Number spiral diagonals

We can make squares of numbers, by starting with 1, and add numbers in a spiral.

It is easier to show than descripe:

square <- matrix(c(21, 22, 23, 24, 25,20,  7,  8,  9, 10,19,  6,  1,  2, 11,18,  5,  4,  3, 12,17, 16, 15, 14, 13), 5,5,byrow=T)
square

##      [,1] [,2] [,3] [,4] [,5]
## [1,]   21   22   23   24   25
## [2,]   20    7    8    9   10
## [3,]   19    6    1    2   11
## [4,]   18    5    4    3   12
## [5,]   17   16   15   14   13

This is a 5 by 5 square. Start at 1, add the 2 to the right of that, and add numbers in a clockwise spiral.

If we look at all the numbers in the diagonals, we can see that the sum of those numbers, is 101. Ie, 21 + 7 + 1 + 3 + 13 + 25 + 9 + 5 + 17 = 101

We are now asked: What is the sum of the numbers in the diagonals in a 1001 by 1001 square?

Looking at an n by n square, where n is uneven, we can see that the north-eastern corner of the square will always be n^2.

We know that the side is n. So the value of the north-western corner will always be n² – (n-1).

Similar the south-western corner is n² – 2(n-1). And the south-eastern is n² – 3(n-1).

The first “square” is a special case. But I can now calculate the corners in all the squares with unequal n.

The sum of those corners is:

n² + n² -(n-1) + n² – 2(n-1) + n² – 3(n-1)

Which reduces to:

4*n² – 6n + 6

If I calculate that for all unequal n from 3 to 1001, sum them, and add 1, I have my answer.

library(purrr)
answer <- seq(3,1001,by=2) %>%
  map(function(x) 4*x*x - 6*x + 6) %>%
  unlist() %>%
  sum() 
answer <- answer + 1

Lessons learned

Not really.

This problem is basically about making a sudoku solver, that can be automated. If you don’t know what a sudoku is – google it and come back.

I have solved a LOT of sudokus. Manually. There are several techniques, but some of them are not that easy to code. And of course I’ll need a method that guarantees a solution.

Fortunately a brute force solution exists:

https://en.wikipedia.org/wiki/Sudoku_solving_algorithms

It is called backtracking. The process is to look at all the empty cells.

In the first one – insert the lowest allowed value, and progress to the next empty cell. Insert the lowest allowed value in it. Continue until you reach a cell with no allowed values. When you do that, go back to the previous cell, and insert the next-lowest allowed value. And then go back (forward?) to the next cell. If the cell you went back to also have no allowed values, you go one step futher back. Continue until you reach the last cell in the puzzle.

So. We’ll need a few things.

First of all we’ll need a matrix in which to keep the sudoku.

I’m going with a matrix rather than a dataframe after reading the R Inferno (https://www.burns-stat.com/pages/Tutor/R_inferno.pdf), that suggests that there is no reason to work with a dataframe if a matrix will do.

I’ll need a couple, or three (actually more like five) functions to figure out what values are allowd for a given cell in a given sudoku.

And I’ll need a way to keep track of which cells that are “frozen”, and which cells should be filled out.

Lets begin by making a matrix to keep the puzzle in. Project Euler provides us with a puzzle and the matching solution. And a file with 50 puzzles that needs to be solved.

Lets read in that file.

problems <- read.csv("p096_sudoku.txt",header=FALSE, stringsAsFactors = FALSE)

The structure is rather simple. 500 rows. First one row with an identifier, eg “Grid 01”. And then 9 rows with just the rows of digits, 0 for the empty cells.

We’ll need to pick out first the rows 1 to 10, then rows 11 to 20 etc.

Lets make a dataframe, and read it in:

start <- seq(1,500,by=10)
end <- seq(10,500,by=10)
prob_df <- data.frame(id=character(),problem=character(), stringsAsFactors = FALSE)
for(i in 1:50){
  prob_df[i,2] <- paste(problems[start[i]:end[i],][-1], collapse="")
}
prob_df[1,2]

## [1] "003020600900305001001806400008102900700000008006708200002609500800203009005010300"

Nice and simple, 50 rows with an 81 character long string of digits. This is the first puzzle in the file. It is also the puzzle where we are provided with a solution.

We now need to convert that to a matrix.

Or – we don’t need to. There are certainly techniques that would allow me to handle it as-is. But I think it is more intuitive to get it into a matrix.

So, lets do that.

mat <- matrix(as.numeric(unlist(strsplit(prob_df[1,2],""))),9,9,byrow=TRUE)  
mat

##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
##  [1,]    0    0    3    0    2    0    6    0    0
##  [2,]    9    0    0    3    0    5    0    0    1
##  [3,]    0    0    1    8    0    6    4    0    0
##  [4,]    0    0    8    1    0    2    9    0    0
##  [5,]    7    0    0    0    0    0    0    0    8
##  [6,]    0    0    6    7    0    8    2    0    0
##  [7,]    0    0    2    6    0    9    5    0    0
##  [8,]    8    0    0    2    0    3    0    0    9
##  [9,]    0    0    5    0    1    0    3    0    0

Neat. Split the string in individual characters. Unlist it, and cast it to numeric. Then pass it all to matrix(), noting that we want 9 rows and 9 columns. And that the matrix should be filled by row.

Depending on where you count from, the first empty cell is 1,2. Given the values that are already filled in, what values are allowed in that cell?

First, what values are allowed based on the row? From inspection it is obvious that only the values 1, 4, 5, 6, 7, 8 and 9 are allowed. Lets write a function for that:

rowall <- function(x,mat){
  setdiff(1:9, mat[x,])
}
rowall(1,mat)

## [1] 1 4 5 7 8 9

The function takes a row-number, in this example 1, and a matrix. mat[x,] returns the values in that row. And setdiff returns the values that are in 1:9 but not in the list of values already in the row.

More or less the exact same thing for the column:

# returnerer tilladte værdier i kolonne y i en matrix mat
colall <- function(y,mat){
  setdiff(1:9, mat[,y])
}
colall(2, mat)

## [1] 1 2 3 4 5 6 7 8 9

What about the frame, the 3×3 set of numbers?

I need a subset of the matrix. For our example, cell 1,2, I need the columns 1, 2 and 3. And the rows 1, 2 and 3. Or a vector 1:3. Thankfully there is symmetry, so column 1 gives the same vector as row 1. I need a function that returns 1:3 when I pass it 1, 2 or 3. 4:6 for the values 4, 5 or 6. and 7:9 for the values 7, 8 or 9.

My good colleague Henrik has tried to solve the same problem using Python. And he suggested that the way to do it is by integer division. In this way:

framelookup <- list(a=c(1,2,3),b=c(4,5,6),c=c(7,8,9))

getint <- function(x){
  x <- (x -1)%/%3+1
  unlist(framelookup[x])
}

• Take the value x.
• Substract 1 and divide the result by 3.
• Then add 1. For x = 1,2 or 3, the result is 1, for x = 4, 5 or 6, the result is 2. And for 7 through 9, we get 3.
• We can then use that to look up the vector we need in a list, and return the unlisted number.

At first I had done it with three if-statements. This solution is a bit more elegant, and Henrik thinks that if-statements should be outlawed. So in the interest of keeping the peace in our office, I’m going with that.

Now I can write a function that returns the values allowed in a given cell in a given matrix, given the values that are already filled out in the frame that cell is in:

fraall <- function(x,y,mat){
  res <- mat[getint(x),getint(y)]
  setdiff(1:9, res)
}

The function takes the coordinates of the cell, and a matrix. Based on that, getint() returns the vectors describing the frame, and saves that subframe in res. Then I use the setdiff() function in the same way I’ve done previously.

That gives me all the constraints. What values are allowed based on row, column and frame. The intersection between these three vectors gives me the allowed values for the cell. It is now simple to write a function that returns the values allowed for a given cell in a given matrix:

allowed <- function(x,y,mat){
  res <- intersect(rowall(x,mat), colall(y,mat))
  res <- intersect(res, fraall(x,y,mat))
  return(res)
}

Actually this is all I need for solving quite a lot of sudokus. Go through all the empty cells. Figure out what values are allowed. If only one value is allowed, plot it into the cell. Repeat until there are no more empty cells where only one value is allowed. A lot of sudokus can be solved with only that method.

The ones that cannot be solved in this way can be solved by the backtracking algorithm. But that is slow, so it might be a good idea to simplify the those puzzles by filling out what can be filled out with this method first.
So let’s write a function that does exactly that.

First I’ll need a list of the empty cells. The which() function can help me.

test <- which(mat ==0, arr.ind=T)
nrow(test)

## [1] 49

Which parts of the matrix is equal to 0? and arr.ind tells that what we want returned (when which is used on an array) is the array indeces. There are 56 0’es in the puzzle.

The matrix test has this structure:

head(test)

##      row col
## [1,]   1   1
## [2,]   3   1
## [3,]   4   1
## [4,]   6   1
## [5,]   7   1
## [6,]   9   1

The first pair of values is 2,1, the next is 4,1 etc. And if I take each one of those, looks at the allowed values for those positions in the matrix, and, if there is only one value allowed, place that value in that position, I get a step closer to the solution.

This does exactly that:

for(i in 1:nrow(test)){
  all <- allowed(test[i,1],test[i,2],mat)
  if(length(all)==1){
    mat[test[i,1],test[i,2]] <- all
  }
}
nrow(which(mat ==0, arr.ind=T))

## [1] 43

And, hey presto, 6 empty cells were filled. The logic is: Find the list of allowed values based on the values in the list of empty cells. If the length is 1, set the current cell to that value.

Now I need to get a new list of empty cells:

test <- which(mat ==0, arr.ind=T)

And I could do it again:

for(i in 1:nrow(test)){
  all <- allowed(test[i,1],test[i,2],mat)
  if(length(all)==1){
    mat[test[i,1],test[i,2]] <- all
  }
}
nrow(which(mat ==0, arr.ind=T))

## [1] 29

14 previously empty cells filled!

Lets make a function that does that again and again, until there is no more improvement.

We’ll start by writing a function that does it one time:

singlevalues <- function(mat){
  test <- which(mat ==0, arr.ind=T)
  for(i in 1:nrow(test)){
    all <- allowed(test[i,1],test[i,2],mat)
  if(length(all)==1){
    mat[test[i,1],test[i,2]] <- all
  }
  }
  return(mat)
}

Then I need to run that several times. When the number of empty cells after running it is the same as the number of empty cells before, stop. No more values can be filled in.

repsinglevalues <- function(mat){
  oldlen <- 1
  newlen <- 0
  while(oldlen > newlen){
  oldlen <- length(which(mat == 0))
  mat <- singlevalues(mat)
  newlen <- length(which(mat == 0))
  if(newlen==0) break()
  }
  return(mat)
}

Take a matrix. Set oldlen to 1, and newlen to 0. As long ans oldlen is larger than newlen, do this:
Set oldlen to the number of 0’es in the matrix. Run the singlevalues function from above. Find out how many empty cells, or zeroes there are now, and set newlen to that. If newlen is equal to 0, stop everything, otherwise repeat. And finally return the changed matrix.

Lets test it:

repsinglevalues(mat)

##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
##  [1,]    4    8    3    9    2    1    6    5    7
##  [2,]    9    6    7    3    4    5    8    2    1
##  [3,]    2    5    1    8    7    6    4    9    3
##  [4,]    5    4    8    1    3    2    9    7    6
##  [5,]    7    2    9    5    6    4    1    3    8
##  [6,]    1    3    6    7    9    8    2    4    5
##  [7,]    3    7    2    6    8    9    5    1    4
##  [8,]    8    1    4    2    5    3    7    6    9
##  [9,]    6    9    5    4    1    7    3    8    2

Yeah! The sudoku is solved!

But not every puzzle can be solved in that way. Now for the brutish way…

First of all, lets get a fresh copy of the original puzzle read in. And a fresh list of empyt cells.

All righty! Nu kan jeg løse en del sudokuer alene ved denne metode. Så er der resten…

mat <- matrix(as.numeric(unlist(strsplit(prob_df[1,2],""))),9,9,byrow=TRUE)  
test <- which(mat ==0, arr.ind=T)
nrow(test)

## [1] 49

Back to the 49 empty cells.

The algorithm is as follows.

• Set i <- 1
• Get the row and column indeces from test[i,1] and test[i,2] respectively.
• Identify the list of allowed values for that cell.
• Set the value of the cell to the lowest of the allowed values.
• Increment i <- i + 1.
• Get the row and column indeces again.
• Get the list of allowed values for that cell. If there are no allowed values, decrement i <- i -1.
• If there are allowed values, set the value of the cell to the lowest of the allowed values.
• When we return to a cell that already has a value, we should not set it to the lowest allowed value. We should set it to the next lowest. That is the lowest of the allowed values, excluding the value that was already there.
• If we get to a cell with no allowed values (excluding the one that might already be there), the value of the cell should be set to 0, and i should be decrementet.

This function does that.

solve <- function(mat){
  i <- 1
  test <- which(mat ==0, arr.ind=T)
  while(nrow(which(mat==0,arr.ind=T))>0){
    allowedvalues <- allowed(test[i,1],test[i,2],mat)
    currentvalue <- mat[test[i,1],test[i,2]]
    allowedvalues <- allowedvalues[allowedvalues>currentvalue]
    if(length(allowedvalues)==0){
      mat[test[i,1],test[i,2]] <- 0
      i <- i -1
    } else{
      mat[test[i,1],test[i,2]] <- min(allowedvalues)
      i <- i + 1
    }
  }
  mat
  }

The function takes a matrix. It sets the counter/pointer i to 1.

Then a test-matrix is generated, containing all the positions that are empty.

While the number of empty places in the matrix is larger than zero do this:

• Find the allowed values.
• Find the current value in the cell.
• Remove all allowed values that are smaller than the current value. In that way, taking the minimum of the remaining allowed values, will always give us the next lowest value. And if the original value was 0, we will get the lowest.
• If there are no allowed values, set the value of the cell to zero, and decrement the counter.
• If there are allowed values, set the value of the cell to the lowest of the remaining allowed values, and increment the counter.
• End by returning the now filled matrix.

Does it work?

solve(mat)

##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
##  [1,]    4    8    3    9    2    1    6    5    7
##  [2,]    9    6    7    3    4    5    8    2    1
##  [3,]    2    5    1    8    7    6    4    9    3
##  [4,]    5    4    8    1    3    2    9    7    6
##  [5,]    7    2    9    5    6    4    1    3    8
##  [6,]    1    3    6    7    9    8    2    4    5
##  [7,]    3    7    2    6    8    9    5    1    4
##  [8,]    8    1    4    2    5    3    7    6    9
##  [9,]    6    9    5    4    1    7    3    8    2

It does! Qap’la!

I now have two functions that solves sudokus. solve(), that bruteforces it and will always get a solution. And
repsinglevalues() that uses logic, gets a solution sometimes, but not always.

I also have a dataframe with 50 sudokus that I need to solve.

And what was the task again? Take the three first digits in the first row. Concatenate them to a single three digit number. Do that for all 50 puzzles. Add them all up. That should be the answer.

answer <- 0
for(i in 1:50){
  mat <- matrix(as.numeric(unlist(strsplit(prob_df[i,2],""))),9,9,byrow=TRUE)  
  mat <- repsinglevalues(mat)
  mat <- solve(mat)
  answer <<- answer + as.numeric(paste(mat[1,1:3],collapse=""))
}

• Set a variable answer to zero.
• Pick out the sudokus one by one.
• Run the logic-solving function on it.
• Then run the brute-force function on the result of that.
• Pick out the three first digits in row 1, collapse them, cast as numeric, and add to the answer-variable.

Done!

Lessons learned

1. Errors in the logic are a bit difficult to locate. I forgot to take into account that the list of allowed values in the brute-force solution does not include the value that is already in the cell. That made the function run for eternity. Or something pretty close. Had I not made that mistake, I would finished this problem a day earlier.
2. Technically the lesson has not been learned yet. But below there is a note ot speed. Adding just one extra value to the puzzle reduces the time it takes to bruteforce the solution significantly.
3. Set-functions, here specifically setdiff(), are neat!

Notes on speed

OK. I needed a function for returning a vector for getting the relevant frame in the matrix. As I mentioned, my colleague Henrik thought it should be done wit integer division rather than if-statements.

This was the way I originally did it:

ifgetint <- function(x){
if(x %in% c(1:3)){
  res <- c(1:3)
} else if(x %in% c(4:6)){
  res <- c(4:6)
  } else {  
      res <- c(7:9)
  }
  return(res)
}

But what way is faster? Lets find out:

library(microbenchmark)
mbm <- microbenchmark(ifgetint, getint, times=1000)
mbm

## Unit: nanoseconds
##      expr min lq   mean median uq   max neval cld
##  ifgetint  50 55 60.824     56 58  3906  1000   a
##    getint  52 57 78.630     59 60 17763  1000   a

Actually my original way of doing it was a little bit faster.

library(ggplot2)
autoplot(mbm)

So. I found two methods for solving sudokus. It would be interesting to compare them. I can’t do that on just any sudoku. I need to use one that can be solved with both methods. Luckily the first problem in the set can do just that.

mat <- matrix(as.numeric(unlist(strsplit(prob_df[1,2],""))),9,9,byrow=TRUE)  
solve(mat)

##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
##  [1,]    4    8    3    9    2    1    6    5    7
##  [2,]    9    6    7    3    4    5    8    2    1
##  [3,]    2    5    1    8    7    6    4    9    3
##  [4,]    5    4    8    1    3    2    9    7    6
##  [5,]    7    2    9    5    6    4    1    3    8
##  [6,]    1    3    6    7    9    8    2    4    5
##  [7,]    3    7    2    6    8    9    5    1    4
##  [8,]    8    1    4    2    5    3    7    6    9
##  [9,]    6    9    5    4    1    7    3    8    2

repsinglevalues(mat)

##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
##  [1,]    4    8    3    9    2    1    6    5    7
##  [2,]    9    6    7    3    4    5    8    2    1
##  [3,]    2    5    1    8    7    6    4    9    3
##  [4,]    5    4    8    1    3    2    9    7    6
##  [5,]    7    2    9    5    6    4    1    3    8
##  [6,]    1    3    6    7    9    8    2    4    5
##  [7,]    3    7    2    6    8    9    5    1    4
##  [8,]    8    1    4    2    5    3    7    6    9
##  [9,]    6    9    5    4    1    7    3    8    2

mbm <- microbenchmark(solve(mat),repsinglevalues(mat), times=100)
mbm

## Unit: milliseconds
##                  expr       min        lq      mean    median        uq
##            solve(mat) 47.681675 52.437511 55.053319 53.852274 54.621338
##  repsinglevalues(mat)  6.637329  7.001639  7.631094  7.157844  7.393044
##        max neval cld
##  171.67914   100   b
##   11.03237   100  a

No surprise there. The logical method is much faster. Something like 8 times faster.

autoplot(mbm)

How much of a difference does it make when part of the sudoku has been solved already? By inspection (ie, trial and error) I find sudoku number 11. The original puzzle has 53 empty cells. By filling out what can be filled out using logic, that is reduced to 52 empty cells.

How much faster is it to solve sudoku number 11, when an additional cell has been filled out using logic?
mat is the original sudoku. redmat is the sudoku partially solved by logic.

i <- 11
mat <- matrix(as.numeric(unlist(strsplit(prob_df[i,2],""))),9,9,byrow=TRUE)  
redmat <- repsinglevalues(mat)

mbm <- microbenchmark(solve(mat),solve(redmat), times=10)
mbm

## Unit: milliseconds
##           expr      min       lq     mean   median       uq      max neval
##     solve(mat) 767.3325 777.2788 794.4597 781.4264 791.7815 911.4314    10
##  solve(redmat) 544.3811 546.2505 574.9207 549.2904 553.6324 690.5734    10
##  cld
##    b
##   a

Wow! That really makes a difference!! Filling in just a single extra value using logic cuts about one third of the time it takes to solve the sudoku.

autoplot(mbm)

Just a short note to help me remember the melt()-function.

Lets create some messy data:

ID <- 1:10
T1 <- runif(10,10,20)
T2 <- runif(10,20,30)
T3 <- runif(10,30,40)
df <- data.frame(ID=ID, T1=T1, T2=T2, T3=T3)

This is heavily inspired by a practical problem a student came to us with. There is 10 different patients, at time = T1, there is a certain value measured on the patient. At time = T2 the same value is measured. And again at time = T3, where T1<T2<T3.

We would now like to plot the development of those values as a function of time (or just T1,T2 and T3).

How to do that?

Using reshape2, and making sure we have the tidyverse packages loaded:

library(tidyverse)
library(reshape2)

clean <- melt(df, id.vars=c("ID"), value.names=c("T1", "T2", "T3"))
head(clean)

##   ID variable    value
## 1  1       T1 13.93662
## 2  2       T1 10.30468
## 3  3       T1 18.14351
## 4  4       T1 17.58294
## 5  5       T1 18.13877
## 6  6       T1 10.21993

Nice, we now have tidy data.

Note that melt() also takes a variable “variable.name”, in case we have a different sort of mess.

Now it is easy to plot:

plot <- ggplot(clean, aes(x=variable, y=value, group=ID)) +
  geom_line()
plot

Neat.

Project Euler – problem 100

Back to the hopeless examples of probabilities from school.
In a bag there are 15 black balls and six white ones. Project Euler talks about discs, math-teachers has always used balls as examples, and they where always white and black. So I’ll stick with that.

It you draw two balls from the bag, there is a 50/50 chance of drawing 2 black balls:

(15/21)*(14/20)

## [1] 0.5

I’m told that the next set of balls in the bag with that property, is 85 black balls and 35 white ones:
(85/120)

(85/120)*(84/119)

## [1] 0.5

Find the mix of black and white balls, that gives a probability of 50/50 of drawing 2 black balls, given that there should be more than 10¹² = 1000000000000 balls in the rather large bag.

That should be straight-forward.
Lets call the number of black balls b and the number of white balls w. And lets define the total number of balls in the back as n=w+b
The probability of drawing two black balls is:

(b/n)((b-1)/(n-1)) = ½

n = w + b > 10¹²

Two equations with two unknowns.
The probability can be rearranged:

(b/n)((b-1)/(n-1)) = ½ <=>

b(b-1) / n(n-1) = ½ <=>

(b² – b) / (n² – n) = ½ <=>

b² – b = ½(n² – n) <=>

2b² – 2b = n² – n <=>

2b² – 2b – n² + n = 0

Hm. Maybe it is not that simple after all. First of all I don’t know if n is 100000000000 or 100000000001. That actually makes a pretty big difference:

1000000000001**2 - 1000000000000**2

## [1] 1.999978e+12

Second of all, I need to find integer solutions. An analytical solution might not give integer results. And I can’t have one third of a ball in the bag.

Googling “finding integer solutions to equations” give, as the first result, a link to the wikipedia article on “Diophantine equations”.
Which apparently are equations that should have integer solutions.

All right, a couple of the problems I’ve tackled earlier, and quite a lot of Project Euler problems I’ve given up on appears to be about solving these Diophantine equations.

So. Nice. The last link of the wikipedia page is to https://www.alpertron.com.ar/QUAD.HTM.
I should probably read up on the methods. But that will have to wait.

The point is, that this Diophantine equation can be solved by:

b~n+1~ = 3b~n~ + 2n~n~ -2

n~n+1~ = 4b~n~ + 3n~n~ -3

The idea is that we have a solution (b~n~, n~n~). And these two equations allows us to calculate the next solution, (b~n+1~, n~n+1~)

Lets try that, we was given that (15,21) was a solution. The next should be (85,120). Do we get that?

b <- 15
n <- 21
b_n <- 3*b + 2*n -2
n_n <- 4*b + 3*n -3
print(paste(b_n, n_n, sep=","))

## [1] "85,120"

Qap’la, it works. Nice. Now I just need to run through this until n~n+1~ gets above 10¹².

b <- 15
n <- 21
while(n<10**12){
  b_n <- 3*b + 2*n -2
  n_n <- 4*b + 3*n -3
  b <- b_n
  n <- n_n
}
answer <- b

Lessons learned:

1. Solving Diophantine equations is at the heart of a lot of these problems. I’ve learned a new tools to handle them!
2. If you want to subscript stuff in RMarkdown, you place a ~ on each side of what you want subscripted.

Other stuff to note: Maybe it is time someone wrote a new solver for Diophantine equations. The one I found is 19 years old. Something to do in Shiny perhaps?

Problem 80 from Project Euler.

The problem tells us that if the square root of a natural number is not an integer, it is irrational.
Project Euler also claims that it is well known. I did not know it.

We are then told that the square root of 2 is 1.4142135623730950… And that the digital sum of the first 100 digits is 475.

The task is now to take the first 100 natural numbers. And find the total of the digital sums for the first 100 digits for all the irrational square roots.

Lets begin by figuring out how to handle that many digits. R does not support more than around 15 places after the decimal point.

The library Rmpfr can handle arbitrary precision:

library(Rmpfr)

## Loading required package: gmp

## 
## Attaching package: 'gmp'

## The following objects are masked from 'package:base':
## 
##     %*%, apply, crossprod, matrix, tcrossprod

## C code of R package 'Rmpfr': GMP using 64 bits per limb

## 
## Attaching package: 'Rmpfr'

## The following objects are masked from 'package:stats':
## 
##     dbinom, dnorm, dpois, pnorm

## The following objects are masked from 'package:base':
## 
##     cbind, pmax, pmin, rbind

a <- sqrt(mpfr(2,500))

The variable a now contains the square root of 2 with a precision of 500 bytes. I’m not quite sure how many decimal places that actually translates to. But testing the following code allows me to conclude with confidence that it is at least 100.

A thing to note here is, that

a <- mpfr(sqrt(2),500)

and

a <- sqrt(mpfr(2,500))

are not equal. In the first exampel sqrt(2) is evaluated before saving the value with the high precision. Start by converting the number 2 to a high precision representation, before doing math on it.

Next is writing a function that will return the digital sum of the first 100 digits of a number.

digitsum <- function(x){
  s <- 0
  for(i in 1:100){
    s <- s + floor(x)
    x <- (x - floor(x))*10
  }
  s
}

First s is initialized to 0. Then floor(x) gives us the first digit in x. We add that to s, and subtract it from x, and multiply by 10. Repeat that 100 times, and you get the sum of the first 100 digits in x.

Let us test that it works. Project Euler told us what the result for sqrt(2) is:

digitsum(a)

## 1 'mpfr' number of precision  500   bits 
## [1] 475

Nice, the correct result (not that that guarantees that I’ve done everything correctly).

Now, lets find all the irrational square roots we need to look at:

library(purrr)
t <- 1:100
s <- t %>%
  keep(~as.logical(sqrt(.x)%%1))

I need to practice this way of coding a bit more. t contains the first 100 natural numbers. I pass that to the keep()-function, and the predicate function takes the square root of each number, take the modulus 1, and convert it to a logical value. If the modulus of the square root is 0, the square root is an integer, and 0 i false. So we’re keeping all the non-integer squareroots.

Now I’ll convert all the natural numbers to the mpfr-class. The next line takes the square root. The third line calculate the digitalsum. And the final line gives us the result.

s <- mpfr(s,500)
r <- sqrt(s)
r <- digitsum(r)
sum(r)

## 1 'mpfr' number of precision  500   bits 
## [1] Censored

Lessons learned:
Rmpfr allows us to work with (more or less) arbitrary precision.
But we need to convert numbers to the relevant class before doing math on it.

Given a set of data, where some values indicate that they are the same as a previous value, how to replace them with the correct value.

Eg, this dataframe:

(m <- data.frame(i=c(1:10,NA), t=c("lorem", "do", "do", "Do", "ipsum", "do", "Do", "(do)", "dolor", NA, "test"), stringsAsFactors=F))

##     i     t
## 1   1 lorem
## 2   2    do
## 3   3    do
## 4   4    Do
## 5   5 ipsum
## 6   6    do
## 7   7    Do
## 8   8  (do)
## 9   9 dolor
## 10 10  <NA>
## 11 NA  test

How to replace the first three “do”s with “lorem” and the next set of “do”s with “ipsum”

Using fill() from the tidyr package is straight forward. It takes a vector, locates all NA, and replaces them with the last, non-NA value.
Simple enough, change all the variations of “do” to NA, run fill(). Done.
One problem, there might be NAs in the dataset, that we do not want to affect.
Solution – there might be a more elegant one, but this works:

1. Change the NAs to something that do not occur in the data
2. Change to variations of “do” to NA
3. Use the fill()-function
4. Change the NAs from step 1 back to NA

library(tidyr)
rpl <- "replacement"
m[is.na(m$t),]$t <- rpl
doset <- c("do", "Do", "(do)")
m[(m$t %in% doset),]$t <- NA

m <- m %>% fill(t)
m[(m$t == rpl),]$t <- NA
m

##     i     t
## 1   1 lorem
## 2   2 lorem
## 3   3 lorem
## 4   4 lorem
## 5   5 ipsum
## 6   6 ipsum
## 7   7 ipsum
## 8   8 ipsum
## 9   9 dolor
## 10 10  <NA>
## 11 NA  test

Done!

Oh, and by the way, this is my first post generated directly from RStudio!

Og et par andre steder. Noget af det jeg bruger en del tid på, både privat og professionelt, er R. Som i det statistiske program R.

Jeg skal derfor snart have taget et kig på denne side:

Og få publiceret hvad jeg alligevel nusser rundt med her og andre steder.

So – you want to hide some rows on a worksheet, based on the value in a cell. Or more than one.
Here’s how to do that with VBA
Find the last row of the range that you want to apply the hiding to.
Get a range of rows, in this case starting at A7, and ending at “LastRow”.
For each value in that range, if the value i column A is equal to the value in cell G1 (that is Cells(1,7), And the value three columns over, eg in colum D (that is c.Offset(0,3)), is equal to the value in cell G2 (Cells(2,7), then set the entire row to be hidden, else set it to be shown.


Private Sub Worksheet_Change(ByVal Target As Range)
Dim LastRow As Long, c As Range
Application.EnableEvents = False
LastRow = Application.WorksheetFunction.CountA(Range("A7:A100000")) + 6
On Error Resume Next
For Each c In Range("A7:A" & LastRow)
If (c.Value = Cells(1, 7).Value And c.Offset(0, 3).Value = Cells(2, 7).Value) Then
c.EntireRow.Hidden = False
Else
c.EntireRow.Hidden = True
End If
Next
On Error GoTo 0
Application.EnableEvents = True
End Sub


Project Euler is a pretty good way to exercise your programming muscles. I tend to think that the hardest part is usually the math.

So when I figure one of them out in minutes, I’m pretty happy about it.

library(gtools)
library(numbers)
pandig <- function(n){
x <- permutations(n,n)
x <- apply(x,1, function(x) paste(x, collapse=""))
return(x)
}
y <- as.numeric(pandig(7))
max(y[isPrime(y)])

The surprising thing is that largest pandigital prime does not begin with either 8 or 9.

These days, when at least one crypto currency has tanked completely, Bitcoin is coming under increasing pressure from authorities, revelations that there are organized manipulation of the trading etc etc etc.

I am reminded of something that happened in the Netherlands in 1637.

Read about it here. I am continually baffled by our inability to learn from history.