Base 4 Computer
(Karnaugh Maps of Base 4 Addition)
 

Here is a Base-4 truth table for the addition of two 2-bit integers. It might be possible to reduce the number of clock cycles that are required for all math operations for variable length numeric operations. A base of 256 (8 bit integers) might generate a significant increase in the speed for math operations, along with a organizing effect on the use and storage of numeric values. The fixed length binary math operations are very efficient, however going to a variable length number system will require a different approach to the storage and on the hardware instruction set. Going to a single segmented variable length base integer unit will permit software development greater latitude in information handling.

A
B,C
D,E
X
Y,Z
 
= overflow from previous addition
= 2 bits of Register 1
= 2 bits of Register 2
= overflow bit
= 2 bits of Accumulator
 
	A  B C  D E  X Y Z 
1	0  0 0  0 0  0 0 0
2	0  0 0  0 1  0 0 1
3	0  0 0  1 0  0 1 0
4	0  0 0  1 1  0 1 1

5	0  0 1  0 0  0 0 1
6	0  0 1  0 1  0 1 0
7	0  0 1  1 0  0 1 1
8	0  0 1  1 1  1 0 0

9	0  1 0  0 0  0 1 0
10	0  1 0  0 1  0 1 1
11	0  1 0  1 0  1 0 0
12	0  1 0  1 1  1 0 1

13	0  1 1  0 0  0 1 1
14	0  1 1  0 1  1 0 0
15	0  1 1  1 0  1 0 1
16	0  1 1  1 1  1 1 0

17	1  0 0  0 0  0 0 1
18	1  0 0  0 1  0 1 0
19	1  0 0  1 0  0 1 1
20	1  0 0  1 1  1 0 0

21	1  0 1  0 0  0 1 0
22	1  0 1  0 1  0 1 1
23	1  0 1  1 0  1 0 0
24	1  0 1  1 1  1 0 1

25	1  1 0  0 0  0 1 1
26	1  1 0  0 1  1 0 0
27	1  1 0  1 0  1 0 1
28	1  1 0  1 1  1 1 0

29	1  1 1  0 0  1 0 0
30	1  1 1  0 1  1 0 1
31	1  1 1  1 0  1 1 0
32	1  1 1  1 1  1 1 1
X (A = 0)
 

BC

DE
  00 01 11 10
00 0 0 0 0
01 0 0 1 0
11 0 1 1 1
10 0 0 1 1
X = BCE + DEC + BD

 

 

X (A = 1)
 

 BC

DE
  00 01 11 10
00 0 0 1 0
01 0 0 1 1
11 1 1 1 1
10 0 1 1 1
X = BC + DE + BE + BD + CD

 

X = BD + CDE + BCE + ADE + ACD + ABE + ABC

Y (A = 0)
 

 BC

DE
  00 01 11 10
00 0 0 0 1
01 0 1 1 1
11 1 0 1 0
10 1 1 0 0
Y = B'C'D + DEB' + CD'E + BCD' + BC'D'

 

 

Y (A = 1)
 

 BC

DE
  00 01 11 10
00 0 1 0 0
01 1 1 1 0
11 0 0 1 1
10 1 0 0 1
Y = C'DE' + B'CD' + C'D'E + CD'E + BDE
Z (A = 0)
 

 BC

DE
  00 01 11 10
00 0 1 1 0
01 1 0 0 1
11 1 0 0 1
10 0 1 1 0
Z = C'E + CE'

 

 

Z (A = 1)
 

 BC

DE
  00 01 11 10
00 1 0 0 1
01 0 1 1 0
11 0 1 1 0
10 1 0 0 1
Z = CE + E'C