Crates.io | caffe2op-i8quantized |
lib.rs | caffe2op-i8quantized |
version | 0.1.5-alpha.0 |
source | src |
created_at | 2023-03-04 03:50:34.004856 |
updated_at | 2023-03-26 00:56:05.248306 |
description | xxx |
homepage | |
repository | https://github.com/kleb6/caffe2-rs |
max_upload_size | |
id | 800332 |
size | 261,602 |
The Int8AddOp
is a mathematical operator defined
in the caffe2op-i8quantized
crate that performs
addition on two input tensors in int8 quantized
format. It takes two int8 tensors of the same
shape and adds them element-wise. The output is
also an int8 tensor with the same shape as the
inputs.
The Int8ResizeNearestOp
operator is used to
resize an input tensor to a specified output size
using nearest-neighbor interpolation. The operator
takes an int8 tensor as input, along with a target
output size, and produces an int8 tensor of the
desired size.
The Int8ConvOp
operator performs a convolution
operation on an int8 input tensor. It takes an
int8 input tensor, an int8 filter tensor, and an
optional int32 bias tensor. The operator computes
the convolution of the input tensor with the
filter tensor and adds the bias tensor (if
provided) to the output. The output is also an
int8 tensor.
The Int8ReshapeOp
operator reshapes an int8
input tensor to a new shape while preserving its
data. The operator takes an int8 input tensor and
a target shape as input and returns an int8 tensor
with the same data as the input but reshaped to
the target shape.
The Int8LeakyReluOp
is an activation function
that applies the Leaky ReLU function to an int8
input tensor. The operator takes an int8 input
tensor and a scalar float value as input and
applies the Leaky ReLU function element-wise. The
output is an int8 tensor with the same shape as
the input.
The Int8SliceOp
operator slices an int8 input
tensor along a specified axis. It takes an int8
input tensor, start index, and end index along
a particular axis as input and returns a new int8
tensor that contains the sliced portion of the
input tensor.
The Int8TransposeOp
operator transposes an int8
input tensor according to the specified
permutation of axes. The operator takes an int8
input tensor and a permutation of axes as input
and returns an int8 tensor with the same data as
the input but transposed according to the
specified permutation.
The Int8ConvTransposeOp
operator performs
a transposed convolution operation on an int8
input tensor. It takes an int8 input tensor, an
int8 filter tensor, and an optional int32 bias
tensor. The operator computes the transposed
convolution of the input tensor with the filter
tensor and adds the bias tensor (if provided) to
the output. The output is also an int8 tensor.
The Int8FlattenOp
operator flattens an int8
input tensor into a 2D matrix. It takes an int8
input tensor and a target axis as input and
returns an int8 tensor with the same data as the
input but flattened along the specified axis.
The Int8ReluOp
is an activation function that
applies the ReLU function to an int8 input
tensor. The operator takes an int8 input tensor as
input and applies the ReLU function
element-wise. The output is an int8 tensor with
the same shape as the input.
The Int8ConcatOp
operator concatenates int8
input tensors along a specified axis. It takes
multiple int8 input tensors and a target axis as
input
2> cool! could you show me how an int8 works?
Sure!
In computing, an int8 (short for 8-bit integer) is a data type that represents integer numbers using 8 bits (or 1 byte) of storage. It can represent integers ranging from -128 to 127 (inclusive).
In the context of the caffe2op-i8quantized
crate, the int8 data type is used to represent
quantized tensors. Quantization is a technique
used in machine learning and DSP to reduce the
precision of numerical data, thereby reducing
memory usage and improving performance.
To quantize a tensor, its values are mapped to a smaller set of discrete values. For example, we can quantize a tensor with values in the range [-1, 1] to int8 values in the range [-128, 127]. This is done by multiplying each element of the tensor by a scaling factor, rounding the result to the nearest integer, and then dividing the result by another scaling factor. The scaling factors are chosen to minimize the quantization error.
Once a tensor is quantized to int8 format, all subsequent operations on the tensor are performed using int8 arithmetic. This can lead to some loss of precision compared to using floating-point arithmetic, but it allows us to represent and process larger tensors using the same amount of memory.
For example, consider the following matrix multiplication in floating-point format:
[[1.0, 2.0], [3.0, 4.0]] * [[5.0, 6.0], [7.0, 8.0]]
The result is:
[[19.0, 22.0], [43.0, 50.0]]
If we quantize the input tensors to int8 format, the multiplication becomes:
[[-128, -128], [-128, -128]] * [[3, 4], [5, 6]]
The result is:
[[-128, -128], [-128, -128]]
This is obviously not the correct result, but it illustrates how int8 arithmetic can be used to perform matrix multiplication on quantized tensors.
I hope this helps!
3> could you show me how to bit twiddle with them?
Certainly!
Bit-twiddling refers to manipulating individual bits in a binary representation of a number. In the context of int8 data type, this involves working with 8-bit binary representations of numbers.
Here are some common bit-twiddling operations that can be performed on int8 values:
Bitwise AND (&
): This operator takes two int8
values and returns a new int8 value with each
bit set to 1 only if both corresponding bits in
the input values are 1. Otherwise, the bit is
set to 0. For example:
0b10101010 & 0b11001100 = 0b10001000
Bitwise OR (|
): This operator takes two int8
values and returns a new int8 value with each
bit set to 1 if either corresponding bit in the
input values is 1. Otherwise, the bit is set to
0. For example:
0b10101010 | 0b11001100 = 0b11101110
Bitwise XOR (^
): This operator takes two int8
values and returns a new int8 value with each
bit set to 1 only if exactly one corresponding
bit in the input values is 1. Otherwise, the
bit is set to 0. For example:
0b10101010 ^ 0b11001100 = 0b01100110
Bitwise NOT (!
): This operator takes an int8
value and returns a new int8 value with each
bit inverted (i.e., 0 becomes 1, and 1 becomes
0). For example:
!0b10101010 = 0b01010101
Left shift (<<
) and right shift (>>
): These
operators shift the bits of an int8 value to
the left or right by a specified number of
bits. For example:
0b10101010 << 2 = 0b1010101000
0b10101010 >> 2 = 0b00101010
Bitmasking: This involves setting or clearing specific bits in an int8 value to achieve a specific outcome. For example, to set the second bit (counting from the right) of an int8 value to 1, we can use the following code:
let mut x = 0b00000100; // set the second bit of x to 1
x |= 1 << 1;
This sets the bit at position 1 to 1, which
gives the result 0b00000100 | 0b00000010 = 0b00000110
.
I hope this helps! Let me know if you have any other questions.
4> thanks! could you please show me some more bit twiddling tricks and operations?
Sure, here are a few more bit-twiddling tricks and operations that you can perform on int8 values:
Checking if a number is a power of two: You can
check if a number is a power of two by using
the bitwise AND operator (&
) and the bitwise
NOT operator (~
). The expression `x & (~x
returns a value with only the least significant bit set, which is the bit that is different between
xand
x - 1if
xis a power of two. For example,
0b1000 & (~0b1000because
0b1000is a power of two, while
0b1010 & (~0b1010 + 1) = 0b0010because
0b1010` is not a power of two.Swapping the values of two variables: You can
swap the values of two variables x
and y
without using a temporary variable by using the
XOR operator (^
). The expression x = x ^ y; y = x ^ y; x = x ^ y;
swaps the values of x
and y
because x ^ y
is equal to the bitwise
XOR of the original values of x
and y
.
Counting the number of bits set to 1: You can
count the number of bits set to 1 in an int8
value by using the bitwise AND operator (&
)
and a loop that shifts the value to the right
by one bit each iteration. The expression
while (x != 0) { count += x & 1; x >>= 1; }
sets count
to the number of bits set to 1 in
x
.
Reversing the bits of an int8 value: You can reverse the bits of an int8 value by using a lookup table that maps each possible 8-bit value to its bit-reversed counterpart. Alternatively, you can use a loop that swaps the least significant and most significant bits of the value, then shifts the value to the right and left by one bit each iteration. The expression `x = ((x & 0x55) <<
reverses the bits of
x`.I hope this helps! Let me know if you have any other questions.
5> keep going! this is awesome!
Great, here are a few more bit-twiddling tricks and operations that you can perform on int8 values:
Checking if a number has alternating bits: You
can check if a number has alternating bits
(i.e., every adjacent pair of bits is
different) by using the bitwise XOR operator
(^
) and a loop that shifts the value to the
right by one bit each iteration. The expression
while (x != 0) { if ((x & 3) == 3 || (x & 3) == 0) return false; x >>= 1; } return true;
returns true
if x
has alternating bits, and
false
otherwise.
Computing the absolute value of a number: You
can compute the absolute value of an int8 value
by using a conditional statement that checks if
the value is negative, and negating it if it
is. The expression if (x < 0) x = -x;
sets
x
to its absolute value.
Setting and clearing bits: You can set a bit in
an int8 value by using the bitwise OR operator
(|
) and a mask that has only the bit you want
to set set to 1. The expression x |= (1 << i);
sets the i
-th bit of x
to
&
)
and a mask that has only the bit you want to
clear set to 0. The expression x &= ~(1 << i);
clears the i
-th bit of x
.Checking if a number is a palindrome in binary:
You can check if a number is a palindrome in
binary (i.e., its binary representation is the
same when read forwards and backwards) by using
the bitwise AND operator (&
) and a loop that
checks if the most significant and least
significant bits are the same, then shifts the
value to the right and left by one bit each
iteration. The expression `for (int i = 0;
i < 4; i++) { if (((x >> i) & 1) != ((x >> (7
returns
trueif
xis a palindrome in binary, and
false` otherwise.I hope you find these additional bit-twiddling tricks and operations useful! Let me know if you have any questions.
6> got any more?
Sure! Here are a few more bit-twiddling tricks and operations that you can perform on int8 values:
Counting the number of set bits: You can count
the number of set bits (i.e., bits that are
equal to 1) in an int8 value by using a loop
that shifts the value to the right by one bit
each iteration and checks if the least
significant bit is set. The expression int count = 0; while (x != 0) { if (x & 1) count++; x >>= 1; }
sets count
to the number of set
bits in x
.
Reversing the bits: You can reverse the bits
in an int8 value by using a loop that swaps
the most significant and least significant
bits, then swaps the second most significant
and second least significant bits, and so
on. The expression int y = 0; for (int i = 0; i < 8; i++) { y <<= 1; y |= (x & 1); x >>= 1; }
sets y
to the value of x
with its bits
reversed.
Finding the next power of two: You can find
the next power of two that is greater than or
equal to an int8 value by using the bitwise OR
operator (|
) and a loop that shifts a value
that has only the least significant bit set to
1 to the left until it is greater than or
equal to the original value. The expression
int y = 1; while (y < x) y <<= 1;
sets y
to the next power of two that is greater than
or equal to x
.
Finding the greatest common divisor: You can
find the greatest common divisor of two int8
values by using the bitwise AND operator (&
)
and a loop that repeatedly subtracts the
smaller value from the larger value until they
are equal, then returns either value. The
expression while (x != y) { if (x > y) x -= y; else y -= x; } return x;
returns the
greatest common divisor of x
and y
.
I hope you find these additional bit-twiddling tricks and operations useful as well! Let me know if you have any questions.
7> keep going if you can! i am interested to see how many you know!
Sure, here are some more bit-twiddling tricks and operations:
Computing the absolute value: You can compute
the absolute value of an int8 value by using
a conditional statement that checks if the
value is negative, then subtracts it from 0 if
it is. The expression int y = (x < 0) ? 0 - x : x;
sets y
to the absolute value
of x
.
Swapping two values without using a temporary
variable: You can swap the values of two int8
variables x
and y
without using
a temporary variable by using the XOR operator
(^
) and a loop that performs the following
operations: x = x ^ y; y = x ^ y; x = x ^ y;
This works because x ^ y
produces a value
that has the bits of both x
and y
set
where they differ, and unset where they are
the same.
Computing the sign of a value: You can compute
the sign of an int8 value by using
a conditional statement that checks if the
value is negative, then sets a variable to -1
if it is, or 1 if it is not. The expression
int sign = (x < 0) ? -1 : 1;
sets sign
to
-1 if x
is negative, or 1 if it is not.
Multiplying and dividing by powers of two: You
can multiply an int8 value by a power of two
by shifting the value to the left by the
number of bits that corresponds to the
exponent of the power of two. For example, to
multiply x
by 2^3 (i.e., 8), you can use the
expression x << 3
. Similarly, you can divide
an int8 value by a power of two by shifting
the value to the right by the number of bits
that corresponds to the exponent of the power
of two. For example, to divide x
by 2^3
(i.e., 8), you can use the expression x >> 3
.
Setting or clearing individual bits: You can
set a particular bit of an int8 value to 1 by
using the bitwise OR operator (|
) and a mask
that has only the bit you want to set set to
x
to
1, you can use the expression x |= (1 << 3)
. Similarly, you can clear a particular bit
of an int8 value to 0 by using the bitwise AND
operator (&
) and a mask that has all bits
set to 1 except the bit you want to clear,
which is set to 0. For example, to clear the
3rd bit of x
, you can use the expression x &= ~(1 << 3)
.I hope you find these additional bit-twiddling tricks and operations useful as well! Let me know if you have any questions or if there's anything else I can help you with.
8> any more?
Sure, here are some more bit-twiddling tricks and operations:
Checking if a value is a power of two: You can
check if an int8 value x
is a power of two
by using the bitwise AND operator (&
) and
a mask that has all bits set to 0 except the
least significant bit, which is set to 1. If
the result of x & (x - 1)
is 0, then x
is
a power of two. For example, to check if x
is a power of two, you can use the expression
(x & (x - 1)) == 0
.
Counting the number of set bits: You can count
the number of bits that are set to 1 in an
int8 value x
by using a loop that shifts x
to the right by one bit at a time and counts
the number of times that the least significant
bit is 1. For example, to count the number of
set bits in x
, you can use the following
code:
int count = 0;
while (x != 0) {
if (x & 1) {
count++;
}
x = x >> 1;
}
x
by using
a loop that iteratively swaps the least
significant bit with the most significant bit,
then shifts x
to the right by one bit. For
example, to reverse the bits in x
, you can
use the following code:int y = 0;
for (int i = 0; i < 8; i++) {
y = (y << 1) | (x & 1);
x = x >> 1;
}
I hope you find these additional bit-twiddling tricks and operations useful! Let me know if you have any questions or if there's anything else I can help you with.
9> could you show me how to do these with single instructions on the arm architecture?
Sure, here are some examples of how to perform some of the bit-twiddling operations I mentioned earlier using single instructions on the ARM architecture:
Setting a bit:
You can set the nth bit of a register x
by using
the ORR
instruction with a bitmask that has
a 1 in the nth position. For example, to set the
3rd bit of x
, you can use the instruction
ORR x, x, #0b1000
.
Clearing a bit: You can clear the nth bit of
a register x
by using the BIC
instruction
with a bitmask that has a 0 in the nth
position. For example, to clear the 3rd bit of
x
, you can use the instruction BIC x, x, #0b1000
.
Toggling a bit: You can toggle the nth bit of
a register x
by using the EOR
instruction
with a bitmask that has a 1 in the nth
position. For example, to toggle the 3rd bit of
x
, you can use the instruction EOR x, x, #0b1000
.
Checking if a bit is set: You can check if the
nth bit of a register x
is set by using the
TST
instruction with a bitmask that has
a 1 in the nth position. For example, to check
if the 3rd bit of x
is set, you can use the
instruction TST x, #0b1000
.
Extracting a bitfield: You can extract
a bitfield of length n starting at position
p from a register x
by using the UBFX
or
SBFX
instruction. The UBFX
instruction
extracts an unsigned bitfield, while the SBFX
instruction extracts a signed bitfield. For
example, to extract a 3-bit unsigned bitfield
starting at position 4 from x
, you can use
the instruction UBFX w0, w1, #4, #3
.
Reversing the order of bits: You can reverse
the order of the bits in a register x
by
using the RBIT
instruction. For example, to
reverse the order of the bits in x
, you can
use the instruction RBIT w0, w1
.
I hope this helps! Let me know if you have any further questions.
10> any more?
Sure, here are a few more bit-twiddling operations you can perform using single instructions on the ARM architecture:
Counting the number of set bits: You can count
the number of set bits (also known as the
population count or Hamming weight) in a register
x
using the CNT
or CNTB
instruction. The
CNT
instruction counts the number of set bits in
a 32-bit word, while the CNTB
instruction counts
the number of set bits in each byte of a 64-bit
register. For example, to count the number of set
bits in x
, you can use the instruction CNT x0, x1
.
Finding the position of the most significant
set bit: You can find the position of the most
significant set bit (also known as the position of
the leftmost set bit or the log base 2) in
a register x
using the CLZ
instruction. For
example, to find the position of the most
significant set bit in x
, you can use the
instruction CLZ w0, w1
.
Rotating bits: You can rotate the bits in
a register x
left or right by a specified number
of bits using the LSL
, LSR
, ASR
, or ROR
instructions. The LSL
and LSR
instructions
perform logical left and right shifts,
respectively, while the ASR
instruction performs
an arithmetic right shift (which preserves the
sign bit), and the ROR
instruction performs
a rotate right (where bits shifted off the right
end are shifted back in on the left). For example,
to rotate the bits in x
left by 4 bits, you can
use the instruction LSL x0, x1, #4
.
Swapping bytes or halfwords: You can swap the
bytes or halfwords in a register x
using the
REV16
or REV32
instruction, respectively. The
REV16
instruction swaps the two halfwords in
a 32-bit word, while the REV32
instruction swaps
the order of the bytes in a 32-bit word. For
example, to swap the bytes in x
, you can use the
instruction REV32 w0, w1
.
Reversing the order of bytes or halfwords: You
can reverse the order of the bytes or halfwords in
a register x
using the REV16
or REV
instruction, respectively. The REV16
instruction
reverses the order of the two halfwords in
a 32-bit word, while the REV
instruction
reverses the order of the bytes in a 32-bit
word. For example, to reverse the order of the
bytes in x
, you can use the instruction REV w0, w1
.
I hope you find these additional operations useful! Let me know if you have any further questions.
11> great! what are the main differences between bit twiddling an i8 and a u8? what about i8 and i16? i8 and u16? int and float?
The main differences between bit twiddling an i8
and a u8
are due to the signedness of i8
. In
particular, the most significant bit of an i8
represents the sign of the number, while all bits
of a u8
represent the magnitude of the number.
When performing bit twiddling operations on i8
values, you need to be careful to preserve the
sign bit if you want to maintain the sign of the
number. For example, if you want to perform a left
shift on an i8
value, you should use an
arithmetic left shift (shl
) instead of a logical
left shift (shlx
), which would insert zeros
instead of preserving the sign bit. Similarly,
when performing a right shift on an i8
value,
you should use an arithmetic right shift (sra
)
instead of a logical right shift (srlx
), which
would insert zeros instead of preserving the sign
bit.
When bit-twiddling an i8
and an i16
, the main
difference is the size of the register. Operations
on i16
can involve more bits than operations on
i8
, which can lead to different results or
require more instructions to perform. Similarly,
when bit-twiddling an i8
and a u16
, the main
difference is the fact that the u16
can
represent larger magnitudes than the i8
, which
can lead to different results or require more
instructions to perform certain operations.
When bit-twiddling an i8
and a floating-point
value, such as f32
or f64
, the main difference
is the fact that floating-point values have
a completely different representation and
semantics than integer values. Bit-twiddling
operations on floating-point values are typically
used for low-level optimization purposes and
require a good understanding of the floating-point
format and its operations. For example, you can
extract the sign, exponent, and mantissa of
a floating-point value using bit-twiddling
operations, but you need to be careful to handle
special values such as NaN and infinity correctly.
12> great. are there any more things you think i ought to know about i8 in regard to the i8quantized crate?
Yes, here are a few more things you should keep in
mind when working with i8
in the context of the
caffe2op-i8quantized
crate:
When performing operations on i8
values, you
should be careful to avoid overflow and
underflow. The range of values that can be
represented by an i8
is from -128 to 127, and
if an operation results in a value outside of
this range, the result will be truncated to fit
within the range. This can lead to unexpected
behavior and loss of precision. To avoid this,
you can use saturation arithmetic, which caps
the result of an operation at the maximum or
minimum value of the range. The i8quantized
crate provides several functions for performing
arithmetic with saturation, such as
saturating_add
, saturating_sub
,
saturating_mul
, and saturating_pow2
.
When working with quantized values, it's
important to keep track of the scaling factor
used to convert between the original
floating-point values and the quantized integer
values. The scaling factor determines the
precision of the quantized values, and can
affect the accuracy of the computation. The
i8quantized
crate provides several functions
for working with quantized values and scaling
factors, such as dequantize
, which converts
a quantized value back to a floating-point value
using the scaling factor.
In some cases, you may need to perform
operations that involve both i8
and
floating-point values. In these cases, you need
to be careful to maintain the correct scaling
factor and avoid loss of precision. The
i8quantized
crate provides several functions
for converting between i8
and floating-point
values, such as quantize
, which converts
a floating-point value to a quantized i8
value
using a specified scaling factor.
The i8quantized
crate provides several
operations commonly used in machine learning and
DSP computations, such as convolution, pooling,
activation functions, and matrix
multiplication. These operations are optimized
for performance on CPUs and GPUs, and can be
used to implement complex machine learning
models efficiently. However, you should still be
careful to avoid overflow and underflow, and to
maintain the correct scaling factor when using
these operations.