In the ideal world, $A$ sends several inputs $x_1,...,x_\lambda$ to the functionality and $B$ sends input $y$. Functionality sends $f(x_1,y),...,f(x_\lambda,y)$ to $A$ and nothing to $B$.

Suppose that we have a protocol which is secure against semi-honest adversaries to realize this functionality.

In this protocol, $A$ sends encrypted values of $x_i$s to $B$. $B$ computes the encrypted result of each $f(x_i,y)$ and sends them back to $A$. $A$ decrypts these values to know each $f(x_i,y)$.

1. Is it possible for $A$ to choose $x_i$s in a way that this protocol is also secure against malicious $B$? For example, $A$ can choose some $x_i$s equal or with some specific relation in such a way that any deviation from the protocol is detected?

What I am intended to say is something like cut-and-choose but for outputs instead of inputs.

2. More generally, is it necessary to force the malicious party to act honestly is each intermediate step or it is sufficient to force him to output correctly?

In the ideal world, $A$ sends several inputs $x_1,...,x_\lambda$ to the functionality and $B$ sends input $y$. Functionality sends $f(x_1,y),...,f(x_\lambda,y)$ to $A$ and nothing to $B$.

Suppose that we have a protocol which is secure against semi-honest adversaries to realize this functionality.

In this protocol, $A$ sends encrypted values of $x_i$s to $B$. $B$ computes the encrypted result of each $f(x_i,y)$ and sends them back to $A$. $A$ decrypts these values to know each $f(x_i,y)$.

1. Is it possible for $A$ to choose $x_i$s in a way that this protocol is also secure against malicious $B$? For example, $A$ can choose some $x_i$s equal or with some specific relation in such a way that any deviation from the protocol is detected?

What I am intended to say is something like cut-and-choose but for outputs instead of inputs.

2. More generally, is it necessary to force the malicious party to act honestly at each intermediate step or it is sufficient to force him to output correctly?