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By "as easy as each other" I meant reducible to each other with a polynomial factor, as is standard in complexity theory.
by sid0 16y ago
By "as easy as each other" I meant reducible to each other with a polynomial factor, as is standard in complexity theory.
- eru 16y agoYes, that why I prefaced with "To be pedantic". I just felt like pointing out that there are meaningful differences between polynomial problems.