Excel BI - Excel Challenge 730

excel-challenges

excel-formulas

🔰 List the odd numbers in a Square

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 730

Challenge Description

🔰 List the odd numbers in a Square

Solutions

library(tidyverse)
library(readxl)
library(matricks)

path = "Excel/700-799/730/730 Pick the Odd Numbers in a Grid Across Diagonals.xlsx"
input1 = read_excel(path, range = "A2:B3", col_names = FALSE) %>% as.matrix()
input2 = read_excel(path, range = "A5:C7", col_names = FALSE) %>% as.matrix()
input3 = read_excel(path, range = "A9:C11", col_names = FALSE) %>% as.matrix()
input4 = read_excel(path, range = "A13:D16", col_names = FALSE) %>% as.matrix()
input5 = read_excel(path, range = "A18:E22", col_names = FALSE) %>% as.matrix()
test1 = read_excel(path, range = "G2", col_names = FALSE) %>% pull()
test2 = NA
test3 = read_excel(path, range = "G9", col_names = FALSE) %>% pull()
test4 = read_excel(path, range = "G13", col_names = FALSE) %>% pull()
test5 = read_excel(path, range = "G18", col_names = FALSE) %>% pull()

check_diag = function(mat) {
  diag = diag(mat) %>% paste(collapse = "") %>% as.numeric()
  antidiag = antidiag(mat) %>% paste(collapse = "") %>% as.numeric()
  if (all(diag %% 2 == 1) && all(antidiag %% 2 == 1)) {
    return(paste(diag, antidiag, sep = ", "))
  } else if (all(diag %% 2 == 1)) {
    return(diag)
  } else if (all(antidiag %% 2 == 1)) {
    return(antidiag)
  } else {
    return(NA)
  }
}

all.equal(check_diag(input1), test1, check.attributes = FALSE) # TRUE
all.equal(check_diag(input2), test2, check.attributes = FALSE) # TRUE
all.equal(check_diag(input3), test3, check.attributes = FALSE) # TRUE
all.equal(check_diag(input4), test4, check.attributes = FALSE) # TRUE
all.equal(check_diag(input5), test5, check.attributes = FALSE) # TRUE

Logic: Read the workbook ranges needed for the challenge.
Strengths: The code maps the workbook rule into a compact, reproducible pipeline.
Areas for Improvement: The solution assumes the workbook layout and selected ranges remain stable, so any structural change in the sheet would require small adjustments.
Gem: The elegant part is how little code is needed once the correct intermediate representation is chosen.

import pandas as pd

path = "700-799/730/730 Pick the Odd Numbers in a Grid Across Diagonals.xlsx"

input1 = pd.read_excel(path, header=None, usecols="A:B", skiprows=1, nrows=2).to_numpy()
input2 = pd.read_excel(path, header=None, usecols="A:C", skiprows=4, nrows=3).to_numpy()
input3 = pd.read_excel(path, header=None, usecols="A:C", skiprows=8, nrows=3).to_numpy()
input4 = pd.read_excel(path, header=None, usecols="A:D", skiprows=12, nrows=4).to_numpy()
input5 = pd.read_excel(path, header=None, usecols="A:E", skiprows=17, nrows=5).to_numpy()
test1 = str(pd.read_excel(path, header=None, usecols="G", skiprows=1, nrows=1).iloc[0,0])
test2 = ""
test3 = str(pd.read_excel(path, header=None, usecols="G", skiprows=8, nrows=1).iloc[0,0])
test4 = pd.read_excel(path, header=None, usecols="G", skiprows=12, nrows=1).iloc[0,0]
test5 = pd.read_excel(path, header=None, usecols="G", skiprows=17, nrows=1).iloc[0,0]

def print_odd_diagonals(m):
    n = len(m)
    d1 = int(''.join(str(m[i][i]) for i in range(n)))
    d2 = int(''.join(str(m[i][n-1-i]) for i in range(n)))
    odds = [str(x) for x in {d1, d2} if x % 2]
    if not odds:
        return ""
    return ', '.join(odds)


print(print_odd_diagonals(input1) == (test1)) # True
print(print_odd_diagonals(input2) == (test2)) # True
print(print_odd_diagonals(input3) == (test3)) # True
print(print_odd_diagonals(input4) == (test4)) # True
print(print_odd_diagonals(input5) == (test5)) # True

The Python version keeps the algorithm explicit, which helps when the challenge depends on a greedy or iterative rule.

Difficulty Level

Easy / Medium

The business rule is clear, though the workbook still needs a few transformation steps to reach the expected output.