Excel BI - Excel Challenge 651

🔰 Create the vortex grid of numbers (n x n) for a value given in column A.

Illustration for Excel BI - Excel Challenge 651

Challenge Description

🔰 Create the vortex grid of numbers (n x n) for a value given in column A. Samples are given for 2, 4 and 7.

Solutions

R Solution

library(tidyverse)
library(readxl)

path = "Excel/651 Vortex Sequence.xlsx"
test2 = read_excel(path, range = "B2:C3", col_names = F) %>% as.matrix()
test4 = read_excel(path, range = "B6:E9", col_names = F) %>% as.matrix()
test7 = read_excel(path, range = "B12:H18", col_names = F) %>% as.matrix()

spiral_matrix <- function(n) {
  m <- matrix(0, n, n)
  directions <- list(c(-1, 0), c(0, 1), c(1, 0), c(0, -1))
  dir_idx <- 1
  pos <- c(n, 1)
  
  for (i in 1:(n * n)) {
    m[pos[1], pos[2]] <- i
    next_pos <- pos + directions[[dir_idx]]
    
    if (!(all(next_pos >= 1 & next_pos <= n) && m[next_pos[1], next_pos[2]] == 0)) {
      dir_idx <- (dir_idx %% 4) + 1
      next_pos <- pos + directions[[dir_idx]]
    }
    pos <- next_pos
  }
  m
}

all.equal(spiral_matrix(2), test2, check.attributes = F) # TRUE
all.equal(spiral_matrix(4), test4, check.attributes = F) # TRUE
all.equal(spiral_matrix(7), test7, check.attributes = F) # TRUE

R Analysis

• Logic: Read the workbook ranges needed for the challenge; Iterate through the sequence until the rule is satisfied.
• Strengths: The algorithm is explicit about the sequence rule, so the control flow is easy to validate against the prompt.
• 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 non-obvious part is the local rule inside the loop, because that rule determines the whole output.

Python Solution

import pandas as pd
import numpy as np

path = "651 Vortex Sequence.xlsx"
test2 = pd.read_excel(path,  usecols="B:C", skiprows=1,  nrows = 2, header=None).to_numpy()
test4 = pd.read_excel(path,  usecols="B:E", skiprows = 5, nrows = 4, header=None).to_numpy()
test7 = pd.read_excel(path,  usecols="B:H", skiprows = 11, nrows = 7, header=None).to_numpy()

def spiral_matrix(n):
    matrix = np.zeros((n, n), dtype=int)
    directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]
    dir_idx = 0
    pos = [n-1, 0] 
    
    for i in range(1, n * n + 1):
        matrix[pos[0], pos[1]] = i
        next_pos = [pos[0] + directions[dir_idx][0], pos[1] + directions[dir_idx][1]]
        
        if (not (0 <= next_pos[0] < n and 0 <= next_pos[1] < n) or 
            matrix[next_pos[0], next_pos[1]] != 0):
            dir_idx = (dir_idx + 1) % 4
            next_pos = [pos[0] + directions[dir_idx][0], pos[1] + directions[dir_idx][1]]
        pos = next_pos
    
    return matrix

print(np.array_equal(test2, spiral_matrix(2))) # True
print(np.array_equal(test4, spiral_matrix(4))) # True
print(np.array_equal(test7, spiral_matrix(7))) # True

Python

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.

Related Challenges

• Excel Challenge 652
• Excel Challenge 653
• Excel Challenge 654