Another Better Lower Bound for $n=17$ Square Packing

By Gustavo | August 21, 2026

The objective of this work is to refine a recent discovery and demonstrate that $4.5058\ (?)\le s(17)$ by utilizing specific weights. To begin, we must establish the definition of $s(n)$.

Definition: Let $s(n)$ represent the side length of the smallest possible square capable of enclosing $n$ unit squares.

Context and Background

For $n=15$, it is trivial to see that the squares fit in a $4 \times 4$ area, meaning $s(15) \le 4$. However, proving that this is the absolute minimum is quite difficult; this was eventually settled by Erich Friedman in 1999, confirming $s(15)=4$.

Regarding $n=17$, the obvious intuitive enclosing square is $5 \times 5$. However, in 1998, John Bidwell demonstrated a configuration showing that a square of $4.6756\dots$ is sufficient, thus $s(17) \le 4.6756\dots$. This arrangement is fascinating and worth reviewing in the general collection of square packings.

Conversely, Trevor Green established in 2000 that $4.4452\dots \le s(17)$. This left a significant gap in our knowledge: $4.4452\dots \le s(17) \le 4.6756\dots$

The Evolution of the Lower Bound

Recently, Sam Burns (utilizing ChatGPT 5.6 Sol) proposed an improved lower bound. While I believe the logic is sound, there is always a possibility of a missed corner case in the code or proof. Consequently, I have marked these newer values with a $(?)$.

My primary critique of Burns' work was the lack of a high-quality graphic. By refining the program, I have pushed the lower bound further to $4.5058$. The current state of the bound is: $4.4452\dots \le 4.4811\ (?)\ \le 4.5058\ (?)\ \le s(17) \le 4.6756\dots$

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Analysis of Trevor Green's Bound

Green's proof for $\frac{19+40\sqrt{2}}{17} \approx 4.4452\dots \le s(17)$ relies on identifying 16 "unavoidable" points within a square of that side length. Through geometric proof, he showed that any unit square placed within this area must cover at least one of these points. Therefore, attempting to pack 17 unit squares would force at least two squares to overlap at one of these 16 points.

While the analytical definitions were missing, I deduced the following margins based on the formula:

• Left/Right Margin: $0.5$
• Top/Bottom Margin: $\sqrt{2} - 0.5 \approx 0.9142\dots$

This configuration creates a diagonal segment of length $1$, which is ideal for constructing triangles with unavoidable vertices. The grid is $6 \times 4$ with a total size of $2.6168\dots \times 2.4452\dots$. In a symmetrized version, each unit square contains at least 4 points, though some "thicker" points count double.

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Analysis of Sam Burns' Bound

Burns' approach to prove $4.4811\ (?)\ \le s(17)$ utilizes 268 points with varying weights, totaling a weight of $16.9476$. The verification process involves a Python program that tests a finite set of directions for "almost-unit" squares.

# Conceptual logic of the verification program
for square in almost_unit_squares(size=0.9973):
    if sum_of_weights(square) < 1.0003:
        return "Bound potentially invalid"
return "Bound verified"

If the program confirms the weight sum is always $\ge 1.0003$, the bound holds. However, the $0.5$ margin seems too narrow compared to other examples. I noticed non-zero weights at grid position (1, 11) and its symmetric counterparts; I suspect these might be unnecessary.

Visualization Details: To render the points as circles, the radius is determined by: $r = \sqrt{\text{weight}^{1/\gamma}} \times \text{scale}$

• $\gamma = 1.0$: Area is proportional to weight (small weights are too tiny).
• $\gamma = 2.0$: Better visibility for smaller weights.
• scale = 0.07: Optimized for my display.

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The New Bound: $4.5058\ (?)$

My strategy involved experimenting with different internal grid sizes and margins. Since the origin of Burns' weight selection was unclear, I employed a more systematic approach:

• Use Linear Programming to determine weights for fixed sizes.
• Apply a mix of brute force search and intuition to optimize the grid.
• Round weights into clean fractions for aesthetic and practical purposes.

After extensive iteration, I achieved a lower bound of $4.5058\ (?)$, utilizing 168 points. (Further technical specifics will be detailed in a subsequent post).

Additional References: