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The rings are interlocked so as not to become misplaced. A puzzle ring is a jewelry ring made up of multiple interconnected bands, which is a type of mechanical puzzle most likely developed as an elaboration of the European gimmal ring. [1] The puzzle ring is also sometimes called a "Turkish wedding ring" or "harem ring."
A baguenaudier Diagrammatic representation of a four-ring baguenaudier A metal version of the puzzle. Baguenaudier (pronounced; French for "time-waster"), [1] also known as the Chinese rings, Cardan's suspension, Cardano's rings, Devil's needle or five pillars puzzle, is a disentanglement puzzle featuring a loop which must be disentangled from a sequence of rings on interlinked pillars. [1]
Most puzzle solvers try to solve such puzzles by mechanical manipulation, but some branches of mathematics can be used to create a model of disentanglement puzzles. Applying a configuration space with a topological framework is an analytical method to gain insight into the properties and solution of some disentanglement puzzles. However, some ...
The puzzle has 12 panels interconnected with nylon wires in a 2 × 6 rectangular shape, measuring approximately 4.25 inches (10.5 cm) by 13 inches (32 cm). The goal of the game is the same as for Rubik's Magic, which is to fold the puzzle from a 2 × 6 rectangular shape into a W-like shape with a certain tile arrangement.
The ant on a rubber rope is a mathematical puzzle with a solution that appears counterintuitive or paradoxical. It is sometimes given as a worm, or inchworm, on a rubber or elastic band, but the principles of the puzzle remain the same. The details of the puzzle can vary, [1] [2] but a typical form is as follows:
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The band runs straight through certain cubes, but bends 90° in others, creating a specific sequence of straight and bent connections. The cubelets can rotate freely. The aim of the puzzle is to arrange the chain in such a way that they will form a 3×3×3 or 4×4×4 cube. While all commercially available 3×3×3 cubes appear to be identical in ...
The puzzle can be generalised to n glasses instead of four. For two glasses it is trivially solved in one turn by inverting either glass. For three glasses there is a two-turn algorithm. For five or more glasses there is no algorithm that guarantees the bell will ring in a finite number of turns. [2]