# R Solution for Excel Puzzles

**Numbers around us - Medium**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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Puzzles no. 514–518

### Puzzles

Author: ExcelBI

All files (xlsx with puzzle and R with solution) for each and every puzzle are available on my Github. Enjoy.

### Puzzle #514

Today we are searching for treasure. Exactly what we need is to find highest 2×2 submatrix within 5×5 matrix. Manually it would take maybe a minute, but we need to code it somehow. Not the easiest task but managable. Lets look at this problem.

#### Loading libraries and data

library(tidyverse) library(readxl) path = "Excel/514 Sub Grid Maximum Sum.xlsx" input = read_excel(path, range = "B2:F6", col_names = F) %>% as.matrix() test = read_excel(path, range = "H1:H4") %>% arrange(`Answer Expected`)

#### Transformation

indices = expand.grid(i = 1:(nrow(input) - 1), j = 1:(ncol(input) - 1)) results = indices %>% pmap(function(i, j) { sub_mat = input[i:(i + 1), j:(j + 1)] list(matrix = sub_mat, sum = sum(sub_mat, na.rm = TRUE)) }) max_sum = max(map_dbl(results, "sum")) max_subs = keep(results, ~ .x$sum == max_sum) max_subs_str = map_chr(max_subs, ~ paste(apply(.x$matrix, 1, paste, collapse = ", "), collapse = " ; ")) %>% tibble(`Answer Expected` = .) %>% arrange(`Answer Expected`)

#### Validation

identical(max_subs_str, test) #> [1] TRUE

### Puzzle #515

Sometimes data notation is exactly like in “matryoshka” doll from Russia. It is concatenated, then some rows are bind together and it is not the last level of depth. But if something is wrapped, can be unfolded again, just as our data. Check it out.

#### Loading libraries and data

library(tidyverse) library(readxl) path = "Excel/515 Normalization of Data.xlsx" input = read_excel(path, range = "A2:B7") test = read_excel(path, range = "D2:F20")

#### Transformation

result = input %>% separate_rows(Data, sep = "(?=[A-Z])") %>% separate(Data, into = c("Name", "Seq"), sep = ":") %>% separate_rows(Seq, sep = ",") %>% filter(!is.na(Seq)) %>% mutate(Seq = as.numeric(Seq), Name = trimws(Name)) %>% select(Seq, Name, State) %>% arrange(Seq)

#### Validation

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

### Puzzle #516

Sometimes we are checking different types and differently named numbers. Today we need to find out what is the smallest possible number (if possible at all), which product of digits give us our number. Diving deep into properties of numbers is what we like the most. Lets find it out.

#### Loading libraries and data

library(tidyverse) library(readxl) path = "Excel/516 Product of Digits of Result is Equal to Number.xlsx" input = read_excel(path, range = "A1:A10") test = read_excel(path, range = "B1:B10") %>% mutate(`Answer Expected` = as.character(`Answer Expected`))

#### Transformation

find_smallest_number_with_digit_product = function(n) { if (n == 0) return(10) if (n == 1) return(1) factors = c() for (i in 9:2) { while (n %% i == 0) { factors = c(factors, i) n = n / i } } if (n > 1) return("NP") return(paste(sort(factors), collapse = "")) } result = input %>% mutate(`Answer Expected` = map_chr(Number, find_smallest_number_with_digit_product)) %>% select(2)

#### Validation

identical(result, test) # [1] TRUE

### Puzzle #517

Again we are fighting with power of exponentiation, but little bit another way. We do not need to calculate squares or cubes, we just need to add some numbers together to get squares. Just like laying down domino, we have to find order of numbers in which each consecutive pair gives square as sum. There is more than two solution for this task, but I managed to get one we have in possible answers. Easy math in not easy task… Let’s do it.

#### Loading libraries and data

library(tidyverse) library(readxl) library(combinat) path = "Excel/517 Arrange Numbers to Form Square Chains.xlsx" input = read_excel(path, range = "A1:A10") %>% unlist() test = read_excel(path, range = "B1:B10") %>% unlist()

#### Transformation

is_perfect_square <- function(x) { sqrt_x <- sqrt(x) sqrt_x == floor(sqrt_x) } is_valid_sequence <- function(nums) { all(map2_lgl(nums[-length(nums)], nums[-1], ~ is_perfect_square(.x + .y))) } find_valid_permutation <- function(nums) { permutations <- permn(nums) valid_perm <- keep(permutations, is_valid_sequence) if (length(valid_perm) > 0) { return(valid_perm[[1]]) } else { return(NULL) } } result = find_valid_permutation(input)

#### Validation

all.equal(unname(result), unname(test)) # [1] TRUE

### Puzzle #518

And nice “almost” real-life task. We need to rank students based on the grades. But we have American grades, and they are not making this task easy, because they are letters, not numbers. So we will use benefit of factors this time. And we need to ignore F’s into ranking. Let’s go.

#### Loading libraries and data

library(tidyverse) library(readxl) path = "Excel/518 Rank Students.xlsx" input = read_excel(path, range = "A1:B20") test = read_excel(path, range = "C1:C20")

#### Transformation

input$Grades <- factor(input$Grades, levels = c("A+", "A", "A-", "B+", "B", "B-", "C+", "C", "C-", "D+", "D", "D-"), ordered = TRUE) result <- input %>% mutate(rank = ifelse(Grades == "F", NA, as.numeric(dense_rank(Grades))))

#### Validation

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

Feel free to comment, share and contact me with advices, questions and your ideas how to improve anything. Contact me on Linkedin if you wish as well.

On my Github repo there are also solutions for the same puzzles in Python. Check it out!

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