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Comparison of Wang tiling solvers

12 décembre 2018 | Catégories: sage, slabbe spkg, math | View Comments

During the last year, I have written a Python module to deal with Wang tiles containing about 4K lines of code including doctests and documentation.

It can be installed like this:

sage -pip install slabbe


It can be used like this:

sage: from slabbe import WangTileSet
sage: tiles = [(2,4,2,1), (2,2,2,0), (1,1,3,1), (1,2,3,2), (3,1,3,3),
....: (0,1,3,1), (0,0,0,1), (3,1,0,2), (0,2,1,2), (1,2,1,4), (3,3,1,2)]
sage: T0 = WangTileSet([map(str,t) for t in tiles])
sage: T0.tikz(ncolumns=11).pdf()


The module on wang tiles contains a class WangTileSolver which contains three reductions of the Wang tiling problem the first using MILP solvers, the second using SAT solvers and the third using Knuth's dancing links.

Here is one example of a tiling found using the dancing links reduction:

sage: %time tiling = T0.solver(10,10).solve(solver='dancing_links')
CPU times: user 36 ms, sys: 12 ms, total: 48 ms
Wall time: 65.5 ms
sage: tiling.tikz().pdf()


All these reductions now allow me to compare the efficiency of various types of solvers restricted to the Wang tiling type of problems. Here is the list of solvers that I often use.

List of solvers
Solver Description
'Gurobi' MILP solver
'GLPK' MILP solver
'PPL' MILP solver
'LP' a SAT solver using a reduction to LP
'cryptominisat' SAT solver
'picosat' SAT solver
'glucose' SAT solver

In this recent work on the substitutive structure of Jeandel-Rao tilings, I introduced various Wang tile sets $T_i$ for $i\in\{0,1,\dots,12\}$. In this blog post, we will concentrate on the 11 Wang tile set $T_0$ introduced by Jeandel and Rao as well as $T_2$ containing 20 tiles and $T_3$ containing 24 tiles.

Tiling a n x n square

The most natural question to ask is to find valid Wang tilings of $n\times n$ square with given Wang tiles. Below is the time spent by each mentionned solvers to find a valid tiling of a $n\times n$ square in less than 10 seconds for each of the three wang tile sets $T_0$, $T_2$ and $T_3$.

We remark that MILP solvers are slower. Dancing links can solve 20x20 squares with Jeandel Rao tiles $T_0$ and SAT solvers are performing very well with Glucose being the best as it can find a 55x55 tiling with Jeandel-Rao tiles $T_0$ in less than 10 seconds.

Finding all dominoes allowing a surrounding of given radius

One thing that is often needed in my research is to enumerate all horizontal and vertical dominoes that allow a given surrounding radius. This is a difficult question in general as deciding if a given tile set admits a tiling of the infinite plane is undecidable. But in some cases, the information we get from the dominoes admitting a surrounding of radius 1, 2, 3 or 4 is enough to conclude that the tiling can be desubstituted for instance. This is why we need to answer this question as fast as possible.

Below is the comparison in the time taken by each solver to compute all vertical and horizontal dominoes allowing a surrounding of radius 1, 2 and 3 (in less than 1000 seconds for each execution).

What is surprising at first is that the solvers that performed well in the first $n\times n$ square experience are not the best in the second experiment computing valid dominoes. Dancing links and the MILP solver Gurobi are now the best algorithms to compute all dominoes. They are followed by picosat and cryptominisat and then glucose.

The source code of the above comparisons

The source code of the above comparison can be found in this Jupyter notebook. Note that it depends on the use of Glucose as a Sage optional package (#26361) and on the most recent development version of slabbe optional Sage Package.

Wooden laser-cut Jeandel-Rao tiles

07 septembre 2018 | Catégories: sage, slabbe spkg, math | View Comments

I have been working on Jeandel-Rao tiles lately.

Before the conference Model Sets and Aperiodic Order held in Durham UK (Sep 3-7 2018), I thought it would be a good idea to bring some real tiles at the conference. So I first decided of some conventions to represent the above tiles as topologically closed disk basically using the representation of integers in base 1:

With these shapes, I created a 33 x 19 patch. With 3cm on each side, the patch takes 99cm x 57cm just within the capacity of the laser cut machine (1m x 60 cm):

With the help of David Renault from LaBRI, we went at Coh@bit, the FabLab of Bordeaux University and we laser cut two 3mm thick plywood for a total of 1282 Wang tiles. This is the result:

One may recreate the 33 x 19 tiling as follows (note that I am using Cartesian-like coordinates, so the first list data[0] actually is the first column from bottom to top):

sage: data = [[10, 4, 6, 1, 3, 3, 7, 0, 9, 7, 2, 6, 1, 3, 8, 7, 0, 9, 7],
....:  [4, 5, 6, 1, 8, 10, 4, 0, 9, 3, 8, 7, 0, 9, 7, 5, 0, 9, 3],
....:  [3, 7, 6, 1, 7, 2, 5, 0, 9, 8, 7, 5, 0, 9, 3, 7, 0, 9, 10],
....:  [10, 4, 6, 1, 3, 8, 7, 0, 9, 7, 5, 6, 1, 8, 10, 4, 0, 9, 3],
....:  [2, 5, 6, 1, 8, 7, 5, 0, 9, 3, 7, 6, 1, 7, 2, 5, 0, 9, 8],
....:  [8, 7, 6, 1, 7, 5, 6, 1, 8, 10, 4, 6, 1, 3, 8, 7, 0, 9, 7],
....:  [7, 5, 6, 1, 3, 7, 6, 1, 7, 2, 5, 6, 1, 8, 7, 5, 0, 9, 3],
....:  [3, 7, 6, 1, 10, 4, 6, 1, 3, 8, 7, 6, 1, 7, 5, 6, 1, 8, 10],
....:  [10, 4, 6, 1, 3, 3, 7, 0, 9, 7, 5, 6, 1, 3, 7, 6, 1, 7, 2],
....:  [2, 5, 6, 1, 8, 10, 4, 0, 9, 3, 7, 6, 1, 10, 4, 6, 1, 3, 8],
....:  [8, 7, 6, 1, 7, 5, 5, 0, 9, 10, 4, 6, 1, 3, 3, 7, 0, 9, 7],
....:  [7, 5, 6, 1, 3, 7, 6, 1, 10, 4, 5, 6, 1, 8, 10, 4, 0, 9, 3],
....:  [3, 7, 6, 1, 10, 4, 6, 1, 3, 3, 7, 6, 1, 7, 2, 5, 0, 9, 8],
....:  [10, 4, 6, 1, 3, 3, 7, 0, 9, 10, 4, 6, 1, 3, 8, 7, 0, 9, 7],
....:  [4, 5, 6, 1, 8, 10, 4, 0, 9, 3, 3, 7, 0, 9, 7, 5, 0, 9, 3],
....:  [3, 7, 6, 1, 7, 2, 5, 0, 9, 8, 10, 4, 0, 9, 3, 7, 0, 9, 10],
....:  [10, 4, 6, 1, 3, 8, 7, 0, 9, 7, 5, 5, 0, 9, 10, 4, 0, 9, 3],
....:  [2, 5, 6, 1, 8, 7, 5, 0, 9, 3, 7, 6, 1, 10, 4, 5, 0, 9, 8],
....:  [8, 7, 6, 1, 7, 5, 6, 1, 8, 10, 4, 6, 1, 3, 3, 7, 0, 9, 7],
....:  [7, 5, 6, 1, 3, 7, 6, 1, 7, 2, 5, 6, 1, 8, 10, 4, 0, 9, 3],
....:  [3, 7, 6, 1, 10, 4, 6, 1, 3, 8, 7, 6, 1, 7, 2, 5, 0, 9, 8],
....:  [10, 4, 6, 1, 3, 3, 7, 0, 9, 7, 2, 6, 1, 3, 8, 7, 0, 9, 7],
....:  [4, 5, 6, 1, 8, 10, 4, 0, 9, 3, 8, 7, 0, 9, 7, 5, 0, 9, 3],
....:  [3, 7, 6, 1, 7, 2, 5, 0, 9, 8, 7, 5, 0, 9, 3, 7, 0, 9, 10],
....:  [10, 4, 6, 1, 3, 8, 7, 0, 9, 7, 5, 6, 1, 8, 10, 4, 0, 9, 3],
....:  [3, 3, 7, 0, 9, 7, 5, 0, 9, 3, 7, 6, 1, 7, 2, 5, 0, 9, 8],
....:  [8, 10, 4, 0, 9, 3, 7, 0, 9, 10, 4, 6, 1, 3, 8, 7, 0, 9, 7],
....:  [7, 5, 5, 0, 9, 10, 4, 0, 9, 3, 3, 7, 0, 9, 7, 5, 0, 9, 3],
....:  [3, 7, 6, 1, 10, 4, 5, 0, 9, 8, 10, 4, 0, 9, 3, 7, 0, 9, 10],
....:  [10, 4, 6, 1, 3, 3, 7, 0, 9, 7, 5, 5, 0, 9, 10, 4, 0, 9, 3],
....:  [2, 5, 6, 1, 8, 10, 4, 0, 9, 3, 7, 6, 1, 10, 4, 5, 0, 9, 8],
....:  [8, 7, 6, 1, 7, 5, 5, 0, 9, 10, 4, 6, 1, 3, 3, 7, 0, 9, 7],
....:  [7, 5, 6, 1, 3, 7, 6, 1, 10, 4, 5, 6, 1, 8, 10, 4, 0, 9, 3]]


The above patch have been chosen among 1000 other randomly generated as the closest to the asymptotic frequencies of the tiles in Jeandel-Rao tilings (or at least in the minimal subshift that I describe in the preprint):

sage: from collections import Counter
sage: c = Counter(flatten(data))
sage: tile_count = [c[i] for i in range(11)]


The asymptotic frequencies:

sage: phi = golden_ratio.n()
sage: Linv = [2*phi + 6, 2*phi + 6, 18*phi + 10, 2*phi + 6, 8*phi + 2,
....:      5*phi + 4, 2*phi + 6, 12/5*phi + 14/5, 8*phi + 2,
....:      2*phi + 6, 8*phi + 2]
sage: perfect_proportions = vector([1/a for a in Linv])


Comparison of the number of tiles of each type with the expected frequency:

sage: header_row = ['tile id', 'Asymptotic frequency', 'Expected nb of copies',
....:               'Nb copies in the 33x19 patch']
sage: columns = [range(11), perfect_proportions, vector(perfect_proportions)*33*19, tile_count]
tile id   Asymptotic frequency   Expected nb of copies   Nb copies in the 33x19 patch
+---------+----------------------+-----------------------+------------------------------+
0         0.108271182329550      67.8860313206280        67
1         0.108271182329550      67.8860313206280        65
2         0.0255593590340479     16.0257181143480        16
3         0.108271182329550      67.8860313206280        71
4         0.0669152706817991     41.9558747174880        42
5         0.0827118232955023     51.8603132062800        51
6         0.108271182329550      67.8860313206280        65
7         0.149627093977301      93.8161879237680        95
8         0.0669152706817991     41.9558747174880        44
9         0.108271182329550      67.8860313206280        67
10        0.0669152706817991     41.9558747174880        44


I brought the $33\times19=641$ tiles at the conference and offered to the first 7 persons to find a $7\times 7$ tiling the opportunity to keep the 49 tiles they used. 49 is a good number since the frequency of the lowest tile (with id 2) is about 2% which allows to have at least one copy of each tile in a subset of 49 tiles allowing a solution.

A natural question to ask is how many such $7\times 7$ tilings does there exist? With ticket #25125 that was merged in Sage 8.3 this Spring, it is possible to enumerate and count solutions in parallel with Knuth dancing links algorithm. After the installation of the Sage Optional package slabbe (sage -pip install slabbe), one may compute that there are 152244 solutions.

sage: from slabbe import WangTileSet
sage: tiles = [(2,4,2,1), (2,2,2,0), (1,1,3,1), (1,2,3,2), (3,1,3,3),
....: (0,1,3,1), (0,0,0,1), (3,1,0,2), (0,2,1,2), (1,2,1,4), (3,3,1,2)]
sage: T0 = WangTileSet(tiles)
sage: T0_solver = T0.solver(7,7)
sage: %time T0_solver.number_of_solutions(ncpus=8)
CPU times: user 16 ms, sys: 82.3 ms, total: 98.3 ms
Wall time: 388 ms
152244


One may also get the list of all solutions and print one of them:

sage: %time L = T0_solver.all_solutions(); print(len(L))
152244
CPU times: user 6.46 s, sys: 344 ms, total: 6.8 s
Wall time: 6.82 s
sage: L[0]
A wang tiling of a 7 x 7 rectangle
sage: L[0].table()  # warning: the output is in Cartesian-like coordinates
[[1, 8, 10, 4, 5, 0, 9],
[1, 7, 2, 5, 6, 1, 8],
[1, 3, 8, 7, 6, 1, 7],
[0, 9, 7, 5, 6, 1, 3],
[0, 9, 3, 7, 6, 1, 8],
[1, 8, 10, 4, 6, 1, 7],
[1, 7, 2, 2, 6, 1, 3]]


This is the number of distinct sets of 49 tiles which admits a 7x7 solution:

sage: from collections import Counter
sage: def count_tiles(tiling):
....:     C = Counter(flatten(tiling.table()))
....:     return tuple(C.get(a,0) for a in range(11))
sage: Lfreq = map(count_tiles, L)
sage: Lfreq_count = Counter(Lfreq)
sage: len(Lfreq_count)
83258


Number of other solutions with the same set of 49 tiles:

sage: Counter(Lfreq_count.values())
Counter({1: 49076, 2: 19849, 3: 6313, 4: 3664, 6: 1410, 5: 1341, 7: 705, 8:
293, 9: 159, 14: 116, 10: 104, 12: 97, 18: 44, 11: 26, 15: 24, 13: 10, 17: 8,
22: 6, 32: 6, 16: 3, 28: 2, 19: 1, 21: 1})


How the number of $k\times k$-solutions grows for k from 0 to 9:

sage: [T0.solver(k,k).number_of_solutions() for k in range(10)]
[0, 11, 85, 444, 1723, 9172, 50638, 152244, 262019, 1641695]


Unfortunately, most of those $k\times k$-solutions are not extendable to a tiling of the whole plane. Indeed the number of $k\times k$ patches in the language of the minimal aperiodic subshift that I am able to describe and which is a proper subset of Jeandel-Rao tilings seems, according to some heuristic, to be something like:

[1, 11, 49, 108, 184, 268, 367, 483]


I do not share my (ugly) code for this computation yet, as I will rather share clean code soon when times come. So among the 152244 about only 483 (0.32%) of them are prolongable into a uniformly recurrent tiling of the plane.

Installing Ubuntu 18.04 on Aspire ES 11 ES1-132-C6LG

20 mai 2018 | Catégories: ubuntu | View Comments

I tried to install Ubuntu 18.04 on a Aspire ES 11 ES1-132-C6LG and got troubles because I get the screen "No Bootable Device found" after installation. The first help I found on Google is this posts on itsfoss.com but at step 3, the option "Select an UEFI file as trusted for executing" is unavailable so the proposed solution does not work for me.

Some comments on the same post shows that I am not alone with this problem:

Alex 2 months ago

I cannot “Select an UEFI file”… Maybe Windows 10 configured my BIOS to
reject any other drive (SSD) that has not Windows on it?

Tom 1 month ago

No, you need to install ubuntu without bootloader, mount the partitions
after the installation, modprobe efivars and then install grub-efi


After looking at what grub-uefi and modprobe mean, I found this discussion (April 2017, Ubuntu 17.04) and this discussion (Dec. 2016) on askubuntu and this discussion (Dec 2016) on community acer very close to my problem. In the end, the most simple (no need to modprobe anything or install grub-efi) and usefull one (it worked) was this discussion (Jan 2017) on community acer especially the two posts by Spektro37 (on page 1, the second on page 2) that I recopy and adapt a little bit below.

1. Install Ubuntu;

2. Boot using Ubuntu USB and select "Try without installing";

3. Launch Gparted to get the EFI partition address. In my case it's /dev/mmcblk0p1;

4. Open the Terminal (ctrl + alt + T);

5. In the Terminal execute the following to create a directory in the media folder and to mount the EFI partition to that folder:

sudo mkdir /media/EFI


In my case it would look like:

sudo mkdir /media/EFI
sudo mount /dev/mmcblk0p1 /media/EFI

1. Create the /EFI/Linux/ folder:

sudo mkdir /media/EFI/EFI/Linux

2. Copy all the existing files from the folder that was created during Ubuntu instalation. In my case, the default folder was /EFI/ubuntu/:

sudo ls /media/EFI/EFI/ubuntu
BOOTX64.CSV  fw  fwupx64.efi  grub.cfg  grubx64.efi  mmx64.efi  shimx64.efi


To do so, you can use the following command:

sudo cp -R /media/EFI/EFI/ubuntu/* /media/EFI/EFI/Linux/

3. Copy the BOOTX64.EFI from the BOOT directory to the Linux folder (without it, I confirm it didn't work):

sudo cp /media/EFI/EFI/BOOT/BOOTX64.EFI /media/EFI/EFI/Linux


According to Spectro37, /EFI/Linux/BOOTX64.efi is an hardcoded path for Linux.

Plus de 3000 questions maintenant disponibles en français dans l'interface WebWork

24 mars 2018 | Catégories: webwork | View Comments

Six ans après avoir utilisé et entamé la traduction de WebWork en français dans un cours que j'avais donné à l'Université du Québec à Montréal, je viens de voir sur internet que WebWork est maintenant utilisé à grande échelle en français au Québec.

Le projet Développement de WeBWorK pour le réseau des cégeps francophones a porté ses fruits. Un serveur du Centre collégial de développement de matériel didactique (CCDMD), donne facilement accès à un compte sur la plateforme WeBWorK. Selon le site, plus de 3000 questions sont maintenant traduites en français dans la Banque de problèmes libres (BPL) en français dans un répertoire sur github. De plus, des ressources en ligne sont disponibles notamment sur ce sujet sur le site de MathemaTIC.

Félicitations à toutes celles et ceux qui ont été impliqués dans la poursuite du projet dans les 6 dernières années.