[WIP] add new operator LU factorization #1019

Chuancysun · 2024-04-30T11:28:40Z

Thanks for your contribution and we appreciate it a lot. 🚀🚀

1. Motivation

add floating point operator LU factorization

2. Modification

add implementation of floating point LU factorization

3. Test Report

3.1 Modification Details

3.1.1 Accuracy Acceptance Standard

For static threshold standard details, see: MLU-OPS™ Accuracy Acceptance Standard.

static threshold
- diff1
  - float32 mlu diff1 <= 1e-5
  - [*] float32 mlu diff1 <= 3e-3
  - float16 mlu diff1 <= 3e-3
- diff2
  - float32 mlu diff2 <= 1e-5
  - [* ] float32 mlu diff2 <= 3e-3
  - float16 mlu diff2 <= 3e-3
- diff3
  - mlu diff3 == 0
  - mlu diff3_1 == 0
  - mlu diff3_2 == 0
dynamic threshold
- diff1: mlu diff1 <= max(baseline diff1 * 10, static threshold)
- diff2: mlu diff2 <= max(baseline diff2 * 10, static threshold)
- diff3: mlu diff3 <= max(baseline diff3 * 10, static threshold)
  - float32, threshold = 1e-5
  - float16, threshold = 1e-3

3.1.2 Operator Scheme checklist

Supported hardware
- [* ] MLU370
- MLU590
Job types
- BLOCK
- UNION1
- UNION2
- UNION4
- [* ] The operator will dynamically select the most suitable task type, for example, UNION8

3.2 Accuracy Test

3.2.1 Accuracy Test

If you have checked the following items, please tick the relevant box.

3.3 Performance Test

Platform：MLU370

----------- case0 -----------
case0
[Op name ]: sgetrf
[Shape ]: input.shape=[256,256], output.shape=[256,256]
[Data type] ]: float32
[MLU Hardware Time ]: 6460 (us)
[MLU Interface Time ]: 15336.7 (us)
[MLU IO Efficiency ]: 0.00026419
[MLU Compute Efficiency ]: 9.90712e-06
[MLU Workspace Size ]: -1 (Bytes)
[MLU Kernel Name(s) ]: {}
[MLU TheoryOps ]: 65536 (Ops)
[MLU TheoryIOs ]: 524288 (Bytes)
[MLU ComputeForce ]: 1.024e+12 (op/s)
[MLU IoBandWidth ]: 307.2 (GB/s)
[GPU Hardware Time ]: -1 (us)
[GPU IO Efficiency ]: -1
[GPU Compute Efficiency ]: -1
[GPU Workspace Size ]: -1 (Bytes)
[Diffs]:
[output]
DIFF1: 1.798500e-04
DIFF2: 7.016698e-04
[^ OK ] ../../test/mlu_op_gtest/pb_gtest/src/zoo/sgetrf/test_case/case0.prototxt
[ OK ] sgetrf/TestSuite.mluOp/0 (36 ms)
[----------] 1 test from sgetrf/TestSuite (36 ms total)

[----------] Global test environment tear-down
[ SUMMARY ] Total 1 cases of 1 op(s).
ALL PASSED.
[==========] 1 test case from 1 test suite ran. (3727 ms total)
[ PASSED ] 1 test case.

3.4 Summary Analysis

Please give a brief overview here, if you want to note and summarize the content.

kernels/sgetrf/add_union1.mlu

PetrelYy · 2024-05-22T09:32:28Z

kernels/sgetrf/add_union1.mlu

+ * permit persons to whom the Software is furnished to do so, subject to
+ * the following conditions:
+ *
+ * The above copyright notice and this permission notice shall be included


补充 proto，docs/bangc-docs/user_guide/9_operators/index.rst 算子说明
可参考 https://github.com/Cambricon/mlu-ops/pull/662/files#diff-7f0a558d8f985a4ebd89cd6674a4bf1a91549ddcc6e708a897f351cb2006f0e8

已添加算子实现方案

mlu_op.h

test/mlu_op_gtest/pb_gtest/src/zoo/sgetrf/sgetrf.cpp

kernels/sgetrf/add_union1.mlu

kernels/sgetrf/trsm_union1.mlu

docs/design_docs/sgetrf/sgetrf.md

ArtIntAI · 2024-05-31T02:55:34Z

kernels/sgetrf/sgetrf_mlu.cpp

+    nb = get_sgetrf_native_nb(m, n);
+
+    float *workspace;
+    cnrtMalloc((void **)&workspace, nb * nb * sizeof(float));


一般workspace是用户自己传进来的，算子是提供getWorkspace的接口来让用户去分配对应的空间，可以参考https://github.com/Cambricon/mlu-ops/blob/master/mlu_op.h#L3716

ArtIntAI · 2024-05-31T02:56:25Z

mlu_op.h

+  typedef enum
+  {
+    MLUOP_STATUS_SUCCESS = 0, /*!< The operation is successfully completed. */
+    MLUOP_STATUS_NOT_INITIALIZED = 1,


建议头文件回退其他修改，只增加自己这次提交的部分。

kernels/sgetrf2/mycnrtMemcpy2D_union1.mlu

mlu_op.h

test/mlu_op_gtest/pb_gtest/src/zoo/sgerf2/sgetrf.cpp

kernels/sgetrf2/myinverse_union1.mlu

kernels/sgetrf2/sgetrf_mlu.cpp

docs/design_docs/sgetrf/sgetrf.md

mlu_op.h

kernels/sgetrf2/sgetrf2.h

kernels/sgetrf2/mycnrtMemcpy2D_union1.mlu

ArtIntAI · 2024-08-14T07:18:52Z

kernels/sgetrf2/myinverse_union1.mlu

+            {
+                temp += mul_result[k];
+            }
+            temp = temp * -1.0 * diag_element;


这里标量计算太慢了，不能向量化调用bangc接口吗

这里是将一个向量中的所有元素中的值相加，文档没有找到能实现此功能的bang函数

reduce_sum或者sumpool可以试试

查看文档后发现这两个函数的功能跟算子的逻辑不符

kernels/sgetrf2/swap_union1.mlu

docs/user_guide/9_operators/index.rst

ArtIntAI · 2024-11-20T12:46:37Z

docs/design_docs/sgetrf/sgetrf.md

+
+### 4.6. **算子接口设计**
+
+mluOpStatus_t mluOp_WIN_API mluOpLUFactorization(mluOpHandle_t handle, const mluOpTensorDescriptor_t input_desc, void *input,const mluOpTensorDescriptor_t output_desc, void *output,const bool mode);


缺少workspace接口的描述

mlu_op.h

ArtIntAI · 2024-11-20T12:50:29Z

kernels/sgetrf2/myinverse_union1.mlu

@@ -0,0 +1,412 @@
+/*************************************************************************
+ * Copyright (C) [2022] by Cambricon, Inc.


copyright的年份不对，改成2024吧

ArtIntAI · 2024-11-20T12:57:31Z

kernels/sgetrf2/sgetrf.cpp

+              x_desc->dim == 2 || x_desc->dim == 3 || x_desc->dim == 4);
+  PARAM_CHECK("mluOpSgetrf2", x_desc->dims[0] > 0);
+  PARAM_CHECK("mluOpSgetrf2", x_desc->dims[1] > 0);
+


缺少必要的阶段性VLOG信息，host上拼接的kenrel前后都加上VLOG(5)，参考

mlu-ops/kernels/abs/abs.cpp

Line 166 in 5ae8c94

VLOG(5) << "kernel Kernel3StagePipelineAbs";

ArtIntAI · 2024-11-20T12:57:41Z

kernels/sgetrf2/sgetrf.cpp

+  mluOpDataType_t dtype = x_desc->dtype;
+  PARAM_CHECK("mluOpSgetrf2", x_desc != NULL);
+
+  PARAM_CHECK("mluOpSgetrf2",


参数检查缺少0元素和large tensor 处理，建议参考下

mlu-ops/kernels/abs/abs.cpp

ArtIntAI · 2024-11-20T13:01:38Z

mlu_op.h

+ *
+ * @par Scale Limitation
+ * - The dimension of input tensor must be either 2, 3 or 4.
+ * Considering the size of the GDRAM, the space occupied by the input matrix should not exceed 7GB.


如果只是gdram空间导致的问题，不是算子本身的限制，可以删除
换句话说，如果gdram空间无限制的话，算子本身是不是没有规模限制？

ArtIntAI · 2024-11-20T13:04:15Z

kernels/sgetrf2/sgetrf.cpp

+  PARAM_CHECK("mluOpSgetrf2", x_desc->dims[1] > 0);
+
+  /* sgetrf参数转换*/
+  int m, n, batch = 1;


这里用了int，对于单个维度超过int max，这里会出现overflow异常

单维度规模如果超过int max会在调用算子时就会报空间不足的错误

ArtIntAI · 2024-11-21T02:16:56Z

mlu_op.h

+ *
+ * @par Data Layout
+ * - The supported combinations of data types are shown below:
+ * - size_t(size)


这里data layout也标记成None吧，workspace接口不做额外说明

ArtIntAI · 2024-11-21T02:18:34Z

mlu_op.h

+ * - size_t(size)
+ *
+ * @par Scale Limitation
+ * - The dimension of input tensor must be either 2, 3 or 4.


这里的限制也标记成None吧

ArtIntAI · 2024-11-21T02:19:03Z

mlu_op.h

+ *
+ * @par Data Type
+ * - The supported combinations of data types are shown below:
+ *   - float(x) - float(y)


不也支持complex吗？

ArtIntAI · 2024-11-21T02:28:05Z

mlu_op.h

+ * Considering the size of the GDRAM, the space occupied by the input matrix should not exceed 7GB.
+ *
+ * @par API Dependency
+ * - The allocated extra workspace should be passed to ::mluOpSgetrf2 to perform the LU operation.


Before calling this function to perform ::mluOpRoiPointPool3d, you need to

get the size of workspace by ::mluOpGetRoiPointPool3dWorkspaceSize.
这里的API依赖描述类似这样，这里和workspace接口的一样了！

shunshen93 · 2024-11-21T02:08:47Z

mlu_op.h

+ * - None.
+ */
+mluOpStatus_t MLUOP_WIN_API
+mluOpGetSgetrf2WorkspaceSize(mluOpHandle_t handle, const mluOpTensorDescriptor_t input_desc, size_t *workspace_size);


参数名和函数定义一致吧。函数定义的地方是x_desc，这里是input_desc

mlu_op.h

test/mlu_op_gtest/pb_gtest/src/zoo/sgetrf2/sgetrf.cpp

shunshen93 · 2024-11-21T04:01:07Z

docs/design_docs/sgetrf/sgetrf.md

+| Input_desc  | 矩阵描述符         | 输入      |                      |                  |                                               |
+| Input       | 输入矩阵           | 输入/输出 | float、complex float | shape[batch,M,N] | 所占空间不超过7GB，且单维度不得超过整型数大小 |
+| output_desc | 矩阵描述符         | 输入      |                      |                  |                                               |
+| output      | 输入矩阵           | 输入/输出 | float、complex float | shape[batch,M,N] | 所占空间不超过7GB，且单维度不得超过整型数大小 |


参考https://pytorch.org/docs/1.13/generated/torch.linalg.lu.html?highlight=lu#torch.linalg.lu
P, L, U = torch.linalg.lu(A)，MLU的输出应该是多个tensor。
目前的输出是P,L,U3个矩阵组合成的1个矩阵，在精度计算的时候，在c++或者python端，将1个tensor拆分成P,L,U3个tensor，计算torch.matmul(P,torch.matmul(L,U))与A的误差。
上面说的在c++或者python端，将1个tensor拆分成P,L,U3个tensor，这部分应该使用MLU实现。

对接之后发现主要是和torch.linalg.lu_factor进行对齐。输出应该是2个tensor，需要将mluOpSgetrf2接口中的ipiv作为device的输出，直接使用data_vector_[2].device_ptr获取，不要在host端new分配空间

shunshen93 · 2024-11-21T07:39:07Z

docs/design_docs/sgetrf/sgetrf.md

+| Input_desc  | 矩阵描述符         | 输入      |                      |                  |                                               |
+| Input       | 输入矩阵           | 输入/输出 | float、complex float | shape[batch,M,N] | 所占空间不超过7GB，且单维度不得超过整型数大小 |
+| output_desc | 矩阵描述符         | 输入      |                      |                  |                                               |
+| output      | 输入矩阵           | 输入/输出 | float、complex float | shape[batch,M,N] | 所占空间不超过7GB，且单维度不得超过整型数大小 |


对接之后发现主要是和torch.linalg.lu_factor进行对齐。输出应该是2个tensor，需要将mluOpSgetrf2接口中的ipiv作为device的输出，直接使用data_vector_[2].device_ptr获取，不要在host端new分配空间

shunshen93 · 2024-11-21T07:42:43Z

test/mlu_op_gtest/pb_gtest/src/zoo/sgetrf2/sgetrf.cpp

+    row = tensor_desc_[0].tensor->dims[2];
+    col = tensor_desc_[0].tensor->dims[3];
+  }
+  std::unique_ptr<int[]> ipiv0(new int[batch * row]);


参考上面的描述，ipiv作为算子的输出，直接使用data_vector_[2].host_ptr获取。对应的测试用例中的mannul_shape_1.json中新增一个output输出，类似与下面这种
{
"inputs": [{"shape": [65536,64], "dtype": "complex_float", "random_distribution": {"uniform": [-5,5]}, "layout":"ARRAY"}],
"outputs": [{"shape": [65536,64], "dtype": "complex_float", "layout": "ARRAY"}，
{"shape": [64], "dtype": "int32", "layout": "ARRAY"}],
"op_params": {"mode": 0},
"proto_params": {"large_tensor": false}
},

// int* ipiv = (int*)data_vector_[2].host_ptr;
std::unique_ptr<int[]> ipiv0(new int[batch * row]);
int *ipiv = ipiv0.get();
不止要获取，还要使用。应该用上面被注释掉的代码，而不是下面这种在host分配空间

shunshen93 · 2024-11-21T08:18:16Z

docs/user_guide/9_operators/index.rst

+---------------
+执行 LU 分解，将一个矩阵分解为一个下三角矩阵（L）和一个上三角矩阵（U），参数``mode``用来指定是否进行选主元操作。
+
+该算子包含7个输入：handle 为操作句柄，x_desc 与 x 分别描述并提供输入矩阵的信息；两个输出：y_desc 与 y 分别描述并存储输出矩阵的信息；此外，还包含一个参数 mode，用于指定是否进行选主元，值为0表示选择非主元模式，ipiv表示置换矩阵，以及一个 workspace 用于临时存储计算过程中的数据。


ipiv也作为输出矩阵，也需要对应的ipiv_desc

kernels/sgetrf2/sgetrf.cpp

ArtIntAI · 2024-11-22T02:40:17Z

kernels/sgetrf2/sgetrf2_native.cpp

+        }
+      }
+    }
+    k_dim.x = dim_x;


是这部分内容来抽取到policyFunc，参考下面的实现
void policyFuncBallQuery(const mluOpHandle_t &handle,
const mluOpTensorDescriptor_t &desc, cnrtDim3_t *k_dim,
cnrtFunctionType_t *k_type) {
size_t cluster_num = mluop::runtime::getClusterLimitCapability(handle);
VLOG(5) << "In current device, cluster_num:" << cluster_num;
size_t core_in_cluster = handle->core_num_per_cluster;
VLOG(5) << "In current device, core_in_cluster:" << core_in_cluster;

size_t total_data_num = desc->total_element_num;

// On a core, a lot of new_xyz data element can be stored; but only one data
// element can be processed at a time. So a cluster can only process four data
// element.
size_t needed_cluster_num =
(total_data_num + core_in_cluster - 1) / core_in_cluster;
*k_type = cnrtFuncTypeUnion1;
k_dim->x = core_in_cluster;
k_dim->y =
needed_cluster_num > cluster_num ? cluster_num : needed_cluster_num;
k_dim->z = 1;
}

AndyQiao0828 · 2024-11-22T02:37:27Z

docs/user_guide/9_operators/index.rst

@@ -793,3 +793,10 @@ mluOpLgamma

 - ``x`` 为输入张量。

+.. Sgetrf2::


.. Sgetrf2::>>>.. _sgetrf2:

AndyQiao0828 · 2024-11-22T02:38:10Z

docs/user_guide/9_operators/index.rst

@@ -793,3 +793,10 @@ mluOpLgamma

 - ``x`` 为输入张量。

+.. Sgetrf2::


这个新增内容在手册的更新历史章节也补充下哈

这个能具体解释一下吗

AndyQiao0828 · 2024-11-22T02:43:18Z

docs/user_guide/9_operators/index.rst

+
+mluOpSgetrf2
+---------------
+执行 LU 分解，将一个矩阵分解为一个下三角矩阵（L）和一个上三角矩阵（U），参数``mode``用来指定是否进行选主元操作。


参数mode用来》》参数 mode 用来

这个是什么意思？

AndyQiao0828 · 2024-11-22T02:47:59Z

docs/user_guide/9_operators/index.rst

+---------------
+执行 LU 分解，将一个矩阵分解为一个下三角矩阵（L）和一个上三角矩阵（U），参数``mode``用来指定是否进行选主元操作。
+
+该算子包含7个输入：handle 为操作句柄，x_desc 与 x 分别描述并提供输入矩阵的信息；两个输出：y_desc 与 y 分别描述并存储输出矩阵的信息；此外，还包含一个参数 mode，用于指定是否进行选主元，值为0表示选择非主元模式，ipiv表示置换矩阵，以及一个 workspace 用于临时存储计算过程中的数据。


有公式的话，需要补充下公式。

Suggested change

该算子包含7个输入：handle 为操作句柄，x_desc 与 x 分别描述并提供输入矩阵的信息；两个输出：y_desc 与 y 分别描述并存储输出矩阵的信息；此外，还包含一个参数 mode，用于指定是否进行选主元，值为0表示选择非主元模式，ipiv表示置换矩阵，以及一个 workspace 用于临时存储计算过程中的数据。

该算子包含7个输入：其中，``handle`` 为操作句柄，``x_desc`` 与 ``x`` 分别描述并提供输入矩阵的信息；两个输出：``y_desc`` 与 ``y`` 分别描述并存储输出矩阵的信息；此外，还包含一个参数 ``mode`` 用于指定是否进行选主元操作，值为0时，表示选择非主元模式，``ipiv`` 表示置换矩阵，以及 ``workspace`` 用于临时存储计算过程中的数据。

》》7个输入中提到了三个（handle，x_desc和x）？mode、ipiv和workspace也是输入吗？

AndyQiao0828 · 2024-11-22T02:49:26Z

mlu_op.h

@@ -14523,6 +14523,132 @@ mluOpLgamma(mluOpHandle_t handle,
            const mluOpTensorDescriptor_t y_desc,
            void *y);

+/*!


这几个接口属于哪个group呢？参照其他补充下哈：
// Group:Lgamma

AndyQiao0828 · 2024-11-22T03:03:52Z

mlu_op.h

+ * INTEGER array, dimension (m);
+ * The pivot indices; row i of the matrix was interchanged with row IPIV(i)


Suggested change

* INTEGER array, dimension (m);

* The pivot indices; row i of the matrix was interchanged with row IPIV(i)

* An integer array, dimension (m);

* The pivot indices; row i of the matrix was interchanged with row IPIV(i).

需要优化，说明ipiv的含义。写成完整的句子：这里dimension跟pivot indices是啥关系。

AndyQiao0828 · 2024-11-22T03:04:30Z

mlu_op.h

+ * INTEGER array, dimension (m);
+ * The pivot indices; row i of the matrix was interchanged with row IPIV(i)
+ *
+ * @param[out] info


在参数解释最前面先说明infor的含义

AndyQiao0828 · 2024-11-22T03:06:09Z

mlu_op.h

+ *                 to solve a system of equations.
+ *
+ * @param[in] mode
+ *   option to perform operation with pivoting/no pivoting versions


Suggested change

* option to perform operation with pivoting/no pivoting versions

* Option to perform the operation with pivoting/no pivoting versions

都有什么mode呢？

已在代码中修改

AndyQiao0828 · 2024-11-22T03:07:04Z

mlu_op.h

+ * - The data layout of y should be MLUOP_LAYOUT_ARRAY.
+ *
+ * @par Scale Limitation
+ * - The dimension of input tensor must be either 2, 3 or 4.


The dimension of input tensor must be 2, 3 or 4.

AndyQiao0828 · 2024-11-22T03:09:48Z

mlu_op.h

+ * @param[out] info
+ *     -     = 0:  successful exit
+ *     -     < 0:  if INFO = -i, the i-th argument had an illegal value
+ *                 or another error occured, such as memory allocation failed.


这是什么意思呢？

shunshen93 · 2024-11-22T08:40:22Z

kernels/sgetrf2/sgetrf.cpp

+    m = x_desc->dims[0];
+    n = x_desc->dims[1];
+  } else if (x_desc->dim == 3) {
+    PARAM_CHECK("mluOpSgetrf2", x_desc->dims[2] > 0);


[0,0]，[1,0,0]，[1,1,0,0]这些情况也要支持，目前测试会在这里被防呆拦住

[0,0]，[1,0,0]，[1,1,0,0]这几种情况有测试过吗？
已[1,0,0]为例，dim=3,x_desc->dims[2]=0，在这里会被防呆拦住。结果需要与gpu的结果保持一致

shunshen93 · 2024-11-22T08:48:48Z

test/mlu_op_gtest/pb_gtest/src/zoo/sgetrf2/test_case/case0.prototxt

@@ -0,0 +1,41 @@
+op_name: "test_sgetrf2"
+input {
+  id: "input"


在590上测试精度超出阈值，且generator产生的case，590精度也会超出阈值

在590上测试精度超出阈值，且generator产生的case，590精度也会超出阈值

能否给出矩阵输入规模等信息及输出？

[22,29,206]，DTYPE_COMPLEX_FLOAT的case会超时

[22,29,206]，DTYPE_COMPLEX_FLOAT的case会超时

请问[22,29,206]对应矩阵规模的[batch,m,n]/[m,n,batch]吗？我分别测试了这两种情况都没有问题

是跑的complex类型吗？

已复现并修改代码

op_name: "test_xgetrf" input { id: "input" shape: { dims: 1 dims: 22 dims: 129 dims: 206 } layout: LAYOUT_ARRAY dtype: DTYPE_COMPLEX_FLOAT random_data: { seed: 25 upper_bound: 10.0 lower_bound: -10.0 distribution: UNIFORM } } output { id: "output" shape: { dims: 1 dims: 22 dims: 129 dims: 206 } layout: LAYOUT_ARRAY dtype: DTYPE_COMPLEX_FLOAT } output { id: "output2" shape { dims: 1 dims: 22 dims: 129 } layout: LAYOUT_ARRAY dtype: DTYPE_INT32 } xgetrf_param{ mode: 0 } test_param: { error_func: DIFF1 error_func: DIFF2 error_threshold: 0.003 error_threshold: 0.003 baseline_device: CPU }

测试下这个case

这个case已复现并修复，请问mlu端怎么关闭nan/inf的检查呢？包含nan/inf的case的测试标准是什么呢？

shunshen93 · 2024-11-22T09:32:47Z

kernels/sgetrf2/sgetrf2.h

+#define c_one 1.0
+#define c_neg_one -1.0
+#define taskType 8
+#define MAX_M_SIZE_COMPLEX 1024


加下注释说明MAX_M_SIZE，MAX_M_SIZE1，MAX_M_SIZE_COMPLEX是什么意思

shunshen93 · 2024-11-22T09:46:57Z

kernels/sgetrf2/scal_ger_union1.mlu

+  float *i_extra_vec_nram = i_shared_y + N * N;
+  float *i_temp = i_extra_vec_nram + N * N;
+  float *i_L1 = i_temp + MAX_M_SIZE_COMPLEX + N + N * 2;
+  float *i_temp_L1 = i_L1 + (MAX_NRAM_SIZE / 16 - 4 * 1024);


为什么是MAX_NRAM_SIZE / 16 - 4 * 1024，加下注释说明

shunshen93 · 2024-11-22T09:47:33Z

kernels/sgetrf2/scal_ger_union1.mlu

+  int gbj, STEP;
+  float *r_shared_y = NRAM_BUFFER;
+  float *r_extra_vec = SRAM_BUFFER;
+  float *r_extra_vec_nram = r_shared_y + N * N;


设计文档中写，存储每次分解时的第一行元素是NN,为什么是NN呢？

ArtIntAI · 2024-11-25T06:42:05Z

另外针对用户感知到的一些tensor信息，如下所列，支持的做下测试，不支持的可以参考下其他算子做好参数拦截
1.large tensor(tensor单个维度超过2G num, tensor的所有维度乘积超过2G num)
2. inplace，输入和输出tensor地址一致
3. stride，如果不支持做好参数检查报错
4. 广播，如果不支持做好参数检查报错
5. 输入和输出包含nan/inf时精度是否和GPU精度对齐
6. 输入tensor是0元素，某个维度是0

ArtIntAI · 2024-11-27T12:33:03Z

kernels/sgetrf2/sgetrf2_native.cpp

+
+  if (dtype == MLUOP_DTYPE_COMPLEX_FLOAT) {
+    if (batch > 1) {
+      k_type = CNRT_FUNC_TYPE_UNION8;


板卡上不一定有这个类型，建议参考这里进行设置：

mlu-ops/kernels/dynamic_point_to_voxel/dynamic_point_to_voxel_backward/dynamic_point_to_voxel_backward.cpp

Line 191 in 5ae8c94

*k_type = mluop::runtime::getJobLimitCapabilityCnrtFuncType(handle);

mlu-ops/kernels/fft/c2c_fft/c2c_fft_host.cpp

Line 1668 in 5ae8c94

int task_type = mluop::runtime::getJobLimitCapability(handle);

ArtIntAI · 2024-11-27T12:42:29Z

kernels/sgetrf2/sgetrf.cpp

+    transpose(handle, MLUOP_DTYPE_COMPLEX_FLOAT, batch, m, n, (float *)x,
+              (float *)y, handle->queue);
+  } else {
+    cnrtMemcpy((float *)y, (float *)x, batch * m * n * sizeof(float),


不建议使用cnrtMemcpy和cnrtMemset，cnrtQueueSync，会对上层使用mlu_graph有问题
建议cnrtMemcpy使用片上的__memcpy来替换
cnrtMemset使用片上设置数据来替换
cnrtQueueSync可以去掉，对于同一个queue来说，queue内的kernel调用（使用<<<>>>）是串行的

shunshen93 · 2024-12-03T06:43:05Z

docs/design_docs/xgetrf/xgetrf.md

+| output      | 输入矩阵           | 输入/输出 | float、complex float | shape[batch,M,N] | 所占空间不超过7GB，且单维度不得超过整型数大小 |
+| mode        | 模式pivot/no pivot | 输入      | int                  |                  |                                               |
+| dipiv_desc  | 矩阵描述符         | 输入      |                      |                  |                                               |
+| dipiv       | 置换矩阵           | 输入      | int*                 | shape[M]         |                                               |


dipiv也是输出？shape是[batch, M]?

shunshen93 · 2024-12-06T08:48:35Z

kernels/xgetrf/xgetrf_mlu.cpp

+        *k_type = CNRT_FUNC_TYPE_UNION1;
+        dim_x = handle->core_num_per_cluster * 1;
+      } else if (taskType == 8) {
+        *k_type = CNRT_FUNC_TYPE_UNION8;


不要写死Unoin8，因为可能其它板卡没有8个cluster。使用mluop::runtime::getJobLimitCapabilityCnrtFuncType(handle)获取

Chuancysun mentioned this pull request May 20, 2024

【新算子】- linalg.lu 算子开发 #1007

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