Re: [gcj] about problem query of death Distributed GCJ 2017

2017-06-05 Thread Reynaldo Gil-Pons 
OK, I didn't think anyone would like to review code... This is the overall 
explanation: first each node tries to sum values on an interval, and send 
answer to node 0. When a node notices it's wrong, sends a different info to 
node 0. Then node 0 solves the rest of the problem. If node 0 was the one that 
failed, then node 1 solves the rest of the problem...


code (hope not too messy):


#include 
#include 
#include 
#include "query_of_death.h"
#include 
#include 
using namespace std;
// Receive(int)
// GetLL(int)
// Send(int)
// PutLL(int)
// MyNodeId()
// NumberOfNodes()
typedef long long i64;

const int MOD = 1 << 20;
int main() {
int n = GetLength();
int OFFSET = 1e6;
int node = MyNodeId();
int nnodes = NumberOfNodes();
assert(nnodes == 100);
int bad = -1;
int sum = 0;
srand(1);
for(int i = OFFSET * node ; i < min(n, OFFSET * (node + 1)) ; i++){

  int cur = GetValue(i);
  
  for(int j = 0; j < 33; j++){
  int tmp = GetValue(i);
  if(tmp != cur){
bad = i;
break;
}
}
if(bad != -1){
  break;
  }
sum += cur;

  }
  if(node == 0){
  
  if(bad != -1){
int ans = 0;
for(int i = 1; i < nnodes; i++){
Receive(i);
int cur = GetLL(i);
assert(cur >= 0);
ans += cur;
  }


PutLL(1, -bad - 1);
Send(1);
PutLL(1, ans);
Send(1);


}
else{
  int ans = 0;
  int badi = -1;
  for(int i = 1; i < nnodes; i++){
Receive(i);
int resp = GetLL(i);
if(resp < 0){
  badi = i;
  bad = -(resp + 1);
}else
  ans += resp;
  }
  assert(bad >= 0);
  assert(badi > 0);
  for(int i = badi * OFFSET; i < min(n, (badi + 1) * OFFSET); i++){
if(i == bad)continue;
//assert(i >= 0 and i < n);
int ii = GetValue(i);
ans += ii;
assert(GetValue(i) == ii);
}
//assert(bad >= 0 and bad < n);
ans += GetValue(bad);
cout << ans << endl;

  

  
  
  
  PutLL(1, 0);
  Send(1);
}

}
else{
  if(bad == -1){
PutLL(0, sum);
Send(0);
  }
  else{
PutLL(0, -bad - 1);
Send(0);
}
  
  
  }
  if(node == 1){
  Receive(0);
  int resp = GetLL(0);
  if(resp < 0){
  bad = -(resp + 1);
  Receive(0);
  int ans = GetLL(0);
  for(int i = 0; i < min(n, OFFSET); i++){
if(i == bad)continue;
int ii = GetValue(i);
ans += ii;
assert(GetValue(i) == ii);
}
//assert(bad >= 0 and bad < n);
ans += GetValue(bad);
cout << ans << endl;
}

}
  
  
  return 0;
}


On Monday, June 5, 2017 at 9:48:24 AM UTC-4, Luke Pebody wrote:
> If you are getting WA you have probably made a logical error somewhere. You 
> might have to post your code to know where.
> 
> 
> On 5 Jun 2017 2:35 p.m., "Reynaldo Gil-Pons"  wrote:
> There are 100 computers, 1 microsecond = 1e-6 seconds, so time for reading 
> each number 30 times is: 10 ^ 8 * (0.2 * 10^ -6) / 100 * 30 = 6 seconds, so a 
> little bit above, but I have tested and it passes the time limit, although 
> giving WA, as if the probability reasoning were incorrect...
> 
> 
> 
> 
> On Monday, June 5, 2017 at 12:59:48 AM UTC-4, Xiongqi ZHANG wrote:
> 
> > 10^8 * 0.2 microsecond = 20 seconds while Time Limit is 2 second.
> 
> >
> 
> >
> 
> > Reynaldo Gil-Pons 于2017年6月4日周日 下午7:39写道：
> 
> 
> > 0.2 millisecons to read a single bit? They say 0.2 microseconds. I get WA 
> > when doing what I explained, in 1.7 seconds...
> 
> >
> 
> >
> 
> >
> 
> > On Friday, June 2, 2017 at 12:09:47 PM UTC-4, Luke Pebody wrote:
> 
> >
> 
> > > I would think it can't work on the large case because it takes 0.2 
> > > milliseconds to read a single bit once and therefore 20 seconds to read 
> > > all 100M bits once and 10 minutes to read them 30 times.
> 
> >
> 
> > >
> 
> >
> 
> > >
> 
> >
> 
> > > On 2 Jun 2017 4:56 p.m., "Reynaldo Gil-Pons"  wrote:
> 
> >
> 
> > > In the analisis they explain a solution for the Small case, and point out 
> > > it cannot work for the Large one. When I calculate the probabilities of 
> > > getting WA using 30 checks for each position, and assuming 100 
> > > testcases (cant be more than this right?) I get less than 1 / 1000. But I 
> > > still get WA. I calculate an upper bound on the probability of failure as 
> > > 1 - (1 - 1 / (2 ** 30)) ** number_test_cases. Is there anything wrong?
> 
> >
> 
> > >
> 
> >
> 
> > >
> 
> >
> 
> > >
> 
> >
> 
> > > --
> 
> >

Re: [gcj] about problem query of death Distributed GCJ 2017

2017-06-05 Thread Luke Pebody 
If you are getting WA you have probably made a logical error somewhere. You
might have to post your code to know where.

On 5 Jun 2017 2:35 p.m., "Reynaldo Gil-Pons"  wrote:

There are 100 computers, 1 microsecond = 1e-6 seconds, so time for reading
each number 30 times is: 10 ^ 8 * (0.2 * 10^ -6) / 100 * 30 = 6 seconds, so
a little bit above, but I have tested and it passes the time limit,
although giving WA, as if the probability reasoning were incorrect...

On Monday, June 5, 2017 at 12:59:48 AM UTC-4, Xiongqi ZHANG wrote:
> 10^8 * 0.2 microsecond = 20 seconds while Time Limit is 2 second.
>
>
> Reynaldo Gil-Pons 于2017年6月4日周日 下午7:39写道：
> 0.2 millisecons to read a single bit? They say 0.2 microseconds. I get WA
when doing what I explained, in 1.7 seconds...
>
>
>
> On Friday, June 2, 2017 at 12:09:47 PM UTC-4, Luke Pebody wrote:
>
> > I would think it can't work on the large case because it takes 0.2
milliseconds to read a single bit once and therefore 20 seconds to read all
100M bits once and 10 minutes to read them 30 times.
>
> >
>
> >
>
> > On 2 Jun 2017 4:56 p.m., "Reynaldo Gil-Pons"  wrote:
>
> > In the analisis they explain a solution for the Small case, and point
out it cannot work for the Large one. When I calculate the probabilities of
getting WA using 30 checks for each position, and assuming 100
testcases (cant be more than this right?) I get less than 1 / 1000. But I
still get WA. I calculate an upper bound on the probability of failure as 1
- (1 - 1 / (2 ** 30)) ** number_test_cases. Is there anything wrong?
>
> >
>
> >
>
> >
>
> > --
>
> >
>
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> >
>
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msgid/google-code/11b46b16-82c3-4824-b319-626b05cf9e76%40googlegroups.com.
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> >
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Re: [gcj] about problem query of death Distributed GCJ 2017

2017-06-05 Thread Reynaldo Gil-Pons 
There are 100 computers, 1 microsecond = 1e-6 seconds, so time for reading each 
number 30 times is: 10 ^ 8 * (0.2 * 10^ -6) / 100 * 30 = 6 seconds, so a little 
bit above, but I have tested and it passes the time limit, although giving WA, 
as if the probability reasoning were incorrect... 

On Monday, June 5, 2017 at 12:59:48 AM UTC-4, Xiongqi ZHANG wrote:
> 10^8 * 0.2 microsecond = 20 seconds while Time Limit is 2 second.
> 
> 
> Reynaldo Gil-Pons 于2017年6月4日周日 下午7:39写道：
> 0.2 millisecons to read a single bit? They say 0.2 microseconds. I get WA 
> when doing what I explained, in 1.7 seconds...
> 
> 
> 
> On Friday, June 2, 2017 at 12:09:47 PM UTC-4, Luke Pebody wrote:
> 
> > I would think it can't work on the large case because it takes 0.2 
> > milliseconds to read a single bit once and therefore 20 seconds to read all 
> > 100M bits once and 10 minutes to read them 30 times.
> 
> >
> 
> >
> 
> > On 2 Jun 2017 4:56 p.m., "Reynaldo Gil-Pons"  wrote:
> 
> > In the analisis they explain a solution for the Small case, and point out 
> > it cannot work for the Large one. When I calculate the probabilities of 
> > getting WA using 30 checks for each position, and assuming 100 
> > testcases (cant be more than this right?) I get less than 1 / 1000. But I 
> > still get WA. I calculate an upper bound on the probability of failure as 1 
> > - (1 - 1 / (2 ** 30)) ** number_test_cases. Is there anything wrong?
> 
> >
> 
> >
> 
> >
> 
> > --
> 
> >
> 
> > You received this message because you are subscribed to the Google Groups 
> > "Google Code Jam" group.
> 
> >
> 
> > To unsubscribe from this group and stop receiving emails from it, send an 
> > email to google-code...@googlegroups.com.
> 
> >
> 
> > To post to this group, send email to googl...@googlegroups.com.
> 
> >
> 
> > To view this discussion on the web visit 
> > https://groups.google.com/d/msgid/google-code/11b46b16-82c3-4824-b319-626b05cf9e76%40googlegroups.com.
> 
> >
> 
> > For more options, visit https://groups.google.com/d/optout.
> 
> 
> 
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> 
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