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.