Three Cube Packing for All Dimensions

: Let V n ( d ) denote the least number, such that every collection of n d -cubes with total volume 1 in d -dimensional (Euclidean) space can be packed parallelly into some d -box of volume V n ( d ) . We show that V 3 ( d ) = r 1 − d d if d ≥ 11 and V 3 ( d ) =


Introduction
In 1966 (according to [1], in 1963 according to [2]), Leo Moser spread a collection of 50 problems named "Poorly Formulated Unsolved Problems of Combinatorial Geometry".The collection consisted of only mimeographed copies, and was not fully published in its original form until 1991 in [1].Problem 7 was "What is the smallest number A such that every set of squares of total area 1 can be accommodated in some rectangle of area A?".This problem can also be found in [2][3][4][5].
The problem has been extended to higher dimensions, and has been studied for a specific number of squares (cubes).We reformulate the problem, to distinguish between the number of dimensions and cubes and to clarify it.
Packing of a (finite or infinite) collection of d-dimensional cubes (d-cubes, for short) into a d-dimensional rectangular parallelepiped (d-box, for short) means that the union of the d-cubes is a subset of the d-box and the intersection of the interior of any two d-cubes is the empty set.The packing, in which each edge of any packed d-cube is parallel to an edge of d-box, is called parallel packing.
We denote by V n (d) the least number such that every collection of n d-cubes with the total volume 1 in d-dimensional (Euclidean) space can be packed parallelly into some d-box of volume V n (d).V(d) denotes the maximum of all V n (d), n = 1, 2, 3, . . .Most of the results are for two-dimensional space.In 1967, Moon and Moser [4] proved that 1.2 ≤ V(d) ≤ 2. In 1970, Kleitman and Krieger [6] proved that V(2) ≤ √ 3 < 1.733, and the rectangle with edge lengths 1 and √ 3 is sufficient.Five years later, Kleitman and Krieger [7] again proved that V(2) ≤ 4 √ 6 .
Packing squares into a rectangle is an over half-century-old problem, and even though there are multiple partial results, it remains unresolved.We investigated a modified problem: packing three d-cubes in d-dimensional space.Some results for smaller dimensions are known.We provide the solution for all dimensions d ≥ 2.
, where r is the only solution of the equation , where r is the only solution of the equation 2 , 1 .The maximum volume is achieved by d-cubes with edges x, y, z, such that: Proof.Let x, y, z be the edge lengths of d-cubes in the d-dimensional Euclidean space (d ≥ 2), where 1 > x ≥ y ≥ z > 0 and the total volume Let k and m be real numbers, such that y = kx and z = mx.In the proof, we use three constraints: 1. Three d-cubes can be packed only in two meaningful ways; see Figure 1.
be the function of the volume of the packing, as shown in Figure 1a be the function of the volume of the packing, as shown in Figure 1b.
It is shown as the triangle ABC in Figure 2. The domain is the same for each d ≥ 2.
Our goal is to find the global maximum of G for each dimension (d ≥ 2) and the edge lengths x, y, z for which it occurs. If gives the curve PB: 1 = k(k + m).PB is continuous and divides the triangle ABC, into two regions (see Figure 2): 1.
Region C 1 where 1 ≤ k(k + m) holds.Therefore, W 1 ≤ W 2 and, consequently, Region C 2 where 1 ≥ k(k + m) holds.Therefore, W 1 ≥ W 2 and, consequently, The curve PB belongs to both regions.The point P = ( 2 ) is used in the proofs several times, . For the sake of clarity, the rest of the proof is divided into nine claims.

Proof. ∂W
Assume, for the sake of contradiction, that if d ≥ 8, then Equation (1) holds inside of the region C 2 .We calculate an upper bound of the right-hand side of Equation ( 1) and we obtain 1 Assume, for the sake of contradiction, that if 2 Step 1: From Equation ( 1), it follows that We calculate a lower bound of the right-hand side of Equation ( 2) using We minimize the numerator (k = m = 0.5) and maximize the denominator is successively greater than 0.530, 0.543, 0.584, 0.623, 0.656, 0.684.Therefore, m is at least 0.53.
Step 2: We repeat the calculations analogously to step 1.
Step 3: We repeat the calculations, for the last time.
For 2 ≤ d ≤ 7, the right-hand side of inequality m ≥ 0.73 d +0.58 d +1 ( is successively greater than 0.661, 0.611, 0.627, 0.651, 0.675, 0.697; m ≥ 0.61. Assume, for the sake of contradiction, that if 2 ≤ d ≤ 6, then W AB ≥ W 2 (P) at some denote the right-hand side, and t(k) increases on 2 .Substituting it into W 2 , we obtain W 2 .Therefore, the global maximum of W AP occurs at point P, which also belongs to PB.It implies, together with Claims 2 and 3, that the global maximum of W 2 must occur on PB.

Claim 5.
The global maximum of G does not occur at points P, B, C. Proof.BC is a part of a line k = 1 if m ∈ (0, 1).Substituting it into W 1 , we obtain W 1 (1, m) = m+2 m d +2 .We denote it by W BC (m), m ∈ (0, 1).We show that, for each m 0 ∈ (0, 1), there exists k 0 , such that point (k 0 , m 0 ) ∈ C 1 and 2 , then the line k = 1 − m 3 is a part of C 1 (see the dash-dotted line in Figure 2).Assume, for the sake of contradiction, that and d ≥ 2, then the global maximum of W 1 does not occur on BC.
So, the global maximum of W 1 does not occur on BC.
The previous claims imply that the global maximum of G must occur on interior of PC or PB.
. We denote it by W PB (k), Proof.
Let h(k) denote the left-hand side of Equation ( 4).The function h(k) is increasing and  Proof.Assume, for the sake of contradiction, that if d ≥ 11, then W PB ≥ G(P) at some k ∈ [0.74, 1).Using the endpoints of the interval in W PB we obtain an upper bound of W PB on the interval.The inequality gives 37 87 37 50 After setting d = 11 and d = 12 into the inequality, we see that they do not satisfy it.The left-hand side is always greater than 37 87 , and the right-hand side is decreasing.If d ≥ 13, then the right-hand side is even smaller than 37 87 .We obtain a contradiction.Therefore, if d ≥ 11, then the global maximum of W PB does not occur on [0.74, 1).
Assume, for the sake of contradiction, that if d ≥ 11, then 2 , 0.74 , we obtain: Using the endpoints of the interval in Equation ( 5), we obtain an upper bound of the left-hand side and a lower bound of the right-hand side.Proof.We remove the fractions from Equation ( 5) by multiplying it by k d .The global maximum of G does not occur on BC, so we remove the root k = 1 by dividing the equation by (k − 1), and we obtain: If d is odd, then we divide the equation additionally by (k + 1) 2 .If d = 2, then we divide the equation additionally by (2k + 1).Removing these roots is not necessary, but we reduce the degree of the polynomial in this way.With the help of Sturm's theorem, we prove that Equation ( 6) has exactly one solution on √ 2 2 , 1 .Table 1 shows Sturm chain for d = 5.

Sign at
Sign at 1  , y = z = rx.

Discussion
There are three d-cubes, that is, three variables.In the previous proofs [20][21][22][23], the apparent substitution z d = 1 − x d − y d is used to achieve only two variables.As a result of it, the domain boundaries x d + y d = 1 and x d + 2y d = 1 change as the dimension changes.For example, in [23], the shape of the curve C : x d y d + y 2d − y d + (x 2 − y 2 ) d = 0 is analyzed depending on the dimension, concluding: ". . . the shape of the curve C is similar . ..".The word "similar" (but not the same) is essential.The curve C is also dimension dependent.(The curve C in previous proofs has the same role as our curve PB.) efficiently we can fill a region with identical objects.The study of maximal packing helps us understand the optimal arrangement of shapes in different dimensions.
• Materials science and crystal structures: In materials science, maximal packing is crucial for understanding the arrangement of atoms or molecules in crystalline structures.Crystals exhibit specific packing arrangements (e.g., face-centered cubic, hexagonal close-packed) that maximize the density of particles while maintaining stability.The efficiency of packing affects material properties such as hardness, conductivity, and optical behavior.

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Optimization and efficiency: Maximal packing problems often arise in optimization scenarios.Solving these problems has practical applications in logistics, manufacturing, and resource utilization.• Computational complexity: Determining the optimal packing arrangement can be computationally challenging.Researchers use heuristics, algorithms, and mathematical techniques to approximate solutions.The study of maximal packing contributes to our understanding of computational complexity and algorithmic efficiency.

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In historical context: ancient civilizations (such as the Egyptians and Babylonians) were interested in efficient packing for practical reasons (e.g., storing grain, arranging bricks) and Kepler's conjecture about the densest sphere packing in three dimensions dates back to the 17th century.
In summary, maximal packing plays a vital role in understanding spatial arrangements, optimizing resource usage, and solving complex problems across various disciplines.

Figure 1 .
Figure 1.Two cases of packing three d-cubes.(a) Packing used by W 1 .(b) Packing used by W 2 .

Figure 2 .
Figure 2. The domain of the function G(k, m).

2 .Claim 2 .
The derivatives are never undefined.The equation ∂W 1 ∂k = 0 = ∂W 1 ∂m gives k = m (line segment PC).So, W 1 has no local extremum inside of the region C 1 .The global maximum of W 1 must occur on the boundary of the region C 1 .The global maximum of W 2 must occur on the boundary of the region C 2 .
We denote the polynomial on the left-hand side by p(k).It has one sign change, as the sequence of signs is +, +|−, − (or +|−, − if d = 2).Therefore, according to Descartes' rule of signs, it has exactly one positive real root.If d ≥ 2, then p(2) = 2 d+1 (3d − 5) − 5 is positive.Setting 2 ≤ d ≤ 5 in p 72 100 = 122) − 61 , we see that it is negative.If d ≥ 5, then the exponential term dominates and p 72 100 is decreasing.So, if d ≥ 2, then p 72 100 is negative.Therefore, if d ≥ 2, then the only positive root of W ′ AP = 0 is greater than 0.72.Since W ′ AP 1 2

Claim 7 .
If d ≤ 10, then the global maximum of G must occur on the interior of PB.If d ≥ 11, then W PC has exactly one critical point on PC.

√ 2 2 , 1 .Claim 8 .
Therefore, if d ≥ 11, then W PC has exactly one critical point on PC.If d ≥ 11, then the global maximum of G must occur on the interior of PC.
Claim 3. The global maximum of W 2 does not occur on AB.
therefore Equation (4) holds on ≥ 1.From the first inequality, we obtain √ 2 + 2 d ≤ 2 d/2 + 2, which only holds for d ≥ 11, where h(1) ≥ 1 holds for d ≥ 2. So, if 2 ≤ d ≤ 10, then W PC has no local extremum inside of PC.Therefore, if 2 ≤ d ≤ 10, then the global maximum of W PC must occur at P or at C, but G does not attain the global maximum at P or C. Therefore, if 2 ≤ d ≤ 10, then the global maximum of G must occur on the interior of PB.The function h(k) is increasing and continuous if k > 0; therefore, if d ≥ 11, then Equation (4) has exactly one solution on .95, then the left-hand side is 0, if d ≥ 7.23, then the left-hand side is decreasing.Therefore, if d ≥ 11, then the left-hand side is always negative.This is a contradiction.Hence, if d ≥ 11, then W PB has no local extremum on ≥ 11, then the global maximum of W PB occurs at P or at B, but G does not attain the global maximum at P or at B. Therefore, if d ≥ 11, then the global maximum of G must occur on the interior of PC.If 2 ≤ d ≤ 10, then W PB has exactly one critical point inside PB.

Table 2 .
The table shows real numbers, but fractions are used in the calculation.−|+,+|−,−, −| + |−, i.e., four sign changes.There are three sign changes at p i (1).The difference of these values is the number of real roots on ( ≤ d ≤ 10, we only show the sign patterns and the number of sign changes in Table2.Summary of Sturm chain for 2 ≤ d ≤ 10.