Archive for as.numeric()

Le Monde puzzle [#939bis]

Posted in Books, Kids, R with tags , , , , , , , on December 18, 2015 by xi'an

If you remember the previous post, I had two interpretations about Le Monde mathematical puzzle #639:

Find all integers with less than 11 digits that are perfect squares and can be written as a(a+6), a being an integer.

and:

Find all integers with less than 11 digits that are perfect squares and can be written as x concatenated with (x+6), x being an integer.

I got a nice email from Jamie Owen, from Newcastle, Britain, about an R resolution with a clever code, as opposed to mine!

About the second version of the puzzle, Jamie first creates the vector of concatenations:

 
x = 1:1e5
cats = x * (10^floor(log10(x+6) + 1) +1)+ 6

He then made the function perfect more… perfect:

perfect=function(b){
  a=trunc(sqrt(b))
  any((a:(a+1))^2 == b)
  }

(using a function any() I had not seen before, and then got the collection of solutions as

x = 1:1e5
x[sapply(x * (10^floor(log10(x+6) + 1) +1)+ 6,perfect)]
[1] 15 38

which runs about 25 times faster than my R solution! (And he further designed a 100 times faster version…)

Jamie also proposed an R code for solving the first version of that puzzle:

max = 1e10
squares = (1:floor(sqrt(max)))^2
# possible answers to a(a+6)
a = -1e6:1e6
# which squares have solutions
sols = intersect(a*(a + 6), squares)
# what are they?
f = function(x){
power = floor(floor(log10(x))/2)+1
a = -10^power:10^power
sols = c(x,a[a*(a+6) - x == 0])
names(sols) = c("square", "a1", "a2")
sols
}
sapply(sols,f)
## [,1]
## square 16
## a1 -8
## a2 2

which returns again 2 as the unique positive solution (equivalent to -8, if considering relative integers). A great lesson in efficient R programming, thanks Jamie!

Le Monde puzzle [#939]

Posted in Books, Kids, R with tags , , , , , , on December 11, 2015 by xi'an

A Le Monde mathematical puzzle about special integers:

Find all integers with less than 11 digits that are perfect squares and can be written as a(a+6), a being an integer.

Eleven digits being too much for a brute force exploration of the form `for (t in 1:1e11)`…, some preliminary  analysis is needed, but I could not figure out a reason why there is no solution apart from 2… (I checked up to 1e8!)

Since I had “guessed” the above puzzle from the solution published one week later (!), I checked the quality of my guesswork with my friend Jean-Louis Fouley, who gave me the genuine question, based on a different interpretation of a(a+6):

Find all integers with less than 11 digits that are perfect squares and can be written as x concatenated with (x+6), x being an integer.

This is more open to brute-force R exploration (with some help from stack overflow) since x only has five digits at most!

 
perfect=function(b){
  x=FALSE
  a=trunc(sqrt(b))
  for (i in a:(a+1))
    if (i^2==b) x=TRUE
  return(x)}

for (x in 1:(1e6-1))
  if (perfect(
    as.numeric(paste(c(as.numeric(strsplit(as.character(x), "")[[1]]),
    as.numeric(strsplit(as.character(x+6), "")[[1]])),collapse="")))) 
      print(x)

Which returns

[1] 15
[1] 38

and then crashes for x=99994, because

strsplit(as.character(1e+05), "") 

does not return the six digits of 1e+05 but

[[1]]
[1] "1" "e" "+" "0" "5"

instead. Except for this value of x, no other solution is found using this R code. And for x=99994, y=99994100000 is not a perfect square.