Out signal based on two other our signals

Question

I am trying to write a circuit to prove that a number is within a specified range. I am using the circomlib library : https://github.com/iden3/circomlib/tree/master/circuits comparators.circuit file. My code till now is this :

    template RangeProof(n) {
        assert(n <= 252);
        signal input in; // number to be proved 
        signal input range[2]; //  [lower bound, upper bound]
        signal output out;

    component low = LessEqThan(n);
    component high = GreaterEqThan(n);
    low.in[0] <== in;
    low.in[1] <== range[0];
    low.out === 1;
    high.in[0] <== in;
    high.in[1]<==range[1];
    high.out === 1;
    out <== (low.out + high.out) == 2 ? 1: 0; //this is the line in question
}

So I want to return 1 if true and 0 if false. But that would depend on whether the other two out signals. Any idea how I can do this? Any help is greatly appreciated.

Why do you assert `low.out===1` and `high.out===1`? That means your code will crash upon compiling the witness anytime your input `in` is out of the range. And if I'm not mistaken, the ternary operator `a ? b : c` cannot be applied on signals, only on variables. — Victor Colomb, May 11 '22 at 08:46

score 1 · Accepted Answer · answered May 19 '22 at 02:42

In your code, you can simply do out <== low.out * high.out.

It achieves exactly what you want:

low.out 0 * high.out 0 = out 0
low.out 1 * high.out 0 = out 0
low.out 0 * high.out 1 = out 0
low.out 1 * high.out 1 = out 1

In general though, if you want to assign to a signal based on an 'if statement' which branches into 2 branches, you need to use a Multiplexer. See Mux1 template from circomlib https://github.com/iden3/circomlib/blob/master/circuits/mux1.circom#L33

Out signal based on two other our signals

1 Answers1