How to build a bitwise circuit¶

Bitwise is another built-in circuit type in the circkit framework. We can follow the same steps as when we build an arithmetic circuit to build a bitwise one.

Operations¶

A bitwise circuit supports the following operations:

Operation	Notation	Note
AND	\(\&\)
OR	\(\\|\)
XOR	\(\wedge\)
NOT	\(\sim\)	unary operation
SHL	\(\ll\)	shift left (constant postions)
SHR	\(\gg\)	shift right (constant postions)
ROL	`node.rol(n)`	circular shift left (constant postions)
ROR	`node.ror(n)`	circular shift right (constant postions)
ADD	\(+\)	addition
SUB	\(-\)	subtraction
MUL	\(*\)	multiplication
DIV	\(/\)	division
MOD	\(\%\)	modulo
NEG	\(-\)	negation, unary operation
LUT	`node.lookup_in(-)`	lookup table, value of node is index
RND	`C.RND()()`	\(\text{C}\) is the circuit

Examples¶

To build a bitwise circuit, we have to specify \(\text{word_size}\) which is the bit size of a word (value of a node). We denote \(\text{mask} = 2^\text{word_size} - 1\). Every word in the circuit is automatically ensured not to exceed \(\text{word_size}\) bits by AND-ing with the mask. Below is a simple example.

from circkit.bitwise import BitwiseCircuit

C = BitwiseCircuit(word_size=8)
a = C.add_input("a")
b = C.add_input("b")

x0 = a + b
x1 = a - b + 2
x2 = a * b - 3
x3 = a / b * 5
x4 = a % 3 + b

x5 = x0 << 2
x6 = x1 >> 4
x7 = x2.rol(5)
x8 = x3.ror(6)
x9 = ~x4

x10 = x5 ^ x6
x11 = x7 | x8
x12 = x9 & x10
x13 = -x11

C.add_output(x12)
C.add_output(x13)

inp = [100, 200]
out = C.evaluate(inp)
print("Circuit output:")
print(out)

Circuit output:
[48, 93]

In the advanced examples of this documentation, we build the Simon cipher and Speck cipher based on bitwise circuits to demonstrate the usefulness of this circuit type in applications

from circkit.bitwise import BitwiseCircuit

C = BitwiseCircuit(word_size=8)
a = C.add_input("a")

x0 = a + 257
x1 = a + 1

x2 = a / 3

C.add_output([x0, x1, x2])

inp = [1]
out = C.evaluate(inp)
print("Circuit output:")
print(out)

Circuit output:
[2, 2, 0]

In the above example, \(257\) is represented by \(1\) in a 8-bit word, thus \(x_0 = x_1\). The division only takes the round number, thus \(x_2 = 1/3 = 0\).