Block-stacking problem

The single-wide problem involves having only one block at any given level. In the ideal case of perfectly rectangular blocks, the solution to the single-wide problem is that the maximum overhang is given by ${\textstyle \sum _{i=1}^{N}{\frac {1}{2i}}}$ times the width of a block. This sum is one half of the corresponding partial sum of the harmonic series. Because the harmonic series diverges, the maximal overhang tends to infinity as ${\displaystyle N}$ increases, meaning that it is possible to achieve any arbitrarily large overhang, with sufficient blocks.

The number of blocks required to reach at least ${\displaystyle N}$ block-lengths past the edge of the table is 4, 31, 227, 1674, 12367, 91380, ... (sequence A014537 in the OEIS).[1]

Comparison of the solutions to the single-wide (top) and multi-wide (bottom) block-stacking problem with three blocks

Multi-wide stacks using counterbalancing can give larger overhangs than a single width stack. Even for three blocks, stacking two counterbalanced blocks on top of another block can give an overhang of 1, while the overhang in the simple ideal case is at most 11/12. As Paterson et al. (2007) showed, asymptotically, the maximum overhang that can be achieved by multi-wide stacks is proportional to the cube root of the number of blocks, in contrast to the single-wide case in which the overhang is proportional to the logarithm of the number of blocks. However, it has been shown that in reality this is impossible and the number of blocks that we can move to the right, due to block stress, is not more than a specified number. For example, for a special brick with h = 0.20 m, Young's modulus E = 3000 MPa and density ρ = 1.8×10³ kg/m³ and limiting compressive stress 3 MPa,the approximate value of N will be 853 and the maximum tower height becomes 170 m.[2]

Mike Paterson's method to increase the overhang of 16 blocks of unit width and breadth b to 2√1 + b ² by offsetting the blocks perpendicular to their lengths in a diamond formation [3]