# Ziggy - Can you make a square from 49 polyomino pieces

This puzzle is variant of the puzzle Ziggy - Make a square from 8 polyomino pieces by jaap-scherphuis.

This puzzle is based on the fact that $$1+2+3+\cdots+49=35^2$$. It consists of 49 zigzag polyomino pieces, ranging in size from 1 to 49 squares. The first eight pieces are shown below:

The remaining pieces follow the zigzag pattern of the eight given pieces.

Is it possible to completely tile a 35 by 35 square using all 49 pieces? Rotating and flipping the pieces is allowed.

• Candidate squares: oeis.org/A001110 Commented Nov 9, 2023 at 15:38

I think it is

impossible.

Outline of a proof:

Note: As the devil is in the detail I'm only 50%70% confident this is correct. Especially the last part is still rathera bit squishy. Anyway, just want to put it out there before the question gets cold.

Step 1:

Show that there must be two opposite triangle corners. This is comparatively easy. Count boundary squares (136). There being only 49 tiles there must be many that cover 3 or 4 boundary points and such tiles cut off a triangle corner.

Doing some mildly annoying case bashing we can then show that a triangle corner must contain either the monomino or the L-tromino, so there can only be two of them.

Inspecting the numbers more closely we find that the triangles must cover more than half the boundary squares and must therefore be in opposite corners. The number of leftover odd-sized tiles is at most 4 and they are not short ones.

Using checkerboard parity we can also easily show that the smaller triangle must be a clean stack of all 4n+1-sized or all 4n-1-sized tiles. (The other triangle could in principle have a few composite outer "onion shells" but this can be ruled out using a 4-way colouring (121212 // 343434 // 121212 ...) of the board.)

Step 2:

With the triangular caps established mentally extend the stripe pattern of the caps to the entire board, i.e. split it into pairs of diagonals (or "ziggies" I suppose). Domino positions on the board are now split into two equal size groups "on grid", the domino lies entirely inside one ziggy, and "off grid", the domino straddles two ziggies.

Let's designate one triangle the "north pole" and the other one the "south pole", the min diagonal separating them would be the "equator". Let us number northern ziggies 1,2,... starting from the equator and southern ziggies -1,-2,... also starting from the equator.

Assuming we have managed to cover the hexagon left between the triangle caps. Cut the covering tiles into dominoes. This is unambiguous for the even sized tiles, for the handful if any odd-sized tiles we cut off a Monomino at one end and cut the rest into dominoes. Note that for any tile regardless its orientation either all its dominoes are on-grid or all off-grid.

Next, consider the dominoes and monominoes covering a given ziggy k. Off-grid dominoes can straddle either northwards N_k or southwards S_k. We have N_k = S_k+1 and N_k + MN_k= S_k + MS_k +/- 1 where the + is valid on the Southern hemisphere and the - in the northern hemisphere and MN_k and MS_k are the numbers of monominoes on the northern an southern parts of the ziggy. By convention also N_-1 = S_1 at the equator and N_max = S_min = 0 at the triangle edges. Also, recall from step 1 that there are at most 4 monominoes total.

These equations mandate a small but nonzero number of off-grid dominoes. Except for at most 4 ziggy boundaries the delta in straddlers across this and neighbouring boundaries is 1. The tiles that were cut into the off-grid dominoes contribute either a stack of one domino per layer crossing or all dominoes to the same layer crossing. In a fully or predominantly grid parallel orientation there are not enough short tiles available. In the orthogonal orientation there are not enough "free" (meaning non boundary touching) tiles available and we cannot attach the tiles to the boundary because of symmetry w.r.t. the equator this would cut another triangle (FIXME: Can the monominoes screw us here?).

UPDATE

An improved version of the last argument: An orthogonal off-grid tile that spans several ziggies will split each of the S_k,N_k... in between in two subquantities because it acts as a barrier. They are governed by similar equations as above except that the +/-1 offset is retained in only one of the subequations. And here is the twist: Which side retains it switches at the equator. This crossing over makes it impossible to stack orthogonal offgrid for the purpose of cancelling the cumulative +/-1 offsets.

• The next in the sequence is $288$ pieces that sum to $204^2$. Like Jack's previous question, these are even numbers. Is it possible in this case? Or are $8$ pieces that sum to $6^2$ unique? (Apart from the trivial case of $1$). Commented Nov 12, 2023 at 0:12
• @WeatherVane not sure. Most of the proof should carry over since the relevant ratios converge (288/204 ~> sqrt 2 if you forgive the super sloppy notation). What we lose is the global imbalance in the 4-way colouring. My gut feeling would be that 8/6 is the only solvable but that's just my gut. Commented Nov 12, 2023 at 2:37

A near miss (omits 2 and 8) via integer linear programming with a binary decision variable for each possible placement of a polyomino:

$$\begin{matrix}&16&28&28&26&26&24&24&22&22&6&6&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15&11&11&7&7&3&3\\&16&16&28&28&26&26&24&24&22&22&6&6&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15&11&11&7&7&3\\&.&16&16&28&28&26&26&24&24&22&22&6&6&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15&11&11&7&7\\&20&20&16&16&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15&11&11&7\\&.&20&20&16&16&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15&11&11\\&18&18&20&20&16&16&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15&11\\&46&18&18&20&20&16&16&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15&15\\&46&46&18&18&20&20&16&16&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19&15\\&.&46&46&18&18&20&20&16&.&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19&19\\&12&12&46&46&18&18&20&20&4&4&28&28&26&26&24&24&22&22&14&14&47&47&43&43&39&39&35&35&31&31&27&27&23&23&19\\&49&12&12&46&46&18&18&20&20&4&4&28&28&26&26&24&24&22&22&30&30&47&47&43&43&39&39&35&35&31&31&27&27&23&23\\&49&49&12&12&46&46&18&18&20&20&48&48&28&28&26&26&24&24&32&32&30&30&47&47&43&43&39&39&35&35&31&31&27&27&23\\&45&49&49&12&12&46&46&18&18&20&20&48&48&28&28&26&26&34&34&32&32&30&30&47&47&43&43&39&39&35&35&31&31&27&27\\&45&45&49&49&12&12&46&46&18&18&44&44&48&48&28&28&36&36&34&34&32&32&30&30&47&47&43&43&39&39&35&35&31&31&27\\&41&45&45&49&49&12&12&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43&39&39&35&35&31&31\\&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43&39&39&35&35&31\\&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43&39&39&35&35\\&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43&39&39&35\\&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43&39&39\\&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43&39\\&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43&43\\&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47&43\\&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47&47\\&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30&47\\&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&30&30\\&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32&.\\&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&32&32\\&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34&.\\&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&46&42&42&44&44&48&48&38&38&36&36&34&34\\&13&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&46&.&42&42&44&44&48&48&38&38&36&36&.\\&9&13&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&10&10&42&42&44&44&48&48&38&38&36&36\\&9&9&13&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&10&10&42&42&44&44&48&48&38&38&.\\&5&9&9&13&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&10&10&42&42&44&44&48&48&38&38\\&5&5&9&9&13&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&10&10&42&42&44&44&48&48&.\\&1&5&5&9&9&13&13&17&17&21&21&25&25&29&29&33&33&37&37&41&41&45&45&49&49&40&40&10&10&42&42&44&44&48&48\\\end{matrix}$$

Another near miss (omits 4 and 6):

$$\begin{matrix}&2&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13&13&9&9&5&5&1\\&2&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13&13&9&9&5&5\\&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13&13&9&9&5\\&8&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13&13&9&9\\&8&8&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13&13&9\\&.&8&8&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13&13\\&16&16&8&8&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17&13\\&.&16&16&8&.&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17&17\\&14&14&16&16&10&10&20&20&18&18&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21&17\\&.&14&14&16&16&10&10&20&20&18&.&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21&21\\&12&12&14&14&16&16&10&10&20&20&49&49&28&28&26&26&24&24&22&22&48&48&45&45&41&41&37&37&33&33&29&29&25&25&21\\&47&12&12&14&14&16&16&10&10&20&20&49&49&28&28&26&26&24&24&30&30&48&48&45&45&41&41&37&37&33&33&29&29&25&25\\&47&47&12&12&14&14&16&16&10&10&46&46&49&49&28&28&26&26&32&32&30&30&48&48&45&45&41&41&37&37&33&33&29&29&25\\&43&47&47&12&12&14&14&16&16&44&44&46&46&49&49&28&28&34&34&32&32&30&30&48&48&45&45&41&41&37&37&33&33&29&29\\&43&43&47&47&12&12&14&14&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41&41&37&37&33&33&29\\&39&43&43&47&47&12&12&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41&41&37&37&33&33\\&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41&41&37&37&33\\&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41&41&37&37\\&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41&41&37\\&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41&41\\&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45&41\\&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45&45\\&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48&45\\&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&48&48\\&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30&.\\&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&30&30\\&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32&.\\&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&32&32\\&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34&.\\&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&34&34\\&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36&.\\&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&36&36\\&7&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49&.\\&3&7&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49&49\\&3&3&7&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&38&38&40&40&42&42&44&44&46&46&49\\\end{matrix}$$

A nearer miss (omits 8):

$$\begin{matrix}&2&14&49&49&.&38&.&30&32&32&48&48&44&44&41&41&37&37&33&33&.&28&25&25&21&21&17&17&13&13&9&9&5&5&.\\&2&14&14&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17&13&13&9&9&5&5\\&42&42&14&14&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17&13&13&9&9&5\\&29&42&42&14&14&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17&13&13&9&9\\&29&29&42&42&14&14&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17&13&13&9\\&.&29&29&42&42&14&14&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17&13&13\\&12&12&29&29&42&42&14&14&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17&13\\&45&12&12&29&29&42&42&14&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17&17\\&45&45&12&12&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21&17\\&.&45&45&12&12&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21&21\\&4&4&45&45&12&12&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25&21\\&47&4&4&45&45&12&12&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25&25\\&47&47&46&46&45&45&40&40&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28&25\\&43&47&47&46&46&45&45&40&40&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28&28\\&43&43&47&47&46&46&45&45&40&40&29&29&42&42&36&36&49&49&38&38&30&30&32&32&48&48&44&44&41&41&37&37&33&33&28\\&39&43&43&47&47&46&46&45&45&40&40&29&29&42&42&36&36&49&49&38&38&30&26&32&32&48&48&44&44&41&41&37&37&33&33\\&39&39&43&43&47&47&46&46&45&45&40&40&29&29&42&42&36&36&49&49&38&38&26&26&1&20&48&48&44&44&41&41&37&37&33\\&35&39&39&43&43&47&47&46&46&45&45&40&40&29&29&42&42&36&36&49&49&38&38&26&26&20&20&48&48&44&44&41&41&37&37\\&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&42&42&36&36&49&49&38&38&26&26&20&20&48&48&44&44&41&41&37\\&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&42&42&36&36&49&49&38&22&26&26&20&20&48&48&44&44&41&41\\&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&42&42&36&36&49&49&22&22&26&26&20&20&48&48&44&44&41\\&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&42&42&36&36&49&49&22&22&26&26&20&20&48&48&44&44\\&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&42&42&36&36&49&49&22&22&26&26&20&20&48&48&.\\&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&36&36&49&49&22&22&26&26&20&20&48&48\\&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&36&36&49&16&22&22&26&26&20&20&.\\&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&36&18&16&16&22&22&26&26&20&20\\&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&18&18&16&16&22&22&26&26&20\\&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&18&18&16&16&22&22&26&26\\&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&18&18&16&16&22&22&26\\&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&45&45&40&40&34&34&24&24&18&18&16&16&22&22\\&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&10&10&40&40&34&34&24&24&18&18&16&16&22\\&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&10&10&40&40&34&34&24&24&18&18&16&16\\&7&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&10&10&6&6&34&34&24&24&18&18&16\\&3&7&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&10&10&6&6&34&34&24&24&18&18\\&3&3&7&7&11&11&15&15&19&19&23&23&27&27&31&31&35&35&39&39&43&43&47&47&46&46&10&10&6&6&34&34&24&24&18\\\end{matrix}$$

• Still in progress. I'm trying to minimize the number of uncovered cells. Commented Nov 9, 2023 at 4:54
• @WillOctagonGibson IIRC I did an exhaustive search for this at the time and did not find a solution. It would be nice to see it confirmed. Commented Nov 9, 2023 at 5:39
• Actually, I probably only started the search. I don't think I completed an exhaustive search because I remember trying to think up a proof for why there would be no solutions. Commented Nov 9, 2023 at 13:56

Billions of near misses...

I arranged the odd sized pieces as per the pic and then tried to tile the even pieces into the blue area. After searching a negligible fraction of the remaining search space (and preferring placing bigger pieces earlier) my tiling program has placed 23 of the 24 even pieces around half a billion times in four hours. This is of course not any sort of proof, but normally when this happens it's time to dust off the real math skills and try to find a disproof.

• I had also tried forcing this pattern and restricting the even polyominoes to have positive slope. The resulting integer linear programming solver eventually returned a positive lower bound for the number of uncovered cells, meaning that a tiling is impossible under those restrictions. Commented Nov 10, 2023 at 14:07