X-Git-Url: https://git.distorted.org.uk/~mdw/catacomb/blobdiff_plain/ceb8d344ecb165be756a77ac8527c870b36945b2..HEAD:/math/gfreduce.c

diff --git a/math/gfreduce.c b/math/gfreduce.c
index 37703a43..f29c1048 100644
--- a/math/gfreduce.c
+++ b/math/gfreduce.c
@@ -49,14 +49,15 @@ DA_DECL(instr_v, gfreduce_instr);
  * sense fast.  Here, we do efficient reduction modulo sparse polynomials.
  * (It works for arbitrary polynomials, but isn't efficient for dense ones.)
  *
- * Suppose that %$p(x) = x^n + p'(x) = \sum_{0\le i<n} p_i x^i$%, hopefully
- * with only a few other %$p_i \ne 0$%.  We're going to compile %$p$% into a
- * sequence of instructions which can be used to perform reduction modulo
- * %$p$%.  The important observation is that %$x^n \equiv p' \pmod p$%.
+ * Suppose that %$p = x^n + p'$% where %$p' = \sum_{0\le i<n} p_i x^i$%,
+ * hopefully with only a few %$p_i \ne 0$%.  We're going to compile %$p$%
+ * into a sequence of instructions which can be used to perform reduction
+ * modulo %$p$%.  The important observation is that
+ * %$x^n \equiv p' \pmod p$%.
  *
  * Suppose we're working with %$w$%-bit words; let %$n = N w + n'$% with
  * %$0 \le n' < w$%.  Let %$u(x)$% be some arbitrary polynomial.  Write
- * %$u = z x^k + u'$% with %$\deg u' < k \ge n$%; then a reduction step uses
+ * %$u = z x^k + u'$% with %$\deg u' < k \ge n$%.  Then a reduction step uses
  * that %$u \equiv u' + z p' x^{k-n} \pmod p$%: the right hand side has
  * degree %$\max \{ \deg u', k + \deg p' - n + \deg z \} < \deg u$%, so this
  * makes progress towards a complete reduction.
@@ -296,7 +297,7 @@ void gfreduce_destroy(gfreduce *r)
 
 /* --- @gfreduce_dump@ --- *
  *
- * Arguments:	@gfreduce *r@ = structure to dump
+ * Arguments:	@const gfreduce *r@ = structure to dump
  *		@FILE *fp@ = file to dump on
  *
  * Returns:	---
@@ -304,7 +305,7 @@ void gfreduce_destroy(gfreduce *r)
  * Use:		Dumps a reduction context.
  */
 
-void gfreduce_dump(gfreduce *r, FILE *fp)
+void gfreduce_dump(const gfreduce *r, FILE *fp)
 {
   size_t i;
 
@@ -326,7 +327,7 @@ void gfreduce_dump(gfreduce *r, FILE *fp)
 
 /* --- @gfreduce_do@ --- *
  *
- * Arguments:	@gfreduce *r@ = reduction context
+ * Arguments:	@const gfreduce *r@ = reduction context
  *		@mp *d@ = destination
  *		@mp *x@ = source
  *
@@ -349,7 +350,7 @@ static void run(const gfreduce_instr *i, const gfreduce_instr *il,
   }
 }
 
-mp *gfreduce_do(gfreduce *r, mp *d, mp *x)
+mp *gfreduce_do(const gfreduce *r, mp *d, mp *x)
 {
   mpw *v, *vl;
   const gfreduce_instr *il;
@@ -391,20 +392,26 @@ mp *gfreduce_do(gfreduce *r, mp *d, mp *x)
 
 /* --- @gfreduce_sqrt@ --- *
  *
- * Arguments:	@gfreduce *r@ = pointer to reduction context
+ * Arguments:	@const gfreduce *r@ = pointer to reduction context
  *		@mp *d@ = destination
  *		@mp *x@ = some polynomial
  *
  * Returns:	The square root of @x@ modulo @r->p@, or null.
  */
 
-mp *gfreduce_sqrt(gfreduce *r, mp *d, mp *x)
+mp *gfreduce_sqrt(const gfreduce *r, mp *d, mp *x)
 {
   mp *y = MP_COPY(x);
   mp *z, *spare = MP_NEW;
   unsigned long m = mp_bits(r->p) - 1;
   unsigned long i;
 
+  /* --- This is pretty easy --- *
+   *
+   * Note that %$x = x^{2^m}$%; therefore %$(x^{2^{m-1}})^2 = x^{2^m} = x$%,
+   * so %$x^{2^{m-1}}$% is the square root we seek.
+   */
+
   for (i = 0; i < m - 1; i++) {
     mp *t = gf_sqr(spare, y);
     spare = y;
@@ -423,14 +430,17 @@ mp *gfreduce_sqrt(gfreduce *r, mp *d, mp *x)
 
 /* --- @gfreduce_trace@ --- *
  *
- * Arguments:	@gfreduce *r@ = pointer to reduction context
+ * Arguments:	@const gfreduce *r@ = pointer to reduction context
  *		@mp *x@ = some polynomial
  *
  * Returns:	The trace of @x@. (%$\Tr(x)=x + x^2 + \cdots + x^{2^{m-1}}$%
- *		if %$x \in \gf{2^m}$%).
+ *		if %$x \in \gf{2^m}$%).  Since the trace is invariant under
+ *		the Frobenius automorphism (i.e., %$\Tr(x)^2 = \Tr(x)$%), it
+ *		must be an element of the base field, i.e., %$\gf{2}$%, and
+ *		we only need a single bit to represent it.
  */
 
-int gfreduce_trace(gfreduce *r, mp *x)
+int gfreduce_trace(const gfreduce *r, mp *x)
 {
   mp *y = MP_COPY(x);
   mp *spare = MP_NEW;
@@ -452,7 +462,7 @@ int gfreduce_trace(gfreduce *r, mp *x)
 
 /* --- @gfreduce_halftrace@ --- *
  *
- * Arguments:	@gfreduce *r@ = pointer to reduction context
+ * Arguments:	@const gfreduce *r@ = pointer to reduction context
  *		@mp *d@ = destination
  *		@mp *x@ = some polynomial
  *
@@ -461,7 +471,7 @@ int gfreduce_trace(gfreduce *r, mp *x)
  *		if %$x \in \gf{2^m}$% with %$m$% odd).
  */
 
-mp *gfreduce_halftrace(gfreduce *r, mp *d, mp *x)
+mp *gfreduce_halftrace(const gfreduce *r, mp *d, mp *x)
 {
   mp *y = MP_COPY(x);
   mp *spare = MP_NEW;
@@ -484,27 +494,82 @@ mp *gfreduce_halftrace(gfreduce *r, mp *d, mp *x)
 
 /* --- @gfreduce_quadsolve@ --- *
  *
- * Arguments:	@gfreduce *r@ = pointer to reduction context
+ * Arguments:	@const gfreduce *r@ = pointer to reduction context
  *		@mp *d@ = destination
  *		@mp *x@ = some polynomial
  *
- * Returns:	A polynomial @y@ such that %$y^2 + y = x$%, or null.
+ * Returns:	A polynomial @z@ such that %$z^2 + z = x$%, or null.
+ *
+ * Use:		Solves quadratic equations in a field with characteristic 2.
+ *		Suppose we have an equation %$y^2 + A y + B = 0$% where
+ *		%$A \ne 0$%.  (If %$A = 0$% then %$y = \sqrt{B}$% and you
+ *		want @gfreduce_sqrt@ instead.)  Use this function to solve
+ *		%$z^2 + z = B/A^2$%; then set %$y = A z$%, since
+ *		%$y^2 + y = A^2 z^2 + A^2 z = A^2 (z^2 + z) = B$% as
+ *		required.
+ *
+ *		The two roots are %$z$% and %$z + 1$%; this function always
+ *		returns the one with zero scalar coefficient.
  */
 
-mp *gfreduce_quadsolve(gfreduce *r, mp *d, mp *x)
+mp *gfreduce_quadsolve(const gfreduce *r, mp *d, mp *x)
 {
   unsigned long m = mp_bits(r->p) - 1;
   mp *t;
 
+  /* --- About the solutions --- *
+   *
+   * Factor %$z^2 + z = z (z + 1)$%.  Therefore, if %$z^2 + z = x$% and
+   * %$z' = z + 1$% then %$z'^2 + z' = z^2 + 1 + z + 1 = z^2 + z$%, so
+   * %$z + 1$% is the other solution.
+   *
+   * A solution exists if and only if %$\Tr(x) = 0$%.  To see the `only if'
+   * implication, recall that the trace function is linear, and hence
+   * $%\Tr(z^2 + z) = \Tr(z)^2 + \Tr(z) = \Tr(z) + \Tr(z) = 0$%.  The `if'
+   * direction will be proven using explicit constructions captured in the
+   * code below.
+   */
+
   MP_COPY(x);
-  if (m & 1)
+  if (m & 1) {
+
+    /* --- A short-cut for fields with odd degree ---
+     *
+     * The method below works in all binary fields, but there's a quicker way
+     * which works whenever the degree is odd.  The half-trace is
+     * %$z = \sum_{0\le i\le (m-1)/2} x^{2^{2i}}$%.  Then %$z^2 + z = {}$%
+     * %$\sum_{0\le i\le (m-1)/2} (x^{2^{2i}} + x^{2^{2i+1}}) = {}$%
+     * %$\Tr(x) + x^{2^m} = \Tr(x) + x$%.  This therefore gives us the
+     * solution we want whenever %$\Tr(x) = 0$%.
+     */
+
     d = gfreduce_halftrace(r, d, x);
-  else {
+  } else {
     mp *z, *w, *rho = MP_NEW;
     mp *spare = MP_NEW;
     grand *fr = fibrand_create(0);
     unsigned long i;
 
+    /* --- Unpicking the magic --- *
+     *
+     * Choose %$\rho \inr \gf{2^m}$% with %$\Tr(\rho) = 1$%.  Let
+     * %$z = \sum_{0\le i<m} \rho^{2^i} \sum_{0\le j<i} x^{2^j} = {}$%
+     * %$\sum_{1\le i<m} \rho^{2^i} (x + \sum_{1\le j<i} x^{2^j} = {}$%
+     * %$\rho^2 x + \rho^4 (x + x^2) + \rho^8 (x + x^2 + x^4) + \cdots + {}$%
+     * %$\rho^{2^{m-1}} (x + x^2 + x^{2^{m-2}})$%.  Then %$z^2 = {}$%
+     * %$\sum_{0\le i<m} \rho^{2^{i+1}} \sum_{0\le j<i} x^{2^{j+1}} = {}$%
+     * %$\sum_{1\le i\le m} \rho^{2^i} \sum_{1\le j<i} x^{2^j} = {}$%
+     * %$\sum_{1\le i<m} \rho^{2^i} \sum_{1\le j<i} x^{2^j} + {}$%
+     * %$\rho^{2^m} \sum_{1\le j<m} x^{2^j}$%; and, somewhat miraculously,
+     * %$z^2 + z = \sum_{1\le i<m} \rho^{2^i} x + {}$%
+     * %$\rho \sum_{1\le i<m} x^{2^i} = x (\Tr(\rho) + \rho) + {}$%
+     * %$\rho (\Tr(x) + x) = x \Tr(\rho) + \rho \Tr(x)$%.  Again,
+     * this gives us the root we want whenever %$\Tr(x) = 0$%.
+     *
+     * The loop below calculates %$w = \Tr(\rho)$% and %$z$% simultaneously,
+     * since the same powers of %$\rho$% are wanted in both calculations.
+     */
+
     for (;;) {
       rho = mprand(rho, m, fr, 0);
       z = MP_ZERO;
@@ -529,6 +594,12 @@ mp *gfreduce_quadsolve(gfreduce *r, mp *d, mp *x)
     d = z;
   }
 
+  /* --- Check that we calculated the right answer --- *
+   *
+   * It should be correct; if it's not then maybe the ring we're working in
+   * isn't really a field.
+   */
+
   t = gf_sqr(MP_NEW, d); t = gfreduce_do(r, t, t); t = gf_add(t, t, d);
   if (!MP_EQ(t, x)) {
     MP_DROP(d);
@@ -536,13 +607,20 @@ mp *gfreduce_quadsolve(gfreduce *r, mp *d, mp *x)
   }
   MP_DROP(t);
   MP_DROP(x);
+
+  /* --- Pick a canonical root --- *
+   *
+   * The two roots are %$z$% and %$z + 1$%; pick the one with a zero
+   * scalar coefficient just for consistency's sake.
+   */
+
   if (d) d->v[0] &= ~(mpw)1;
   return (d);
 }
 
 /* --- @gfreduce_exp@ --- *
  *
- * Arguments:	@gfreduce *gr@ = pointer to reduction context
+ * Arguments:	@const gfreduce *gr@ = pointer to reduction context
  *		@mp *d@ = fake destination
  *		@mp *a@ = base
  *		@mp *e@ = exponent
@@ -550,7 +628,7 @@ mp *gfreduce_quadsolve(gfreduce *r, mp *d, mp *x)
  * Returns:	Result, %$a^e \bmod m$%.
  */
 
-mp *gfreduce_exp(gfreduce *gr, mp *d, mp *a, mp *e)
+mp *gfreduce_exp(const gfreduce *gr, mp *d, mp *a, mp *e)
 {
   mp *x = MP_ONE;
   mp *spare = (e->f & MP_BURN) ? MP_NEWSEC : MP_NEW;