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Pascal's triangle/Puzzle: Difference between revisions
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=={{header|C++}}==
{{trans|C}}
<lang cpp>#include <iostream>
#include <iomanip>
inline int sign(int i) {
return i < 0 ? -1 : i > 0;
}
inline int& E(int *x, int row, int col) {
return x[row * (row + 1) / 2 + col];
}
int iter(int *v, int *diff) {
// 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 (auto i = 1u; i < 5u; i++)
for (auto j = 0u; 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 (auto i = 0u; i < 4u; i++)
for (auto j = 0u; 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);
// do feedback, check if we are done
uint sum;
int e = 0;
for (auto i = sum = 0u; i < 15u; 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;
}
void show(int *x) {
for (auto i = 0u; i < 5u; i++)
for (auto j = 0u; j <= i; j++)
std::cout << std::setw(4u) << *(x++) << (j < i ? ' ' : '\n');
}
int main() {
int v[15] = { 0 }, diff[15] = { 0 };
for (auto i = 1u, s = 1u; s; i++) {
s = iter(v, diff);
std::cout << "pass " << i << ": " << s << std::endl;
}
show(v);
return 0;
}</lang>
=={{header|Clojure}}==
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