From 477ed667b1a9bc7f73850f45449666a8907ecaed Mon Sep 17 00:00:00 2001
From: Sebastiano Tronto <sebastiano@tronto.net>
Date: Wed, 20 Dec 2023 16:32:41 +0100
Subject: Added solution for 20

---
 2023/20/20b.c | 143 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 143 insertions(+)
 create mode 100644 2023/20/20b.c

(limited to '2023/20/20b.c')

diff --git a/2023/20/20b.c b/2023/20/20b.c
new file mode 100644
index 0000000..dc0d6b3
--- /dev/null
+++ b/2023/20/20b.c
@@ -0,0 +1,143 @@
+/*
+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;
+}
-- 
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