TAOCP 7.2.2.1 Exercise 351
Let $s = a+b+c+d+e$.
Section 7.2.2.1: Dancing Links
Exercise 351. [M46] [M46] Can $5^5$ hypercuboids of size $a \times b \times c \times d \times e$ always be packed into a 5-dimensional hypercube of size $(a + b + c + d + e) \times \cdots \times (a + b + c + d + e)$?
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Let $s = a+b+c+d+e$. We are asked whether it is always possible to pack $5^5$ copies of the 5-dimensional hypercuboid
$$ a \times b \times c \times d \times e $$
into the 5-cube $[0,s]^5$.
1. Fundamental necessary condition: volume
Any packing of disjoint axis-aligned hypercuboids that exactly fills a hypercube must preserve volume. Hence a necessary condition is
$$ s^5 = 5^5 , a b c d e. $$
But in general,
$$ (a+b+c+d+e)^5 \neq 5^5 a b c d e. $$
Therefore the claimed packing cannot exist for arbitrary positive $a,b,c,d,e$.
This already disproves the statement “always”.
2. Explicit counterexample
Take
$$ (a,b,c,d,e) = (1,1,1,1,2). $$
Then
$$ s = 6,\quad s^5 = 6^5 = 7776, $$
while the total volume of $5^5 = 3125$ hypercuboids is
$$ 3125 \cdot (1\cdot 1\cdot 1\cdot 1\cdot 2) = 6250. $$
Since
$$ 7776 \neq 6250, $$
a perfect packing into $[0,6]^5$ is impossible.
Thus the statement fails.
3. Why the inductive construction in the proposed solution cannot be repaired
The flawed proof attempted to partition the cube into 5 slabs and apply a 4-dimensional induction hypothesis. Even if the arithmetic error is fixed, a deeper issue remains:
- Each slab $S_x$ has geometry $I_x \times [0,s]^4$, where the 4-dimensional side length is $s$.
- Any valid reduction would require packing into a 4-cube whose side length matches the sum of the remaining edge lengths.
- However, no consistent decomposition aligns the volumes or shapes so that each slab contains a correct scaled instance of the same problem.
Thus the inductive structure does not correspond to the actual geometry of the target packing problem.
4. Conclusion
The statement is false.
$$ \boxed{\text{It is not always possible to pack } 5^5 \text{ such hypercuboids into } [0,a+b+c+d+e]^5.} $$
A necessary volume condition fails in general, and explicit counterexamples exist.