[dv, otbn] Adust the `scramble_test` to tolerate the right amount of false positives
Signed-off-by: Douglas Reis <doreis@lowrisc.org>
diff --git a/sw/device/tests/otbn_mem_scramble_test.c b/sw/device/tests/otbn_mem_scramble_test.c
index 39532b8..31bc85e 100644
--- a/sw/device/tests/otbn_mem_scramble_test.c
+++ b/sw/device/tests/otbn_mem_scramble_test.c
@@ -22,19 +22,59 @@
uint32_t offset_bytes, const void *src,
size_t len_bytes);
-/**
- * Number of distinct addresses to check in each IMEM and DMEM.
- */
-static const int kNumAddrs = 50;
+enum {
+ /**
+ * Number of distinct addresses to check in each IMEM and DMEM.
+ */
+ kNumAddrs = 50,
-/**
- * Minimum number of expected integrity errors in IMEM and DMEM after
- * re-scrambling.
- *
- * Note that there is a non-zero chance for every word that the integrity bits
- * after re-scrambling are still valid.
- */
-static const int kNumIntgErrorsThreshold = kNumAddrs - 1;
+ /**
+ * Minimum number of expected integrity errors in IMEM and DMEM after
+ * re-scrambling.
+ * Note that there are `2^32` valid code words and that each non-valid code
+ * word triggers an error. Therefore, the probability that a random 39-bit
+ * word triggers an error is: `(2^39 - 2^32)/ 2^39 = 127/128`. Then the
+ * probability that all `kNumAddrs` triggers an errors is
+ * `(127/128)^kNumAddrs` after re-scrambling.
+ *
+ * The Generic formula:
+ * (w-i)
+ * 127
+ * Pr(i) = -------- x (w choose i)
+ * w
+ * 128
+ * Where:
+ * w = The number of words tested.
+ * i = The number of words that may not generate errors.
+ * Pr(i) = Probability that i words will not generate an ECC error.
+ *
+ * So for i in (0..4):
+ *
+ * ``` Python
+ * from math import comb
+ * w = 50
+ * t = 0
+ * for i in range(5):
+ * p = ((127**(w-i))/(128**w)) * comb(w,i)
+ * t += p
+ * print(f'Pr({i}): { round(p, 4)},\tsum{{Pr(0-{i})}}: {round(t, 6)}')
+ * ```
+ * ```
+ * Pr(0): 0.6756, sum{Pr(0-0)}: 0.675597
+ * Pr(1): 0.266, sum{Pr(0-1)}: 0.94158
+ * Pr(2): 0.0513, sum{Pr(0-2)}: 0.992891
+ * Pr(3): 0.0065, sum{Pr(0-3)}: 0.999356
+ * Pr(4): 0.0006, sum{Pr(0-4)}: 0.999954
+ * ```
+ * So by choosing `(kNumAddrs - 2) = 48` as the threshold we will have a
+ * probability of `1 - 0.992891 = 0.71%` that this test will fail randomly due
+ * to ECC errors not being generated. That seems a reasonable number.
+ */
+ kNumIntgErrorsThreshold = kNumAddrs - 2,
+};
+static_assert(kNumAddrs == 50,
+ "kNumAddrs changed, so kEccErrorProbability should be "
+ "computed again");
static volatile bool has_irq_fired;
