# R Solution for Excel Puzzles

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Puzzles no. 349–353

### Puzzles

Author: ExcelBI

Puzzles:

#349: content file

#350: content file

#351: content file

#352: content file

#353: content file

Let dig in those numbers.

### Puzzle #349

In this puzzle we have to translate “normal” (Arabic) numbers into Roman numerals. With one twist… Historically roman numerals has its limit (3899 if I recall properly), and we have some bigger numbers, so M (thousands) would be un-naturally repeated. Lets do it.

#### Load libraries and data

library(tidyverse) library(readxl) input = read_excel("Numbers to Roman.xlsx", range = "A1:A10") test = read_excel("Numbers to Roman.xlsx", range = "B1:B10")

#### Transformation

to_roman <- function(number) { if (!is.numeric(number) || number <= 0 || number != as.integer(number)) { return(NA) } roman_symbols <- c("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I") arabic_values <- c(1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1) numeral <- "" for (i in seq_along(roman_symbols)) { while (number >= arabic_values[i]) { numeral <- paste0(numeral, roman_symbols[i]) number <- number - arabic_values[i] } } return(numeral) } result = input %>% mutate(Roman = map_chr(Number, to_roman))

#### Validation

identical(result$Roman, test$Roman) #> [1] TRUE

### Puzzle #350

Today we have insolite numbers. Wait what? What kind of number our host dug up from voids of maths? Insolite numbers are such numbers that are divisible by both sum of squares of digits and product of squares of digits. Just another interesting property of number. Can we extract them from given column? Why not.

#### Load libraries and data

library(tidyverse) library(readxl) input = read_excel("Insolite Number.xlsx", range = "A1:A10") %>% janitor::clean_names() test = read_excel("Insolite Number.xlsx", range = "B1:B6") %>% janitor::clean_names()

#### Transformation

is_insolite = function(number) { digits = strsplit(as.character(number), "")[[1]] digits = as.numeric(digits) sum_of_squares = sum(digits^2) product_of_squares = prod(digits^2) div_by_sum = number %% sum_of_squares == 0 div_by_prod = number %% product_of_squares == 0 return(div_by_sum & div_by_prod) } result = input %>% mutate(answer_expected = map(numbers, is_insolite)) %>% filter(answer_expected == TRUE) %>% select(numbers)

#### Verification

identical(result$numbers, test$answer_expected) #> [1] TRUE

### Puzzle #351

Mixing words again, but not all. What we have to do is mix or rather change order of words into reverse one in part of sentence. We get two numbers that point to start and end of reversed section. Results looks little bit like Yoda’s talks.

#### Load libraries and data

library(tidyverse) library(readxl) input = read_excel("Reverse Words between Positions.xlsx", range = "A1:C9") test = read_excel("Reverse Words between Positions.xlsx", range = "D1:D9")

#### Transform

reverse_words <- function(text, start_pos, end_pos) { words <- str_split(text, " ")[[1]] words[start_pos:end_pos] <- rev(words[start_pos:end_pos]) words <- words[!is.na(words)] paste(words, collapse = " ") } result = input %>% mutate(reversed = pmap_chr(list(text = Sentence, start_pos = `Word No1`, end_pos = `Word No2`), reverse_words)) %>% select(reversed)

#### Validation

identical(test$`Answer Expected`, result$reversed) #> [1] TRUE

### Puzzle #352

There are more numbers with interesting properties and today we have interprimes. How do we describe them? Numbers that are equally distant from neighbouring primes or if it will be easier to understand, this number is a mean of two consecutive primes

I’ve done this two ways: one with dedicated function, one more in base R syntax involved.

#### Load libraries and data

library(tidyverse) library(readxl) library(numbers) input = read_excel("Excel/352 Interprime Numbers.xlsx", range = "A1:A10") %>% janitor::clean_names() test = read_excel("Excel/352 Interprime Numbers.xlsx", range = "B1:B6") %>% janitor::clean_names()

#### Transformation

# Version using ready numbers package functions is_interprime <- function(n) { if (!is.integer(n)) {return(FALSE)} prev_prime <- previousPrime(n) next_prime <- nextPrime(n) return(n == (prev_prime + next_prime) / 2) } # version with own functions (we are gonna use only is_prime from numbers package) next_prime = function(n) { repeat { n = n + 1 if (isPrime(n)) {return(n)} } } previous_prime = function(n) { while (n > 2) { n = n - 1 if (isPrime(n)) {return(n)} } } is_interprime_2 <- function(n) { if (!is.integer(n)) {return(FALSE)} prev_prime <- previous_prime(n) next_prime <- next_prime(n) return(n == (prev_prime + next_prime) / 2) } result = input %>% mutate(number = as.integer(number)) %>% mutate(is_interprime = map_lgl(number, is_interprime), is_interprime_2 = map_lgl(number, is_interprime_2)) %>% filter(is_interprime) %>% bind_cols(test)

#### Validation

identical(as.numeric(result$number), result$answer_expected) #> [1] TRUE

### Puzzle #353

We lost some letters, and there are some holes in it. We have to find what letters are missing in every sequence and give only missing ones.

Try it.

#### Loading library and data

library(tidyverse) library(readxl) input = read_excel("Excel/353 Missing Letters.xlsx", range = "A1:A10") test = read_excel("Excel/353 Missing Letters.xlsx", range = "B1:B10")

#### Transformation

result = input %>% mutate(let = strsplit(Letters, ", ")) %>% mutate(let = map(let, sort)) %>% mutate(min = map(let, min), max = map(let, max), min_index = map_int(min, ~ which(letters == .x)), max_index = map_int(max, ~ which(letters == .x)), seq = map2(min_index, max_index, ~ letters[.x:.y]), diff = map2(seq, let, ~ setdiff(.x, .y)), answer = map_chr(diff, ~ paste(.x, collapse = ", ")) ) %>% select(Letters, answer) %>% mutate(answer = ifelse(answer == "", NA, answer))

#### Validation

identical(result$answer, test$`Expected Answer`) # [1] TRUE

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