1 files changed, 51 insertions, 21 deletions

diff --git a/13.md b/13.md
index e3662a5..2c60929 100644
--- a/13.md
+++ b/13.md
@@ -10,6 +10,8 @@ This NIP defines a way to generate and interpret Proof of Work for nostr notes.
 
 `difficulty` is defined to be the number of leading zero bits in the `NIP-01` id. For example, an id of `000000000e9d97a1ab09fc381030b346cdd7a142ad57e6df0b46dc9bef6c7e2d` has a difficulty of `36` with `36` leading 0 bits.
 
+`002f...` is `0000 0000 0010 1111...` in binary, which has 10 leading zeroes. Do not forget to count leading zeroes for hex digits <= `7`.
+
 Mining
 ------
 
@@ -36,7 +38,7 @@ Example mined note
     [
       "nonce",
       "776797",
-      "20"
+      "21"
     ]
   ],
   "content": "It's just me mining my own business",
@@ -47,33 +49,61 @@ Example mined note
 Validating
 ----------
 
-Here is some reference C code for calculating the difficulty (aka number of leading zero bits) in a nostr note id:
+Here is some reference C code for calculating the difficulty (aka number of leading zero bits) in a nostr event id:
 
 ```c
-int zero_bits(unsigned char b)
-{
-        int n = 0;
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int countLeadingZeroes(const char *hex) {
+    int count = 0;
+
+    for (int i = 0; i < strlen(hex); i++) {
+        int nibble = (int)strtol((char[]){hex[i], '\0'}, NULL, 16);
+        if (nibble == 0) {
+            count += 4;
+        } else {
+            count += __builtin_clz(nibble) - 28;
+            break;
+        }
+    }
 
-        if (b == 0)
-                return 8;
+    return count;
+}
+
+int main(int argc, char *argv[]) {
+    if (argc != 2) {
+        fprintf(stderr, "Usage: %s <hex_string>\n", argv[0]);
+        return 1;
+    }
 
-        while (b >>= 1)
-                n++;
+    const char *hex_string = argv[1];
+    int result = countLeadingZeroes(hex_string);
+    printf("Leading zeroes in hex string %s: %d\n", hex_string, result);
 
-        return 7-n;
+    return 0;
 }
+```
 
-/* find the number of leading zero bits in a hash */
-int count_leading_zero_bits(unsigned char *hash)
-{
-        int bits, total, i;
-        for (i = 0, total = 0; i < 32; i++) {
-                bits = zero_bits(hash[i]);
-                total += bits;
-                if (bits != 8)
-                        break;
-        }
-        return total;
+Here is some JavaScript code for doing the same thing:
+
+```javascript
+// hex should be a hexadecimal string (with no 0x prefix)
+function countLeadingZeroes(hex) {
+  let count = 0;
+
+  for (let i = 0; i < hex.length; i++) {
+    const nibble = parseInt(hex[i], 16);
+    if (nibble === 0) {
+      count += 4;
+    } else {
+      count += Math.clz32(nibble) - 28;
+      break;
+    }
+  }
+
+  return count;
 }
 ```