+static inline int log2i(int n)
+{
+ int log2 = 0;
+
+ for (; n > 1; n >>= 1)
+ log2++;
+
+ return log2;
+}
+
+static inline int exp2i(int n)
+{
+ return 1 << n;
+}
+
+/*
+ * Estimate the number of bisect steps left (after the current step)
+ *
+ * For any x between 0 included and 2^n excluded, the probability for
+ * n - 1 steps left looks like:
+ *
+ * P(2^n + x) == (2^n - x) / (2^n + x)
+ *
+ * and P(2^n + x) < 0.5 means 2^n < 3x
+ */
+int estimate_bisect_steps(int all)
+{
+ int n, x, e;
+
+ if (all < 3)
+ return 0;
+
+ n = log2i(all);
+ e = exp2i(n);
+ x = all - e;
+
+ return (e < 3 * x) ? n : n - 1;
+}