def dencrypt(M,bs,d,n):
a=[]
c=""
i=0
end=len(M[0])
while i<end:
if(i!=end-1):
c=make(str(Pow.resMod(int(M[0][i]),d,n)),bs)
else:
c=make(str(Pow.resMod(int(M[0][i]),d,n)),M[1])
a.append(c)
i+=1
return a
def encrypt(M,bs,e,n):
f=make(str(M),3)
i=0
batches=split(f,bs)[0]
tail=split(f,bs)[1]
while i<len(batches):
batches[i]=str(Pow.resMod(int(batches[i]),e,n))
i+=1
return [batches, tail]
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""Generate pseudo random numbers. Should not be used for security purposes."""
from '__go__/math/rand' import Uint32, Seed
from '__go__/math' import Pow
from '__go__/time' import Now
BPF = 53 # Number of bits in a float
RECIP_BPF = Pow(2, -BPF)
# TODO: The random byte generator currently uses math.rand.Uint32 to generate
# 4 bytes at a time. We should use math.rand.Read to generate the correct
# number of bytes needed. This can be changed once there is a way to
# allocate the needed []byte for Read from python and cast it to a list of
# integers once it is filled.
def _gorandom(nbytes):
byte_arr = []
while len(byte_arr) < nbytes:
i = Uint32()
byte_arr.append(i & 0xff)
byte_arr.append(i >> 8 & 0xff)
byte_arr.append(i >> 16 & 0xff)
byte_arr.append(i >> 24 & 0xff)
def pow(x, y):
return Pow(float(x), float(y))
def main():
# LIS.run()
# MAX_MAX.run()
Pow.run(4, 8)
def findCoprime(K):
n=findCoHelp(K,K,Pow.pow(10,100,10))
while(gcd(n,K)!=1 or n==0):
n-=1
return n
