Number Spiral

Problem Overview

Learning Goals

After solving this problem, you will be able to:

• Recognize diagonal layer patterns in 2D grids
• Derive O(1) formulas for coordinate-based calculations
• Handle alternating direction patterns based on parity
• Convert between 1-indexed coordinates and mathematical formulas

Problem Statement

Problem: A number spiral is an infinite grid where positive integers are written in diagonal layers. Given row y and column x, find the value at position (y, x).

The spiral is constructed as follows:

 1   2   9  10  25 ...
 4   3   8  11  24
 5   6   7  12  23
16  15  14  13  22
17  18  19  20  21
...

Input:

• Line 1: Number of test cases t (1 <= t <= 10^5)
• Lines 2 to t+1: Two integers y and x (1 <= y, x <= 10^9)

Output:

• For each test case, print the value at position (y, x)

Constraints:

• 1 <= t <= 10^5
• 1 <= y, x <= 10^9

Example

Input:
3
2 3
1 1
4 2

Output:
8
1
15

Explanation:

• Position (2, 3): Looking at the grid, row 2 column 3 has value 8
• Position (1, 1): Top-left corner is always 1
• Position (4, 2): Row 4 column 2 has value 15

Intuition: How to Think About This Problem

Pattern Recognition

Key Question: How are numbers arranged in this spiral? What determines the value at any position?

The spiral fills numbers along diagonal layers. Each layer k forms an L-shape starting from position (k, 1) or (1, k). The crucial insight is that the layer number is determined by max(y, x), and within each layer, numbers flow in alternating directions (down/right for odd layers, up/left for even layers).

Breaking Down the Problem

1. What layer is position (y, x) in? Layer k = max(y, x)
2. What’s the starting number of layer k? The corner value at the start of layer k
3. Where in the layer is (y, x)? Count steps from the layer’s starting corner

Analogies

Think of this like floors in a building where each floor has an L-shaped hallway. Floor k starts at room (k-1)^2 + 1 and wraps around in alternating directions. To find a specific room, you determine which floor it’s on, then count how far along the hallway it is.

Visual: Layer Structure

Layer 1:  (1,1)                    Numbers: 1
Layer 2:  (1,2)-(2,2)-(2,1)        Numbers: 2,3,4
Layer 3:  (3,1)-(3,2)-(3,3)        Numbers: 5,6,7
          (2,3)-(1,3)                       8,9
Layer 4:  (1,4)-(2,4)-(3,4)-(4,4)  Numbers: 10,11,12,13
          (4,3)-(4,2)-(4,1)                 14,15,16

Key Observation: Layer k contains numbers from (k-1)^2 + 1 to k^2.

Solution: O(1) Mathematical Formula

Key Insight

The Trick: Every position (y, x) lies on layer k = max(y, x). The layer’s corner value and direction alternation (based on k’s parity) let us compute the answer directly.

Layer Properties

The Formula Derivation

Step 1: Identify the layer

k = max(y, x)

Step 2: Determine if we’re on the row part or column part of the L-shape

• If y >= x: We’re on the vertical part of the L
• If x > y: We’re on the horizontal part of the L

Step 3: Calculate position based on layer parity

For odd layers (k is odd):

• Numbers go DOWN the column, then RIGHT along the row
• At (k, 1): value = (k-1)^2 + 1
• At (k, k): value = (k-1)^2 + k (the corner)
• At (1, k): value = k^2

For even layers (k is even):

• Numbers go RIGHT along the row, then DOWN the column
• At (1, k): value = (k-1)^2 + 1
• At (k, k): value = (k-1)^2 + k (the corner)
• At (k, 1): value = k^2

Algorithm

1. Compute k = max(y, x) to identify the layer
2. If k is odd:
   • If y >= x: answer = (k-1)^2 + y
   • Else: answer = k^2 - x + 1
3. If k is even:
   • If y >= x: answer = k^2 - y + 1
   • Else: answer = (k-1)^2 + x

Dry Run Example

Example 1: Input (y=2, x=3)

Step 1: Find the layer
  k = max(2, 3) = 3

Step 2: Determine parity
  k = 3 is ODD

Step 3: Determine position type
  x > y, so we're on the horizontal part (going up column k)

Step 4: Apply formula
  answer = k^2 - y + 1 = 9 - 2 + 1 = 8

Verify: Grid shows (2,3) = 8

Example 2: Input (y=4, x=2)

Step 1: k = max(4, 2) = 4

Step 2: k = 4 is EVEN

Step 3: y >= x, so we're on the horizontal part (row k, going left)

Step 4: Apply formula
  answer = k^2 - x + 1 = 16 - 2 + 1 = 15

Verify: Grid shows (4,2) = 15

Example 3: Input (y=1, x=1)

Step 1: k = max(1, 1) = 1

Step 2: k = 1 is ODD

Step 3: y >= x

Step 4: answer = (k-1)^2 + y = 0 + 1 = 1

Verify: Origin is always 1

Summary of Formulas:

k = max(y, x)

If k is ODD:
  - If y >= x: answer = (k-1)^2 + y     # vertical part
  - If x > y:  answer = k^2 - y + 1     # horizontal part

If k is EVEN:
  - If y >= x: answer = k^2 - x + 1     # horizontal part
  - If x > y:  answer = (k-1)^2 + x     # vertical part

Visual Diagram

Layer k determines the "ring" of the spiral:

k=1     k=2       k=3         k=4
 1    | 2  9   | 2  9     | 2  9  10
      | 4  3   | 4  3  8  | 4  3   8  11
               | 5  6  7  | 5  6   7  12
                          |16 15  14  13

Odd layers (1,3,5...): Fill DOWN then UP
Even layers (2,4,6...): Fill RIGHT then DOWN

Code

Python Solution:

def solve(y: int, x: int) -> int:
  """
  Find value at position (y, x) in the number spiral.

  Time: O(1)
  Space: O(1)
  """
  k = max(y, x)  # Layer number

  if k % 2 == 1:  # Odd layer
    if y >= x:
      return (k - 1) * (k - 1) + y
    else:
      return k * k - y + 1
  else:  # Even layer
    if y >= x:
      return k * k - x + 1
    else:
      return (k - 1) * (k - 1) + x


def main():
  t = int(input())
  results = []
  for _ in range(t):
    y, x = map(int, input().split())
    results.append(solve(y, x))
  print('\n'.join(map(str, results)))


if __name__ == "__main__":
  main()

Complexity

Common Mistakes

Mistake 1: Forgetting Direction Alternation

# WRONG - Using same formula for all layers
def solve_wrong(y, x):
  k = max(y, x)
  if y >= x:
    return (k - 1) * (k - 1) + y  # Wrong for even layers!
  else:
    return k * k - y + 1

Problem: The spiral alternates direction between odd and even layers. Using the same formula ignores this.

Fix: Check k % 2 and use the appropriate formula for each parity.

Mistake 2: Using 0-indexed Coordinates

# WRONG - Treating input as 0-indexed
def solve_wrong(y, x):
  y += 1  # DON'T do this!
  x += 1
  k = max(y, x)
  # ...

Problem: The problem uses 1-indexed coordinates. Adding 1 shifts all answers incorrectly.

Fix: Use the coordinates as given; they are already 1-indexed.

Mistake 3: Integer Overflow

Problem: When y, x can be up to 10^9, k^2 can be up to 10^18, exceeding 32-bit int range.

Fix: Use long long for all calculations involving k^2.

Mistake 4: Wrong Boundary Condition

# WRONG - Using > instead of >=
if y > x:  # Should be y >= x
  return (k - 1) * (k - 1) + y

Problem: When y == x (diagonal), the position belongs to a specific part of the L-shape depending on direction.

Fix: The condition should be y >= x for odd layers and the same for even layers.

Edge Cases

When to Use This Pattern

Use This Approach When:

• The problem involves a 2D grid with a predictable filling pattern
• You need O(1) lookup for coordinates in a spiral or diagonal pattern
• Layer-based decomposition reveals a mathematical structure
• Values follow a sequential ordering along some path

Don’t Use When:

• The spiral has irregular or input-dependent patterns
• You need to traverse or modify the actual grid
• The pattern doesn’t have a closed-form mathematical description

Pattern Recognition Checklist:

• Values arranged in concentric layers or rings? -> Layer-based formula
• Pattern alternates based on layer/ring number? -> Check parity (odd/even)
• Need value at arbitrary coordinate? -> Derive O(1) closed-form formula
• Sequential filling along a path? -> Identify path direction and starting points

Related Problems

Similar Difficulty

Harder (Do These After)

Key Takeaways

1. The Core Idea: The spiral can be decomposed into layers where max(y, x) determines the layer, and parity determines the direction.

2. Mathematical Insight: Each layer k contains numbers from (k-1)^2 + 1 to k^2, with 2k-1 cells.

3. Direction Matters: Odd layers fill down-then-right; even layers fill right-then-down. Getting this wrong breaks the solution.

4. Pattern: This is a coordinate-to-value mapping problem solvable with O(1) formulas once you understand the structure.

Practice Checklist

Before moving on, make sure you can:

• Draw the first 5 layers of the spiral from memory
• Derive the formula for odd and even layers independently
• Explain why max(y, x) determines the layer
• Handle the test cases without running the code
• Implement in your preferred language in under 5 minutes

Additional Resources

• CSES Number Spiral - Spiral pattern queries
• Spiral Matrix patterns on CP-Algorithms