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Updated SM3中的FF2和GG2函数 (markdown)

Sun Yimin 2023-09-26 08:16:22 +08:00
parent 7e24148498
commit 4aa13d1054
1 changed files with 2 additions and 0 deletions
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SM3中的FF2和GG2函数.md

@ -62,6 +62,7 @@ $(X \land Y) \bigoplus (X \land Z) \bigoplus (Y \land Z)$
$\lnot((X \land Y \land \lnot Z) \lor (X \land Z \land \lnot Y)) \land (Y \land Z)$
= $\lnot(X \land Y \land \lnot Z) \land \lnot(X \land Z \land \lnot Y) \land (Y \land Z)$
= $(\lnot X \lor \lnot Y \lor Z) \land (\lnot X \lor \lnot Z \lor Y) \land (Y \land Z)$
= $(\lnot X \lor (\lnot X \land \lnot Z) \lor (\lnot X \land Y) \lor (\lnot X \land \lnot Y) \lor (\lnot Z \land \lnot Y) \lor (Z \land \lnot X) \lor (Z \land Y)) \land (Y \land Z)$
= $(Y \land Z)$
$((X \land Y \land \lnot Z) \lor (X \land Z \land \lnot Y)) \land (\lnot Y \lor \lnot Z)$
@ -74,4 +75,5 @@ $(X \land Y \land \lnot Z) \lor (X \land Z \land \lnot Y) \lor (Y \land Z)$
相关知识：
* $A \bigoplus B = (\lnot A \land B) \lor (A \land \lnot B) $
* [Boolean algebra](https://en.wikipedia.org/wiki/Boolean_algebra)
* [布尔代数运算律](https://baike.baidu.com/item/%E5%B8%83%E5%B0%94%E4%BB%A3%E6%95%B0%E8%BF%90%E7%AE%97%E5%BE%8B/22804079)