Joshua Moerman
7 years ago
commit
4ceb2dbc70
4 changed files with 2294 additions and 0 deletions
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Tilings of the plane with triangles ... |
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... of equal area, bounded perimenter and all non-congruent. Here, I try to |
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implement the procedure from a paper by Andrey Kupavskii, János Pach and Gábor |
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Tardos. It is not correct. I interpret generic as random, is probably fine for |
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this gimmick. [ArXiv article](https://arxiv.org/pdf/1712.03118.pdf) |
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#include <random> |
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#include <cmath> |
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#include <queue> |
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#include <iostream> |
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using namespace std; |
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using ld = long double; |
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struct pt { |
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ld x = 0, y = 0; |
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}; |
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constexpr pt zero = pt{0, 0}; |
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pt operator+(pt a, pt b) { |
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a.x += b.x; |
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a.y += b.y; |
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return a; |
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} |
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pt operator-(pt a, pt b) { |
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a.x -= b.x; |
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a.y -= b.y; |
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return a; |
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} |
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pt operator*(ld s, pt p) { |
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p.x *= s; |
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p.y *= s; |
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return p; |
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} |
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ld dot(pt a, pt b) { |
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return a.x * b.x + a.y * b.y; |
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} |
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ld normsqr(pt a) { |
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return dot(a, a); |
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} |
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ld lensqr(pt a, pt b) { |
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return normsqr(b - a); |
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} |
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ld norm(pt a) { |
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return sqrt(normsqr(a)); |
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} |
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struct cone { |
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pt p1, d1; |
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pt p2, d2; |
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}; |
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struct cone_d { |
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ld d; |
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cone c; |
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cone_d(cone const & co) : c(co) { |
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// TODO: proper point line segment distance
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d = min(normsqr(c.p1), normsqr(c.p2)); |
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} |
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}; |
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bool operator<(cone_d const & lhs, cone_d const & rhs) { |
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return lhs.d > rhs.d; |
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} |
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// two points and a direction from p2
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// returns x such that
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ld unit_triangle(pt p1, pt p2, pt d2) { |
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const auto d1 = p1 - p2; |
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// from https://en.wikipedia.org/wiki/Triangle#Using_coordinates
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// 1 = area = 1/2 * abs(x * d2.x * d1.y - x * d1.x * d2.y)
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// 2 = x * abs(d2.x * d1.y - d1.x * d2.y)
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// 2 / abs(d2.x * d1.y - d1.x * d2.y) = x
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const auto x = 2.0 / abs(d2.x * d1.y - d1.x * d2.y); |
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return x; |
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} |
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int main() { |
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mt19937 gen(random_device{}()); |
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// This is used for the starting triangles
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uniform_real_distribution<ld> dist(0.8, 1.6); |
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uniform_real_distribution<ld> unif01(0, 1); |
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// the three starting rays
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pt c1 = {0, 1}; |
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pt c2 = { sqrt(3) / 2.0, -0.5}; |
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pt c3 = {-sqrt(3) / 2.0, -0.5}; |
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// and the random triangles attached to them
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pt p12a = dist(gen) * c1; |
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pt p12b = unit_triangle(p12a, {0, 0}, c2) * c2; |
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cone c12{p12a, c1, p12b, c2}; |
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pt p23a = dist(gen) * c2; |
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pt p23b = unit_triangle(p23a, {0, 0}, c3) * c3; |
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cone c23{p23a, c2, p23b, c3}; |
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pt p31a = dist(gen) * c3; |
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pt p31b = unit_triangle(p31a, {0, 0}, c1) * c1; |
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cone c31{p31a, c3, p31b, c1}; |
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// we keep track of all lines in the construction
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vector<pair<pt, pt>> rays; |
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rays.emplace_back(zero, c1); |
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rays.emplace_back(zero, c2); |
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rays.emplace_back(zero, c3); |
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vector<pair<pt, pt>> lines; |
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lines.emplace_back(p12a, p12b); |
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lines.emplace_back(p23a, p23b); |
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lines.emplace_back(p31a, p31b); |
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// I use a priority queue to pick the closest cone
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priority_queue<cone_d> work; |
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work.push(c12); |
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work.push(c23); |
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work.push(c31); |
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// we pick the closest cone to expand
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while(!work.empty() && lines.size() < 2000) { |
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auto c = work.top().c; |
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work.pop(); |
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// one quadrant is enough for me
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if((c.p1.x < -4 && c.p2.x < -4) || (c.p1.y < -4 && c.p2.y < -4)) |
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continue; |
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// split or not?
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if (lensqr(c.p1, c.p2) > 16) { |
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// choose midpoint on c.p2 - c.p1
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// such that both are length >= 2
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// choose direction such that we get two cones
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// The two pow(-) functions skew the distribution,
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// you can artistically choose something here.
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const auto l = sqrt(lensqr(c.p1, c.p2)); |
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const auto wiggle = l - 4; |
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const auto r0 = unif01(gen); |
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const auto r = 2.0 / l + pow(r0, 1.5) * wiggle / l; |
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const auto pm = (1.0 - r) * c.p1 + r * c.p2; |
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const auto r20 = unif01(gen); |
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const auto r2 = 0.001 + 0.998 * pow(r20, 1.2); |
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const auto dm = r2 * c.d1 + (1.0 - r2) * c.d2; |
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rays.emplace_back(pm, dm); |
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cone c1{c.p1, c.d1, pm, dm}; |
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cone c2{pm, dm, c.p2, c.d2}; |
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work.push(c1); |
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work.push(c2); |
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} else { |
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// try the two triangles and see which one fits the bill
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const auto x1 = unit_triangle(c.p2, c.p1, c.d1); |
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const auto x2 = unit_triangle(c.p1, c.p2, c.d2); |
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const auto p1b = c.p1 + x1 * c.d1; |
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const auto p2b = c.p2 + x2 * c.d2; |
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// Here I do not follow the paper correctly. I should pick the
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// new cone with angles at most bla bla bla. Instead, I choose
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// the one with a shorter edge, sounds fine to me. I reintroduce
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// some randomness for artistic reasons. Increase 20 for wild
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// results.
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if (lensqr(c.p1, p2b) + 20*unif01(gen) < lensqr(p1b, c.p2) + 20*unif01(gen)) { |
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cone c2{c.p1, c.d1, p2b, c.d2}; |
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work.push(c2); |
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lines.emplace_back(c.p1, p2b); |
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} else { |
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cone c2{p1b, c.d1, c.p2, c.d2}; |
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work.push(c2); |
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lines.emplace_back(p1b, c.p2); |
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} |
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} |
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} |
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// the print an svg attribute
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const auto p = [](auto str, auto v) { |
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cout << ' ' << str << "=\"" << v << "\""; |
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}; |
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// I roughly estimate the maximal distance from zero
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ld max_l = 0; |
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shuffle(lines.begin(), lines.end(), gen); |
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cout << "<svg>\n"; |
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for(auto && l : lines) { |
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max_l = 0.5 * max_l + 0.5 * max(max_l, max(norm(l.first), norm(l.second))); |
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cout << "<line"; |
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p("x1", l.first.x); |
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p("y1", l.first.y); |
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p("x2", l.second.x); |
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p("y2", l.second.y); |
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p("style", "stroke:#000;stroke-width:0.1"); |
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cout << "/>\n"; |
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} |
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for(auto && l : rays) { |
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const auto p1 = l.first; |
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const auto l_left = max(ld(1.0), max_l - norm(p1)); |
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const auto p2 = p1 + (l_left / norm(l.second)) * l.second; |
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cout << "<line"; |
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p("x1", p1.x); |
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p("y1", p1.y); |
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p("x2", p2.x); |
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p("y2", p2.y); |
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p("style", "stroke:#000;stroke-width:0.1"); |
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cout << "/>\n"; |
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} |
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cout << "</svg>" << endl; |
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} |
After Width: | Height: | Size: 997 KiB |
After Width: | Height: | Size: 196 KiB |
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