def exponentiate(A,i): """Calculate A**i""" if i==0: return 1 if i==1: return A return exponentiate(A,i/2)*exponentiate(A,i/2+(i%2))
This method allows computing Ai in a number of multiplications that is logarithmic in i. We will show how to implement each multiplication in polynomial space by use of recursion: each computation will utilize polynomial memory, and the depth of the recursion will be polynomial in log(i), so the total memory requirements are polynomial.
Consider how multiplication is usually performed. The following code multiplies A by B, writing out the result in C. A, B and C are stored as arrays, where A[x] is the bit of A whose value is 2x. (By convention, if x is larger than the bit-length of A, we take A[x] to be equal to 0.) As we are only interested in what happens up to bit n, this code will compute C only up to bit n. In essence, the code performs "mod 2n+1" multiplication.
def mult_mod(A,B,C,n): """Return in C the value of (A*B) % 2**(n+1)""" acc=0 for counter in range(n+1): acc/=2 k=counter for j in range(counter+1): acc+=A[j]*B[k] k-=1 C[counter]=acc % 2
In this code, acc is used both to compute the next digit and the carry. The carry itself is logarithmic in n, so can be stored.
For our application, A, B an C may be too large to store, so we will not store them. Specifically, we will only execute the line "C[counter]=acc % 2" when counter equals n.
The actual code will therefore look more like this:
def mult_digit(A,B,n): """Return the n'th digit of A*B""" acc=0 for counter in range(n+1): acc/=2 k=counter for j in range(counter+1): acc+=A[j]*B[k] k-=1 return acc % 2
Combining both of the above ideas together, we get our finalized code. Here, A is treated as an array, but, for better readability, n and i are treated as integers.
def exp_digit(n,A,i): """A function that calculates the n'th digit of A**i""" if i==0: return (n==0) if i==1: return A[n] # Zero, if n>bit-length of A. acc=0 for counter in range(n+1): acc/=2 k=counter for j in range(counter+1): acc+=exp_digit(j,A,i/2)*exp_digit(k,A,i/2+(i%2)) k-=1 return acc % 2
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