<p>Under believable hardness assumptions, it is possible to construct pseudo-random generators that are strong enough to deduce all of the assertions P=BPP, P=RP, P=co-RP (and thus, P=ZPP). You can find the details in</p>

<p>Madhu Sudan, Luca Trevisan, Salil Vadhan. Pseudorandom generators without the XOR Lemma.</p>

CC BY-SA 3.0
2013-08-07T05:05:57.880Z
2013-08-07T05:05:57.880Z
Mahdi Cheraghchi
<p>1- what are the Consequences of ZPP=P ? </p>

<p>2- I know that if the advice from Adleman's theorem could be found in polynomial time then BPP=P , and this is likely to happen (I think that there is other evidence that P=BPP). what about evidence that P=ZPP without this result ? is there any other evidences ? </p>

<cc.complexity-theory>
ZPP vs. P status
Fayez Abdlrazaq Deab
# ZPP vs. P status

(Asked by: username_0 on 2013-08-06)

1- what are the Consequences of ZPP=P ?

2- I know that if the advice from Adleman's theorem could be found in polynomial time then BPP=P , and this is likely to happen (I think that there is other evidence that P=BPP). what about evidence that P=ZPP without this result ? is there any other evidences ?

**Tags:** cc.complexity-theory

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## 1 answers

### Answer by username_1 on 2013-08-07. Score: 6 | Accepted answer

Under believable hardness assumptions, it is possible to construct pseudo-random generators that are strong enough to deduce all of the assertions P=BPP, P=RP, P=co-RP (and thus, P=ZPP). You can find the details in

Madhu Sudan, Luca Trevisan, Salil Vadhan. Pseudorandom generators without the XOR Lemma.


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