 - 10n + 5 + 10n -5 = 5n^2$ immortal trees, as desired. From the SubClaim, in any $3k \times 3k$ grid, there will be $5k^2$ immortal trees, which is stronger than majestic, as desired in Claim.

OG! My first C3! The answer is $5.(674)^2$ Proof that this can be achieved: We define $move$ to be a turn of the $gardener$ followed by that of the $Lumber Jack$ Lemma: In a $3 \times 3$ grid, with the same rules, the gardener can ensure at least $5$ $majestic$ $trees$. Proof: LET THE GRID( WHERE SQUARES ARE REPLACED BY ALPHABETS) BE A B C D E F G H I . Let $H(A)$ denote the height of the tree on the square $A$, define $H(B) \cdots H(H)$ similarly. Now the gardener would always chooses the square $E$. Consider the $sum(S)$ of the heights of the trees in squares $A,B,C,D,E$. In each turn Lumber Jack can decrease $S$ maximum by 4, while the gardener ensures that on his turn that $S$ is increased by 5. Thus in all $S$ is increased by at least 1 in each move. So in $,$ moves(where $k \geq 1000$), $S = H(A)+H(B)+H(C)+H(D)+H(E) \geq m$, Thus one of $H(A),H(B),H(C),H(D),H(E) \geq \frac{m}{5}.$ WLOG let it be $H(A)$, Hence The tree on the square $A$ is now majestic(since $m$ can be arbitrarily large), thus call tree $A$ as $amar$(i.e one who can't die = immortal), Now similarly consider$5$ squares other than $A$, and do moves, then do the same process considering other 5 five squares not including the squares which have turned amar already , then again and again perform the moves till we have 4 remaining squares and $5$ $amar$ trees. Note: that the lumberjack cannot reduce a previously turned $amar$ tree back to a $non- amar tree$ as we can ensure insufficient number of moves after the tree gets $amar$ to do so.( In the sense that We ensure that an $amar$ tree gets so large, that it is greater than the combined length of the latter turned $amar$ trees) (For an example: We ensure the first $amar$ tree is of length $10^{10^5}$, second of length $10^{10^4}$, third of length $10^{10^3}$, fourth of $10^{10^2}$ and $10^{10^1}$ units now note the if the $ith$ $amar-tree$ has turned of the length $10^{10^i}$, then the subsequent turned $amar-trees$ are of the length $10^{10^{i-1}}$ $\cdots$ $10^{10^1}$. Note that we need a maximum of $5x$ to make a $amar- tree$ of length $x$, thus, after the $ith$ tree turned $amar$, there at maximum of $5(10^{10^{i-1}} +\cdots 10^{10^1}$moves performed which is way less than $10^{10^i}$ Note that the $2022 \times 2022$ grid can be divided into $674^2$ $3 \times 3$ grids. and thus we have $5.674^2$ $amar$ trees, we can also note that the lumberjack cannot reduce a previously turned $amar$ tree back to a $non- amar$ $tree$ by the same logic here as well. And of course an $amar$ is a majestic tree Proof that $5.(674)^2$ is optimal: Strategy that lumberjack uses: same as the above posts.

I request if anyone can review please please please please please please

The answer is $\frac{5}{9} \cdot 2022^2=2271380$. Label the squares as $(x,y)$ where $1 \leq x,y \leq 2022$. Mark all cells $(x,y)$ where $x,y \equiv \pm 1 \pmod{3}$ red, and all other cells green. In every turn, the gardener can grow at most four trees on red squares, since the gardener cannot grow two trees on different squares $(x_1,y_1),(x_2,y_2)$ where $x_1 \equiv x_2$ and $y_1 \equiv y_2 \pmod{3}$. Assume the lumberjack picks the following strategy: In every turn, choose four red squares, which includes all the red squares with trees grown on that turn. Then clearly, all trees on red squares can never have heights greater than $1$, so the upper bound is established. Now we consider the $3 \times 3$ grids consisting of squares $(3m+i,3n+j)$ where $1 \leq i,j \leq 3$ and $0 \leq m,n \leq 673$. Label these grids $A_1, \cdots A_{674^2}$ in some order. We claim the following is true for any $3 \times 3$ grid and $1 \leq i \leq 5$: For any positive integer $k$, if the gardener repeatedly chooses the centre cell, there will eventually be at least $i$ trees in the grid with heights all $\geq k$. We prove this by induction on $i$. The claim is obviously true for $i=1$: The total height of the trees increase by at least $5$ each turn, so the total height is unbounded, so height of the tallest tree must also be unbounded. For $2 \geq i \geq 5$, if the claim is true for $i-1$, then consider the total height of the remaining $9-(i-1)$ trees in the grid. This total height must increase by at least $5-(i-1)>0$ each turn, so the height of the tallest tree in those remaining trees must eventually exceed $k$. So the claim is proved. Next, for any positive integer $k$, let $f(k)$ denote the smallest positive integer $n$ such that, after repeatedly choosing the centre cell for $n$ turns, there must exist at least $5$ trees with height $ \geq k$. Define $g_i(k)$ for $i \in \mathbb{Z}^+$ recursively as follows: $g_1(k)=f(k)$ and $g_{i+1}(k)=f \left(k + \sum_{j=1}^i g_{j}(k) \right)$. The gardener can perform the following procedure: Choose the centre square of $A_{674^2}$ for $g_{674^2}(10^6)$ turns, then the centre square of $A_{674^2-1}$ for $g_{674^2-1}(10^6)$ turns, and so on, finally choosing the centre square of $A_1$ for $g_1(10^6)$ turns. Note that by definition there must exist $5$ trees in $A_1$ with heights $\geq 10^6$ at the end of the procedure. For any $1 < i \leq 674^2$, there must be $5$ trees in $A_i$ with heights at least $10^6+\sum_{j=1}^{i-1} g_{j}(10^6)$ after the centre square of $A_i$ has been chosen for $g_{i}(10^6)$ turns. Then, exactly $\sum_{j=1}^{i-1} g_{j}(10^6)$ more turns pass by. The lumberjack can now decrease the heights of the $5$ tallest trees in $A_i$ by at most $\sum_{j=1}^{i-1} g_{j}(10^6)$ each, leaving each tree with heights at least $10^6$ at the end of the procedure, so we are done.

short solution to the problem. (I actually solved in like 15 minutes lmao) claim: $K\ge (5/9)\cdot 2022^2$ proof: consider when gardener choose $674^2$ square with $(3i+2, 3j+2$), (taking $674^2$ turn). So height of all trees increase by $1$. here, lumberjack only can decrement at most $4\cdot 674^2 = (4/9)\cdot 2022^2$, completes the proof. If $K=(5/9)\cdot 2022^2$, then we can color the square like the following (shown below is $6\times 6$), so that colored square never be majestic. (even with height $2$. This follows by every $3\times 3$ square contains $4$ colored square.) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1., xmax = 7., ymin = -1., ymax = 7.; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); fill((0.,1.)--(0.,3.)--(1.,3.)--(1.,2.)--(3.,2.)--(3.,3.)--(4.,3.)--(4.,2.)--(5.,2.)--(6.,2.)--(6.,1.)--cycle, red); fill((0.,6.)--(0.,4.)--(6.,4.)--(6.,5.)--(4.,5.)--(4.,6.)--(3.,6.)--(3.,5.)--(1.,5.)--(1.,6.)--cycle, red); /* draw figures */ draw((0.,0.)--(0.,2.), linewidth(2.) + zzttqq); draw((0.,2.)--(1.,2.), linewidth(2.) + zzttqq); draw((1.,2.)--(1.,1.), linewidth(2.) + zzttqq); draw((1.,1.)--(3.,1.), linewidth(2.) + zzttqq); draw((3.,1.)--(3.,2.), linewidth(2.) + zzttqq); draw((3.,2.)--(4.,2.), linewidth(2.) + zzttqq); draw((4.,2.)--(4.,1.), linewidth(2.) + zzttqq); draw((4.,1.)--(5.,1.), linewidth(2.) + zzttqq); draw((5.,1.)--(6.,1.), linewidth(2.) + zzttqq); draw((6.,1.)--(6.,0.), linewidth(2.) + zzttqq); draw((6.,0.)--(0.,0.), linewidth(2.) + zzttqq); draw((0.,6.)--(0.,4.), linewidth(2.) + zzttqq); draw((0.,4.)--(6.,4.), linewidth(2.) + zzttqq); draw((6.,4.)--(6.,5.), linewidth(2.) + zzttqq); draw((6.,5.)--(4.,5.), linewidth(2.) + zzttqq); draw((4.,5.)--(4.,6.), linewidth(2.) + zzttqq); draw((4.,6.)--(3.,6.), linewidth(2.) + zzttqq); draw((3.,6.)--(3.,5.), linewidth(2.) + zzttqq); draw((3.,5.)--(1.,5.), linewidth(2.) + zzttqq); draw((1.,5.)--(1.,6.), linewidth(2.) + zzttqq); draw((1.,6.)--(0.,6.), linewidth(2.) + zzttqq); draw((0.,0.)--(0.,2.), linewidth(2.) + zzttqq); draw((0.,2.)--(1.,2.), linewidth(2.) + zzttqq); draw((1.,2.)--(1.,1.), linewidth(2.) + zzttqq); draw((1.,1.)--(3.,1.), linewidth(2.) + zzttqq); draw((3.,1.)--(3.,2.), linewidth(2.) + zzttqq); draw((3.,2.)--(4.,2.), linewidth(2.) + zzttqq); draw((4.,2.)--(4.,1.), linewidth(2.) + zzttqq); draw((4.,1.)--(5.,1.), linewidth(2.) + zzttqq); draw((5.,1.)--(6.,1.), linewidth(2.) + zzttqq); draw((6.,1.)--(6.,0.), linewidth(2.) + zzttqq); draw((6.,0.)--(0.,0.), linewidth(2.) + zzttqq); draw((0.,6.)--(0.,0.), linewidth(2.)); draw((1.,6.)--(1.,0.), linewidth(2.)); draw((2.,0.)--(2.,6.), linewidth(2.)); draw((3.,6.)--(3.,0.), linewidth(2.)); draw((4.,0.)--(4.,6.), linewidth(2.)); draw((5.,6.)--(5.,0.), linewidth(2.)); draw((6.,0.)--(6.,6.), linewidth(2.)); draw((0.,6.)--(0.44,6.), linewidth(2.)); draw((0.,6.)--(6.,6.), linewidth(2.)); draw((6.,5.)--(0.,5.), linewidth(2.)); draw((0.,4.)--(6.,4.), linewidth(2.)); draw((6.,3.)--(0.,3.), linewidth(2.)); draw((0.,2.)--(6.,2.), linewidth(2.)); draw((6.,1.)--(0.,1.), linewidth(2.)); draw((0.,0.)--(6.,0.), linewidth(2.)); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]