tasks a-s

Task/Pascals-triangle-Puzzle/C/pascals-triangle-puzzle-1.c

@@ -0,0 +1,39 @@
+/* Pascal's pyramid solver
+ *
+ *               [top]
+ *            [   ] [   ]
+ *         [mid] [   ] [   ]
+ *      [   ] [   ] [   ] [   ]
+ *   [ x ] [ a ] [ y ] [ b ] [ z ]
+ *             x + z = y
+ *
+ * This solution makes use of a little bit of mathematical observation,
+ * such as the fact that top = 4(a+b) + 7(x+z) and mid = 2x + 2a + z.
+ */
+
+#include <stdio.h>
+#include <math.h>
+
+void pascal(int a, int b, int mid, int top, int* x, int* y, int* z)
+{
+    double ytemp = (top - 4 * (a + b)) / 7.;
+    if(fmod(ytemp, 1.) >= 0.0001)
+    {
+        x = 0;
+        return;
+    }
+    *y = ytemp;
+    *x = mid - 2 * a - *y;
+    *z = *y - *x;
+}
+int main()
+{
+    int a = 11, b = 4, mid = 40, top = 151;
+    int x, y, z;
+    pascal(a, b, mid, top, &x, &y, &z);
+    if(x != 0)
+        printf("x: %d, y: %d, z: %d\n", x, y, z);
+    else printf("No solution\n");
+
+    return 0;
+}

Task/Pascals-triangle-Puzzle/C/pascals-triangle-puzzle-2.c

@@ -0,0 +1,73 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+void show(int *x) {
+	int i, j;
+
+	for (i = 0; i < 5; i++)
+		for (j = 0; j <= i; j++)
+			printf("%4d%c", *(x++), j < i ? ' ' : '\n');
+}
+
+inline int sign(int i)
+{
+	return i < 0 ? -1 : i > 0;
+}
+
+int iter(int *v, int *diff) {
+	int sum, i, j, e = 0;
+
+#	define E(x, row, col) x[(row) * ((row) + 1) / 2 + (col)]
+	/* enforce boundary conditions */
+	E(v, 0, 0) = 151;
+	E(v, 2, 0) = 40;
+	E(v, 4, 1) = 11;
+	E(v, 4, 3) = 4;
+
+	/* calculate difference from equilibrium */
+	for (i = 1; i < 5; i++) {
+		for (j = 0; j <= i; j++) {
+			E(diff, i, j) = 0;
+			if (j < i)
+				E(diff, i, j) += E(v, i - 1, j) -
+						 E(v, i, j + 1) -
+						 E(v, i, j);
+			if (j)
+				E(diff, i, j) += E(v, i - 1, j - 1) -
+						 E(v, i, j - 1) -
+						 E(v, i, j);
+		}
+	}
+
+	for (i = 0; i < 4; i++)
+		for (j = 0; j < i; j++)
+			E(diff, i, j) += E(v, i + 1, j) +
+					 E(v, i + 1, j + 1) -
+					 E(v, i, j);
+
+	E(diff, 4, 2) += E(v, 4, 0) + E(v, 4, 4) - E(v, 4, 2);
+#	undef E
+
+	/* Do feedback, check if we are done. */
+	for (i = sum = 0; i < 15; i++) {
+		sum += !!sign(e = diff[i]);
+
+		/* 1/5-ish feedback strength on average.  These numbers are highly
+		   magical, depending on nodes' connectivities. */
+		if (e >= 4 || e <= -4) 		v[i] += e/5;
+		else if (rand() < RAND_MAX/4)	v[i] += sign(e);
+	}
+	return sum;
+}
+
+int main() {
+	int v[15] = { 0 }, diff[15] = { 0 }, i, s;
+
+	for (i = s = 1; s; i++) {
+		s = iter(v, diff);
+		printf("pass %d: %d\n", i, s);
+	}
+	show(v);
+
+	return 0;
+}

Task/Pascals-triangle-Puzzle/C/pascals-triangle-puzzle-3.c

@@ -0,0 +1,13 @@
+pass 1: 12
+pass 2: 12
+pass 3: 14
+pass 4: 14
+...
+pass 113: 4
+pass 114: 7
+pass 115: 0
+ 151
+  81   70
+  40   41   29
+  16   24   17   12
+   5   11   13    4    8