Comments (4)
补充TODO
- 找个开源json库
from autograd.
找助教确认的事情
- 是否需要类型检查, 举个张量形状不一致的例子
A<16, 32>[i, j] = A<15, 32>[i, j] + B<16, 32>[i, k] * C<32, 32>[k, j];
- 是否可以读run.h(函数签名那里可以偷懒一下吗?
- 对于这两个例子的区别, 这样理解是否正确?
// A[i][j]=A[i][j]+B[i][k]*B[k][j]
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++){
for(int k =0; k < l; k++){
A[i][j]+=B[i][k]*C[k][j];
}
// A[i][j]=B[i][k]*B[k][j]
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++){
A[i][j]=0;
for(int k =0; k < l; k++)
A[i][j]+=B[i][k]*C[k][j];
}
}
- 和助教确认一下目前对爱因斯坦求和规范的理解是否正确(见上)
from autograd.
找助教确认的事情
- 是否需要类型检查, 举个张量形状不一致的例子
A<16, 32>[i, j] = A<15, 32>[i, j] + B<16, 32>[i, k] * C<32, 32>[k, j];
- 是否可以读run.h(函数签名那里可以偷懒一下吗?
- 对于这两个例子的区别, 这样理解是否正确?
// A[i][j]=A[i][j]+B[i][k]*B[k][j] for(int i = 0; i < n; i++) for(int j = 0; j < m; j++){ for(int k =0; k < l; k++){ A[i][j]+=B[i][k]*C[k][j]; }// A[i][j]=B[i][k]*B[k][j] for(int i = 0; i < n; i++) for(int j = 0; j < m; j++){ A[i][j]=0; for(int k =0; k < l; k++) A[i][j]+=B[i][k]*C[k][j]; } }
- 和助教确认一下目前对爱因斯坦求和规范的理解是否正确(见上)
- 不需要
- 不可
注意, 问过了助教, 助教的理解和我们的完全不一样, 这里不是严格的爱因斯坦求和规则
每一个statement一定是一个严格的loop nest, kernel就是最内层的语句
from autograd.
补充TODO
- 找个开源json库
https://github.com/nlohmann/json
from autograd.
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from autograd.