X-Git-Url: https://git.distorted.org.uk/~mdw/catacomb/blobdiff_plain/8e663045cd18e14e99affd3f904170dc686fd76c..eea933cdf237601a44e13006cfcfd3aad7a08349:/math/gfreduce.c

diff --git a/math/gfreduce.c b/math/gfreduce.c
index 96e0cedf..83ffeb10 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.
@@ -79,9 +80,12 @@ struct gen {
   unsigned f;				/* Flags */
 #define f_lsr 1u			/*   Overflow from previous word */
 #define f_load 2u			/*   Outstanding @LOAD@ */
+#define f_fip 4u			/*   Final-pass offset is set */
   instr_v iv;				/* Instruction vector */
+  size_t fip;				/* Offset for final-pass reduction */
   size_t w;				/* Currently loaded target word */
   size_t wi;				/* Left-shifts for current word */
+  gfreduce *r;				/* Reduction context pointer */
 };
 
 #define INSTR(g_, op_, arg_) do {					\
@@ -97,6 +101,24 @@ struct gen {
 
 static void emit_load(struct gen *g, size_t w)
 {
+  /* --- If this is not the low-order word then note final-pass start --- *
+   *
+   * Once we've eliminated the whole high-degree words, there will possibly
+   * remain a few high-degree bits.  We can further reduce the subject
+   * polynomial by subtracting an appropriate multiple of %$p'$%, but if we
+   * do this naively we'll end up addressing `low-order' words beyond the
+   * bottom of our input.  We solve this problem by storing an alternative
+   * start position for this final pass (which works because we scan bits
+   * right-to-left).
+   */
+
+  if (!(g->f & f_fip) && w < g->r->lim) {
+    g->fip = DA_LEN(&g->iv);
+    g->f |= f_fip;
+  }
+
+  /* --- Actually emit the instruction --- */
+
   INSTR(g, GFRI_LOAD, w);
   g->f |= f_load;
   g->w = w;
@@ -160,6 +182,7 @@ void gfreduce_create(gfreduce *r, mp *p)
 
   /* --- Sort out the easy stuff --- */
 
+  g.r = r;
   d = mp_bits(p); assert(d); d--;
   r->lim = d/MPW_BITS;
   dw = d%MPW_BITS;
@@ -234,9 +257,20 @@ void gfreduce_create(gfreduce *r, mp *p)
     INSTR(&g, GFRI_STORE, g.w);
   }
 
+  /* --- Copy the instruction vector.
+   *
+   * If we've not set a final-pass offset yet then now would be an excellent
+   * time.  Obviously it should be right at the end, because there's nothing
+   * for a final pass to do.
+   */
+
   r->in = DA_LEN(&g.iv);
   r->iv = xmalloc(r->in * sizeof(gfreduce_instr));
   memcpy(r->iv, DA(&g.iv), r->in * sizeof(gfreduce_instr));
+
+  if (!(g.f & f_fip)) g.fip = DA_LEN(&g.iv);
+  r->fiv = r->iv + g.fip;
+
   DA_DESTROY(&g.iv);
 }
 
@@ -244,6 +278,7 @@ void gfreduce_create(gfreduce *r, mp *p)
 
 #undef f_lsr
 #undef f_load
+#undef f_fip
 
 /* --- @gfreduce_destroy@ --- *
  *
@@ -279,11 +314,15 @@ void gfreduce_dump(gfreduce *r, FILE *fp)
 	  (unsigned long)r->lim, (unsigned long)r->mask);
   for (i = 0; i < r->in; i++) {
     static const char *opname[] = { "load", "lsl", "lsr", "store" };
+    if (&r->iv[i] == r->fiv)
+      fputs("final:\n", fp);
     assert(r->iv[i].op < N(opname));
     fprintf(fp, "  %s %lu\n",
 	    opname[r->iv[i].op],
 	    (unsigned long)r->iv[i].arg);
   }
+  if (&r->iv[i] == r->fiv)
+    fputs("final:\n", fp);
 }
 
 /* --- @gfreduce_do@ --- *
@@ -340,7 +379,7 @@ mp *gfreduce_do(gfreduce *r, mp *d, mp *x)
       while (*vl & r->mask) {
 	z = *vl & r->mask;
 	*vl &= ~r->mask;
-	run(r->iv, il, vl, z);
+	run(r->fiv, il, vl, z);
       }
     }
   }
@@ -367,6 +406,12 @@ mp *gfreduce_sqrt(gfreduce *r, mp *d, mp *x)
   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;
@@ -389,7 +434,10 @@ mp *gfreduce_sqrt(gfreduce *r, mp *d, mp *x)
  *		@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)
@@ -450,7 +498,18 @@ mp *gfreduce_halftrace(gfreduce *r, mp *d, mp *x)
  *		@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)
@@ -458,15 +517,55 @@ mp *gfreduce_quadsolve(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
+     * %$z = \sum_{0\le i<m} \rho^{2^i} \sum_{0\le j<i} x^{2^j} = {}$%
+     * %$\rho^2 x + \rho^4 (x + x^2) + \rho^8 (x + x^2 + x^4) + \cdots + {}$%
+     * %$\rho^{2^{m-1}} (x + x^2 + x^{2^{m-2}})$%.  Then %$z^2 = {}$%
+     * %$\sum_{0\le i<m} \rho^{2^{i+1}} \sum_{0\le j<i} x^{2^{j+1}} = {}$%
+     * %$\sum_{1\le i\le m} \rho^{2^i} \sum_{1\le j\le i} x^{2^j}$% and,
+     * somewhat miraculously, %$z^2 + z = \sum_{0\le i<m} \rho^{2^i} x + {}$%
+     * %$\rho \sum_{1\le i<m} x^{2^i} = x \Tr(\rho) + \rho \Tr(x)$%.  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;
@@ -491,6 +590,12 @@ 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);
@@ -498,6 +603,13 @@ 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);
 }
@@ -539,8 +651,6 @@ mp *gfreduce_exp(gfreduce *gr, mp *d, mp *a, mp *e)
 
 #ifdef TEST_RIG
 
-#define MP(x) mp_readstring(MP_NEW, #x, 0, 0)
-
 static int vreduce(dstr *v)
 {
   mp *d = *(mp **)v[0].buf;