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If A guesses the correct password pw = pw', then A can obtain x1 correctly.
In this scenario, if A guesses the correct password pw = pw', then A succeeds.
If A guesses an invalid password pw ≠ pw', then z can be treated as a random number in Z n *.
U a computes (3) e i ⊕ f i = [ h (ID i || x s ) ⊕ r i ] ⊕ f i = [ h (ID i || x s ) ⊕ h (PW i ⊕ K i ) ⊕ f i ] ⊕ f i = [ h (ID i || x s ) ⊕ h (PW i ⊕ K i ) ] to obtain [ h (ID i || x s ) ⊕ h (PW i ⊕ K i ) ]. U a guesses PW a as user's possible password and computes M1 a = [ e i ⊕ f i ] ⊕ h(PW a ⊕ K i ).
But if M 3 a = M 3, then it provides U a with the exact password PW i of C i. U a computes (4) e i ⊕ f i = [ h (ID i || x s ) ⊕ r i ] ⊕ f i = [ h (ID i || x s ) ⊕ h (PW i ⊕ K i ) ⊕ f i ] ⊕ f i = [ h (ID i || x s ) ⊕ h (PW i ⊕ K i ) ] to obtain [ h (ID i || x s ) ⊕ h (PW i ⊕ K i ) ]. U a guesses PW a as user's possible password and computes M1 a = [ e i ⊕ f i ] ⊕ h(PW a ⊕ K i ).
Similar(55)
So take a guess.
(Just a guess).
Take a guess.
Mr. Graifman has a guess.
Ms. Huffstetler hazarded a guess.
I will venture a guess.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com