Timeline for Why lower bounds for boolean Circuits does not imply arithmetic circuits lower bounds
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2012 at 16:45 | vote | accept | Somebody | ||
| Nov 30, 2012 at 6:03 | history | edited | Joshua Grochow | CC BY-SA 3.0 |
Added general statement
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| Nov 30, 2012 at 5:55 | comment | added | Joshua Grochow | @Klim: No. The issue is that a circuit over C can still use irrational (or even non-real) constants, which you still can't take "mod 2." | |
| Nov 30, 2012 at 5:27 | comment | added | Klim | Does it mean that proving some kind of theorem like von-division(i.e. that you do not need to divide by two) will imply circuit lower bounds over C? | |
| Nov 30, 2012 at 4:00 | comment | added | Joshua Grochow | So that when you take things mod 2 $b$ has an inverse mod 2, i.e. $a/b \in \mathbb{Q}$ becomes $ab^{-1} \pmod{2}$ and the latter is well-defined. | |
| Nov 30, 2012 at 3:31 | comment | added | Suresh Venkat | what is the significance of $b$ even ? | |
| Nov 30, 2012 at 2:02 | history | answered | Joshua Grochow | CC BY-SA 3.0 |