Puzzle description:

The problem of polyomino rectification is to fill a rectangle with copies of a single polyomino. For some of the polyominoes the solution with the mininmal number of pieces needs a quite large rectangle and is surprisingly difficult.

For the P-heptomino considered here the minimal rectangle has size 14x14 and consists of 28 identical P-heptominoes.

Each heptomino consists of seven solid cubies. So you also can try to built cuboids. Try to use some of the P-heptominoes to fill the following boxes:

3x3x7, 3x4x7, 3x5x7, ...

4x4x7, 4x5x7, 4x6x7, ...

5x5x7, 4x6x7, 4x7x7, ...

With more pieces you can fill any box having one side divisible by 7.

Source: https://www.math.uni-bielefeld.de/~sillke/PENTA/qu7-p3

If you can read German: Mehr Informationen gibt es in https://welt-der-geduldspiele.blogspot.com/search?q=3D-Druck

Puzzle difficulty:

The 14x14 square is of advanced difficulty.

If you do not succeed immediately, the tray contains a spare place for one heptomino.

Printing instructions:

Print the tray and 18 pieces of the G-hexomino.

Prints are fine with no supports.

I printed the puzzle in PLA at 0.1 mm layer height.

Print Settings

Printer Brand:

Prusa

Printer: 

Mini

Filament: Prusament PLA black, orange

Category: Puzzles