Liu Hui's π algorithm was invented by Liu Hui (fl. 3rd century), a mathematician of the state of Cao Wei. Before his time, the ratio of the circumference of a circle to its diameter was often taken experimentally as three in China, while Zhang Heng (78–139) rendered it as 3.1724 (from the proportion of the celestial circle to the diameter of the earth, 92/29) or as π ≈ 10 ≈ 3.162 {\displaystyle \pi \approx {\sqrt {10}}\approx 3.162} . Liu Hui was not satisfied with this value. He commented that it was too large and overshot the mark. Another mathematician Wang Fan (219–257) provided π ≈ 142/45 ≈ 3.156. All these empirical π values were accurate to two digits (i.e. one decimal place). Liu Hui was the first Chinese mathematician to provide a rigorous algorithm for calculation of π to any accuracy. Liu Hui's own calculation with a 96-gon provided an accuracy of five digits i.e. π ≈ 3.1416. Liu Hui remarked in his commentary to The Nine Chapters on the Mathematical Art, that the ratio of the circumference of an inscribed hexagon to the diameter of the circle was three, hence π must be greater than three. He went on to provide a detailed step-by-step description of an iterative algorithm to calculate π to any required accuracy based on bisecting polygons; he calculated π to between 3.141024 and 3.142708 with a 96-gon; he suggested that 3.14 was a good enough approximation, and expressed π as 157/50; he admitted that this number was a bit small. Later he invented a quick method to improve on it, and obtained π ≈ 3.1416 with only a 96-gon, a level of accuracy comparable to that from a 1536-gon. His most important contribution in this area was his simple iterative π algorithm.
Area of a circle
Liu Hui argued:
"Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the finer we cut, the smaller the loss with respect to the area of circle, thus with further cut after cut, the area of the resulting polygon will coincide and become one with the circle; there will be no loss". This is essentially equivalent to:
lim N → ∞ area of N -gon = area of circle . {\displaystyle \lim _{N\to \infty }{\text{area of }}N{\text{-gon}}={\text{area of circle}}.\,}
Further, Liu Hui proved that the area of a circle is half of its circumference multiplied by its radius. He said: "Between a polygon and a circle, there is excess radius. Multiply the excess radius by a side of the polygon. The resulting area exceeds the boundary of the circle". In the diagram d = excess radius. Multiplying d by one side results in oblong ABCD which exceeds the boundary of the circle. If a side of the polygon is small (i.e. there is a very large number of sides), then the excess radius will be small, hence excess area will be small. As in the diagram, when N → ∞, d → 0, and ABCD → 0. "Multiply the side of a polygon by its radius, and the area doubles; hence multiply half the circumference by the radius to yield the area of circle". When N → ∞, half the circumference of the N-gon approaches a semicircle, thus half a circumference of a circle multiplied by its radius equals the area of the circle. Liu Hui did not explain in detail this deduction. However, it is self-evident by using Liu Hui's "in-out complement principle" which he provided elsewhere in The Nine Chapters on the Mathematical Art: Cut up a geometric shape into parts, rearrange the parts to form another shape, the area of the two shapes will be identical. Thus rearranging the six green triangles, three blue triangles and three red triangles into a rectangle with width = 3L, and height R shows that the area of the dodecagon = 3RL. In general, multiplying half of the circumference of a N-gon by its radius yields the area of a 2N-gon. Liu Hui used this result repetitively in his π algorithm.
Liu Hui's π inequality
Liu Hui proved an inequality involving π by considering the area of inscribed polygons with N and 2N sides. In the diagram, the yellow area represents the area of an N-gon, denoted by A N {\displaystyle A_{N}} , and the yellow area plus the green area represents the area of a 2N-gon, denoted by A 2 N {\displaystyle A_{2N}} . Therefore, the green area represents the difference between the areas of the 2N-gon and the N-gon:
D 2 N = A 2 N − A N . {\displaystyle D_{2N}=A_{2N}-A_{N}.}
The red area is equal to the green area, and so is also D 2 N {\displaystyle D_{2N}} . So
Yellow area + green area + red area = A 2 N + D 2 N . {\displaystyle A_{2N}+D_{2N}.}
Let A C {\displaystyle A_{C}} represent the area of the circle. Then
A 2 N < A C < A 2 N + D 2 N . {\displaystyle A_{2N}<A_{C}<A_{2N}+D_{2N}.}
If the radius of the circle is taken to be 1, then we have Liu Hui's π inequality:
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