Dork Scratchings

    I was given a new version of Dobble for Xmas called Dobble Connect.  Dobble consists of stack of cards with pictures on it where each pair of cards has one match, and each pair of symbols appears on one card.  This is a new version which has 91 symbols in total, 10 symbols per card, and 91 cards.  The fingerprint of the original Dobble was (57 symbols, 8 symbols, 57 cards). An additional innovation in Dobble Connect is that the cards are hexagonal and grouped into a colour per player.  Players build a tiling on the tabletop, placing cards next to each other as soon as they spot a matching symbol between their top card and one of the cards already placed.  The first player to get four of their own colour in a row wins. How is it possible for each pair of cards to have one and only one symbol in common, and for each pair of symbols to appear on one and only one card?  Does that have something to do with the strange number 91?  The answer is "yes" and it turns out this is o

Here's a nice theorem by Gauss The (2D) shoelace theorem Suppose $\mathbf{x_1},\mathbf{x_2},...,\mathbf{x_n}$ are ordered vertices of a polygon.  Let $x_i, y_i$ be the coordinates of $\mathbf{x_i}$ and define $\mathbf{x_{n+1}}$ to be $\mathbf{x_1}$.   Then the area of the polygon is plus or minus $$ \frac{1}{2}\sum_i{x_iy_{i+1} - x_{i+1}y_i} $$ This formula gives rise to the name of the theorem as shown by the illustration below Proof We're going start off by proving the simple case where the polygon is a triangle.  The first thing to notice is that if you take any two adjacent vertices, .e.g. $\mathbf{x_1}$ and $\mathbf{x_3}$, they form a triangle with the origin.  This has an area equal to half of the parallelpiped formed by the two vertices (treated as vectors from the origin). And the area of the parallelpiped is plus or minus $det(\mathbf{x_1}\vert \mathbf{x_3})$, with the sign being positive if the columns taken in order produce an anti-clockwise motion around the origin.

 Photo taken in the Outer Hebrides by my brother-in-law (The doctor one, not the Flat Earth one) This photo was taken a few days ago by my brother-in-law Alex from the Outer Hebrides where he's off galavanting at the moment. You can clearly see that half the ship is missing indicating either a) it's sinking b) the Earth is round Going with (b) for the time being we can actually come up with a pretty good guesstimate for the size of the Earth just from this picture. Let's guess that the ship is 10km away. (That's not likely to be accurate but it's certainly more than 1km - the width of Lake Windermere - and less than 100km - the distance from Portsmouth to Le Havre). So $l=10000$. Now let's assume that the bottom 10m of the ship is missing from view (again, not likely to be accurate but good enough for an order of magnitude calculation). So $h=10$. Now chuck it into this equation which is easily worked out with a bit of trig $$ r=\frac{l^2}{2h} $$ and, hey presto

Why does $\sqrt{2} \ne \frac{7}{5}$?  Well, if it did, we could draw a square of side 5 and diagonal 7 Not in proportion! Then, by removing 5 from the diagonal we could create a second square of side 2 and diagonal 3, which would mean $\sqrt{2} = \frac{3}{2}$, which is a  different fraction to $\frac{7}{5}$. Actually, we can easily generalize this argument to show that if $$ \sqrt{2} = \frac{a}{b} $$ for some whole numbers $a$ and $b$, then $$ \sqrt{2} = \frac{c}{d} $$ for some smaller whole numbers $c \lt a$ and $d \lt b$.  Repeating this argument over and over leads us to the conclusion that $\sqrt{2}$ is a whole number itself!  This is clearly incorrect and leads us to the conclusion that we can't in fact write $\sqrt{2}$ as a fraction. This is not a particularly modern way to prove the existence of irrational (non-fraction) numbers, but it is - supposedly - how the ancient Greeks originally did it!  Unfortunately the person who discovered this fact, Hippasus of Metapontum , fo

Theorem of the week: The Hairy Ball Theorem This says you can't comb a hairy ball without introducing discontinuities such as partings or whorls, unless there's a bald spot.  (There's a more mathematical statement below under the heading "theorem".)  The proof is from An Extremely Short Proof of the Hairy Ball Theorem, by P McGrath , but I've put it into my own words, completely removed all maths notation, and added pictures to make it as accessible as possible.  In addition to being extremely short, it's extremely elegant, and somewhat reminiscent of the Ham Sandwich Theorem . Theorem It is not possible to impose a continuous vector field onto a sphere, such that the vectors are all tangential to the surface, unless the field is zero somewhere Proof Let's assume the sphere does have a continuous, tangential, everywhere non-zero vector field, and attempt to derive a contradiction. Draw a small circle around a point p.  Do one lap around

Is Our Universe Finite? A while ago I drew the pictures above to try to understand current ideas about the size of the universe. The diagrams are based on some pictures I saw in the book " Our Mathematical Universe ". The diagrams show two dimensional slices of four dimensional spacetime. The blue stuff is "inflationary material" which expands at an enormous rate. The current theory of inflation states that universes like ours form as bubbles in the inflationary material as some of the inflationary material changes phase and "evaporates" out as non-inflationary material. An important point is that the sides of this bubble are moving away from each other way too fast for anything - even light - to travel from one side to the other.  The 1st diagram illustrates the point that in this model there is room for more, far more, than one universe. The yellow region in the 2nd and 3rd diagrams is what is known as a light cone. The point in the middle

Credit: Teodomiro, wikipedia.org The ancient Greeks were obsessed with ruler and compass constructions, and they had a lot of successes.  They bisected angles, constructed pentagons, and much more.  One thing that eluded them was finding a general method for trisecting an angle.  Although they could trisect certain angles, e.g. $90^{\circ}$, they tried in vain to come up with a general recipe given just three starting points. It turns out that trisecting an angle is in general impossible, but the proof that this is the case had to wait for some mathematics developed by Galois.  In this post I'll give a proof using polynomials, fields , and vector spaces . Most impossibility proofs work the same way.  First you identify some property which remains invariant with each step, and then you show that the property would need to change to get to the final state.  In this case the property is very abstract.... The Invariant Property Let $\mathbb{F}_0$ be the minimal subfield of

Our holiday cottage in Lindos has this wonderful mosaic in the floor. It's made from black and white pebbles each about 2cm in diameter which gently massage your feet as you walk over them. In the doorway the same pebbles write out the date 1908. Looking at a design like this I can't help trying to reverse engineer it.  The first thing you notice is that the edges of the white quadrilaterals share a straight line with the edges of the black ones.  In fact all the shapes other than the concentric circles are formed by intersecting chords of the outer circle. The next observation is that these chords come in parallel pairs which are tangent to the inner circle on opposite sides.  These observations are enough to generate a recipe: Draw the outer circle Mark every 20 degrees to split into 18 equal parts Draw a chord between point n and point n+8 Repeat for n = 0...8 Draw the inner circle I had a go at this using Handwrite Pro on my phone. I ended up using a slig

I discovered the card game Dobble a few days ago whilst visiting relatives in Germany.  There are 57 cards with 8 symbols on each.  To play the most basic version of the game you deal the cards face down between the players such that one card remains (if there are an odd number of players you may need to hold back more than 1 card).  Then you turn face up the remaining card (or one of the remaining cards) and each person turns their top card face up. The first person to spot a symbol on their own card that matches one on the left over card shouts out the name of that symbol (e.g. "car!") and gets to move their card to the top of the shared stack; the others move their top card to the bottom of their own stack.  The winner is the person who gets rid of their cards first. At first sight this doesn't seem very mathematical.  However, after a while I noticed something odd about the cards: every card shares exactly one symbol with each other card in the pack.  I thou

Connecting Geometry and Algebra Matrix determinants have the following strange definition that seems to have been pulled out of thin air: $$ det(A) = \sum_{\sigma}{sign(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)}} $$ where $A = (a_{ij})$ is a real $n\times n$ matrix $\sigma$ ranges over all permutations of $\{1,...,n\}$ $sign(\sigma)$ is $+1$ if $\sigma$ is a product of an even number of transpositions and $-1$ otherwise$^\dagger$ However, in the geometric world the definition is far more intuitive: The determinant of A is the volume of the parallelepiped formed by its columns, multiplied by minus one if these have the opposite handedness to the unit vectors. Why are these two definitions the same?  To begin to answer this we need to first define elementary matrices and then show that every square matrix can be written as a product of these. Definition The elementary matrices are $E_{i, j}$ for $1 \le i,j \le n$ $E_{i,\lambda}$ for every real $\lambda$ and $1 \l

Theorem of the week This week's theorem of the week is the fundamental theorem of algebra , and the picture is the proof! Theorem Every non degree-zero polynomial $p(z) = a_nz^n + ...+ a_1z +  a_0$ has a root in $\mathbb{C}$. Picture proof To see how the picture proves this, write $z$ as $Re^{i\theta}$, then for all $k$ $$ z^k = R^ke^{k i \theta} $$ So for sufficiently large $R$ the $a_nz^n$ term dwarfs all the others and so the image of $\{z\in\mathbb{C}: \lvert z \rvert = R\}$ must go around the origin $n$ times, like the rubber band in the photo.  But when $R= 0$ the image is just $\{a_0\}$ which goes around the origin zero times.  So, for some $0 < r < R$ the image of $\{z\in\mathbb{C}: \lvert z \rvert = r\}$ must cross the origin.  QED. Less handwavy proof In order to obtain a contradiction assume $p(z)$ has no zeros.  Then $\frac{z^{n-1}}{p(z)}$ is everywhere differentiable, which in turn means that its closed loop integrals are zero$^{(\dagger)}$.

A proof that the net force due to pressure on a (fully or partially) submerged object is equal to the weight of the water displaced. The facts/approximations used were pressure is isotropic density is constant gravitational force is constant And the method of proof was to apply the Divergence Theorem. POSTSCRIPT I was just wondering why pressure is isotropic when I found this post and  I realised that the scratchings above actually prove it!  Imagine a blob of water submerged in water: the above proof shows that the net force on it is zero, but only if pressure is isotropic !

I heard somewhere or another the claim that soap film forms a "surface of zero mean curvature".  I wasn't sure exactly what that meant until I read this book which gave me the tools to understand what that means, and prove it.  It turns out to be a simple consequence of Gauss's Divergence theorem.  In tensor notation Gauss's Divergence theorem states $$ \int_{\Omega}\nabla_iT^idV = \int_{\partial{\Omega}}T_iN^idA $$ where $\Omega$ is some volume of space $\nabla_i$ is the covariant derivative along the $i$th coordinate $z^i$ $\delta{\Omega}$ is the surface of the volume $T^i$ is any single index tensor defined over the whole volume $N^i$ is the unit vector normal to the surface in contravariant form  This works in any number of dimensions, so if you take surface embedded in 3 dimensional euclidean space $\vec{z} = \vec{z}(s^1,s^2)$ and cut it, then the theorem tells us $$ \int_{S}\nabla_{\alpha}T^{\alpha}dA = \int_{\partial{S}}T_{\alpha}n^{\alph

In More Mathematical Puzzles and Diversions , Martin Gardner makes a passing reference to the Ham Sandwich Theorem.  It goes like this Any 3 shapes in 3 dimensional space can be simultaneously bisected by a single plane So imagine you have two roughly cut pieces of bread and a slice of ham, then you can always cut the sandwich in half such that each half has exactly half of each piece of bread and half of the ham, no matter how roughly strewn the pieces are. According to Gardner the generalized version has been proved by Tukey and Stone: any n shapes in $\mathbb{R}^n$ can be simultaneously halved by a single $n-1$ dimensional hyperplane.  But I thought I'd have a go at proving it myself in the 3D case, just for kicks. First observe that there are at least enough degrees of freedom to make it not impossible .  A plane (other than one going through the origin) can be described by the equation $k_xx+k_yy+k_zz = 1$ for some $\boldsymbol{k} \neq \boldsymbol{0}$ so there are 3 p

Photo taken by me at 18:00 BST May 20, 2018 I was out and about in Cambridge and I saw a modern sundial on the side of a building in Tennis Court road .  It occurred to me then that I should be able to build a sundial which tells you both the time - in GMT - and the date.  (Although you do need to know whether it is before or after the summer solstice!).  In fact all you need is to know one out of compass bearing, time, date and you should be able to work out the other two! To run it you need to do the following copy the text into sundial.py and chmod +x sundial.py install pre-requisite packages: sudo apt install python-numpy python-matplotlib run it:  ./sundial.py The result is a printout like the one shown.  If you want to adapt the picture for your locale just edit the parameters passed to plot_fixed_lat_long() . When I printed out my first sundial I was surprised to discover that the trajectory of the shadow is straight on the equinoxes.  After a day of pondering