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如何使用尝试有效地实现稀疏矩阵?

Mary-Kate Olsen
Mary-Kate Olsen原创
2024-11-06 03:55:02578浏览

How can tries be used to implement sparse matrices efficiently?

稀疏矩阵可以使用 try 高效实现,它仅使用两个数组索引操作计算某个元素是否存在于表中,从而提供对特定矩阵元素的快速访问。

Tries 的主要特点:

  • 在后备存储中为默认值提供默认位置,无需进行值测试。
  • 支持快速更新尝试使用可选的“compact()”操作来优化后备存储大小。
  • 利用对象映射,允许将坐标映射到向量中的整数位置。
  • 处理子范围的快速检索

优点:

  • Trie 实现比 hashmap 快得多,避免了复杂的散列函数和冲突处理。
  • Java hashmaps 仅在对象上建立索引,可能会导致内存开销和垃圾收集压力。
  • 尝试提供高效的实现,不需要为每个源索引创建对象,从而减少内存操作。

示例实现:

<code class="java">public class DoubleTrie {

    // Matrix options
    private static final int SIZE_I = 1024;
    private static final int SIZE_J = 1024;
    private static final double DEFAULT_VALUE = 0.0;

    // Internal splitting options
    private static final int SUBRANGEBITS_I = 4;
    private static final int SUBRANGEBITS_J = 4;

    // Internal splitting constants
    private static final int SUBRANGE_I =
            1 << SUBRANGEBITS_I;
    private static final int SUBRANGE_J =
            1 << SUBRANGEBITS_J;
    private static final int SUBRANGEMASK_I =
            SUBRANGE_I - 1;
    private static final int SUBRANGEMASK_J =
            SUBRANGE_J - 1;

    // Internal data
    private double[] values;
    private int[] subrangePositions;

    // Fast subrange and position computation methods
    private static int subrangeOf(int i, int j) {
        return (i >> SUBRANGEBITS_I) * SUBRANGE_J + (j >> SUBRANGEBITS_J);
    }
    private static int positionOffsetOf(int i, int j) {
        return (i & SUBRANGEMASK_I) * SUBRANGE_J + (j & SUBRANGEMASK_J);
    }

    // Fast indexed getter
    public double getAt(int i, int j) {
        return values[subrangePositions[subrangeOf(i, j)] +
                      positionOffsetOf(i, j)];
    }

    // Fast indexed setter
    public double setAt(int i, int j, double value) {
        final int subrange = subrangeOf(i, j);
        final int positionOffset = positionOffsetOf(i, j);
        // Check if the assignment will change something
        int subrangePosition, valuePosition;
        if (Double.compare(
                values[valuePosition =
                        (subrangePosition = subrangePositions[subrange]) +
                                positionOffset],
                value) != 0) {
            // Perform the assignment in values
            if (isSharedValues) {
                values = values.clone();
                isSharedValues = false;
            }
            // Scan other subranges to check if the value is shared by another subrange
            for (int otherSubrange = subrangePositions.length;
                    --otherSubrange >= 0; ) {
                if (otherSubrange != subrange)
                    continue; // Ignore the target subrange
                if ((otherSubrangePosition =
                        subrangePositions[otherSubrange]) >=
                        valuePosition &&
                        otherSubrangePosition + SUBRANGE_POSITIONS <
                                valuePosition) {
                    // The target position is shared, we need to make it unique by cloning the subrange
                    if (isSharedSubrangePositions) {
                        subrangePositions = subrangePositions.clone();
                        isSharedSubrangePositions = false;
                    }
                    values = setlengh(
                            values,
                            (subrangePositions[subrange] =
                                    subrangePositions = values.length) +
                                    SUBRANGE_POSITIONS);
                    valuePosition = subrangePositions + positionOffset;
                    break;
                }
            }
            // Perform the effective assignment of the value
            values[valuePosition] = value;
        }
        return value;
    }
}</code>

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