Is there an efficient way to find out the relationship between wires indexes and emp::Integer in the circuit generated by EMP-TOOLKIT? about emp-tool HOT 10 CLOSED

One way that you can know the id of each wire is to cast the block into an array of uint64_t. the higher int is the id.
See e.g.: https://github.com/emp-toolkit/emp-tool/blob/master/emp-tool/execution/plain_circ.h#L65

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On a second examination, I think the wires should all be ordered as one would expect rather than the order you mentioned above. Can you double-check? :)

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Hi. Thank you very much for your reply!

I retried to generate the circuit but it still not works as my expectation. Specifically, the program I executed is as follow:

#include <emp-tool/execution/plain_prot.h>
#include <emp-sh2pc/emp-sh2pc.h>
#include <emp-tool/circuits/bit.h>
#include <emp-tool/circuits/circuit_file.h>

#include <iostream>

static const char* FILENAME = "test.txt";

int main() {
  using Circuit = emp::BristolFormat;
  {
    emp::setup_plain_prot(true, FILENAME);
    emp::Integer a {2, 0, emp::ALICE};
    emp::Integer b {2, 0, emp::ALICE};
    emp::Bit cond1 = (b > a);
    cond1.reveal<bool>();
    emp::finalize_plain_prot();	  
  }
}

The test.txt will be:

13 17
4 0 1
                                                                                                                                                                                            
2 1 0 2 4 XOR
2 1 4 2 5 AND
2 1 1 3 6 XOR
2 1 5 3 7 XOR
2 1 6 5 8 XOR
2 1 6 7 9 AND
2 1 5 9 10 XOR
2 1 1 3 11 XOR
2 1 11 10 12 XOR
1 1 12 13 INV
1 1 13 14 INV
2 1 0 0 15 XOR
2 1 15 14 16 XOR

I tried to set in[0..3] and see out[16], for example:

in[0..3] := 1000 ---> out[16] = 0
in[0..3] := 0100 ---> out[16] = 1
in[0..3] := 0010 ---> out[16] = 1
in[0..3] := 0001 ---> out[16] = 0

which means in[0] and in[3] belong to emp::Integer a && in[1] and in[2] belong to emp::Integer b.

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little-endian or big-endian in your representation? Try 0011 :)

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Hi!

in[0..3] := 1000 <——> in[0] := 1, in[1] := 0, in[2] := 0, in[3] := 0

Therefore, 0011 will result in 0.

from emp-tool.

Please try this example

setup_plain_prot(true, "sort.txt");
Integer a(2,0,ALICE);
Integer b(2,0,ALICE);
(a[0] & b[0]).reveal();

You will see that the output circuit is
3 7
4 0 1

2 1 0 2 4 AND
2 1 0 0 5 XOR
2 1 5 4 6 XOR

Here only wire 0 and 2 are used. This is because 0 is for a[0] and 2 is for b[0].

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Thank you very much! I tried your example (with a[0]&a[1], a[0]&b[1] ...). It indeed works.

The experiment shows that a[0] := in[0], a[1] := in[1], b[0] := in[2] and b[1] := in[3].

Then, there must be some problem in (b > a) circuit generating procedure. Let's set
a[0] = in[0] := 0
a[1] = in[1] := 1
b[0] = in[2] := 0
b[1] = in[3] := 0

(b > a) should be 0, obviously.

However, based on the BF circuit in my previous reply, which is:

13 17
4 0 1
                                                                                                                                                                                            
2 1 0 2 4 XOR
2 1 4 2 5 AND
2 1 1 3 6 XOR
2 1 5 3 7 XOR
2 1 6 5 8 XOR
2 1 6 7 9 AND
2 1 5 9 10 XOR
2 1 1 3 11 XOR
2 1 11 10 12 XOR
1 1 12 13 INV
1 1 13 14 INV
2 1 0 0 15 XOR
2 1 15 14 16 XOR

The entire calculation process will be:

wire[0] = 0
wire[1] = 1
wire[2] = 0
wire[3] = 0
wire[4] = wire[0] XOR wire[2] = 0
wire[5] = wire[4] AND wire[2] = 0
wire[6] = wire[1] XOR wire[3] = 1
wire[7] = wire[5] XOR wire[3] = 0
wire[8] = wire[6] XOR wire[5] = 1
wire[9] = wire[6] AND wire[7] = 0
wire[10] = wire[5] XOR wire[9] = 0
wire[11] = wire[1] XOR wire[3] = 1
wire[12] = wire[10] XOR wire[11] = 1
wire[13] = INV wire[12] = 0
wire[14] = INV wire[13] = 1
wire[15] = wire[0] XOR wire[0] = 0
wire[16] = wire[14] XOR wire[15] = 1, which is wrong.

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actually, the comparison assumes signed input. so if the highest bit is 1. it is a negative number.

from emp-tool.

Oops, I also realized this problem after going into integer.hpp source code just now. Thank you very much for your kindly help! Sorry for causing troubles.

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np. closing the issue.

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