Monthly Archives: June 2014

The Goldbach’s Comet

Every even integer greater than 2 can be expressed as the sum of two primes (Christian Goldbach, 1742)

The point cloud known as Goldbach’s Comet represents the amount of different ways (y axis) an even number (x axis) can be writen as sum of two prime numbers. In this plot, x axis is between 2 and 50.000 (stars are just ornaments):

Goldbach

Mathematicians are still waiting for a proof of this conjecture. This is the code for drawing this plot:

require(schoolmath)
require(utils)
library(plyr)
n=50000
data(primlist)
primes=as.data.frame(t(combn(primlist[primlist>2 & primlist<n-2], 2)))
primes$V3=primes$V1+primes$V2
primes2=count(primes, "V3")
primes2=primes2[primes2$V3<=n,]
stars=cbind(runif(50, min=-n*0.05, max=n), runif(200, min=-n*0.001, max=max(primes2$freq)))
plot.new()
par(mai = rep(0, 4), bg = "gray12")
plot(NA,type="n", xlim=c(-n*0.05,n), xaxs="i", ylim=c(-n*0.001,max(primes2$freq)))
points(stars, col = "blue4", cex=.7, pch=16)
points(stars, col = "blue", cex=.3, pch=16)
points(stars, col = "gray75", cex=.1, pch=16)
apply(primes2, 1, function(x) points(x=x[1],y=x[2], col = if (runif(1)&gt;x[1]/n) {"white"} else {sample(colours(),1)}, cex=.1, pch=16))

The Ikeda’s Galaxy

Chaos is the score upon which reality is written (Henry Miller)

Nonlinear dynamical systems are an enormous seam of amazing images. The Ikeda Map is an example of strange attractor which represents the movement of particles under the rules of certain differential equations.

I have drawn the trajectories followed by of 200 particles under the 2D-Ikeda Map with the same tecnique I used in this previous post, resulting this nice galaxy:

ikeda0.918

Wold you like to create your own galaxies? Here you have the code:

u=0.918 #Parameter between 0 and 1
n=200 #Number of particles
m=40 #Number of iterations
ikeda=data.frame(it=1,x1=runif(n, min = -40, max = 40), y1=runif(n, min = -40, max = 40))
ikeda$x2=1+u*(ikeda$x1*cos(0.4-6/(1+ikeda$x1^2+ikeda$y1^2))-ikeda$y1*sin(0.4-6/(1+ikeda$x1^2+ikeda$y1^2)))
ikeda$y2=  u*(ikeda$x1*sin(0.4-6/(1+ikeda$x1^2+ikeda$y1^2))+ikeda$y1*cos(0.4-6/(1+ikeda$x1^2+ikeda$y1^2)))
for (k in 1:m)
{
df=as.data.frame(cbind(rep(k+1,n),
ikeda[ikeda$it==k,]$x2,
ikeda[ikeda$it==k,]$y2,
1+u*(ikeda[ikeda$it==k,]$x2*cos(0.4-6/(1+ikeda[ikeda$it==k,]$x2^2+ikeda[ikeda$it==k,]$y2^2))-ikeda[ikeda$it==k,]$y2*sin(0.4-6/(1+ikeda[ikeda$it==k,]$x2^2+ikeda[ikeda$it==k,]$y2^2))),
u*(ikeda[ikeda$it==k,]$x2*sin(0.4-6/(1+ikeda[ikeda$it==k,]$x2^2+ikeda[ikeda$it==k,]$y2^2))+ikeda[ikeda$it==k,]$y2*cos(0.4-6/(1+ikeda[ikeda$it==k,]$x2^2+ikeda[ikeda$it==k,]$y2^2)))))
names(df)=names(ikeda)
ikeda=rbind(df, ikeda)
}
plot.new()
par(mai = rep(0, 4), bg = "gray12")
plot(c(0,0),type="n", xlim=c(-35, 35), ylim=c(-35,35))
apply(ikeda, 1, function(x) lines(x=c(x[2],x[4]), y=c(x[3],x[5]), col = paste("gray", as.character(min(round(jitter(x[1]*80/(m-1)+(20*m-100)/(m-1), amount=5)), 100)), sep = ""), lwd=0.1))

The Gilbreath’s Conjecture

317 is a prime, not because we think so, or because our minds are shaped in one way rather than another, but because it is so, because mathematical reality is built that way (G.H. Hardy)

In 1958, the mathematician and magician Norman L. Gilbreath presented a disconcerting hypothesis conceived in the back of a napkin. Gilbreath wrote first prime numbers in a row. In the next rows, he wrote the difference  in absolute value of consecutive values of previous row. Beginning with first 20 primes in the first row, he obtained something like this:Gilbreath01

The conjecture is easy: except for the first one, all elements of the first column are 1. So far, no one has demonstrated this hypothesis. In fact, according to mathematician Richard Guy, it seems unlikely that we will see a demonstration of Gilbreath’s conjecture in the near future, though probably this conjecture is true.

In the previous chart, I coloured zeros in white, ones in violet and rest of numbers in gold. The conjecture says that except for the first element, the first column is entirely violet. Following you can see the coloured chart for first 20, 40, 60 and 80 prime numbers:

Gilbreath02

It is nice how zeros create triangular patterns similar to patterns created by cellular automata. This is the chart for first 200 primes:Gilbreath03

How much time will it take to demonstrate this simple conjecture? Who knows. Meanwhile, you can draw triangles with this code:

library(ggplot2)
create.gilbreath=function(n)
{
require(reshape)
require(schoolmath)
data(primlist)
gilbreath=t(matrix(primlist[2:(n+1)], n, n))
for (i in 2:n) {gilbreath[i,]=c(eval(parse(text=paste(c(paste("abs(diff(",collapse=""), "gilbreath[i-1,]", paste("))",collapse="")),collapse=""))),rep(NA,1))}
na.omit(melt(t(gilbreath)))
}
opt=theme(legend.position="none",
panel.background = element_blank(),
panel.grid = element_blank(),
axis.ticks=element_blank(),
axis.title=element_blank(),
axis.text =element_blank())
gilbreath=create.gilbreath(20)
gilbreath$value1=cut(gilbreath$value, breaks=c(-Inf,1:2,Inf), right = FALSE)
ggplot(gilbreath, aes(x=X1, y=X2)) +
geom_tile(aes(fill = value1), colour="grey") +
scale_fill_manual(values = c("white", "darkviolet", "gold"))+
geom_text(label=gilbreath$value, size=8)+
scale_y_reverse()+opt

The Three Little Pigs

Jesse, you asked me if I was in the meth business or the money business. Neither. I’m in the empire business (Walter White in Breaking Bad)

The game of pig has simple rules but complex strategies. It was described for the first time in 1945  by a magician called John Scarne. Playing the pig game is easy: each turn, a player repeatedly rolls a die until either a 1 is rolled or the player decides to hold:

  • If the player rolls a 1, they score nothing and it becomes the next player’s turn
  • If the player rolls any other number, it is added to their turn total and the player’s turn continues
  • If a player chooses to hold, their turn total is added to their score, and it becomes the next player’s turn

The first player who reach at least 100 points is the winner. For example: you obtain a 3 and then decide to roll again, obtaining a 1. Your score is zero in this turn. Next player gets the sequence 3-4-6 and decides to hold, obtaining a score of 13 points in this turn.

Despite its simplicity, the pig game has a very complex and against-intuition optimal strategy. It was calculated in 2004 by Todd W. Neller and Clifton Presser from Gettysburg College of Pennsilvania with the help of computers.

To illustrate the game, I simulated three players (pigs) playing the pig game with three different strategies:

  • The Coward pig, who only rolls the die a small number of times in every turn
  • The Risky pig, who rolls the die a more times than the coward one
  • The Ambitious pig, who tries to obtain in every turn more points than two others

I simulated several scenarios.

  • Some favorable scenarios for Coward pig:

pigs2Coward pigs4Coward

In first scenario, the Coward pig rolls the die between 1 and 5 times each round and wins if the Risky pig asumes an excessive level of risk (rolling each time between 10 and 15 times). Trying to obtain more than the Coward is a bad option for the Ambitious pig. Simulating this scenario 100 times gives victory to Coward a 51% of times (25% to Risky and 24% to Ambitious).

Second scenario puts closer Coward and Risky pigs (first one rolls the die between 4 and 7 times  each round and second one between 6 and 9 times). Coward wins 54% of times (34% Risky and only 12% Ambitious).

Being coward seems to be a good strategy when you play against a reckless or when you are just a bit more conservative than a Risky one.

  • Some favorable scenarios for Risky pig:

pigs5RiskyPigs1Risky

Rolling the die between 4 and 6 times each round seems to be a good option, even more when you are playing against a extremely conservative player who rolls no more than 3 times each time. Simulating 100 times these previous scenarios gives victory to Risky pig a 58% of times in first the case in which Coward rolls allways 1 and Risky 6 times each round (0% for Coward and only 42% form Ambitious) and 66% of times in the second one (only 5% to Coward and 29% to Ambitious).

Being Risky is a good strategy when you play against a chicken.

  • Some favorable scenarios for Ambitious pig:

pigs3Ambitious pigs6ambitious

The Ambitious pig wins when two others turn into extremely coward and risky pigs as can be seen in the first scenario in which Ambitious wins 65% of the times (31% for Coward and 4% for Risky). Ambitious pig also wins when two others get closer and hit the die a small number of times (2 rolls the Coward and 4 rolls the Risky). In this scenario the Ambitious wins 58% of times (5% for Coward and 37% for Risky). By the way, these two scenarios sound very unreal.

Being ambitious seems to be dangerous but works well when you play against a crazy and a chicken or against very conservative players.

From my point of view, this is a good example to experiment with simulations, game strategies and xkcd style graphics.

The code:

require(ggplot2)
require(extrafont)
#Number of hits for Coward
CowardLower=2
CowardUpper=2
#Number of hits for Risky
RiskyLower=4
RiskyUpper=4
game=data.frame(ROUND=0, part.p1=0, part.p2=0, part.p3=0, Coward=0, Risky=0, Ambitious=0)
while(max(game$Coward)<100 & max(game$Risky)<100 & max(game$Ambitious)<100)
{
#Coward Little Pig
p1=sample(1:6,sample(CowardLower:CowardUpper,1), replace=TRUE)
s1=min(min(p1-1),1)*sum(p1)
#Risky Little Pig
p2=sample(1:6,sample(RiskyLower:RiskyUpper,1), replace=TRUE)
s2=min(min(p2-1),1)*sum(p2)
#Ambitious Little Pig
s3=0
repeat {
p3=sample(1:6,1)
s3=(p3+s3)*min(min(p3-1),1)
if (p3==1|s3>max(s1,s2)) break
}
game[nrow(game)+1,]=c(max(game$ROUND)+1,s1,s2,s3,max(game$Coward)+s1,max(game$Risky)+s2,max(game$Ambitious)+s3)
}
opts=theme(
panel.background = element_rect(fill="darkolivegreen1"),
panel.border = element_rect(colour="black", fill=NA),
axis.line = element_line(size = 0.5, colour = "black"),
axis.ticks = element_line(colour="black"),
panel.grid.major = element_line(colour="white", linetype = 1),
panel.grid.minor = element_blank(),
axis.text.y = element_text(colour="black"),
axis.text.x = element_text(colour="black"),
text = element_text(size=25, family="xkcd"),
legend.key = element_blank(),
legend.position = c(.2,.75),
legend.background = element_blank(),
plot.title = element_text(size = 50)
)
ggplot(game, mapping=aes(x=game$ROUND, y=game$Coward)) +
geom_line(color="red", size=1.5) +
geom_line(aes(x=game$ROUND, y=game$Risky), color="blue", size=1.5) +
geom_line(aes(x=game$ROUND, y=game$Ambitious), color="green4", size=1.5) +
geom_point(aes(x=game$ROUND, y=game$Coward, colour="c1"), size=5.5) +
geom_point(aes(x=game$ROUND, y=game$Risky, colour="c2"), size=5.5) +
geom_point(aes(x=game$ROUND, y=game$Ambitious, colour="c3"), size=5.5) +
ggtitle("THE THREE LITTLE PIGS") +
xlab("ROUND") + ylab("SCORING") +
geom_text(aes(max(game$ROUND), max(max(game$Coward, game$Risky, game$Ambitious)), hjust=1.2, family="xkcd", label="WINNER!"), size=10)+
geom_hline(yintercept=100, linetype=2, size=1)+
scale_y_continuous(breaks=seq(0, max(max(game$Coward, game$Risky, game$Ambitious))+10, 10))+
scale_x_continuous(breaks=seq(0, max(game$ROUND), 1))+
scale_colour_manual("",
labels = c(paste("Coward: ", CowardLower, "-", CowardUpper, " hits", sep = ""), paste("Risky: ", RiskyLower, "-", RiskyUpper, " hits", sep = ""), "Ambitious"),
breaks = c("c1", "c2", "c3"),
values = c("red", "blue", "green4"))+ opts

floweR

It is the time you have wasted for your rose that makes your rose so important (Antoine de Saint-Exupéry, The Little Prince)

Yesterday I found a package called circular and I could not suppress to do this:

rose

Make your own floweRs:

require(circular)
for (i in 1:8)
{
  rose.diag(circular(runif(5000, 0, 2*pi)), 
            bins = (25-round(abs(jitter(0, amount=2*i)))), 
            axes=FALSE,
            border=rgb(255,50+204*((i-1)/8),50+204*((i-1)/8), alpha=255, max=255),
            ticks = FALSE,
            col=rgb(250,0+204*((i-1)/8),0+204*((i-1)/8), alpha=255, max=255), 
            control.circle=circle.control(lty=0), 
            shrink=0.22+(i-1)*(2-0.22)/8)
  par(new=TRUE)  
}
for (i in 1:150)
  {
  q = runif(1)*pi*2
  r = sqrt(runif(1))
  x = (0.18*r)* cos(q)
  y = (0.18*r)* sin(q)
  points(x, y, 
         col = rgb(255 ,sample(80:120, 1), 0, alpha= 95, max=255),
         bg = rgb(255 ,sample(100:200, 1), 0, alpha= 95, max=255), pch=21, cex=2)
  }