Correctness of ideal protocol of MPC in presence of malicious adversary

In presence of malicious adversaries in ideal protocol of Multi Party Computation, how can honest parties get the correct output? i.e., corrupted parties modify their inputs and give to the trusted party. Trusted Party computes the function by taking honest party inputs and modified inputs of corrupted parties which results in a wrong output.

• Are you assuming honest majority? Also, what do you mean by modified inputs? For example, if the malicious party's input is $5$ and they instead send $0$? There isn't anything you can do in that case. Jul 15 '15 at 11:43
• Assume honest majority.In example, malicious party send 0 instead of 5. In that case, trusted party computes the function using 0 as one of the input which results in wrong output.Trusted party sends the output(wrong) to honest party i.e., it violates the correctness property of MPC. Jul 16 '15 at 6:04
• @preethi : $\:$ How does that violate "the correctness property of MPC"? $\;\;\;\;$
– user991
Jul 16 '15 at 13:31

The traditional MPC definition of correctness has no notion of correctness on the inputs. The traditional MPC correctness property deals with the output, i.e., the protocol is correct if $y$ where $y=f(x_1,x_2,\dots,x_n)$ is guaranteed to be output. What the $x_1,\dots,x_n$ values are is completely up to the inputting party.