### Vector Displays The Sketchpad, developed at MIT’s Lincoln Lab,...
Below you can see a picture of a a IBM 709 along with the included ...
The resulting line on the raster screen will be something like this...
The **Newton-Raphson** method is an iterative root-finding algorith...
This iterative method of finding square roots has been known since ...
Finding the square root of a number $V$ is equivalent to using Newt...
I think $G_n+1$ is a misprint for $G_{n+1}$.
Here is a simple implementation of the Bresenham’s algorithm in Jav...
### Floating Point Numbers You can think of floating point numbe...
Given that $x$ is the fraction and $a$ is the characteristic: $$\s... In _Figure 3_ it would be better to say that the circle is circumsc... There is a long history of computing more and more precise approxim... Using a rapidly converging series: \pi = \sqrt{12}\left(1-\frac{1... Professor John von Neumann was one of the people that expressed int... ### Modern computation of π In the early 20th century the Indian m... Of course the second identity immediately implies the first by sett... The following three algorithms don’t have much in common except for two things: They were implemented between 1949 and 1965— during the so-called ﬁrst generation of comput- ing hardware—and they are good examples of rigorous thinking. First, I explain Bresenham’s algorithm and apply it to a fundamental process of computer graphics—drawing a line on a raster screen. Second, although there are sever- al ways to calculate a square root, the square root algorithm is interesting because it takes advantage of hardware design. And, finally, while the calculation of π goes back more than 2,200 years, Machin’s algorithm from 1706 is of importance to the history of computing. Bresenham’s algorithm Cathode-ray tube (CRT) displays for com- puters were developed principally along two lines, vector and raster, until the 1970s, when memory became inexpensive. Initially, the advantage of a vector display was that it could draw a line by passing its end points to analog circuitry, which moved an electron beam across the CRT. To draw a line on a raster display, on the other hand, required calculating the loca- tions of every point that the line passed through and illuminating each one. Raster technology won out when it became economically feasible to draw a display in a dedicated memory and refresh the CRT from it. Even though raster computation is more com- plicated, vector display time varied with the complexity of the diagram, making realistic animation, for example, impossible. In the 1950s, the IBM 704 and 709 comput- ers included a CRT display with an addressable raster of 1,024 × 1,024 dots. Once illuminated, a dot would persist on the screen for about two seconds. The CPU—excessively slow by today’s standards—controlled the display, which had no refresh buffer. The 704, for example, had a 12-microsecond cycle, meaning that the fastest instruction executed in 24 microseconds. Floating-point calculations were much slower; a ﬂoating divide instruction took 18 cycles, or 216 microseconds. Each dot was activated by the program’s building a 36-bit word containing the X- and Y- address of the dot, with the X-address in bits 8–17 and the Y-address in bits 26–35 and “copy- ing” the word to the CRT (see Figure 1). Obviously, the fastest way to refresh the screen was to create a word for each illuminated dot and to set up a “copy” loop. However, the maxi- mum memory size for the 704 was 32,000 words, including the program, so prestoring all the dots wasn’t always feasible. In actual use, the visible CRT was a “slave” to another unit in a black box with a shutterless 35-mm camera. The general purpose of this arrangement was to enable the camera to take a picture of the master screen by illuminating dots only once. Nevertheless, the process of calculat- ing the positions of dots was so slow that only a few dots could be displayed before the ﬁrst one began to fade. There was a demonstration pro- gram—a tic-tac-toe game—for the display, and even this simple picture ﬂickered badly. It was in this context, around 1960, that computer scientist Jack Bresenham 1 consid- ered how to minimize the calculation time for the dots that make up a line. For a line with a slope between 0 and 1, the Y-difference between consecutive points on a straight line is less than or equal to 1. For the line between X1,Y1 and X2,Y2, such that 0 ≤∆Y ≤∆X and X1 < X2, the X-addresses of the dots are X1, X1 + 1, X1 + 2, , X2. If X = X2 – X1 and Y = Y2 – Y1, then for each step in X, the val- ues of Y are Y1, Y1 +∆Y/X, Y1 + 2Y/X, … ,Y2; but raster point addresses are integers, so we 10 IEEE Annals of the History of Computing 1058-6180/02/17.00 © 2002 IEEE Three Early Algorithms Keith S. Reid-Green Three well-known algorithms, implemented separately within a span of 20 years, demonstrate the use of mathematics developed hundreds of years ago. Figure 1. This 36-bit word activates a dot of a raster display. S 1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435 X-address Y-address use Y1, int(Y1 +∆Y/X), int(Y1 + 2Y/X), and so forth, without rounding. In the following implementation of Bresenham’s algorithm, Ratio and Delta are real numbers; all other variables are integers. Note that only addition and subtraction are used in the For … Next loop. DeltaX = X2 X1 DeltaY = Y2 Y1 Ratio = DeltaY / DeltaX Delta = Ratio 0.5 j = Y1 For i = X1 to X2 Call Dot (i, j) If Delta > 0 Then Delta = Delta 1.0 j = j + 1 End If Delta = Delta + Ratio Next i The For … Next loop is executed once per dot, and so this part of the code should be minimized. This loop would have taken about 60 cycles, or 720 microseconds. Because the raster size of the master CRT was 3.366 × 3.366 inches, there were about 300 dots per inch. Thus it took about 0.25 seconds to plot a line one inch long. Note that the algorithm can be improved by avoiding floating-point opera- tions entirely: DeltaX = X2 X1´ All integer variables DeltaY = Y2 Y1 Dy2 = 2 x DeltaY Ratio = Dy2 2 x DeltaX Delta = Dy2 DeltaX j = Y1 For i = X1 to X2 Call Dot (i, j) If Delta > 0 Then Delta = Delta + Ratio j = j + 1 Else Delta = Delta + Dy2 End If Next i This algorithm improves the loop’s speed by about 15 percent. Generalizing the algorithm to plot any line requires determining which X-value is the smaller, determining if the slope is 1 or > 1, and requires using X or Y, respectively, as the indexing value. A special case is if X1 = X2, but this is easy to deal with. A square root algorithm From the very beginning of computing, mathematical subroutines such as sine, cosine, logarithm, A X , square root, and so on were avail- able to programmers. The importance of opti- mizing their speed is obvious because most mathematical programs used them, and a single program could call them thousands of times. In the early days of minicomputers, hardware costs were kept as low as possible so most mini- computers did not have ﬂoating-point hard- ware. Instead, ﬂoating-point operations were accomplished in subroutines. At the same time, functions like SQRT were done in software, so a square root function with ﬂoating-point soft- ware could be slow. The usual way to ﬁnd a square root was to use a Newton’s iteration: where V is the value for which the square root is to be found, and G1 is the first approxima- tion (or ﬁrst guess) of the square root. The cal- culation is repeated until the difference between G n + 1 and G n is small enough, that is, until G 2 n+1 is approximately equal to V. Given that V can range over the set of posi- tive numbers, a good ﬁrst approximation of G 1 is not easy to ﬁnd, and of course the better the first guess, the faster the iteration converges. However, for values of V less than 1, conver- gence is fast. Here’s where knowledge of the structure of a floating-point number comes in handy. A ﬂoating-point number mimics scientiﬁc nota- tion—for example, we write 12.35 as 0.1235 × 10 2 . The power of 10, called the characteristic, is stored as a signed integer along with the frac- tion and its sign. This structure suggests that we can calculate the square root of the number by dividing the characteristic by 2 and ﬁnding the square root of the fraction. Floating-point numbers are normalized, that is, the first digit to the right of the point is never zero (unless the value of the number is zero). In a binary floating-point number, this means that the first digit to the right of the binary point is always a 1, thus the value of the fraction is at least 0.5. So the square root of a binary floating-point number 0.11000 … × 2 3 (in decimal, 6.0) can be found by halving the characteristic and finding the square root of 0.11000 …, except that the characteristic is an GG GV G nn n n + =− 1 2 2 October–December 2002 11 odd number. This is easily remedied by adding 1 to the characteristic if it is odd and shifting the fraction one place to the right, thereby dividing it by two, yielding 0.011000 … × 2 4 , or 0.375 × 16—still 6.0. The square root of the characteristic is 2 2 , and we can ﬁnd the square root of the fraction in four Newton’s iterations to an error of not more than 0.000002, using G 1 = 0.7. (It’s necessary to set the characteristic to 2 0 so that the fraction V is between 0.25 and 1.) Furthermore, with a little effort, we can determine a single calculation that develops four iterations without a loop: and thus and so on. Finally, the resulting square root G 4 must be multiplied by 10 to the power of the original characteristic divided by 2, to yield the square root of the original number V. Machin’s algorithm; computation of π The pursuit of more and more digits of π may start with a reference in the Old Testament, 2 to zero decimal places; that is, π= 3. The Babylonians got by nicely with π=3.125 (well, strictly speaking, 3 1/8, because decimal fractions were unknown to the Babylonians). The ancient Egyptians used π=4(8/9) 2 , which is about 3.1605, not really any better than the Babylonian estimate. It was Archimedes (287–212 BCE) who start- ed the contest of ﬁnding more and more digits of π by inscribing and circumscribing a regular polygon about a circle. This was a big step, because it provided both an upper and a lower bound for π, and the more sides to the polygon, the tighter the bounds became. The area of a cir- cle is πr 2 , so the area of a circle of unit radius is π. For a hexagon, we see in Figure 2 that the inscribed polygon consists of six equilateral triangles with sides of length 1. Each has an area of the square root of 3 divided by 4, so the area of the inscribed hexagon is a bit more than 2.598. For a circumscribed hexagon (see Figure 3) the height of each triangle is 1, so the area is slightly greater than 0.57735, giving the area of the circumscribed hexagon as about 3.4641. Archimedes used these numbers to bound π. He wrote the equivalent of 2.598 < π < 3.464 which suggests that π is about 3.03. This isn’t very good, except that Archimedes went on to use the same approach with 96-sided polygons and reﬁned his bounds to 3.14084 < π < 3.142858 although he had to use proper fractions: 3 10/71 < π < 3 1/7 This is more like it—the average of the bounds is 3.14185; accurate to three decimals. GG GV G G GV G V G GV G 31 1 2 1 1 1 2 1 2 1 1 2 1 2 2 2 2 2 =− GG GV G 32 2 2 2 2 =− GG GV G 21 1 2 1 2 =− 12 IEEE Annals of the History of Computing Three Early Algorithms R=1 Figure 2. An inscribed hexagon decomposed into six equilateral triangles. R=1 Figure 3. A circumscribed hexagon. For almost 2,000 years, reﬁnements of π as cal- culated in the Western world were done by increasing the number of sides of the polygons. Finally, with the advent of the calculus in the 17th century came new ways to find π. Notable was John Machin, a London professor of astronomy, who in 1706 used the difference between two arctangents. It was this method that scientists used on ENIAC 3 in 1949 at the University of Pennsylvania, setting a new record of 2,037 digits: π / 4 = 4 tan – 1 (1/5) – tan – 1 (1/239) We know that π / 4 radians is 45 degrees, and that the angle whose tangent is 1 is 45 degrees. Unfortunately, we can’t calculate tan – 1 (1) using a series because the arctangent series only con- verges for values less than 1—hence Machin’s equation. It is easy to test the equation: Debug.Print 4# * (4# * Atn(1# / 5#) Atn(1# / 239#)) gives the result 3.14159265358979, exactly right as far as it goes, but not a proof. To show that tan – 1 (1) = 4 tan – 1 (1/5) – tan – 1 (1/239) we need two trigonometric identities: tan(2a) = (2 tan(a)) / (1 – tan 2 (a)) (1) and tan(a + b) = (tan(a) + tan(b)) / (1 – tan(a) tan(b)) (2) If we let x = atn(1/5), then using Equation 1 we get tan(2x) = (2 tan x) / (1 – tan 2 (x)) = 5/12 and tan(4x) = (2 tan(2x)) / (1 – tan 2 (2x)) = 120/119 but we want the result to be 1. We can ﬁnd the difference from Equation 2: tan(4x – π / 4) = (tan(4x) – 1) / (1 + tan(4x)) = 1/239 hence tan(4 atn(1/5) – atn(1/239)) resolves to 1. Machin found π / 4 to 100 places using two arc- tangent series: π / 4 = 4 (1/5 – 1 / (3 * 5 3 ) + 1 / (5 * 5 5 ) – …) – (1/239 – 1 / (3 * 239 3 ) + 1 / (5 * 239 5 ) – …) and this is how it was done on ENIAC to 2,037 places. Finding a very high precision result is a nontrivial programming exercise, but that’s another matter. Acknowledgments I would like to acknowledge the assistance of Michael Greenwald, Dick Painter, and Dennis Quardt in the preparation of this article. References and notes 1. W.M. Newman and R.F. Sproull, Principles of Inter- active Computer Graphics, 2nd ed., McGraw-Hill, New York, 1979. Jack Bresenham is currently a professor of computer science at Winthrop Uni- versity. He retired from IBM as a senior technical staff member in 1987 after 27 years. He is best known for fast line and circle algorithms. 2. 1 Kings 23, Authorized (King James) Version. The verse reads: “And he made a molten sea, ten cubits from the one brim to the other: it was round all about, and his height was ﬁve cubits: and a line of thirty cubits did compass it round about.” The “molten sea” is a brass vat. Its diameter is ten cubits, thus its radius is ﬁve cubits. The given circumference—the line of 30 cubits that “did compass it round about”—implies that π = 3, because the circumference is computed as 2πr, from 2π * 5 = 30. 3. P. Beckmann, A History of Pi, Dorset Press, New York, 1989. Keith S. Reid-Green retired in 1999 after 42 years as a com- puter scientist. He began as an IBM 704 programmer in 1957 and is known for pioneering work in data reduction, com- puter-aided manufacturing, and image processing. He is widely published in the history of computing and was an adjunct professor at the University of Utah and at Rutgers, where he taught computer graphics to graduate students. He holds bachelor’s degrees in mathematics and physics and a master’s degree in computer science. Readers may contact Keith Reid-Green at 23 Point Court, Lawrenceville, NJ 08648, email KReid- Green@msn.com. October–December 2002 13 Thanks Jadrian, I went ahead and corrected it. Finding the square root of a number V is equivalent to using Newton’s method to find the root of f(x) = x^2 - V. Newton’s method finds a root of f(x) from a guess x_n by approximating f(x) as its tangent line at x_n : f(x_n) + f’(x_n)(x - x_n). To get x_{n+1} you find the root of the tangent:$$ f(x_n) + f’(x_n)(x - x_n) = 0 f’(x_n)x = f’(x_n)x_n - f(x_n) x = x_n - \frac{f(x_n)}{f’(x_n)}$$For f(x) = x^2 - V you get:$$x_{n+1} = x_n - \frac{x_n^2 - V}{2x_n}$$This iterative method of finding square roots has been known since Babylonian times. This method is also known as Heron's method, after the first-century Greek mathematician Hero of Alexandria who gave the first explicit description of the method in his AD 60 work: Metrica. ### Modern computation of π In the early 20th century the Indian mathematician Srinivasa Ramanujan found several rapidly converging infinite series of π, including:$${\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}$$This series computes a further eight decimal places of π with each term. Ramanujan series are the basis of a lot of the fastest algorithms used to compute Pi. The current record was set in 2020 by Timothy Mullican, who computed 50 trillion digits of π. He wrote about it [here](https://blog.timothymullican.com/calculating-pi-my-attempt-breaking-pi-record) Below you can see a picture of a a IBM 709 along with the included CRT display. The IBM 709 was an early vacuum tube powered computer introduced by IBM in 1958. The IBM 709 had 32,768 words of 36-bit memory, could execute 42,000 add instructions per second and could multiply two 36-bit integers at a rate of 5000 per second. ![IBM 709](https://i.imgur.com/80AKAlk.jpg) Here is a simple implementation of the Bresenham’s algorithm in JavaScript.  let draw_line = (x0, y0, x1, y1) => { // Calculate "deltas" of the line (difference between two ending points) let dx = x1 - x0; let dy = y1 - y0; // Calculate the line equation based on deltas let D = (2 * dy) - dx; let y = y0; // Draw the line based on arguments provided for (let x = x0; x < x1; x++) { // Place pixel on the raster display pixel(x, y); if (D >= 0) { y = y + 1; D = D - 2 * dx; } D = D + 2 * dy; } };  You can find a full tutorial along with an interactive demo [here](http://www.javascriptteacher.com/bresenham-line-drawing-algorithm.html) I think G_n+1 is a misprint for G_{n+1}. Of course the second identity immediately implies the first by setting b = a. In _Figure 3_ it would be better to say that the circle is circumscribed by the hexagon. “A circumscribed hexagon” implies, rather, _Figure 2_ where the hexagon is circumscribed by a circle. The resulting line on the raster screen will be something like this: ![](https://i.imgur.com/YWZDeYZ.png) The basic intuition is that on each loop you have to decide whether you are increasing the y coordinate by 1 or not. ### Vector Displays The Sketchpad, developed at MIT’s Lincoln Lab, is an example of an early computer that made use of a vector displays. ![sketchpad](https://i.imgflip.com/3r3v47.gif) ### Raster Displays Here is an example of a raster display - a slow motion recording of a CRT displaying Super Mario Bros. on a NES: ![raster](https://i.imgur.com/kknDRng.gif) There is a long history of computing more and more precise approximations of π. Broadly you can divide it into 2 eras: before and after electronic digital computers. Throughout the centuries many mathematicians worked on developing new methods of computing π. The 16th century German-Dutch mathematician Ludolph van Ceulen computed the first 35 decimal places of π with a 262 sided polygon. He was so proud of his achievement that he had the digits inscribed on his tombstone. ![Ludolph van Ceulen tombstone](https://i.imgur.com/37uFixl.png) *Ludolph van Ceulen's tombstone* The longest expansion of π before the advent of digital computing was achieved by the English amateur mathematician William Shanks in the late 19th century. He spent over 20 years calculating π to 527 places. To put it into perspective, in order to calculate the circumference of the the observable universe, 93 billion light-years in diameter, to a precision of less than one Planck length you need an approximation of π to 62 decimal places. ### Floating Point Numbers You can think of floating point numbers as being written in *scientific notation*. A floating point number has 3 parts: - **sign**: indicates whether the number is positive or negative - **significant or fraction**: The significant digits of the number - **exponent or characteristic**: indicates how large (or small) the number is In the example below I’m using a biased exponent - which was popularized by the IBM 704. This was made a standard by the IEEE Standard for Floating-Point Arithmetic (IEEE 754). In this paper the author uses signed exponents. Example Let’s say we wanted to store 2019 as a floating point number: - Decimal Representation: 2019 - Binary Representation: 11111100011 - Scientific notation:  2.019 \times 10^{3}  - Binary Scientific notation: 1.1111100011 \times 2^{10} - Double Precision Raw Binary: ![](https://i.imgur.com/sa4lgWK.png) As you can see in the raw binary, the exponent is stored as 10000001001, which is 1033: 10 (the actual binary exponent) + 1023 (the bias, which allows you to store the exponent as an unsigned integer). This means that the exponent range is −1022 to +1023 (exponents of −1023 (all 0s) and +1024 (all 1s) are reserved for special numbers.) Given that x is the fraction and a is the characteristic:$$\sqrt{x \times 10^a} \Leftrightarrow (x \times 10^a)^\frac{1}{2} \Leftrightarrow \sqrt{x} \times 10^\frac{a}{2} It's Newton–*Raphson*; there's no *L* in Raphson's name. The **Newton-Raphson** method is an iterative root-finding algorithm which produces successively better approximations to the zeroes of a real-valued function. Here is a representation of a few iterations of the method ![](https://media.giphy.com/media/j2GcSDWz93j3rn3QIY/giphy.gif) ### Convergence The **Newton-Raphson** method doesn’t always converge. Sometimes it can fail: - Derivative may be zero at the root - Function may fail to be continuously differentiable - Choosing a bad starting point, i.e. one that lies outside the range of guaranteed convergence Using a rapidly converging series: $\pi = \sqrt{12}\left(1-\frac{1}{3.3}+\frac{1}{3.5^{2}}-\frac{1}{3.7^{2}}+\ldots \right)$ Madhava of Sangamagrama (Modern day Kerala, India) computed the the correct value of $\pi$ to 11 decimal digits as early as the period (c. 1340 – c. 1425) . The value of $\pi$ correct to 13 decimals, $\pi=3.1415926535898$, the most accurate since the 5th century, is sometimes attributed to Madhava. https://en.wikipedia.org/wiki/Madhava_of_Sangamagrama Professor John von Neumann was one of the people that expressed interest in having the ENIAC compute the value of π to many decimal places. Von Neumann was in fact interested in obtaining a statistical measure of the randomness of the distribution of the digits.