jacquesattack

gen.fibmorse = function(start = c(0), depth = 1){
  # check inputs
  if(class(start) != "numeric"){
    print("Input not a numeric vector!")
    stop();
  }
  for(val in start){
    if(!(val %in% c(0,1)) ){
      print(paste("Encountered non 0/1 value in vector:",val))
      stop()

jacquesattack

mse = function(v1,v2){
  se = (v2 - v1)^2
  mse = mean(se)
  return(mse)
}

# See http://en.wikipedia.org/wiki/Forecast_skill for info on forecasting skill. 
# One thing I don't like about the vanilla method is that the domain is (-Inf,1] which is kind of hard to understand.
# The Rossignol method, or Rossignol's Modified Forecasting Skill (RMFS) changes this to be [-1,1] which is easier to understand. Use it by setting type="rossignol".
fskill = function(data,type="vanilla"){

jacquesattack

# Calibrated p-values
# -------------------
# See https://stat.duke.edu/courses/Spring10/sta122/Labs/Lab6.pdf
# and http://www.uv.es/sestio/TechRep/tr14-03.pdf
alpha.p = function(p){
  val = 1 + (-exp(1)*p*log(p))^-1
  return(val^-1)
}

jacquesattack

prev.value = function(str){
  if(str == "01") return("0")
  if(str %in% c("0","10")) return("1")
  return(NULL)
}



tokenize = function(str,prev.tokens=c("")){
  if(str == "") return("");

jacquesattack

# simulate pvalue fallacy
# See https://stat.duke.edu/courses/Spring10/sta122/Labs/Lab6.pdf
# and https://stat.duke.edu/~berger/applet2/applet.pdf
sim.pval = function(pi = 0.5, theta.input = 1, sigma.input = 1, n = 100, max = 1000){
  alpha = 0
  beta = 0
  while(alpha + beta < max){
    rand = runif(1)
    if(rand > pi){
      theta = 0

jacquesattack

f1 = function(n){
  x = double(n)
  for(i in 1:n) x[i] = runif(1)
  return(x)
}

f2 = function(n){
  x = c()
  for(i in 1:n) x = c(x,runif(1))

jacquesattack

curve(dnorm(x,0,1),-5,5)
curve(dnorm(x,1,1),-5,5,add=TRUE)
arrows(0,0.2,1,0.2)
abline(v=0,lty=2)
abline(v=1,lty=2)

jacquesattack

## solve stauf painting puzzle
get.move = function(val){
 move = c(0,0,0,0,0,0,0,0,0)
 if(val == 1) move = c(1,1,0,1,1,0,0,0,0)
 if(val == 2) move = c(1,1,1,0,0,0,0,0,0)
 if(val == 3) move = c(0,1,1,0,1,1,0,0,0)
 if(val == 4) move = c(1,0,0,1,0,0,1,0,0)
 if(val == 5) move = c(0,1,0,1,1,1,0,1,0)
 if(val == 6) move = c(0,0,1,0,0,1,0,0,1)
 if(val == 7) move = c(0,0,0,1,1,0,1,1,0)

jacquesattack

# https://www.quantamagazine.org/20150722-solution-information-from-randomness/

n = 1000
vals = runif(n) # values shown by opened hand
g = runif(n) # g values
highest = sample(c(TRUE,FALSE),n,replace=TRUE) # was the value in the opened hand the higher value of the two hands?
# one of two things happens
# 1) g is less than the value you chose, you stay, and you are right
# 2) g is greater than the value you chose, you switched, and you were right
# other cases are failures (add 0 to s) so don't care about them since they are recoverable via n-s

jacquesattack

f = function(expr){
  x = c(13,26,2,4,6,3,9,12,8,10,5,15,18,14,7,21,24,16)
  failure = FALSE
  for(i in seq_along(x)){
    result = eval(expr)
    if(length(result) != 1) next
    print(sprintf("%s, %s, %s",i,x[i],result))
    if(result != x[i]){
      failure = TRUE
      break