A unit square can be covered by a single disk of area π/2. Let us call the ratio of the square’s area to that of the covering disks (i.e. the sum of the areas of the disks) the

efficiencyof the covering, so that in the base case with one disk the efficiency is 2/π≈63.66%. Say that a covering isefficientif its efficiency exceeds this value. If we use a honeycomb (hexagonal) grid of 22 equal disks in alternating rows of four and five disks, we get a covering efficiency of 24/11π≈69.45%; so efficient coverings exist. Allowing disks of different sizes, how few are needed to cover the square efficiently? Can it be done with fewer than 22 disks?

**Answer**

If different circle sizes are allowed, we may reach an efficiency >76.3% with just five disks:

With 13 disks the maximum efficiency is already >80.4%. It is enough to replace each “corner disk” above with three disks with approximately half the radius:

If we adjust the configuration in each corner (allowing the most external disks to overlap with the larger disk) the maximum efficiency with 13 disk is ≥82.5%:

**Attribution***Source : Link , Question Author : John Bentin , Answer Author : Jack D’Aurizio*