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[Operator] Fix symbolic broadcasting (hidet-org#131)
The previous implementation is incorrect when dealing with a pair of dimensions that are both symbolic. Minimal example: import hidet if __name__ == "__main__": x = hidet.symbol(["n"]) y = hidet.symbol(["m"]) z = x + y print(x.shape, y.shape, z.shape) # before: (n,) (m,) (m,)
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# Licensed under the Apache License, Version 2.0 (the "License"); | ||
# you may not use this file except in compliance with the License. | ||
# You may obtain a copy of the License at | ||
# | ||
# http://www.apache.org/licenses/LICENSE-2.0 | ||
# | ||
# Unless required by applicable law or agreed to in writing, software | ||
# distributed under the License is distributed on an "AS IS" BASIS, | ||
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. | ||
# See the License for the specific language governing permissions and | ||
# limitations under the License. | ||
from collections import namedtuple | ||
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import hidet | ||
from hidet.ir.expr import if_then_else | ||
from hidet.transforms.rule_based_simplifier import RuleBasedSimplifier | ||
from hidet.utils import repeat_until_converge | ||
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def test_rule_based_simplify(): | ||
n = hidet.symbol_var("n") | ||
m = hidet.symbol_var("m") | ||
testcase = namedtuple("testcase", ["expr", "expected"]) | ||
cases = [ | ||
testcase(expr=(n + 1) - (1 + n), expected=0), | ||
testcase(expr=if_then_else(n > 0, 1, 1), expected=1), | ||
testcase(expr=(n + m - m) - (m + n - m), expected=0), | ||
testcase(expr=(n + m) - (m + n), expected=0), | ||
testcase(expr=n / n, expected=1), | ||
] | ||
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simp = RuleBasedSimplifier() | ||
for expr, expected in cases: | ||
assert repeat_until_converge(simp, expr) == expected |