Although there is some documentation related to the zone free function of Arrangement_2 module, it is not mentioned in any example files and the usage is not obvious.
Assuming that I have an arrangement of points and line segments based on CGAL::Arr_linear_traits_2, I want to print out all faces visited when walking along a given Segment_2. How can I do that?
You need to use the "assign" function:
void segment_intersect(Arrangement_2 &arr, Segment_2 &c)
{
std::vector<CGAL::Object> zone_elems;
Arrangement_2::Face_handle face;
CGAL::zone(arr, c, std::back_inserter(zone_elems));
for ( int i = 0; i < (int)zone_elems.size(); ++i )
{
if ( assign(face, zone_elems[i]) )
//print the face index...
}
}
The usage actually is quite obvious. To get all elements intersected, this code is enough:
void segment_intersect(Arrangement_2 &arr, Segment_2 &c)
{
std::vector<CGAL::Object> zone_elems;
CGAL::zone(arr, c, std::back_inserter(zone_elems));
}
I have yet to find out how to extract faces out of the vector.
Related
It is my first attempt to implement recursion with CUDA. The goal is to extract all the combinations from a set of chars "12345" using the power of CUDA to parallelize dynamically the task. Here is my kernel:
__device__ char route[31] = { "_________________________"};
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth) {
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
printf("%s\n", route);
int o = 0;
int newlen = 0;
for (int i = 0; i<6; ++i)
{
if (i != threadIdx.x)
{
route[i+x-o] = init[i];
newlen++;
}
else
{
o = 1;
}
}
Recursive<<<1,newlen>>>(depth + 1);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0);
}
The idea is to exclude 1 item (the item corresponding to the threadIdx) in each different thread. In each recursive call, using the variable depth, it works over a different base (variable x) on the route device variable.
I expect the kernel prompts something like:
2345_____________________
1345_____________________
1245_____________________
1234_____________________
2345_345_________________
2345_245_________________
2345_234_________________
2345_345__45_____________
2345_345__35_____________
2345_345__34_____________
..
2345_245__45_____________
..
But it prompts ...
·_____________
·_____________
·_____________
·_____________
·_____________
·2345
·2345
·2345
·2345
...
What I´m doing wrong?
What I´m doing wrong?
I may not articulate every problem with your code, but these items should get you a lot closer.
I recommend providing a complete example. In my view it is basically required by Stack Overflow, see item 1 here, note use of the word "must". Your example is missing any host code, including the original kernel call. It's only a few extra lines of code, why not include it? Sure, in this case, I can deduce what the call must have been, but why not just include it? Anyway, based on the output you indicated, it seems fairly evident the launch configuration of the host launch would have to be <<<1,1>>>.
This doesn't seem to be logical to me:
I expect the kernel prompts something like:
2345_____________________
The very first thing your kernel does is print out the route variable, before making any changes to it, so I would expect _____________________. However we can "fix" this by moving the printout to the end of the kernel.
You may be confused about what a __device__ variable is. It is a global variable, and there is only one copy of it. Therefore, when you modify it in your kernel code, every thread, in every kernel, is attempting to modify the same global variable, at the same time. That cannot possibly have orderly results, in any thread-parallel environment. I chose to "fix" this by making a local copy for each thread to work on.
You have an off-by-1 error, as well as an extent error in this loop:
for (int i = 0; i<6; ++i)
The off-by-1 error is due to the fact that you are iterating over 6 possible items (that is, i can reach a value of 5) but there are only 5 items in your init variable (the 6th item being a null terminator. The correct indexing starts out over 0-4 (with one of those being skipped). On subsequent iteration depths, its necessary to reduce this indexing extent by 1. Note that I've chosen to fix the first error here by increasing the length of init. There are other ways to fix, of course. My method inserts an extra _ between depths in the result.
You assume that at each iteration depth, the correct choice of items is the same, and in the same order, i.e. init. However this is not the case. At each depth, the choices of items must be selected not from the unchanging init variable, but from the choices passed from previous depth. Therefore we need a local, per-thread copy of init also.
A few other comments about CUDA Dynamic Parallelism (CDP). When passing pointers to data from one kernel scope to a child scope, local space pointers cannot be used. Therefore I allocate for the local copy of route from the heap, so it can be passed to child kernels. init can be deduced from route, so we can use an ordinary local variable for myinit.
You're going to quickly hit some dynamic parallelism (and perhaps memory) limits here if you continue this. I believe the total number of kernel launches for this is 5^5, which is 3125 (I'm doing this quickly, I may be mistaken). CDP has a pending launch limit of 2000 kernels by default. We're not hitting this here according to what I see, but you'll run into that sooner or later if you increase the depth or width of this operation. Furthermore, in-kernel allocations from the device heap are by default limited to 8KB. I don't seem to be hitting that limit, but probably I am, so my design should probably be modified to fix that.
Finally, in-kernel printf output is limited to the size of a particular buffer. If this technique is not already hitting that limit, it will soon if you increase the width or depth.
Here is a worked example, attempting to address the various items above. I'm not claiming it is defect free, but I think the output is closer to your expectations. Note that due to character limits on SO answers, I've truncated/excerpted some of the output.
$ cat t1639.cu
#include <stdio.h>
__device__ char route[31] = { "_________________________"};
__device__ char init[7] = { "12345_" };
__global__ void Recursive(int depth, const char *oroute) {
char *nroute = (char *)malloc(31);
char myinit[7];
if (depth == 0) memcpy(myinit, init, 6);
else memcpy(myinit, oroute+(depth-1)*6, 6);
myinit[6] = 0;
if (nroute == NULL) {printf("oops\n"); return;}
memcpy(nroute, oroute, 30);
nroute[30] = 0;
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
//printf("%s\n", nroute);
int o = 0;
int newlen = 0;
for (int i = 0; i<(6-depth); ++i)
{
if (i != threadIdx.x)
{
nroute[i+x-o] = myinit[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", nroute);
Recursive<<<1,newlen>>>(depth + 1, nroute);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0, route);
}
int main(){
RecursiveCount<<<1,1>>>();
cudaDeviceSynchronize();
}
$ nvcc -o t1639 t1639.cu -rdc=true -lcudadevrt -arch=sm_70
$ cuda-memcheck ./t1639
========= CUDA-MEMCHECK
2345_____________________
1345_____________________
1245_____________________
1235_____________________
1234_____________________
2345__345________________
2345__245________________
2345__235________________
2345__234________________
2345__2345_______________
2345__345___45___________
2345__345___35___________
2345__345___34___________
2345__345___345__________
2345__345___45____5______
2345__345___45____4______
2345__345___45____45_____
2345__345___45____5______
2345__345___45____5_____5
2345__345___45____4______
2345__345___45____4_____4
2345__345___45____45____5
2345__345___45____45____4
2345__345___35____5______
2345__345___35____3______
2345__345___35____35_____
2345__345___35____5______
2345__345___35____5_____5
2345__345___35____3______
2345__345___35____3_____3
2345__345___35____35____5
2345__345___35____35____3
2345__345___34____4______
2345__345___34____3______
2345__345___34____34_____
2345__345___34____4______
2345__345___34____4_____4
2345__345___34____3______
2345__345___34____3_____3
2345__345___34____34____4
2345__345___34____34____3
2345__345___345___45_____
2345__345___345___35_____
2345__345___345___34_____
2345__345___345___45____5
2345__345___345___45____4
2345__345___345___35____5
2345__345___345___35____3
2345__345___345___34____4
2345__345___345___34____3
2345__245___45___________
2345__245___25___________
2345__245___24___________
2345__245___245__________
2345__245___45____5______
2345__245___45____4______
2345__245___45____45_____
2345__245___45____5______
2345__245___45____5_____5
2345__245___45____4______
2345__245___45____4_____4
2345__245___45____45____5
2345__245___45____45____4
2345__245___25____5______
2345__245___25____2______
2345__245___25____25_____
2345__245___25____5______
2345__245___25____5_____5
2345__245___25____2______
2345__245___25____2_____2
2345__245___25____25____5
2345__245___25____25____2
2345__245___24____4______
2345__245___24____2______
2345__245___24____24_____
2345__245___24____4______
2345__245___24____4_____4
2345__245___24____2______
2345__245___24____2_____2
2345__245___24____24____4
2345__245___24____24____2
2345__245___245___45_____
2345__245___245___25_____
2345__245___245___24_____
2345__245___245___45____5
2345__245___245___45____4
2345__245___245___25____5
2345__245___245___25____2
2345__245___245___24____4
2345__245___245___24____2
2345__235___35___________
2345__235___25___________
2345__235___23___________
2345__235___235__________
2345__235___35____5______
2345__235___35____3______
2345__235___35____35_____
2345__235___35____5______
2345__235___35____5_____5
2345__235___35____3______
2345__235___35____3_____3
2345__235___35____35____5
2345__235___35____35____3
2345__235___25____5______
2345__235___25____2______
2345__235___25____25_____
2345__235___25____5______
2345__235___25____5_____5
2345__235___25____2______
2345__235___25____2_____2
2345__235___25____25____5
2345__235___25____25____2
2345__235___23____3______
2345__235___23____2______
2345__235___23____23_____
2345__235___23____3______
2345__235___23____3_____3
2345__235___23____2______
2345__235___23____2_____2
2345__235___23____23____3
2345__235___23____23____2
2345__235___235___35_____
2345__235___235___25_____
2345__235___235___23_____
2345__235___235___35____5
2345__235___235___35____3
2345__235___235___25____5
2345__235___235___25____2
2345__235___235___23____3
2345__235___235___23____2
2345__234___34___________
2345__234___24___________
2345__234___23___________
2345__234___234__________
2345__234___34____4______
2345__234___34____3______
2345__234___34____34_____
2345__234___34____4______
2345__234___34____4_____4
2345__234___34____3______
2345__234___34____3_____3
2345__234___34____34____4
2345__234___34____34____3
2345__234___24____4______
2345__234___24____2______
2345__234___24____24_____
2345__234___24____4______
2345__234___24____4_____4
2345__234___24____2______
2345__234___24____2_____2
2345__234___24____24____4
2345__234___24____24____2
2345__234___23____3______
2345__234___23____2______
2345__234___23____23_____
2345__234___23____3______
2345__234___23____3_____3
2345__234___23____2______
2345__234___23____2_____2
2345__234___23____23____3
2345__234___23____23____2
2345__234___234___34_____
2345__234___234___24_____
2345__234___234___23_____
2345__234___234___34____4
2345__234___234___34____3
2345__234___234___24____4
2345__234___234___24____2
2345__234___234___23____3
2345__234___234___23____2
2345__2345__345__________
2345__2345__245__________
2345__2345__235__________
2345__2345__234__________
2345__2345__345___45_____
2345__2345__345___35_____
2345__2345__345___34_____
2345__2345__345___45____5
2345__2345__345___45____4
2345__2345__345___35____5
2345__2345__345___35____3
2345__2345__345___34____4
2345__2345__345___34____3
2345__2345__245___45_____
2345__2345__245___25_____
2345__2345__245___24_____
2345__2345__245___45____5
2345__2345__245___45____4
2345__2345__245___25____5
2345__2345__245___25____2
2345__2345__245___24____4
2345__2345__245___24____2
2345__2345__235___35_____
2345__2345__235___25_____
2345__2345__235___23_____
2345__2345__235___35____5
2345__2345__235___35____3
2345__2345__235___25____5
2345__2345__235___25____2
2345__2345__235___23____3
2345__2345__235___23____2
2345__2345__234___34_____
2345__2345__234___24_____
2345__2345__234___23_____
2345__2345__234___34____4
2345__2345__234___34____3
2345__2345__234___24____4
2345__2345__234___24____2
2345__2345__234___23____3
2345__2345__234___23____2
1345__345________________
1345__145________________
1345__135________________
1345__134________________
1345__1345_______________
1345__345___45___________
1345__345___35___________
1345__345___34___________
1345__345___345__________
1345__345___45____5______
1345__345___45____4______
1345__345___45____45_____
1345__345___45____5______
1345__345___45____5_____5
1345__345___45____4______
1345__345___45____4_____4
1345__345___45____45____5
1345__345___45____45____4
1345__345___35____5______
1345__345___35____3______
1345__345___35____35_____
1345__345___35____5______
1345__345___35____5_____5
1345__345___35____3______
1345__345___35____3_____3
1345__345___35____35____5
1345__345___35____35____3
1345__345___34____4______
1345__345___34____3______
1345__345___34____34_____
1345__345___34____4______
1345__345___34____4_____4
1345__345___34____3______
1345__345___34____3_____3
1345__345___34____34____4
1345__345___34____34____3
1345__345___345___45_____
1345__345___345___35_____
1345__345___345___34_____
1345__345___345___45____5
1345__345___345___45____4
1345__345___345___35____5
1345__345___345___35____3
1345__345___345___34____4
1345__345___345___34____3
1345__145___45___________
1345__145___15___________
1345__145___14___________
1345__145___145__________
1345__145___45____5______
1345__145___45____4______
1345__145___45____45_____
1345__145___45____5______
1345__145___45____5_____5
1345__145___45____4______
1345__145___45____4_____4
1345__145___45____45____5
1345__145___45____45____4
1345__145___15____5______
1345__145___15____1______
1345__145___15____15_____
1345__145___15____5______
1345__145___15____5_____5
1345__145___15____1______
1345__145___15____1_____1
1345__145___15____15____5
1345__145___15____15____1
1345__145___14____4______
1345__145___14____1______
1345__145___14____14_____
1345__145___14____4______
1345__145___14____4_____4
1345__145___14____1______
1345__145___14____1_____1
1345__145___14____14____4
1345__145___14____14____1
1345__145___145___45_____
1345__145___145___15_____
1345__145___145___14_____
1345__145___145___45____5
1345__145___145___45____4
1345__145___145___15____5
1345__145___145___15____1
1345__145___145___14____4
1345__145___145___14____1
1345__135___35___________
1345__135___15___________
1345__135___13___________
1345__135___135__________
1345__135___35____5______
1345__135___35____3______
1345__135___35____35_____
1345__135___35____5______
1345__135___35____5_____5
1345__135___35____3______
1345__135___35____3_____3
1345__135___35____35____5
1345__135___35____35____3
1345__135___15____5______
1345__135___15____1______
1345__135___15____15_____
1345__135___15____5______
1345__135___15____5_____5
1345__135___15____1______
1345__135___15____1_____1
1345__135___15____15____5
1345__135___15____15____1
1345__135___13____3______
1345__135___13____1______
1345__135___13____13_____
1345__135___13____3______
1345__135___13____3_____3
1345__135___13____1______
1345__135___13____1_____1
1345__135___13____13____3
1345__135___13____13____1
1345__135___135___35_____
1345__135___135___15_____
1345__135___135___13_____
1345__135___135___35____5
1345__135___135___35____3
1345__135___135___15____5
1345__135___135___15____1
1345__135___135___13____3
1345__135___135___13____1
1345__134___34___________
1345__134___14___________
1345__134___13___________
1345__134___134__________
1345__134___34____4______
1345__134___34____3______
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1345__134___34____4______
1345__134___34____4_____4
1345__134___34____3______
1345__134___34____3_____3
1345__134___34____34____4
1345__134___34____34____3
1345__134___14____4______
1345__134___14____1______
1345__134___14____14_____
1345__134___14____4______
1345__134___14____4_____4
1345__134___14____1______
1345__134___14____1_____1
1345__134___14____14____4
1345__134___14____14____1
1345__134___13____3______
1345__134___13____1______
1345__134___13____13_____
1345__134___13____3______
1345__134___13____3_____3
1345__134___13____1______
1345__134___13____1_____1
1345__134___13____13____3
1345__134___13____13____1
1345__134___134___34_____
1345__134___134___14_____
1345__134___134___13_____
1345__134___134___34____4
1345__134___134___34____3
1345__134___134___14____4
1345__134___134___14____1
1345__134___134___13____3
1345__134___134___13____1
1345__1345__345__________
1345__1345__145__________
1345__1345__135__________
1345__1345__134__________
1345__1345__345___45_____
1345__1345__345___35_____
1345__1345__345___34_____
1345__1345__345___45____5
1345__1345__345___45____4
1345__1345__345___35____5
1345__1345__345___35____3
1345__1345__345___34____4
1345__1345__345___34____3
1345__1345__145___45_____
1345__1345__145___15_____
1345__1345__145___14_____
1345__1345__145___45____5
1345__1345__145___45____4
1345__1345__145___15____5
1345__1345__145___15____1
1345__1345__145___14____4
1345__1345__145___14____1
1345__1345__135___35_____
1345__1345__135___15_____
1345__1345__135___13_____
1345__1345__135___35____5
1345__1345__135___35____3
1345__1345__135___15____5
1345__1345__135___15____1
1345__1345__135___13____3
1345__1345__135___13____1
1345__1345__134___34_____
1345__1345__134___14_____
1345__1345__134___13_____
1345__1345__134___34____4
1345__1345__134___34____3
1345__1345__134___14____4
1345__1345__134___14____1
1345__1345__134___13____3
1345__1345__134___13____1
1245__245________________
1245__145________________
1245__125________________
1245__124________________
1245__1245_______________
1245__245___45___________
1245__245___25___________
1245__245___24___________
1245__245___245__________
1245__245___45____5______
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1245__245___45____5______
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1245__245___45____45____5
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1245__245___25____5______
1245__245___25____2______
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1245__245___25____2______
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1245__245___25____25____5
1245__245___25____25____2
1245__245___24____4______
1245__245___24____2______
1245__245___24____24_____
1245__245___24____4______
1245__245___24____4_____4
1245__245___24____2______
1245__245___24____2_____2
1245__245___24____24____4
1245__245___24____24____2
1245__245___245___45_____
1245__245___245___25_____
1245__245___245___24_____
1245__245___245___45____5
1245__245___245___45____4
1245__245___245___25____5
1245__245___245___25____2
1245__245___245___24____4
1245__245___245___24____2
1245__145___45___________
1245__145___15___________
1245__145___14___________
1245__145___145__________
1245__145___45____5______
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1245__145___45____45_____
1245__145___45____5______
1245__145___45____5_____5
1245__145___45____4______
...
1235__1235__235___25_____
1235__1235__235___23_____
1235__1235__235___35____5
1235__1235__235___35____3
1235__1235__235___25____5
1235__1235__235___25____2
1235__1235__235___23____3
1235__1235__235___23____2
1235__1235__135___35_____
1235__1235__135___15_____
1235__1235__135___13_____
1235__1235__135___35____5
1235__1235__135___35____3
1235__1235__135___15____5
1235__1235__135___15____1
1235__1235__135___13____3
1235__1235__135___13____1
1235__1235__125___25_____
1235__1235__125___15_____
1235__1235__125___12_____
1235__1235__125___25____5
1235__1235__125___25____2
1235__1235__125___15____5
1235__1235__125___15____1
1235__1235__125___12____2
1235__1235__125___12____1
1235__1235__123___23_____
1235__1235__123___13_____
1235__1235__123___12_____
1235__1235__123___23____3
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========= ERROR SUMMARY: 0 errors
$
The answer given by Robert Crovella is correct at the 5th point, the mistake was in the using of init in every recursive call, but I want to clarify something that can be useful for other beginners with CUDA.
I used this variable because when I tried to launch a child kernel passing a local variable I always got the exception: Error: a pointer to local memory cannot be passed to a launch as an argument.
As I´m C# expert developer I´m not used to using pointers (Ref does the low-level-work for that) so I thought there was no way to do it in CUDA/c programming.
As Robert shows in its code it is possible copying the pointer with memalloc for using it as a referable argument.
Here is a kernel simplified as an example of deep recursion.
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth, const char* route) {
// up to depth 6
if (depth == 5) return;
//declaration for a referable argument (point 6)
char* newroute = (char*)malloc(6);
memcpy(newroute, route, 5);
int o = 0;
int newlen = 0;
for (int i = 0; i < (6 - depth); ++i)
{
if (i != threadIdx.x)
{
newroute[i - o] = route[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", newroute);
Recursive <<<1, newlen>>>(depth + 1, newroute);
}
__global__ void RecursiveCount() {
Recursive <<<1, 5>>>(0, init);
}
I don't add the main call because I´m using ManagedCUDA for C# but as Robert says it can be figured-out how the call RecursiveCount is.
About ending arrays of char with /0 ... sorry but I don't know exactly what is the benefit; this code works fine without them.
I have a problem that I am unwilling to believe hasn't been solved before in Sage.
Given a pair of integers (d,n) as input, I'd like to receive a list (or set, or whatever) of all nondecreasing sequences of length d all of whose entries are no greater than n.
Similarly, I'd like another function which returns all strictly increasing sequences of length d whose entries are no greater than n.
For example, for d = 2 n=3, I'd receive the output:
[[1,2], [1,3], [2,3]]
or
[[1,1], [1,2], [1,3], [2,2], [2,3], [3,3]]
depending on whether I'm using increasing or nondecreasing.
Does anyone know of such a function?
Edit Of course, if there is such a method for nonincreasing or decreasing sequences, I can modify that to fit my purposes. Just something to iterate over sequences
I needed this algorithm too and I finally managed to write one today. I will share the code here, but I only started to learn coding last week, so it is not pretty.
Idea Input=(r,d). Step 1) Create a class "ListAndPosition" that has a list L of arrays Integer[r+1]'s, and an integer q between 0 and r. Step 2) Create a method that receives a ListAndPosition (L,q) and screens sequentially the arrays in L checking if the integer at position q is less than the one at position q+1, if so, it adds a new array at the bottom of the list with that entry ++. When done, the Method calls itself again with the new list and q-1 as input.
The code for Step 1)
import java.util.ArrayList;
public class ListAndPosition {
public static Integer r=5;
public final ArrayList<Integer[]> L;
public int q;
public ListAndPosition(ArrayList<Integer[]> L, int q) {
this.L = L;
this.q = q;
}
public ArrayList<Integer[]> getList(){
return L;
}
public int getPosition() {
return q;
}
public void decreasePosition() {
q--;
}
public void showList() {
for(int i=0;i<L.size();i++){
for(int j=0; j<r+1 ; j++){
System.out.print(""+L.get(i)[j]);
}
System.out.println("");
}
}
}
The code for Step 2)
import java.util.ArrayList;
public class NonDecreasingSeqs {
public static Integer r=5;
public static Integer d=3;
public static void main(String[] args) {
//Creating the first array
Integer[] firstArray;
firstArray = new Integer[r+1];
for(int i=0;i<r;i++){
firstArray[i] = 0;
}
firstArray[r] = d;
//Creating the starting listAndDim
ArrayList<Integer[]> L = new ArrayList<Integer[]>();
L.add(firstArray);
ListAndPosition Lq = new ListAndPosition(L,r-1);
System.out.println(""+nonDecSeqs(Lq).size());
}
public static ArrayList<Integer[]> nonDecSeqs(ListAndPosition Lq){
int iterations = r-1-Lq.getPosition();
System.out.println("How many arrays in the list after "+iterations+" iterations? "+Lq.getList().size());
System.out.print("Should we stop the iteration?");
if(0<Lq.getPosition()){
System.out.println(" No, position = "+Lq.getPosition());
for(int i=0;i<Lq.getList().size();i++){
//Showing particular array
System.out.println("Array of L #"+i+":");
for(int j=0;j<r+1;j++){
System.out.print(""+Lq.getList().get(i)[j]);
}
System.out.print("\nCan it be modified at position "+Lq.getPosition()+"?");
if(Lq.getList().get(i)[Lq.getPosition()]<Lq.getList().get(i)[Lq.getPosition()+1]){
System.out.println(" Yes, "+Lq.getList().get(i)[Lq.getPosition()]+"<"+Lq.getList().get(i)[Lq.getPosition()+1]);
{
Integer[] tempArray = new Integer[r+1];
for(int j=0;j<r+1;j++){
if(j==Lq.getPosition()){
tempArray[j] = new Integer(Lq.getList().get(i)[j])+1;
}
else{
tempArray[j] = new Integer(Lq.getList().get(i)[j]);
}
}
Lq.getList().add(tempArray);
}
System.out.println("New list");Lq.showList();
}
else{
System.out.println(" No, "+Lq.getList().get(i)[Lq.getPosition()]+"="+Lq.getList().get(i)[Lq.getPosition()+1]);
}
}
System.out.print("Old position = "+Lq.getPosition());
Lq.decreasePosition();
System.out.println(", new position = "+Lq.getPosition());
nonDecSeqs(Lq);
}
else{
System.out.println(" Yes, position = "+Lq.getPosition());
}
return Lq.getList();
}
}
Remark: I needed my sequences to start at 0 and end at d.
This is probably not a very good answer to your question. But you could, in principle, use Partitions and the max_slope=-1 argument. Messing around with filtering lists of IntegerVectors sounds equally inefficient and depressing for other reasons.
If this has a canonical name, it might be in the list of sage-combinat functionality, and there is even a base class you could perhaps use for integer lists, which is basically what you are asking about. Maybe you could actually get what you want using IntegerListsLex? Hope this proves helpful.
This question can be solved by using the class "UnorderedTuples" described here:
http://doc.sagemath.org/html/en/reference/combinat/sage/combinat/tuple.html
To return all all nondecreasing sequences with entries between 0 and n-1 of length d, you may type:
UnorderedTuples(range(n),d)
This returns the nondecreasing sequence as a list. I needed an immutable object (because the sequences would become keys of a dictionary). So I used the "tuple" method to turn the lists into tuples:
immutables = []
for s in UnorderedTuples(range(n),d):
immutables.append(tuple(s))
return immutables
And I also wrote a method which picks out only the increasing sequences:
def isIncreasing(list):
for i in range(len(list) - 1):
if list[i] >= list[i+1]:
return false
return true
The method that returns only strictly increasing sequences would look like
immutables = []
for s in UnorderedTuples(range(n),d):
if isIncreasing(s):
immutables.append(tuple(s))
return immutables
I use following dijkstra implementation to calculate all pairs shortest paths in an undirected graph. After calling calculateAllPaths(), dist[i][j] contains shortest path length between i and j (or Integer.MAX_VALUE if no such path available).
The problem is that some vertexes of my graph are removing dynamically and I should recalculate all paths from scratch to update dist matrix. I'm seeking for a solution to optimize update speed by avoiding unnecessary calculations when a vertex removes from my graph. I already search for solution and I now there is some algorithms such as LPA* to do this, but they seem very complicated and I guess a simpler solution may solve my problem.
public static void calculateAllPaths()
{
for(int j=graph.length/2+graph.length%2;j>=0;j--)
{
calculateAllPathsFromSource(j);
}
}
public static void calculateAllPathsFromSource(int s)
{
final boolean visited[] = new boolean[graph.length];
for (int i=0; i<dist.length; i++)
{
if(i == s)
{
continue;
}
//visit next node
int next = -1;
int minDist = Integer.MAX_VALUE;
for (int j=0; j<dist[s].length; j++)
{
if (!visited[j] && dist[s][j] < minDist)
{
next = j;
minDist = dist[s][j];
}
}
if(next == -1)
{
continue;
}
visited[next] = true;
for(int v=0;v<graph.length;v++)
{
if(v == next || graph[next][v] == -1)
{
continue;
}
int md = dist[s][next] + graph[next][v];
if(md < dist[s][v])
{
dist[s][v] = dist[v][s] = md;
}
}
}
}
If you know that vertices are only being removed dynamically, then instead of just storing the best path matrix dist[i][j], you could also store the permutation of each such path. Say, instead of dist[i][j] you make a custom class myBestPathInfo, and the array of an instance of this, say myBestPathInfo[i][j], contain members best distance as well as permutation of the best path. Preferably, the best path permutation is described as an ordered set of some vertex objects, where the latter are of reference type and unique for each vertex (however used in several myBestPathInfo instances). Such objects could include a boolean property isActive (true/false).
Whenever a vertex is removed, you traverse through the best path permutations for each vertex-vertex pair, to make sure no vertex has been deactivated. Finally, only for broken paths (deactivated vertices) do you re-run Dijkstra's algorithm.
Another solution would be to solve the shortest path for all pairs using linear programming (LP) techniques. A removed vertex can be easily implemented as an additional constraint in your program (e.g. flow in <=0 and and flow out of vertex <= 0*), after which the re-solving of the shortest path LP:s can use the previous optimal solution as a feasible basic feasible solution (BFS) in the dual LPs. This property holds since adding a constraint in the primal LP is equivalent to an additional variable in the dual; hence, previously optimal primal BFS will be feasible in dual after additional constraints. (on-the-fly starting on simplex solver for LPs).
Currently the only way I know how to do input in D is with the scanf() function. But god damn it's ugly. You would think that since it's an upgrade from C that they would have fixed that.
I'm looking for a way to do it with a single argument. Currently you have to do:
int foo = 0;
scanf("%i", &foo);
writeln("%i", foo);
But it would look a lot cleaner with a single argument. Something like:
int foo = 0;
scanf(foo);
writeln(foo);
Thanks.
readf("%d", &foo); allows working with std.stdio.File rather than C FILE*
foo = readln().strip().to!int();
For reading entire files with lines formatted in the same way:
int[] numbers = slurp!int("filename", "%d");
There's a really cool user-input module here:
https://github.com/Abscissa/scriptlike/blob/master/src/scriptlike/interact.d
Example code:
if (userInput!bool("Do you want to continue?"))
{
auto outputFolder = pathLocation("Where you do want to place the output?");
auto color = menu!string("What color would you like to use?", ["Blue", "Green"]);
}
auto num = require!(int, "a > 0 && a <= 10")("Enter a number from 1 to 10");
The above answers are great. I just want to add my 2 cents.
I often have the following simple function lying around:
T read(T)()
{
T obj;
readf(" %s", &obj);
return obj;
}
It's generic and pretty handy - it swallows any white space and reads any type you ask. You can use it like this:
auto number = read!int;
auto floating_number = read!float;
// etc.
While I understand that we can detect cycles with the DFS algorithm by detecting back-edges http://cs.wellesley.edu/~cs231/fall01/dfs.pdf. I am not being able to figure out how to output the nodes in the cycle in an efficient and "clean" manner while following the above said method.
Would be gratfeull for some help
Thanks
This is how i did it in my own implementation. It deviates a little bit in the naming conventions from the one used in your PDF but it should be obvious what it does.
All m_* variables are vectors, except m_topoOrder and m_cycle which are stacks.
The nodes of the cycle will be in m_cycle.
The m_onStack keeps track of nodes which are on the recursive call stack.
For a complete description i suggest the book Algorithms by Robert Sedgewick.
void QxDigraph::dfs(int v)
{
m_marked[v] = true;
m_onStack[v] = true;
foreach(int w, m_adj[v]) {
if(hasCycle()) return;
else if(!m_marked[w])
{
m_edgeTo[w] = v;
dfs(w);
}
else if(m_onStack[w])
{
m_cycle.clear();
for(int x=v; x!=w; x = m_edgeTo[x])
m_cycle.push(x);
m_cycle.push(w);
m_cycle.push(v);
}
}
m_onStack[v] = false;
m_topoOrder.push(v);
}