基于其他两个信号的信号

发布于 2025-01-26 11:56:14 字数 857 浏览 2 评论 0 原文

我正在尝试编写一个电路，以证明一个数字在指定的范围内。我正在使用 circomlib library ： comparatorter.circuit.circuit 文件。我的代码是：

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
}

我要返回 1 如果 true 和 0 如果 false 。但这将取决于其他两个信号。我该怎么做？

原文

I am trying to write a circuit to prove that a number is within a specified range. I am using the circomlib library: comparators.circuit file. My code 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
}

I want to return 1 if true and 0 if false. But that would depend on whether the other two out signals. How can I do this?

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梨涡 2025-02-02 11:56:14

在您的代码中，您可以简单地做
out＆lt; == low.out * high.out 。

它可以准确地实现您想要的：

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

不过，如果您想根据“ if语句”分配给两个分支到两个分支的信号，则需要使用多路复用器。参见 mux1 来自Circomlib

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

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