R Solution for Excel Puzzles

Puzzles no. 399–403

Puzzles

Author: ExcelBI

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

Puzzle #399

Today we have been given a list of random strings containing of certain number of letters duplicated. And our task is to count how many of each letters are there and present it as alphabetically pasted string. Sounds nice and it is nice. Let’s go.

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Excel/399 Counter Dictionary.xlsx", range = "A1:A10")
test  = read_excel("Excel/399 Counter Dictionary.xlsx", range = "B1:B10")

Transformation

count_chars = function(string) {
  chars = string %>%
    str_split(., pattern = "") %>%
    unlist() %>%
    tibble(char = .) %>% 
    group_by(char) %>%
    summarise(count = n()) %>%
    ungroup() %>%
    arrange(char) %>%
    unite("char_count", c("char", "count"), sep = ":") %>%
    pull(char_count) %>%
    str_c(collapse = ", ")
  
  return(chars)
}

result = input %>%
  mutate(`Answer Expected` = map_chr(String, count_chars)) %>%
  select(-String)

Validation

identical(result, test)
# [1] TRUE

Puzzle #400

Once again we are playing with coordinates and checking if they form one structure. But this time vertices are mixed and we have some more to do.
In this puzzle I will give you one surprise. Be patient.

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Excel/400 Connected Points_v2.xlsx", range = "A1:D8")
test  = read_excel("Excel/400 Connected Points_v2.xlsx", range = "E1:E8")

Transformation

result = input %>%
  mutate(row = row_number()) %>%
  select(row, everything()) %>%
  pivot_longer(-row, names_to = "col", values_to = "value") %>%
  select(-col) %>%
  na.omit() %>%
  group_by(row) %>%
  separate_rows(value, sep = ", ") %>%
  group_by(row, value) %>%
  summarise(n = n()) %>%
  ungroup() %>%
  select(-value) %>%
  group_by(n, row) %>%
  summarise(count = n()) %>%
  ungroup() %>%
  filter(n == 1) %>%
  mutate(`Answer Expected` = ifelse(count == 2, "Yes", "No")) %>%
  select(`Answer Expected`)

Validation

identical(test, result)
# [1] TRUE

Optimized version

I asked AI chat to optimize my code from above, because I don’t really like when my code is to long without a purpose. So I tried it, and that is really a surprise.

result2 <- input %>%
  mutate(`Answer Expected` = pmap_chr(., ~ {
    unique_values <- na.omit(c(...))
    if (length(unique(unique_values)) == 2) "Yes" else "No"
  })) %>%
  select(`Answer Expected`)


identical(test, result2)
# [1] TRUE

Puzzle #401

I am not using matrices in my daily work often, but I really like puzzles in which I can use them to solve. Today we have to form triangle from string. We have to bend it to size of matrix. Let’s try.

Loading libraries and data

library(tidyverse)
library(readxl)

input1 = read_excel("Excel/401 Make Triangle.xlsx",
                    range = "A2:A2", col_names = F) %>% pull()
input2 = read_excel("Excel/401 Make Triangle.xlsx",
                    range = "A5:A5", col_names = F) %>% pull()
input3 = read_excel("Excel/401 Make Triangle.xlsx",
                    range = "A9:A9", col_names = F) %>% pull()
input4 = read_excel("Excel/401 Make Triangle.xlsx",
                    range = "A14:A14", col_names = F) %>% pull()
input5 = read_excel("Excel/401 Make Triangle.xlsx",
                    range = "A19:A19", col_names = F) %>% pull()

test1 = read_excel("Excel/401 Make Triangle.xlsx",
                   range = "C2:D3", col_names = F) %>% as.matrix(.)
dimnames(test1) = list(NULL, NULL)
test2 = read_excel("Excel/401 Make Triangle.xlsx",
                   range = "C5:D7",col_names = F) %>% as.matrix(.)
dimnames(test2) = list(NULL, NULL)
test3 = read_excel("Excel/401 Make Triangle.xlsx",
                   range = "C9:E12",col_names = F) %>% as.matrix(.)
dimnames(test3) = list(NULL, NULL)
test4 = read_excel("Excel/401 Make Triangle.xlsx",
                   range = "C14:F17", col_names = F) %>% as.matrix(.)
dimnames(test4) = list(NULL, NULL)
test5 = read_excel("Excel/401 Make Triangle.xlsx",
                   range = "C19:G23", col_names = F)  %>% as.matrix(.)
dimnames(test5) = list(NULL, NULL)

Transformation and validation

triangle = function(string) {
  chars = str_split(string, "") %>% unlist()
  nchars = length(chars)
  positions = tibble(row = 1:10) %>%
    mutate(start = cumsum(c(1, row[-5])),
           end = start + row - 1)
  nrow = positions %>%
    mutate(nrow = map2_dbl(start, end, ~ sum(.x <= nchars &
                                               nchars <= .y))) %>%
    filter(nrow == 1) %>%
    pull(row)
  M = matrix(NA, nrow = nrow, ncol = nrow)
  
  for (i in 1:nrow) {
    M[i, 1:i] = chars[positions$start[i]:positions$end[i]]
  }
  
  FM = M %>%
    as_tibble() %>%
    select(where( ~ !all(is.na(.)))) %>%
    as.matrix()
  dimnames(FM) = list(NULL, NULL)
  
  
  return(FM)
}
identical(triangle(input1), test1) # TRUE
identical(triangle(input2), test2) # TRUE
identical(triangle(input3), test3) # TRUE
identical(triangle(input4), test4) # TRUE
identical(triangle(input5), test5) # TRUE

Puzzle #402

One of common topics in our series is of course cyphering. And today we have again some spy level puzzle. We have some phrase and keyword using which we need to code given phrase. Few weeks ago there was puzzle when lacking letters in keyword were taken from coded phrase. Today we are repeating key how many times we need. And there is one more detail, we have to handle spaces as well. Not so simple, but satisfying.

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Excel/402 Vignere Cipher.xlsx", range = "A1:B10")
test  = read_excel("Excel/402 Vignere Cipher.xlsx", range = "C1:C10")

Transformation

code = function(plain_text, key) {
  coding_df = tibble(letters = letters, numbers = 0:25)
  
  plain_text_clean = plain_text %>% 
    str_remove_all(pattern = "\\s") %>% 
    str_split(pattern = "") %>%
    unlist()
  
  key = str_split(key, "") %>% unlist()
  key_full = rep(key, length.out = length(plain_text_clean))

  df = data.frame(plain_text = plain_text_clean, key = key_full) %>%
    left_join(coding_df, by = c("plain_text" = "letters")) %>%
    left_join(coding_df, by = c("key" = "letters")) %>%
    mutate(coded = (numbers.x + numbers.y) %% 26) %>%
    select(coded) %>%
    left_join(coding_df, by = c("coded" = "numbers")) %>%
    pull(letters)
  
  words_starts = str_split(plain_text, " ") %>% 
    unlist() %>% 
    str_length() 
  
  words = list()
  
  for (i in 1:length(words_starts)) {
    if (i == 1) {
      words[[i]] = paste(df[1:words_starts[i]], collapse = "")
    } else {
      words[[i]] = paste(df[(sum(words_starts[1:(i-1)])+1):(sum(words_starts[1:i]))], collapse = "")
    }
  }
  
  words = unlist(words) %>% str_c(collapse = " ")
  
  return(words)
}

result = input %>%
  mutate(`Answer Expected` = map2_chr(`Plain Text`, Keyword, code))

Validation

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

Puzzle #403

We are summarizing some values into year brackets. Usually you do it using crosstab. And our job today is to make crosstab that is not excel crosstab, but should work like it. From R side usually you have to make pivot, but I didn’t. So we have pivot table (another word for crosstab), without using pivot neither in R nor in Excel. How? Look on it.

Loading libraries and data

library(tidyverse)
library(readxl)

input = read_excel("Excel/403 Generate Pivot Table.xlsx", range = "A1:B100")
test  = read_excel("Excel/403 Generate Pivot Table.xlsx", range = "D2:F9")

Transformation

result = input %>%
  add_row(Year = 2024, Value = 0) %>% ## just to have proper year range at the end
  mutate(group = cut(Year, breaks = seq(1989, 2024, 5), labels = FALSE, include.lowest = TRUE)) %>%
  group_by(group) %>%
  summarize(Year = paste0(min(Year), "-", max(Year)), 
            `Sum of Value` = sum(Value)) %>%
  ungroup() %>%
  mutate(`% of Value` = `Sum of Value`/sum(`Sum of Value`)) %>%
  select(-group)

Validation

identical(result, test)
# [1] TRUE

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