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top iTouch hacks

So, you did jailbreak your iTouch and did install some fun or useful stuff via the Install.app … but then, suddenly, the next program on your wish-list fails to install ??!! I know you hate to do drastic things to your iTouch, but sooner or later you’ll have to do it, so why not NOW?

Move the Applications Folder

The problem is that there are two disk partitions (a small one, meant only to host the apple-software and a large one to contain all your music, videos and stuff) and Install.app installs programs in the /Apllications folder on the smaller partition. So, we want to move it to the other partition using a symbolic link trick (as in the wiki-hack post). Here a walkthrough, more details can be found on Koos Kasper’s site.

  • Have BSDsubsystem and OpenSSH installed, so that you can ssh into the iTouch.
  • verify that the second line of the /etc/fstab file reads as below (or edit it if necessary, in my case it was already ok, perhaps this is done during jailbreak?) and reboot the iTouch (if you had to change it)

/dev/disk0s2 /private/var hfs rw 0 2

  • ssh into the iTouch and type in the following commands (to move the folder and make the symbolic link)

cd /
cp -pr Applications /var/root
mv Applications Applications.old
ln -s private/var/root/Applications /Applications

  • reboot the iTouch, ssh into it and remove the old Application-folder to free space

cd /
rm -rf Applications.old

From now on, all (most) new programs are installed on the larger partition. If you reinstall the OpenSSH application (as suggested) make sure to remove on your computer the old key for iTouch.

Stream your Music!

I use the iTouch to read my mail, to read RSS feeds, to administer this blog, to VNC to the home-server and when needed to ssh into the computer at work (running this blog) to restart the apache server. Unless I have to write a lot, there is no need to fire up a computer… But, when someone has a Mac running, I would like to be able to stream the music on my iTouch to hear it loudly. Here’s the procedure, via Rupert Gee’s blog :

  • Have the Auto-Lock set to “Never” in Settings/General
  • Install the UIctl applications (under Utilities)
  • Add a source to Install.app (click on Sources-button lower-right, Edit upper-right and then Add upper-left) http://home.mike.tl/iphone
  • Relaunch Install.app and install FireFlyMediaServer (under Multimedia).
  • Write down the address given during installation to change your password and monitor the Firefly-server (the default root password is ‘dottie’ and so the address should be

http://root:dottie@127.0.0.1:3689

  • Open up UIctl and scoll down to a line saying “org.fireflymediaserver.mt-daapd” and tap on it. Tap on “load-w” and then on “Do It”
  • Now, at the Mac your iTouch should be vusible under Shared in iTunes, click on it and give the password and your music is available!

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mini-sudokube

Via the Arcadian functor I learned of the existence of the Sudokube (picture on the left).

Sudokube is a variation on a Rubik’s Cube in which each face resembles one-ninth of a Sudoku grid: the numbers from one to nine. This makes solving the cube slightly more difficult than a conventional Rubik’s Cube because each number must be in the right place and the centre cubies must also be in the correct orientation.

A much more interesting Sudoku-variation of the cube was invented two weeks ago by one of my freshmen-grouptheory students and was dubbed the mini-sudokube by him. Here’s the story.

At the end of my grouptheory course I let the students write a paper to check whether they have research potential. This year the assignment was to read through the paper mini-sudokus and groups by Carlos Arcos, Gary Brookfield and Mike Krebs, and do something original with it.

Mini-Sudoku is the baby $2 \times 2 $ version of Sudoku. Below a trivial problem and its solution

Of course most of them took the safe road and explained in more detail a result of the paper. Two of them were more original. One used the mini-sudoku solutions to find solutions for 4×4 sudokus, but the most original contribution came from Ibrahim Belkadi who wanted to count all mini-sudokubes. A mini-sudokube is a cube with a mini-sudoku solution on all 6 of its sides BUT NUMBERS CARRY OVER CUBE-EDGES. That is, if we have as the mini-sudoku given by the central square below on the top-face of the cube, then on the 4 side-faces we have already one row filled in.

The problem then is to find out how many compatible solutions there are. It is pretty easy to see that top- and bottom-faces determine all squares of the cube, but clearly not all choices are permitted. For example, with the above top-face fixed there are exactly 4 solutions. Let ${ a,b } = { 1,4 } $ and ${ c,d } = { 2,3 } $ then these four solutions are given below

Putting one of these solutions (or any other) on a $4 \times 4 $-Rubik cube would make a more interesting puzzle, I think. I’ve excused Ibrahim from having to take examination on condition that he writes down what he can prove on his mini-sudokubes by that time. Looking forward to it!

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the modular group and superpotentials (2)

Last time we have that that one can represent (the conjugacy class of) a finite index subgroup of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ by a Farey symbol or by a dessin or by its fundamental domain. Today we will associate a quiver to it.

For example, the modular group itself is represented by the Farey symbol
[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] or by its dessin (the green circle-edge) or by its fundamental domain which is the region of the upper halfplane bounded by the red and blue vertical boundaries. Both the red and blue boundary consist of TWO edges which are identified with each other and are therefore called a and b. These edges carry a natural orientation given by circling counter-clockwise along the boundary of the marked triangle (or clockwise along the boundary of the upper unmarked triangle having $\infty $ as its third vertex). That is the edge a is oriented from $i $ to $0 $ (or from $i $ to $\infty $) and the edge b is oriented from $0 $ to $\rho $ (or from $\infty $ to $\rho $) and the green edge c (which is an inner edge so carries no identifications) from $\rho $ to $i $. That is, the fundamental region consists of two triangles, glued together along their boundary which is the oriented cycle $\vec{abc} $ consistent with the fact that the compactification of $\mathcal{H}/\Gamma $ is the 2-sphere $S^2 = \mathbb{P}^1_{\mathbb{C}} $. Under this identification the triangle-boundary abc can be seen to circle the equator whereas the top triangle gives the upper half sphere and the lower triangle the lower half sphere. Emphasizing the orientation we can depict the triangle-boundary as the quiver

[tex]\xymatrix{i \ar[rd]_a & & \rho \ar[ll]_c \\ & 0 \ar[ru]_b}[/tex]

embedded in the 2-sphere. Note that quiver is just a fancy name for an oriented graph…

Okay, let’s look at the next case, that of the unique index 2 subgroup $\Gamma_2 $ represented by the Farey symbol [tex]\xymatrix{\infty \ar@{-}[r]_{\bullet} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] or the dessin (the two green edges) or by its fundamental domain consisting of the 4 triangles where again the left and right vertical boundaries are to be identified in parts.

That is we have 6 edges on the 2-sphere $\mathcal{H}/\Gamma_2 = S^2 $ all of them oriented by the above rule. So, for example the lower-right triangle is oriented as $\vec{cfb} $. To see how this oriented graph (the quiver) is embedded in $S^2 $ view the big lower region (cdab) as the under hemisphere and the big upper region (abcd) as the upper hemisphere. So, the two green edges together with a and b are the equator and the remaining two yellow edges form the two parts of a bigcircle connecting the north and south pole. That is, the graph are the cut-lines if we cut the sphere in 4 equal parts. The corresponding quiver-picture is

[tex]\xymatrix{& i \ar@/^/[dd]^f \ar@/_/[dd]_e & \\
\rho^2 \ar[ru]^d & & \rho \ar[lu]_c \\
& 0 \ar[lu]^a \ar[ru]_b &}[/tex]

As a mental check, verify that the index 3 subgroup determined by the Farey symbol [tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\circ} & 1 \ar@{-}[r]_{\circ} & \infty}[/tex] , whose fundamental domain with identifications is given on the left, has as its associated quiver picture

[tex]\xymatrix{& & \rho \ar[lld]_d \ar[ld]^f \ar[rd]^e & \\
i \ar[rrd]_a & i+1 \ar[rd]^b & & \omega \ar[ld]^c \\
& & 0 \ar[uu]^h \ar@/^/[uu]^g \ar@/_/[uu]_i &}[/tex]

whereas the index 3 subgroup determined by the Farey symbol [tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{1} & 1 \ar@{-}[r]_{\circ} & \infty}[/tex], whose fundamental domain with identifications is depicted on the right, has as its associated quiver

[tex]\xymatrix{i \ar[rr]^a \ar[dd]^b & & 1 \ar@/^/[ld]^h \ar@/_/[ld]_i \\
& \rho \ar@/^/[lu]^d \ar@/_/[lu]_e \ar[rd]^f & \\
0 \ar[ru]^g & & i+1 \ar[uu]^c}[/tex]

Next time, we will use these quivers to define superpotentials…

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