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Puzzles no. 429–433

### Puzzles

Author: ExcelBI

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

### Puzzle #429

We all (I assume) know the Pythagorean theorem, thanks to which we can calculate sides of triangle. But today we have Pythagorean quadruples, which follows similar rules. Not to be too easy, we get 3 numbers, but without pointing which one is the longest. Let’s check how to do it.

```library(tidyverse)

input = read_excel("Excel/429 Pythagorean Quadruples.xlsx", range = "A1:A10")
test  = read_excel("Excel/429 Pythagorean Quadruples.xlsx", range = "B1:B10")```

#### Transformation

```find_quadr_solution = function(sides) {
numbers = str_split(sides, ", ")[[1]] %>%
as.numeric() %>%
na.omit()

missing1 = sqrt(sum(numbers^2)) # if d side is missing

pot_d = max(numbers)
others = numbers[numbers != pot_d]
missing2 = sqrt(pot_d^2 - sum(others^2)) # if a, b or c side is missing

if (missing1 == floor(missing1)) {
missing = missing1
} else if (missing2 == floor(missing2)) {
missing = missing2
} else {
missing = NA
}

return(missing)
}

result = input %>%
mutate(r = map_dbl(Number, find_quadr_solution)) ```

#### Validation

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

### Puzzle #430

From time to time we are painting with numbers or characters. And this day is today. For better understanding what our host wanted us to do, I’ll better show you this “beam of light made of letters”.

I was tricky but I managed to do it. Check it out.

```library(tidyverse)

#### Transformation

```shift_left <- function(mat, shift_size) {
n_cols <- ncol(mat)
tibble_mat <- as_tibble(mat)
shifted_df <- tibble_mat %>%
pmap_dfr(., ~{
row_values <- c(...)
shifted_values <- c(row_values[(shift_size + 1):length(row_values)], rep(NA, shift_size))
return(as_tibble(t(shifted_values)))
})
as.matrix(shifted_df)
}

for (i in 26:1) {
M <- matrix(NA, nrow = 1, ncol = 52)
M[1, 1:i] <- 1:i
M <- t(apply(M, 1, rev))
M = shift_left(M, 53-2*i)
M[!is.na(M)] <- LETTERS[M[!is.na(M)]]

if (i == 26){
M_final <- M
} else {
M_final <- rbind(M_final, M)
}
}```

#### Export to check in Excel

```mf_df = M_final %>% as.data.frame()

writexl::write_xlsx(mf_df, "430 Excel solution.xlsx")```

Puzzle #431

This time we have very interesting ranking to make. We have to find which region in sales how many times was on which step of podium. Pretty crazy to think, but not really to code. Let’s go.

```library(tidyverse)

input = read_excel("Excel/431 Top 3 Rankings.xlsx", range = "A1:H20")
test  = read_excel("Excel/431 Top 3 Rankings.xlsx", range = "J2:K5")```

#### Transformation

```result = input %>%
pivot_longer(cols = -c(1), names_to = "year", values_to = "result") %>%
mutate(Rank = dense_rank(desc(result)) %>% as.numeric(), .by = year) %>%
filter(Rank <= 3) %>%
summarise(n = n_distinct(year), .by = c("Region", "Rank")) %>%
mutate(check = n == max(n), .by = "Rank") %>%
filter(check) %>%
summarise(Regions = paste(Region, collapse = ", "), .by = "Rank") %>%
arrange(Rank) ```

#### Validation

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

### Puzzle #432

Ciphering is always fun, and this time again we have some words to code. Technique for current task is Bifid cipher. We take letter, then look for their coordinate in coding square, take them, fold it like phylo pastry, squeeze together, get those squeezed numbers and match with coding square again to get encoded letter. Again easier to code than to explain :).

```library(tidyverse)

input = read_excel("Excel/432 Bifid Cipher_Part 1.xlsx", range = "A1:A10")
test  = read_excel("Excel/432 Bifid Cipher_Part 1.xlsx", range = "B1:B10")```

#### Transformation

```create_coding_square <- function() {
Letters = c(letters[1:9], letters[11:26])
df = as.data.frame(matrix(Letters, nrow = 5, byrow = TRUE)) %>%
pivot_longer(cols = everything()) %>%
mutate(column = as.numeric(str_extract(name, "[0-9]+")),
row = rep(1:5,each =  5)) %>%
select(-name)
return(df)
}

bifid_encode = function(text) {
coding_square = create_coding_square()
text = str_replace_all(text, "J", "I")
chars = str_split(text, "")[[1]]

coords = map_dfr(chars, function(char) {
coords = coding_square %>%
filter(value == char) %>%
select(row, column)
return(coords)
})
coords = paste0(coords\$row, coords\$column) %>%
str_split("", simplify = TRUE) %>%
as.numeric() %>%S
matrix(ncol = 2, byrow = TRUE) %>%
as.data.frame()

encoded = coords %>%
left_join(coding_square, by = c("V1" = "row", "V2" = "column")) %>%
pull(value) %>%
paste0(collapse = "")

return(encoded)
}

result = input %>%
mutate(`Answer Expected` = map_chr(`Plain Text`, bifid_encode)) %>%

#### Validation

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

### Puzzle #433

My favourite type of puzzles — text manipulation. We get some kind of notes that should be placed in table columns, but are squeezed in one. But never mind, it is easy to clear this mess. Even in three different way even.

```library(tidyverse)
library(unglue)

input = read_excel("Excel/433 Text Split.xlsx", range = "A1:A20")
test  = read_excel("Excel/433 Text Split.xlsx", range = "C1:G20")```

#### First approach — separate()

```result = input %>%
separate(Text, into = c("Levels", "Names"), sep = " : ") %>%
separate(Levels, into = c("Level1", "Level2", "Level3"), sep = "\\.") %>%
separate(Names, into = c("First Name", "Last Name"), sep = " ") %>%
select(-c(...2))```

#### Second approach — RegEx

```pattern = "(\\d+)(\\.\\d+)?(\\.\\d+)?\\s*:\\s*(\\w+)\\s+(\\w+)"

result = str_match(input\$Text, pattern) %>%
as_tibble() %>%
select(-c(1)) %>%
mutate(across(c("V2", "V3", "V4"), ~ str_replace(.x, pattern = "[:punct:]", replacement = ""))) %>%
setNames(c("Level1", "Level2", "Level3", "First Name", "Last Name"))```

#### Third approach — unglue

```patterns = c("{Level1}.{Level2}.{Level3} : {`First Name`} {`Last Name`}",
"{Level1}.{Level2} : {`First Name`} {`Last Name`}",
"{Level1} : {`First Name`} {`Last Name`}")

result = input %>%
unglue_unnest(Text, patterns = patterns) %>%
select(Level1,Level2,Level3,`First Name` = `X.First.Name.`,`Last Name`= `X.Last.Name.`)```

#### Validation

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

# for each of three solutions :D```

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.

R Solution for Excel Puzzles was originally published in Numbers around us on Medium, where people are continuing the conversation by highlighting and responding to this story.

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