Correct the at most degrees #9
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`L,R,O` are of degree at most `d-1` . `T` is of degree at most `d`. So `P = LR-O=HT` is of degree at most `2(d-1)` and `H` is of degree at most `d-2`.
daira
reviewed
May 29, 2019
| @@ -23,7 +23,7 @@ As we saw in Part V, Alice will typically want to prove she has a satisfying as | |||
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| If Alice has a satisfying assignment it means that, defining :math:`L,R,O,P` as above, there exists a polynomial :math:`H` such that :math:`P=H\cdot T`. In particular, for any :math:`s\in\mathbb{F}_p` we have :math:`P(s)=H(s)\cdot T(s)`. | |||
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| Suppose now that Alice <em>doesn't</em> have a satisfying assignment, but she still constructs :math:`L,R,O,P` as above from some unsatisfying assignment :math:`(c_1,\ldots,c_m)`. Then we are guaranteed that :math:`T` does not divide :math:`P`. This means that for any polynomial :math:`H` of degree at most :math:`d`, :math:`P` and :math:`L,R,O,H` will be different polynomials. Note that :math:`P` and :math:`L,R,O,H` here are both of degree at most :math:`2d`. | |||
| Suppose now that Alice <em>doesn't</em> have a satisfying assignment, but she still constructs :math:`L,R,O,P` as above from some unsatisfying assignment :math:`(c_1,\ldots,c_m)`. Then we are guaranteed that :math:`T` does not divide :math:`P`. This means that for any polynomial :math:`H` of degree at most :math:`d-2`, :math:`P` and :math:`L,R,O,H` will be different polynomials. Note that :math:`P` here is of degree at most :math:`2(d-1)`, :math:`L,R,O` here are of degree at most :math:`d-1` and :math:`H` here is degree at most :math:`d-2`. | |||
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This change looks correct to me.
daira
approved these changes
May 29, 2019
ebfull
approved these changes
May 29, 2019
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L,R,Oare of degree at mostd-1.Tis of degree at mostd.So
P = LR-O=HTis of degree at most2(d-1)andHis of degree at mostd-2.