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reduces the total RTT for revealOblivBoolArray #67

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@weikengchen weikengchen commented Jul 11, 2018

When we reveal data to both parties, we may use revealOblivBoolArray. It passes party=0 to the underlying function. There are many rounds due to underlying implementations.

This commit changes the way revealOblivBoolArray deals with an array to the same way used by revealObliv##tnameArray, which uses party=2 to reveal the result to the second party first, and then uses party=1 to reveal to the first party, rather than uses party=0.

It reduces the total RTT.

I tested the change with the following script:

#include<obliv.h>
#include <obliv.oh>
#include<stdio.h>
#include<stdlib.h>
#include<time.h>

double wallClock()
{
  struct timespec t;
  clock_gettime(CLOCK_REALTIME,&t);
  return t.tv_sec+1e-9*t.tv_nsec;
}

void ocTestUtilTcpOrDie(ProtocolDesc* pd,const char* remote_host,
                        const char* port)
{
  if(!remote_host)
  { if(protocolAcceptTcp2P(pd,port)!=0)
    { fprintf(stderr,"TCP accept failed\n");
      exit(1);
    }
  }
  else 
    if(protocolConnectTcp2P(pd,remote_host,port)!=0) 
    { fprintf(stderr,"TCP connect failed\n");
      exit(1);
    }
}


double lap;

#define MAXN 1000
void readBool(obliv bool* dest, int n, const bool* src,int party)
{
	OblivInputs specs[MAXN];
	int i;
	for(i=0;i<n;++i) setupOblivBool(specs+i,dest+i,src[i]);
	feedOblivInputs(specs,n,party);
}

void gotest(void* vargs)
{
	obliv bool p1_input[1000];
	obliv bool p2_input;

	bool buf[1000];
	readBool(p1_input, 1000, buf, 1);
	readBool(&p2_input, 1, buf, 2);

	fprintf(stderr,"OT time: %lf s\n",wallClock()-lap);

	for(int i = 0; i < 1000; i++){
		p1_input[i] = p1_input[i] ^ p2_input;
	}
        // no party knows

        // original version
        // for(int i = 0; i < 1000; ++i) revealOblivBool(buf + i, p1_input[i], 0);

        // improved version
	revealOblivBoolArray(buf, p1_input, 1000, 0);
}

int main(int argc,char* argv[])
{ 
  ProtocolDesc pd;

  const char* remote_host = (strcmp(argv[2],"--")==0?NULL:argv[2]);
  int i, party = (!remote_host?1:2);

  ocTestUtilTcpOrDie(&pd,remote_host,argv[1]);
  setCurrentParty(&pd,party);

  lap = wallClock();
  execYaoProtocol(&pd, gotest, NULL);
  fprintf(stderr,"Total time: %lf s\n",wallClock()-lap);
  cleanupProtocol(&pd);
  fprintf(stderr,"\n");
  return 0;
}

The original version:

OT time: 0.628520 s
Total time: 22.787672 s

The improved version:

OT time: 0.586414 s
Total time: 0.586716 s

@weikengchen
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weikengchen commented Jul 11, 2018

The example in #66 considers 12345 bools.
I can't help but do an experiment for 12345.

The original version:

OT time: 0.643174 s
Total time: 255.182363 s

The improved version:

OT time: 0.652290 s
Total time: 0.655129 s

For this specific and boring application (1-bit-input from the evaluator, a lot of XOR free gates, 12345-bit-output), the performance is improved by 390x.

The example changes part of the code in the post into the following:

#define MAXN 12345
void readBool(obliv bool* dest, int n, const bool* src,int party)
{
        OblivInputs specs[MAXN];
        int i;
        for(i=0;i<n;++i) setupOblivBool(specs+i,dest+i,src[i]);
        feedOblivInputs(specs,n,party);
}

void gotest(void* vargs)
{
        obliv bool p1_input[12345];
        obliv bool p2_input;

        bool buf[12345];
        readBool(p1_input, 12345, buf, 1);
        readBool(&p2_input, 1, buf, 2);

        fprintf(stderr,"OT time: %lf s\n",wallClock()-lap);

        for(int i = 0; i < 12345; i++){
                p1_input[i] = p1_input[i] ^ p2_input;
        }

        for(int i = 0; i < 12345; ++i) revealOblivBool(buf + i, p1_input[i], 0);
        //revealOblivBoolArray(buf, p1_input, 12345, 0);
}

@weikengchen weikengchen changed the title reduces the RTT for revealOblivBoolArray reduces the total RTT for revealOblivBoolArray Jul 11, 2018
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