library(tidyverse)
library(readxl)
library(combinat)
path = "Excel/800-899/816/816 Penholodigial Numbers.xlsx"
test = read_excel(path, range = "A1:A31", col_types = "numeric") %>% pull()
squares_df = (ceiling(sqrt(123456789)):floor(sqrt(987654321)))^2
squares_df = squares_df[map_lgl(squares_df, ~ setequal(as.integer(str_split(.x, "", simplify = TRUE)), 1:9))]
all.equal(squares_df, test)
# [1] TRUEExcel BI - Excel Challenge 816
excel-challenges
excel-formulas
🔰 List all Penholodigital squares.
Challenge Description
🔰 List all Penholodigital squares. A Penholodigital square is perfect square number that contains every digit from 1 to 9 exactly once, not including zero.
Solutions
- Logic: Read the workbook ranges needed for the challenge; Parse the packed text or string structure.
- 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
import numpy as np
path = "800-899/816/816 Penholodigial Numbers.xlsx"
test_raw = pd.read_excel(path, usecols="A", nrows=31, header=None).iloc[:, 0]
test = [int(x) for x in test_raw if str(x).isdigit()]
squares = [i**2 for i in range(int(np.ceil(np.sqrt(123456789))), int(np.floor(np.sqrt(987654321))) + 1)]
is_pandigital = lambda n: set(str(n)) == set('123456789') and len(str(n)) == 9
squares_df = [sq for sq in squares if is_pandigital(sq)]
result = squares_df == test
print(result)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.