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diff --git a/2023/08/8b.c b/2023/08/8b.c
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+/*
+I hate this stupid problem. If you read the description carefully (which is
+what I did), you would think that the problem is much, much harder than it
+actually is. In reality, the author made many assumptions about the possible
+paths that made is super easy. Let's see a few:
+
+- Different ghost's path do not end up in the same Z-node (this is not super
+  important for the solution).
+- Each ghost has exactly one Z-node in its path (one could think about this
+  when reading the strange remark that "there are as many A-nodes as Z-nodes",
+  which does not look important at first).
+- Each ghosts meets a Z-node exactly once before entering a loop (this is
+  a fundamental and incredibly strong assupmtion).
+- If a ghost encounters a Z-node after X steps, it will encounter it exactly
+  every X steps (like WTF I don't even have to solve a system of congruences?
+  what is this, a problem for babies?)
+
+Fuck.
+
+This code does not even solve the problem by the way, it computes some data
+about the path that each ghost takes. Then you can figure out the solution
+with a pocket calculator, or by hand.
+*/
+#include <inttypes.h>
+#include <math.h>
+#include <stdbool.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+#define N 1000
+
+typedef struct { char last; int next[2]; } node_t;
+typedef struct { int preplen, plen, nz, z[N], zid[N]; } source_t;
+
+int ind(char s[3], char m[][3], int n) {
+        for (int j = 0; j < n; j++)
+                if (s[0] == m[j][0] && s[1] == m[j][1] && s[2] == m[j][2])
+                        return j;
+
+        return -1;
+}
+
+int v[N*N];
+source_t worksource(int i, node_t *nodes, char *dir, int n, int k) {
+        source_t s;
+        int state;
+
+printf("Working source %d\n", i);
+
+        memset(v, 0, sizeof(int) * N*N);
+        for (int j = i, d = 0, l = 1; true; d = (d+1)%k) {
+                j = nodes[j].next[dir[d] == 'R'];
+                state = j*k+d;
+if (nodes[j].last == 'Z') printf("Found Z %d at %d\n", j, l);
+                if (v[state]) {
+                        s.preplen = v[state]-1;
+                        s.plen = l - v[state];
+printf("Stopping at %d. Preperiod: %d, period: %d\n", l, s.preplen, s.plen);
+                        return s;
+                } else v[state] = l++;
+        }
+
+        return s;
+}
+
+int main() {
+        node_t nodes[N];
+        source_t src[N];
+        char *buf, dir[N], line[N], name[N][3], lstr[N][3], rstr[N][3];
+        int k, i, n, m, s;
+
+        k = strlen(fgets(dir, N, stdin)) - 1;
+        fgets(line, N, stdin);
+        for (n = 0; (buf = fgets(line, N, stdin)) != NULL; n++) {
+                memcpy(name[n], buf, 3);
+                nodes[n].last = name[n][2];
+                while (*buf != '(') buf++;
+                memcpy(lstr[n], buf+1, 3);
+                while (*buf != ' ') buf++;
+                memcpy(rstr[n], buf+1, 3);
+        }
+
+        for (i = 0; i < n; i++) {
+                nodes[i].next[0] = ind(lstr[i], name, n);
+                nodes[i].next[1] = ind(rstr[i], name, n);
+        }
+
+printf("%d %d\n", n, k);
+
+        for (i = 0, m = 0; i < n; i++)
+                if (nodes[i].last == 'A')
+                        src[m++] = worksource(i, nodes, dir, n, k);
+
+        return 0;
+}