Examples of net.algart.math.patterns.Pattern.pointCount()

net.algart.math.patterns.Pattern.pointCount()
Returns the number of points in this pattern. This value is always positive (>=1). If the number of points is greater than Long.MAX_VALUE, returns Long.MAX_VALUE.
Warning! This method can work slowly for some forms of large patterns: the required time can be O(N), where N is the number of points (result of this method). In these cases, this method can also throw {@link TooManyPointsInPatternError}or OutOfMemoryError.
There is a guarantee, that if this object implements {@link QuickPointCountPattern} interface,then this method works very quickly (O(1) operations) and without exceptions.
There is a guarantee, that if this object implements {@link DirectPointSetPattern} interface,then the result of this method is not greater than Integer.MAX_VALUE.
Note: if this method returns some value greater than Integer.MAX_VALUE, it means that you cannot use {@link #points()} and {@link #roundedPoints()} methods,because Java Set object cannot contain more than Integer.MAX_VALUE elements. @return the number of {@link Point points} in this pattern. @throws TooManyPointsInPatternError for some forms of large patterns, if the number of points is greater thanInteger.MAX_VALUE or, in some rare situations, is near this limit (OutOfMemoryError can be also thrown instead of this exception). @see #largePointCount() @see #isSurelySinglePoint @see QuickPointCountPattern#isPointCountVeryLarge()

                System.out.println(new TreeSet<Point>(points));
            t1 = System.nanoTime();
            List<Pattern> minkowskiDecomposition = p.minkowskiDecomposition(minimalPointCountForDecomposition);
            t2 = System.nanoTime();
            System.out.printf(Locale.US, "Minkowski decomposition to %d patterns (%.3f ms, %.3f ns / point):%n",
                minkowskiDecomposition.size(), (t2 - t1) * 1e-6, (t2 - t1) / (double)p.pointCount());
            for (Pattern q : minkowskiDecomposition) {
                System.out.println("    " + q);
            }
            System.out.println();

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            List<List<Pattern>> allUnionDecompositions = p.allUnionDecompositions(minimalPointCountForDecomposition);
            t2 = System.nanoTime();
            for (int k = 0, n = allUnionDecompositions.size(); k < n; k++) {
                List<Pattern> unionDecomposition = allUnionDecompositions.get(k);
                System.out.printf(Locale.US, "Union decompositions #%d to %d patterns (%.3f ms, %.3f ns / point):%n",
                    k + 1, unionDecomposition.size(), (t2 - t1) * 1e-6, (t2 - t1) / (double) p.pointCount());
                for (Pattern q : unionDecomposition) {
                    System.out.println("    " + q
                        + (!q.hasMinkowskiDecomposition() ? " " + new TreeSet<IPoint>(q.roundedPoints()) : ""));
                }
                t1 = System.nanoTime();

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                        totalPointCount += s.pointCount();
                }
                t2 = System.nanoTime();
                System.out.printf(Locale.US, "Minkowski decomposition of this union elements to %d patterns, "
                    + "%d total points (%.3f ms, %.3f ns / point)%n%n",
                    totalPatternCount, totalPointCount, (t2 - t1) * 1e-6, (t2 - t1) / (double)p.pointCount());
            }
        }
    }
}

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            points.add(IPoint.valueOf(2000000000 + k, k));
            points.add(IPoint.valueOf(k, 2000000000 + k));
        }
        Pattern p2 = Patterns.newIntegerPattern(points);
        Pattern pSum = Patterns.newMinkowskiSum(p1, p2);
        System.out.println("Large sum: " + pSum + ", contains " + pSum.pointCount() + " points");
    }
}

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