### Archive

Archive for the ‘puzzles’ Category

## Jotto (5-letter Mastermind) in the NAO robot

I would like to show how to code the NAO robot to beat us at Jotto (5-letter Mastermind) with python in Choregraphe. I will employ a brute force technique that does not require any knowledge of the English language, the frequency of its letters, or smart combinations of vowels and consonants to try to minimize the number of attempts. It goes like this:

1. Gather all 5-letter words with no repeated letters in a list.
2. Choose a random word from that list—your guess—, and ask it to be scored ala Mastermind.
3. Filter through the list all words that share the same score with your guess; discard the rest.
4. Go back to step 2 and repeat until the target word is found.

Coding this strategy in python requires only four variables:

• whole_dict: the list with all the words
• step = [x for x in whole_dict]: A copy of whole_dict, which is going to be shortened on each step (hence the name). Note that stating step = whole_dict will change the contents of whole_dict when we change the contents of step — not a good idea.
• guess = random.choice(step): A random choice from the list step.
• score: A string containing the two digits we obtain after scoring the guess. The first digit indicates the number of correct letters in the same position as the target word; the second digit indicates the number of correct letters in the wrong position.
• attempts: optional. The number of attempts at guessing words. For quality control purposes.

At this point, I urge the reader to stop reading the post and try to implement this strategy as a simple script. When done, come back to see how it can be coded in the NAO robot.

## Advanced Problem #18

Another fun problem, as requested by Qinfeng Li in MA598R: Measure Theory

Let $f \in L_2(\mathbb{R})$ so that $g(x)=xf(x) \in L_2(\mathbb{R})$ too.

Show that $f \in L_1(\mathbb{R})$ and $\lVert f \rVert_1^2 \leq 8 \lVert f \rVert_2 \lVert g \rVert_2.$

I got this problem by picking some strictly positive value $\lambda$ and breaking the integral $\int_{\mathbb{R}} \lvert f \rvert\, d\mu$ as follows:

$\begin{array}{rl} \displaystyle{\int_{\mathbb{R}} \lvert f \rvert\, d\mu} &= \displaystyle{\int_{\lvert x \rvert \leq \lambda} \lvert f \rvert\, d\mu + \int_{\lvert x \rvert \geq \lambda} \lvert f \rvert\, d\mu} \\ \\ &= \displaystyle{\int_{-\lambda}^{\lambda} \lvert f \rvert\, d\mu + \int_{\lvert x \rvert \geq \lambda} \tfrac{1}{\lvert x \rvert} \lvert g(x) \rvert\, d\mu} \\ \\ &= \displaystyle{\int_\mathbb{R} \lvert f \rvert \cdot \boldsymbol{1}_{\lvert x \rvert \leq \lambda}\, d\mu + \int_{\mathbb{R}} \lvert g \rvert \cdot \big( \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda}\big) \, d\mu} \\ \\ &\leq \lVert f \rVert_2 \cdot \big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2 + \lVert g \rVert_2 \cdot \big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 \end{array}$

Let us examine now the factors $\big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2$ and $\big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2$ above:

$\big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2 = \sqrt{2\lambda}$

$\big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 = \displaystyle{ \sqrt{ 2\int_\lambda^\infty \dfrac{dx}{x^2}} = \sqrt{ \dfrac{2}{\lambda}} }$

We have thus proven that $f \in L_1(\mathbb{R})$ with $\lVert f \rVert_1 \leq \sqrt{2\lambda} \lVert f \rVert_2 + \sqrt{\tfrac{2}{\lambda}} \lVert g \rVert_2.$ At this point, all you have to do is pick $\lambda = \lVert g \rVert_2 / \lVert f \rVert_2$ (provided the denominator is not zero) and you are done.

Categories: Analysis, puzzles, Teaching

## More on Lindenmayer Systems

September 24, 2013 2 comments

We briefly explored Lindenmayer systems (or L-systems) in an old post: Toying with Basic Fractals. We quickly reviewed this method for creation of an approximation to fractals, and displayed an example (the Koch snowflake) based on tikz libraries.

I would like to show a few more examples of beautiful curves generated with this technique, together with their generating axiom, rules and parameters. Feel free to click on each of the images below to download a larger version.

Note that any coding language with plotting capabilities should be able to tackle this project. I used once again tikz for $\text{\LaTeX}$, but this time with the tikzlibrary lindenmayersystems.

 name : Dragon Curve axiom : X order : 11 step : 5pt angle : 90 rules : X -> X+YF+ Y -> -FX-Y  name : Gosper Space-filling Curve axiom : XF order : 5 step : 2pt angle : 60 rules : XF -> XF+YF++YF-XF--XFXF-YF+ YF -> -XF+YFYF++YF+XF--XF-YF  name : Quadric Koch Island axiom : F+F+F+F order : 4 step : 1pt angle : 90 rules : F -> F+F-F-FF+F+F-F  name : Sierpinski Arrowhead axiom : F order : 8 step : 3.5pt angle : 60 rules : G -> F+G+F F -> G-F-G  name : ? axiom : F+F+F+F order : 4 step : 2pt angle : 90 rules : F -> FF+F+F+F+F+F-F  name : ? axiom : F+F+F+F order : 4 step : 3pt angle : 90 rules : F -> FF+F+F+F+FF 

Would you like to experiment a little with axioms, rules and parameters, and obtain some new pleasant curves with this method? If the mathematical properties of the fractal that they approximate are interesting enough, I bet you could attach your name to them. Like the astronomer that finds through her telescope a new object in the sky, or the zoologist that discover a new species of spider in the forest.

## Sympy should suffice

I have just received a copy of Instant SymPy Starter, by Ronan Lamy—a no-nonsense guide to the main properties of SymPy, the Python library for symbolic mathematics. This short monograph packs everything you should need, with neat examples included, in about 50 pages. Well-worth its money.

To celebrate, I would like to pose a few coding challenges on the use of this library, based on a fun geometric puzzle from cut-the-knot: Rhombus in Circles

Segments $\overline{AB}$ and $\overline{CD}$ are equal. Lines $AB$ and $CD$ intersect at $M.$ Form four circumcircles: $(E)=(ACM), (F)=(ADM), (G)=(BDM), (H)=(BCM).$ Prove that the circumcenters $E, F, G, H$ form a rhombus, with $\angle EFG = \angle AMC.$

Note that if this construction works, it must do so independently of translations, rotations and dilations. We may then assume that $M$ is the origin, that the segments have length one, $A=(2,0), B=(1,0),$ and that for some parameters $a>0, \theta \in (0, \pi),$ it is $C=(a+1) (\cos \theta, \sin\theta), D=a (\cos\theta, \sin\theta).$ We let SymPy take care of the computation of circumcenters:

import sympy
from sympy import *

# Point definitions
M=Point(0,0)
A=Point(2,0)
B=Point(1,0)
a,theta=symbols('a,theta',real=True,positive=True)
C=Point((a+1)*cos(theta),(a+1)*sin(theta))
D=Point(a*cos(theta),a*sin(theta))

#Circumcenters
E=Triangle(A,C,M).circumcenter
F=Triangle(A,D,M).circumcenter
G=Triangle(B,D,M).circumcenter
H=Triangle(B,C,M).circumcenter


Finding that the alternate angles are equal in the quadrilateral $EFGH$ is pretty straightforward:

In [11]: P=Polygon(E,F,G,H)

In [12]: P.angles[E]==P.angles[G]
Out[12]: True

In [13]: P.angles[F]==P.angles[H]
Out[13]: True


To prove it a rhombus, the two sides that coincide on each angle must be equal. This presents us with the first challenge: Note for example that if we naively ask SymPy whether the triangle $\triangle EFG$ is equilateral, we get a False statement:

In [14]: Triangle(E,F,G).is_equilateral()
Out[14]: False

In [15]: F.distance(E)
Out[15]: Abs((a/2 - cos(theta))/sin(theta) - (a - 2*cos(theta) + 1)/(2*sin(theta)))

In [16]: F.distance(G)
Out[16]: sqrt(((a/2 - cos(theta))/sin(theta) - (a - cos(theta))/(2*sin(theta)))**2 + 1/4)


Part of the reason is that we have not indicated anywhere that the parameter theta is to be strictly bounded above by $\pi$ (we did indicate that it must be strictly positive). The other reason is that SymPy does not handle identities well, unless the expressions to be evaluated are perfectly simplified. For example, if we trust the routines of simplification of trigonometric expressions alone, we will not be able to resolve this problem with this technique:

In [17]: trigsimp(F.distance(E)-F.distance(G),deep=True)==0
Out[17]: False


Finding that $\angle EFG = \angle AMC$ with SymPy is not that easy either. This is the second challenge.

How would the reader resolve this situation?

Categories: Geometry, puzzles, sage, Teaching

## Project Euler with Julia

Disclaimer: Project Euler discourages the posting of solutions to their problems, to avoid spoilers. The three solutions I have presented in my blog are to the three first problems (the easiest and most popular), as a means to advertise and encourage my readers to “push the envelope” and go beyond brute-force solutions.

Just for fun, and as a means to learn Julia, I will be attempting problems from the Project Euler coding exclusively in that promising computer language.

The first problem is very simple:

Multiples of 3 and 5
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multip les is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

One easy-to-code way is by dealing with arrays:

julia> sum( filter( x->((x%3==0) | (x%5==0)), [1:999] ) )
233168


A much better way is to obtain it via summation formulas: Note that the sum of all multiples of three between one and 999 is given by

$3+6+\dotsb+999 = 3 (1+2+\dotsb+333) = 3 \frac{333\cdot 334}{2} = 166833.$

Similarly, the sum of all multiples of five between one and 995 is given by

$5+10+\dotsb+995 = 5 (1+2+\dotsb+199) = 5 \frac{199\cdot 200}{2} = 99500.$

Finally, we need to subtract the sum of the multiples of both three and five, since they have been counted twice. This sum is

$15+30+\dotsb+990 = 15 (1+2+\dotsb+66)= 15 \frac{66\cdot 67}{2}= 33165.$

The final sum is then, $166833+99500-33165=233168.$

Categories: puzzles

## Seked

If a pyramid is 250 cubits high and the side of its base 360 cubits long, what is its seked?

Take half of 360; it makes 180. Multiply 250 so as to get 180; it makes $1/2 1/5 1/50$ of a cubit. A cubit is 7 palms. Multiply 7 by $1/2 1/5 1/50:$

$\begin{array}{l|lll} 1&7&&\\ 1/2&3&1/2& \\ 1/5 &1&1/3&1/15\\ 1/50&&1/10&1/25 \end{array}$

The seked is $5 \tfrac{1}{25}$ palms [that is, $(3+1/2) + (1+1/3+1/15) + (1/10+1/25)$].

A’h-mose. The Rhind Papyrus. 33 AD

The image above presents one of the problems included in the Rhind Papyrus, written by the scribe Ahmes (A’h-mose) circa 33 AD. This is a description of all the hieroglyphs, as translated by August Eisenlohr:

The question for the reader, after going carefully over the English translation is: What does seked mean?

Categories: History, puzzles