Hello,

I would like to understand the size of the public key generated with this simple circuit:

```
import time
import numpy as np
import concrete.numpy as cnp
nfeatures = 1000
nclasses = 10
w = np.random.randint(0,16,(nfeatures,10))
X_train = np.random.randint(0,2, (1000, nfeatures)).astype(np.int16)
print(X_train.shape)
cfg = cnp.Configuration(show_graph=True)
@cnp.compiler({"x": "encrypted"})
def g(x):
return x @ w
inputset = X_train[0:4,:]
inpt = X_train[5,:]
print("compiling")
t = time.time()
circuit = g.compile(inputset, configuration=cfg)
print('compiled in ', time.time()-t)
t = time.time()
circuit.keygen()
print("Keygen done in ", time.time()-t)
print(f"Encryption keys: {circuit.size_of_secret_keys} bytes")
print(f"Evaluation keys: {circuit.size_of_bootstrap_keys + circuit.size_of_keyswitch_keys} bytes")
print(f"Inputs: {circuit.size_of_inputs} bytes")
print(f"Outputs: {circuit.size_of_outputs} bytes")
enc = circuit.encrypt(inpt)
t = time.time()
res_enc = circuit.run(enc)
print("Run time : ", time.time()-t)
dec = circuit.decrypt(res_enc)
```

Which gives the following output :

```
Encryption keys: 8200 bytes
Evaluation keys: 0 bytes
Inputs: 8200000 bytes
Outputs: 82000 bytes
```

I understand from PBS paper that no PBS is required as only leveled operations are required for this linear regression, but shouldn’t there be a public key anyway ? Or as no PBS is needed no public key is generated ?

Thanks in advance