Added solution for 20

1 files changed, 143 insertions, 0 deletions

diff --git a/2023/20/20b.c b/2023/20/20b.c
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+/*
+This one is a bit weird. This program works only for my specific input
+(included in this folder). Similarly to number 8, this program outputs
+4 numbers and you have to take the lcm of them.
+
+I solved it this way:
+1. First, using a modified version of the code for part one (graph.c),
+   I printed the graph as a list of adjacency lists.
+2. I then went to https://csacademy.com/app/graph_editor and observed
+   the graph. I noticed that after the button is pressed, the signal
+   is sent to 4 independent parts of the graph. Each of these parts has
+   only one entry point and one exit point. The entry points were not
+   flip-flop nodes.
+3. Finally, I implemented the code to check how long it takes for the
+   graph to go back to the initial state, and I simulated sending a
+   signal to each of the 4 parts independently.
+*/
+
+#include <inttypes.h>
+#include <stdbool.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+#define N 100
+#define ischar(c) (c >= 'a' && c <= 'z')
+
+typedef struct {
+        int nin, nout, in[N], out[N];
+        char name[20], outc[N][20];
+        bool isff, ison, reg[N];
+} node_t;
+
+typedef struct {
+        int i, n;
+        struct {int node; bool hi;} elem[500];
+} queue_t;
+
+char *buf, line[N][N];
+bool rx;
+int b, r, n;
+node_t m[N];
+queue_t q;
+
+void add(int v, bool hi) {
+        q.elem[q.n].node = v;
+        q.elem[q.n].hi = hi;
+        q.n++;
+}
+
+void send(int v, bool hi) {
+        for (int j = 0; j < m[v].nout; j++) {
+                add(m[v].out[j], hi);
+                m[m[v].out[j]].reg[v] = hi;
+        }
+}
+
+int findm(char *name) {
+        for (int k = 0; k < n; k++)
+                if (!strcmp(m[k].name, name))
+                        return k;
+        return n;
+}
+
+void sig(int node, bool high) {
+        q.n = q.i = 0;
+        add(node, high);
+        while (q.i < q.n) {
+                int v = q.elem[q.i].node;
+                bool hi = q.elem[q.i].hi;
+                q.i++;
+
+                if (v == b) {
+                        send(v, hi);
+                } else if (m[v].isff) {
+                        if (!hi)
+                                send(v, m[v].ison = !m[v].ison);
+                } else {
+                        bool allhi = true;
+                        for (int j = 0; j < m[v].nin; j++)
+                                allhi = allhi && m[v].reg[m[v].in[j]];
+                        send(v, !allhi);
+                }
+        }
+}
+
+bool isclean(void) {
+        for (int i = 0; i < n; i++) {
+                if (m[i].isff && m[i].ison)
+                        return false;
+                else
+                        for (int j = 0; j < m[i].nin; j++)
+                                if (m[i].reg[j])
+                                        return false;
+        }
+        return true;
+}
+
+int64_t period(int node) {
+        int64_t npush = 0;
+        do {
+                sig(node, false);
+                npush++;
+        } while (!isclean());
+        return npush;
+}
+
+int main() {
+        for (n = 0; (buf = fgets(line[n], N, stdin)) != NULL; n++) {
+                if (ischar(*buf)) b = n;
+                m[n].isff = *buf == '%';
+                if (!ischar(*buf)) buf++;
+                for (int i = 0; ischar(*buf); m[n].name[i++] = *(buf++)) ;
+                buf += 4;
+                for (int i = 0; *buf != '\n'; i++) {
+                        while (!ischar(*buf)) buf++;
+                        for (int j = 0; ischar(*buf); m[n].outc[i][j++] = *(buf++)) ;
+                }
+        }
+
+        for (int i = 0; i < n; i++) {
+                for (int j = 0; m[i].outc[j][0]; j++) {
+                        int k = findm(m[i].outc[j]);
+                        m[i].out[m[i].nout++] = k;
+                        m[k].in[m[k].nin++] = i;
+                        if (k == n) {
+                                for (int k = 0; m[i].outc[j][k]; k++)
+                                        m[n].name[k] = m[i].outc[j][k];
+                                n++;
+                        }
+                }
+        }
+
+        for (int i = 0; i < n; i++)
+                if (!strcmp(m[i].name, "rx"))
+                        r = i;
+
+        printf("%" PRId64 "\n", period(19));
+        printf("%" PRId64 "\n", period(33));
+        printf("%" PRId64 "\n", period(39));
+        printf("%" PRId64 "\n", period(57));
+        return 0;
+}