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MJD
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Apparently $\leftarrow_\$$This answer is broken according to a report? Seemsfree for anyone to not be the case..edit.

$$\leftarrow_\$$$

Inline in quote:

$\leftarrow_\$$

Display in quote:

$$\leftarrow_\$$$

$$\left|\Pr_{k\leftarrow_\$\ \{0,1\}^n}[\mathcal{A}(1^n, F(k,-))=1] - \Pr_{f\leftarrow_\$\ \{0,1\}^n\to\{0,1\}^n}[\mathcal{A}(1^n, f(-))=1]\right|\leq \mathrm{negl}(n).$$

Apparently $\leftarrow_\$$ is broken according to a report? Seems to not be the case...

$$\leftarrow_\$$$

Inline in quote:

$\leftarrow_\$$

Display in quote:

$$\leftarrow_\$$$

$$\left|\Pr_{k\leftarrow_\$\ \{0,1\}^n}[\mathcal{A}(1^n, F(k,-))=1] - \Pr_{f\leftarrow_\$\ \{0,1\}^n\to\{0,1\}^n}[\mathcal{A}(1^n, f(-))=1]\right|\leq \mathrm{negl}(n).$$

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ComFreek
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Apparently $\leftarrow_\$$ is broken according to a report? Seems to not be the case...

$$\leftarrow_\$$$

Inline in quote:

$\leftarrow_\$$

Display in quote:

$$\leftarrow_\$$$

$$\left|\Pr_{k\leftarrow_\$\ \{0,1\}^n}[\mathcal{A}(1^n, F(k,-))=1] - \Pr_{f\leftarrow_\$\ \{0,1\}^n\to\{0,1\}^n}[\mathcal{A}(1^n, f(-))=1]\right|\leq \mathrm{negl}(n).$$

Apparently $\leftarrow_\$$ is broken according to a report? Seems to not be the case...

Apparently $\leftarrow_\$$ is broken according to a report? Seems to not be the case...

$$\leftarrow_\$$$

Inline in quote:

$\leftarrow_\$$

Display in quote:

$$\leftarrow_\$$$

$$\left|\Pr_{k\leftarrow_\$\ \{0,1\}^n}[\mathcal{A}(1^n, F(k,-))=1] - \Pr_{f\leftarrow_\$\ \{0,1\}^n\to\{0,1\}^n}[\mathcal{A}(1^n, f(-))=1]\right|\leq \mathrm{negl}(n).$$

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SEJPM
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Apparently $\leftarrow_\$$ is broken according to a report? Seems to not be the case...

Post Made Community Wiki by SEJPM