Concrete supports native Python and NumPy operations as much as possible, but not everything in Python or NumPy is available. Therefore, we provide some extensions ourselves to improve your experience.
fhe.univariate(function)
Allows you to wrap any univariate function into a single table lookup:
import numpy as npfrom concrete import fhedefcomplex_univariate_function(x):defper_element(element): result =0for i inrange(element): result += ireturn resultreturn np.vectorize(per_element)(x)@fhe.compiler({"x": "encrypted"})deff(x):return fhe.univariate(complex_univariate_function)(x)inputset = [np.random.randint(0, 5, size=(3, 2))for _ inrange(10)]circuit = f.compile(inputset)sample = np.array([ [0, 4], [2, 1], [3, 0],])assert np.array_equal(circuit.encrypt_run_decrypt(sample), complex_univariate_function(sample))
The wrapped function:
shouldn't have any side effects (e.g., no modification of global state)
should be deterministic (e.g., no random numbers)
should have the same output shape as its input (i.e., output.shape should be the same with input.shape)
each output element should correspond to a single input element (e.g., output[0] should only depend on input[0])
If any of these constraints are violated, the outcome is undefined.
fhe.multivariate(function)
Allows you to wrap any multivariate function into a table lookup:
Currently, only scalars can be used to create arrays.
fhe.zero()
Allows you to create an encrypted scalar zero:
from concrete import fheimport numpy as np@fhe.compiler({"x": "encrypted"})deff(x): z = fhe.zero()return x + zinputset =range(10)circuit = f.compile(inputset)for x inrange(10):assert circuit.encrypt_run_decrypt(x)== x
fhe.zeros(shape)
Allows you to create an encrypted tensor of zeros:
from concrete import fheimport numpy as np@fhe.compiler({"x": "encrypted"})deff(x): z = fhe.zeros((2, 3))return x + zinputset =range(10)circuit = f.compile(inputset)for x inrange(10):assert np.array_equal(circuit.encrypt_run_decrypt(x), np.array([[x, x, x], [x, x, x]]))
fhe.one()
Allows you to create an encrypted scalar one:
from concrete import fheimport numpy as np@fhe.compiler({"x": "encrypted"})deff(x): z = fhe.one()return x + zinputset =range(10)circuit = f.compile(inputset)for x inrange(10):assert circuit.encrypt_run_decrypt(x)== x +1
fhe.ones(shape)
Allows you to create an encrypted tensor of ones:
from concrete import fheimport numpy as np@fhe.compiler({"x": "encrypted"})deff(x): z = fhe.ones((2, 3))return x + zinputset =range(10)circuit = f.compile(inputset)for x inrange(10):assert np.array_equal(circuit.encrypt_run_decrypt(x), np.array([[x, x, x], [x, x, x]]) +1)
fhe.hint(value, **kwargs)
Allows you to hint properties of a value. Imagine you have this circuit:
from concrete import fheimport numpy as np@fhe.compiler({"x": "encrypted"})deff(x,y,z): a = x | y b = y & z c = a ^ breturn cinputset = [ (np.random.randint(0, 2**8), np.random.randint(0, 2**8), np.random.randint(0, 2**8))for _ inrange(3)]circuit = f.compile(inputset)print(circuit)
You'd expect all of a, b, and c to be 8-bits, but because inputset is very small, this code could print:
%0 = x # EncryptedScalar<uint8> ∈ [173, 240]
%1 = y # EncryptedScalar<uint8> ∈ [52, 219]
%2 = z # EncryptedScalar<uint8> ∈ [36, 252]
%3 = bitwise_or(%0, %1) # EncryptedScalar<uint8> ∈ [243, 255]
%4 = bitwise_and(%1, %2) # EncryptedScalar<uint7> ∈ [0, 112]
^^^^^ this can lead to bugs
%5 = bitwise_xor(%3, %4) # EncryptedScalar<uint8> ∈ [131, 255]
return %5
The first solution in these cases should be to use a bigger inputset, but it can still be tricky to solve with the inputset. That's where the hint extension comes into play. Hints are a way to provide extra information to compilation process:
Bit-width hints are for constraining the minimum number of bits in the encoded value. If you hint a value to be 8-bits, it means it should be at least uint8 or int8.
To fix f using hints, you can do:
@fhe.compiler({"x": "encrypted", "y": "encrypted", "z": "encrypted"})deff(x,y,z):# hint that inputs should be considered at least 8-bits x = fhe.hint(x, bit_width=8) y = fhe.hint(y, bit_width=8) z = fhe.hint(z, bit_width=8)# hint that intermediates should be considered at least 8-bits a = fhe.hint(x | y, bit_width=8) b = fhe.hint(y & z, bit_width=8) c = fhe.hint(a ^ b, bit_width=8)return c
Hints are only applied to the value being hinted, and no other value. If you want the hint to be applied to multiple values, you need to hint all of them.
Alternatively, you can use it to make sure a value can store certain integers:
@fhe.compiler({"x": "encrypted", "y": "encrypted"})defis_vectors_same(x,y):assert x.ndim !=1assert y.ndim !=1assertlen(x)==len(y) n =len(x) number_of_same_elements = np.sum(x == y) fhe.hint(number_of_same_elements, can_store=n)# hint that number of same elements can go up to n is_same = number_of_same_elements == nreturn is_same
fhe.relu(value)
Allows you to perform ReLU operation, with the same semantic as x if x >= 0 else 0:
ReLU extension can be converted in two different ways:
With a single TLU on the original bit-width.
With multiple TLUs on smaller bit-widths.
For small bit-widths, the first one is better as it'll have a single TLU on a small bit-width. For big bit-widths, the second one is better as it won't have a TLU on a big bit-width.
The decision between the two can be controlled with relu_on_bits_threshold: int = 7 configuration option:
relu_on_bits_threshold=5 means:
1-bit to 4-bits would be converted using the first way (i.e., using TLU)
5-bits and more would be converted using the second way (i.e., using bits)
There is another option to customize the implementation relu_on_bits_chunk_size: int = 2:
relu_on_bits_chunk_size=4 means:
When using the second implementation:
The input would be split to 4-bit chunks using fhe.bits, and then the ReLU would be applied to those chunks, which are then combined back.
Here is a script showing how execution cost is impacted when changing these values:
from concrete import fheimport numpy as npimport matplotlib.pyplot as pltchunk_sizes = np.array(range(1, 6), dtype=int)bit_widths = np.array(range(5, 17), dtype=int)data = []for bit_width in bit_widths: title =f"{bit_width=}:"print(title)print("-"*len(title)) inputset =range(-2**(bit_width-1), 2**(bit_width-1)) configuration = fhe.Configuration(relu_on_bits_threshold=17) compiler = fhe.Compiler(lambdax: fhe.relu((fhe.relu(x) - (2**(bit_width-2))) *2), {"x": "encrypted"}) circuit = compiler.compile(inputset, configuration)print(f" Complexity: {circuit.complexity} # tlu") data.append((bit_width, 0, circuit.complexity))for chunk_size in chunk_sizes: configuration = fhe.Configuration( relu_on_bits_threshold=1, relu_on_bits_chunk_size=int(chunk_size), ) circuit = compiler.compile(inputset, configuration)print(f" Complexity: {circuit.complexity} # {chunk_size=}") data.append((bit_width, chunk_size, circuit.complexity))print()data = np.array(data)plt.title(f"ReLU using TLU vs using bits")plt.xlabel("Input/Output precision")plt.ylabel("Cost")for i, chunk_size inenumerate([0] +list(chunk_sizes)): costs = [ costfor _, candidate_chunk_size, cost in dataif candidate_chunk_size == chunk_size ]assertlen(costs)==len(bit_widths) label ="Single TLU"if i ==0elsef"Bits extract + multiples {chunk_size +1} bits TLUs" width_bar =0.8/ (len(chunk_sizes)+1)if i ==0: plt.hlines( costs, bit_widths -0.45, bit_widths +0.45, label=label, linestyle="--", )else: plt.bar( np.array(bit_widths) + width_bar * (i - (len(chunk_sizes) +1) /2), height=costs, width=width_bar, label=label, )plt.xticks(bit_widths)plt.legend(loc="upper left")plt.show()
You might need to run the script twice to avoid crashing when plotting.
The script will show the following figure:
The default values of these options are set based on simple circuits. How they affect performance will depend on the circuit, so play around with them to get the most out of this extension.
Conversion with the second method (i.e., using chunks) only works in Native encoding, which is usually selected when all table lookups in the circuit are below or equal to 8 bits.
fhe.if_then_else(condition, x, y)
Allows you to perform ternary if operation, with the same semantic as x if condition else y:
Identity extension can be used to clone an input while changing its bit-width. Imagine you have return x**2, x+100 where x is 2-bits. Because of x+100, x will be assigned 7-bits and x**2 would be more expensive than it needs to be. If return x**2, fhe.identity(x)+100 is used instead, x will be assigned 2-bits as it should and fhe.identity(x) will be assigned 7-bits as necessary.
Identity extension only works in Native encoding, which is usually selected when all table lookups in the circuit are below or equal to 8 bits.
fhe.inputset(...)
Used for creating a random inputset with the given specifications: