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package org.nd4j.linalg.api.ops.executioner;
import lombok.AllArgsConstructor;
import lombok.Data;
import org.nd4j.linalg.api.ndarray.INDArray;
import org.nd4j.linalg.util.ArrayUtil;
import java.util.Arrays;
/**Utility functions for the DefaultOpExecutioner
* @author Alex Black
*/
public class OpExecutionerUtil {
private OpExecutionerUtil() {
}
/** Can we do the op (X = Op(X)) directly on the arrays without breaking X up into 1d tensors first?
* In general, this is possible if the elements of X are contiguous in the buffer, OR if every element
* of X is at position offset+i*elementWiseStride in the buffer
* */
public static boolean canDoOpDirectly(INDArray x) {
if(x.elementWiseStride() < 1)
return false;
if(x.isVector()) return true;
//For a single NDArray all we require is that the elements are contiguous in the buffer or every nth element
//Full buffer -> implies all elements are contiguous (and match)
long l1 = x.lengthLong();
long dl1 = x.data().length();
if(l1 == dl1)
return true;
//Strides are same as a zero offset NDArray -> all elements are contiguous (even if not offset 0)
int[] shape1 = x.shape();
int[] stridesAsInit = (x.ordering()=='c' ? ArrayUtil.calcStrides(shape1) : ArrayUtil.calcStridesFortran(shape1));
boolean stridesSameAsInit = Arrays.equals(x.stride(), stridesAsInit);
return stridesSameAsInit;
}
/** Can we do the transform op (X = Op(X,Y)) directly on the arrays without breaking them up into 1d tensors first? */
public static boolean canDoOpDirectly(INDArray x, INDArray y){
if(x.isVector()) return true;
if(x.ordering() != y.ordering()) return false; //other than vectors, elements in f vs. c NDArrays will never line up
if(x.elementWiseStride() < 1 || y.elementWiseStride() < 1)
return false;
//Full buffer + matching strides -> implies all elements are contiguous (and match)
//Need strides to match, otherwise elements in buffer won't line up (i.e., c vs. f order arrays)
long l1 = x.lengthLong();
long dl1 = x.data().length();
long l2 = y.lengthLong();
long dl2 = y.data().length();
int[] strides1 = x.stride();
int[] strides2 = y.stride();
boolean equalStrides = Arrays.equals(strides1, strides2);
if(l1 == dl1 && l2 == dl2 && equalStrides)
return true;
//Strides match + are same as a zero offset NDArray -> all elements are contiguous (and match)
if(equalStrides) {
int[] shape1 = x.shape();
int[] stridesAsInit = (x.ordering()=='c' ? ArrayUtil.calcStrides(shape1) : ArrayUtil.calcStridesFortran(shape1));
boolean stridesSameAsInit = Arrays.equals(strides1, stridesAsInit);
return stridesSameAsInit;
}
return false;
}
/** Can we do the transform op (Z = Op(X,Y)) directly on the arrays without breaking them up into 1d tensors first? */
public static boolean canDoOpDirectly(INDArray x, INDArray y, INDArray z){
if(x.isVector()) return true;
if(x.ordering() != y.ordering() || x.ordering() != z.ordering() ) return false; //other than vectors, elements in f vs. c NDArrays will never line up
if(x.elementWiseStride() < 1 || y.elementWiseStride() < 1)
return false;
//Full buffer + matching strides -> implies all elements are contiguous (and match)
long l1 = x.lengthLong();
long dl1 = x.data().length();
long l2 = y.lengthLong();
long dl2 = y.data().length();
long l3 = z.lengthLong();
long dl3 = z.data().length();
int[] strides1 = x.stride();
int[] strides2 = y.stride();
int[] strides3 = z.stride();
boolean equalStrides = Arrays.equals(strides1, strides2) && Arrays.equals(strides1,strides3);
if(l1 == dl1 && l2 == dl2 && l3 == dl3 && equalStrides)
return true;
//Strides match + are same as a zero offset NDArray -> all elements are contiguous (and match)
if(equalStrides) {
int[] shape1 = x.shape();
int[] stridesAsInit = (x.ordering() == 'c' ? ArrayUtil.calcStrides(shape1) : ArrayUtil.calcStridesFortran(shape1));
boolean stridesSameAsInit = Arrays.equals(strides1, stridesAsInit);
return stridesSameAsInit;
}
return false;
}
/**
*
* Choose tensor dimension for operations with one argument: x=Op(x) or similar
* When doing some operations in parallel, it is necessary to break up
* operations along a dimension to
* give a set of 1d tensors. The dimension that this is done on is important for performance reasons;
* in summary we want to both minimize the number of tensors
* , but also minimize the separation between
* elements in the buffer (so the resulting operation is efficient - i.e., avoids cache thrashing).
* However, achieving both minimal number
* of tensors and are not always possible.
* @param x NDArray that we want to split
* @return The best dimension to split on
*/
public static int chooseElementWiseTensorDimension(INDArray x) {
if(x.isVector())
return ArrayUtil.argMax(x.shape()); //Execute along the vector
//doing argMin(max(x.stride(i),y.stride(i))) minimizes the maximum
//separation between elements (helps CPU cache) BUT might result in a huge number
//of tiny ops - i.e., addi on NDArrays with shape [5,10^6]
int opAlongDimensionMinStride = ArrayUtil.argMin(x.stride());
//doing argMax on shape gives us smallest number of largest tensors
//but may not be optimal in terms of element separation (for CPU cache etc)
int opAlongDimensionMaxLength = ArrayUtil.argMax(x.shape());
//Edge cases: shapes with 1s in them can have stride of 1 on the dimensions of length 1
if(x.isVector() || x.size(opAlongDimensionMinStride) == 1)
return opAlongDimensionMaxLength;
//Using a heuristic approach here: basically if we get >= 10x as many tensors using the minimum stride
//dimension vs. the maximum size dimension, use the maximum size dimension instead
//The idea is to avoid choosing wrong dimension in cases like shape=[10,10^6]
//Might be able to do better than this with some additional thought
int nOpsAlongMinStride = ArrayUtil.prod(ArrayUtil.removeIndex(x.shape(), opAlongDimensionMinStride));
int nOpsAlongMaxLength = ArrayUtil.prod(ArrayUtil.removeIndex(x.shape(), opAlongDimensionMaxLength));
if(nOpsAlongMinStride <= 10 * nOpsAlongMaxLength)
return opAlongDimensionMinStride;
else
return opAlongDimensionMaxLength;
}
/**
* Choose tensor dimension for operations with 2 arguments: x=Op(x,y) or similar
* @see #chooseElementWiseTensorDimension(INDArray)
*/
public static int chooseElementWiseTensorDimension(INDArray x, INDArray y) {
if(x.isVector())
return ArrayUtil.argMax(x.shape()); //Execute along the vector
//doing argMin(max(x.stride(i),y.stride(i))) minimizes the maximum
//separation between elements (helps CPU cache) BUT might result in a huge number
//of tiny ops - i.e., addi on NDArrays with shape [5,10^6]
int opAlongDimensionMinStride = ArrayUtil.argMinOfMax(x.stride(), y.stride());
//doing argMax on shape gives us smallest number of largest tensors
//but may not be optimal in terms of element separation (for CPU cache etc)
int opAlongDimensionMaxLength = ArrayUtil.argMax(x.shape());
//Edge case: shapes with 1s in them can have stride of 1 on the dimensions of length 1
if(opAlongDimensionMinStride == opAlongDimensionMaxLength || x.size(opAlongDimensionMinStride)==1)
return opAlongDimensionMaxLength;
//Using a heuristic approach here: basically if we get >= 10x as many tensors using the minimum stride
//dimension vs. the maximum size dimension, use the maximum size dimension instead
//The idea is to avoid choosing wrong dimension in cases like shape=[10,10^6]
//Might be able to do better than this with some additional thought
int nOpsAlongMinStride = ArrayUtil.prod(ArrayUtil.removeIndex(x.shape(), opAlongDimensionMinStride));
int nOpsAlongMaxLength = ArrayUtil.prod(ArrayUtil.removeIndex(x.shape(), opAlongDimensionMaxLength));
if(nOpsAlongMinStride <= 10 * nOpsAlongMaxLength)
return opAlongDimensionMinStride;
else return opAlongDimensionMaxLength;
}
/**Choose tensor dimension for operations with 3 arguments: z=Op(x,y) or similar
* @see #chooseElementWiseTensorDimension(INDArray)
*/
public static int chooseElementWiseTensorDimension(INDArray x, INDArray y, INDArray z){
if(x.isVector()) return ArrayUtil.argMax(x.shape());
int opAlongDimensionMinStride = ArrayUtil.argMinOfMax(x.stride(),y.stride(),z.stride());
int opAlongDimensionMaxLength = ArrayUtil.argMax(x.shape());
//Edge case: shapes with 1s in them can have stride of 1 on the dimensions of length 1
if(opAlongDimensionMinStride == opAlongDimensionMaxLength || x.size(opAlongDimensionMinStride)==1)
return opAlongDimensionMaxLength;
int nOpsAlongMinStride = ArrayUtil.prod(ArrayUtil.removeIndex(x.shape(), opAlongDimensionMinStride));
int nOpsAlongMaxLength = ArrayUtil.prod(ArrayUtil.removeIndex(x.shape(), opAlongDimensionMaxLength));
if(nOpsAlongMinStride <= 10 * nOpsAlongMaxLength) return opAlongDimensionMinStride;
else return opAlongDimensionMaxLength;
}
/** Tensor1DStats, used to efficiently iterate through tensors on a matrix (2d NDArray) for element-wise ops
* For example, the offset of each 1d tensor can be calculated using only a single tensorAlongDimension method call,
* hence is potentially faster than approaches requiring multiple tensorAlongDimension calls.
* Note that this can only (generally) be used for 2d NDArrays. For certain 3+d NDArrays, the tensor starts may not
* be in increasing order
*/
public static Tensor1DStats get1DTensorStats(INDArray array, int...dimension) {
int tensorLength = array.size(dimension[0]);
//As per tensorssAlongDimension:
int numTensors = array.tensorssAlongDimension(dimension);
//First tensor always starts with the first element in the NDArray, regardless of dimension
int firstTensorOffset = array.offset();
//Next: Need to work out the separation between the start (first element) of each 1d tensor
int tensorStartSeparation;
int elementWiseStride; //Separation in buffer between elements in the tensor
if(numTensors == 1) {
tensorStartSeparation = -1; //Not applicable
elementWiseStride = array.elementWiseStride();
} else {
INDArray secondTensor = array.tensorAlongDimension(1, dimension);
tensorStartSeparation = secondTensor.offset() - firstTensorOffset;
elementWiseStride = secondTensor.elementWiseStride();
}
return new Tensor1DStats(firstTensorOffset,tensorStartSeparation,
numTensors,tensorLength,elementWiseStride);
}
/** Simple class containing values used for calculating various quantities related to 1d tensors.
* offset of ith tensor: firstTensorOffset + i * tensorStartSeparation
* separation between elements in tensor: elementWiseStride
* number of elements in each 1d tensor: tensorLength
* number of 1d tensors: numTensors
*/
@AllArgsConstructor
@Data
public static class Tensor1DStats {
public final int firstTensorOffset;
public final int tensorStartSeparation;
public final int numTensors;
public final int tensorLength;
public final int elementWiseStride;
}
}
