Digital optical arithmetic

Earl Swartzlander

doi:10.1364/AO.25.003021

Applied Optics
Vol. 25,
Issue 18,
pp. 3021-3032
(1986)
•https://doi.org/10.1364/AO.25.003021

Digital optical arithmetic

Earl Swartzlander

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Earl Swartzlander, "Digital optical arithmetic," Appl. Opt. 25, 3021-3032 (1986)

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Abstract

Since development of the first computing system, there has been a continual quest for faster systems. This is a result of successful accomplishments with a given system which lead to desires to do even more, thereby requiring better performance. In many applications, better performance requires faster arithmetic. Recent progress in optical processing suggests that digital optical arithmetic may eliminate many of today’s performance bottlenecks. This paper provides an introduction to the concepts (and literature) of digital arithmetic. The primary emphasis is on number systems and on algorithms for addition, subtraction, multiplication, and division, since these operations are the basis for current systems.

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Number	Twos complement	Sign magnitude	Ones complement
+¾	011	011	011
+½	010	010	010
+¼	001	001	001
+0	000	000	000
−0	N/A	100	111
−¼	111	101	110
−½	110	110	101
−¾	101	111	100
−1	100	N/A	N/A

Number	Residue mod3	Residue mod4
0	0	0
1	1	1
2	2	2
3	0	3
4	1	0
5	2	1
6	0	2
7	1	3
8	2	0
9	0	1
10	1	2
11	2	3

Add	3 = (0,3)	5 = (2,1)	8 = (2,0)
	+4 = (1,0)	+6 = (0,2)	+2 = (2,2)

	(1,3) = 7	(2,3) = 11	(1,2) = 10
Subtract	3 = (0,3)	10 = (1,2)	8 = (2,0)
	−2 = (2,2)	−6 = (0,2)	−2 = (2,2)

	(1,1) = 1	(1,0) = 4	(0,2) = 6
Multiply	3 = (0,3)	2 = (2,2)	8 = (2,0)
	*3 = (0,3)	*4 = (1,0)	*1 = (1,1)

	(0,1) = 9	(2,0) = 8	(2,0) = 8
Invalid	3 = (0,3)	2 = (2,2)	8 = (2,0)
	+9 = (0,1)	−4 = (1,0)	*6 = (0,2)

	(0,0) = 0	(1,2) = 10	(0,0) = 0

i	a_i	a_i₋₁	Sum of partial products		Operation result
A = ⅝ (= 0.101)				B = ¾ (= 0.110)
0	1	0	0.000	−B	1.010
1	0	1	1.1010	+B	0.0110
2	1	0	0.00110	−B	1.01110
3	0	1	1.101110	+B	0.011110 (= ¹⁵/₃₂)
Thus ⅝ · ¾ = ¹⁵/₃₂
A = −⅝ (= 1.011)				B = ¾ (= 0.110)
0	1	0	0.000	−B	1.010
1	1	1	1.1010	—	1.1010
2	0	1	1.11010	+B	0.10010
3	1	0	0.010010	−B	1.100010 (= −¹⁵/₃₂)
Thus −⅝ · ¾ = −¹⁵/₃₂
A = ⅝ (= 0.101)				B = −¾ (= 1.010)
0	1	0	0.000	−B	0.110
1	0	1	0.0110	+B	1.1010
2	1	0	1.11010	−B	0.10010
3	0	1	0.010010	+B	1.100010 (= −¹⁵/₃₂)
Thus ⅝ · −¾ = −¹⁵/₃₂
A = −⅝ (= 1.011)				B = −¾ (= 1.010)
0	1	0	0.000	−B	0.110
1	1	1	0.0110	—	0.0110
2	0	1	0.00110	+B	1.01110
3	1	0	1.101110	−B	0.011110 (= ¹⁵/₃₂)
Thus −⅝ · −¾ = ¹⁵/₃₂

R⁽⁰⁾ = dividend 0.0101 (←⁵/₁₆)
D = divisor 0.0110 (←⅜)
R⁽⁰⁾ > 0 → q₁ = 1	2R⁽⁰⁾ →	0.1010
	− D →	+1.1010

	2R⁽¹⁾ →	0.0100
R⁽¹⁾ > 0 → q₂ = 1	2R⁽¹⁾ →	0.1000
	− D →	+1.1010

	2R⁽²⁾ →	0.0010
R⁽²⁾ > 0 → q₃ = 1	2R⁽²⁾ →	0.0100
	− D →	+1.1010

	2R⁽³⁾ →	1.1110
R⁽³⁾ > 0 → q₄ = −1	2R⁽³⁾ →	1.1100
	+ D →	+0.0110

	2R⁽⁴⁾ →	0.0010
Q = 0.111 − 1 (signed digit form)
Q = P − N, where P = 0.1110 and N = 0.0001
P →	0.1110
−N →	+1.1111
Q →	0.1101 = ¹³/₁₆

Dividend = ⅝
Divisor = ¾
	Error
Y⁽⁰⁾ = 1.457107 − 0.75 = 0.707107	0.040440
Y⁽¹⁾ = 2 × 0.707107 (1 − 0.75 × 0.707107) = 0.664214	0.002453
Y⁽²⁾ = 2 × 0.664214 (1 − 0.75 × 0.664214) = 0.666658	0.000009
Q = 2 × 0.625 × 0.666658 = 0.833323

Abstract

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Applied Optics