Created
December 30, 2019 04:01
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Multiplicative inverse mod 2^w, without Euclid or Bezout.
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(defun multiplicative-inverse (x bitwidth) | |
"Iteratively finds the multiplicative inverse of x mod 2^bitwidth. | |
The induction step uses the fact that, for a given inverse x^-1 | |
with (x * x^-1) mod 2^w == 1, (x * x^-1) mod 2^{w - 1} == 1 as | |
well. We start with the trivial inverse mod 2, and increase the | |
bitwidth iteratively: given inv_{w - 1} = x^-1 mod 2^{w - 1}, | |
x^-1 mod 2^w is either inv_{w - 1}, or inv_{w - 1} + 2^{w - 1}." | |
(assert (oddp x)) | |
(let ((inverse 1)) | |
;; For any odd x, (x * 1) mod 2 == 1, so | |
;; we begin the search at mod (1 << 2) == mod 4. | |
(loop for i from 2 upto bitwidth | |
do (let ((mask (1- (ash 1 i)))) | |
(setf inverse (logior inverse | |
(if (= 1 (logand (* inverse x) mask)) | |
0 | |
(ash 1 (1- i))))) | |
(assert (= 1 (logand (* inverse x) mask)))) | |
finally | |
(assert (= 1 (mod (* inverse x) (ash 1 bitwidth)))) | |
(return inverse)))) |
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