I use the boost::hana to_map function to remove duplicates from boost::hana tuple of types. See it at the compiler explorer. The code works very well but compiles very long (~10s). I wonder if there exist a faster solution that is compatible with boost::hana tuple.
#include <boost/hana/map.hpp>
#include <boost/hana/pair.hpp>
#include <boost/hana/type.hpp>
#include <boost/hana/basic_tuple.hpp>
#include <boost/hana/size.hpp>
using namespace boost::hana;
constexpr auto to_type_pair = [](auto x) { return make_pair(typeid_(x), x); };
template <class Tuple>
constexpr auto remove_duplicate_types(Tuple tuple)
{
return values(to_map(transform(tuple, to_type_pair)));
}
int main(){
auto tuple = make_basic_tuple(
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30
, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40
, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50
, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60
, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70
);
auto noDuplicatesTuple = remove_duplicate_types(tuple);
// Should return 1 since there is only one distinct type in the tuple
return size(noDuplicatesTuple);
}
I haven't run any benchmarks, but your example does not appear to take 10 seconds on Compiler Explorer. However, I can explain why it is a relatively slow solution, and suggest an alternative that assumes you are only interested getting a unique list of types and not retaining any run-time information in your result.
Creating large tuples and/or instantiating function templates that have large tuples in their prototypes are expensive compile-time operations.
Just your call to transform instantiates a lambda for each element which in turn instantiates pair. The input/output of this call are both large tuples.
The call to to_map makes an empty map and recursively calls insert for each element each time making a new map, but in this simple case the intermediate result will always be hana::map<int>. I'm willing to bet that this is exploding your compile-times if your actual use case is non-trivial. (It was certainly an issue when we were implementing hana::map so we made hana::make_map avoid this since it has all of its inputs up front).
All of this, and there is a significant penalty for these large function types being used in run-time code. You might notice a difference if you wrapped the operations in decltype and only used the resulting type.
Alternatively, using raw template metaprogramming can sometimes yield performance results over function template based metaprogramming. Here is an example for your use case:
#include <boost/hana/basic_tuple.hpp>
#include <boost/mp11/algorithm.hpp>
namespace hana = boost::hana;
using namespace boost::mp11;
int main() {
auto tuple = hana::make_basic_tuple(
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30
, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40
, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50
, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60
, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70
);
hana::basic_tuple<int> no_dups = mp_unique<std::decay_t<decltype(tuple)>>{};
}
https://godbolt.org/z/EnTWf6
The following program prints out a shuffled deck of cards (as integers):
#include <array>
#include <algorithm>
#include <random>
#include <iostream>
typedef unsigned int card;
typedef std::array<card, 52> deck;
auto shuffled_deck(){
deck d = {};
std::iota(d.begin(), d.end(), 0);
std::shuffle(d.begin(), d.end(), std::default_random_engine());
return d;
}
int main(){
for(auto& i: shuffled_deck()) std::cout << i << ", ";
}
Compiled with g++ -std=c++17 the program runs and prints:
18, 34, 27, 46, 11, 3, 12, 19, 33, 21, 41, 13, 36, 49, 40, 22, 8, 9, 28, 2, 6, 30, 50, 24, 37, 32, 35, 4, 15, 45, 47, 43, 14, 44, 20, 23, 29, 7, 31, 51, 26, 10, 42, 48, 0, 38, 5, 16, 17, 1, 25, 39,
This is great, but intuition tells me that this deck could be created at compile-time, so I make the shuffled_deck method constexpr
constexpr auto shuffled_deck(){
deck d = {};
std::iota(d.begin(), d.end(), 0); // Error! Iota isn't constexpr!
std::shuffle(d.begin(), d.end(), std::default_random_engine());
return d;
}
Compiling with g++ -std=c++17 gives you compilation error saying that std::iota is not constexpr. My question is why? Surely std::iota is determinable at compile-time. Is the standard library just lagging behind on this feature?
This should be proposed to be added to the standard. As it stands now it just isn't.
I wrote a program in MSVS 2015, but I need to run it in MSVS 2013.
I get the error
"Error 1 error C2661: 'std::vector>::vector' :
no overloaded function takes 21
arguments \vmwfil04\students$\1302273\visual studio
2013\projects\dartsc++2013\dartsc++2013\gui.h 22 1"
This problem is affecting all of my vectors I've created before runtime.
What could be causing this?
offending code:
vector<int> Double{ 0, 40, 2, 36, 8, 26, 12, 20, 30, 4, 34, 6, 38, 14, 32, 16, 22, 28, 18, 24, 10 };
vector<int> Normal{ 0, 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12, 5 };
vector<int> Treble{ 0, 60, 3, 54, 12, 39, 18, 30, 45, 6, 39, 9, 57, 21, 48, 24, 33, 42, 27, 36, 15 };
vector<int> Bull { 0, 25, 50};
Support for these list-initialisers was new in VS 2015. It's not present in VS 2013. So you can't do it.
You'll have to take the old-fashioned, C++03 approach instead.
I believe this is a bug in Visual Studio 2013 as it does support list initializers (2013 specific documentation of the functionality). Try enclosing the brackets in a parenthesis per this answer.
e.g. vector<int> Double({ 0, 40, 2, 36, 8, 26, 12, 20, 30, 4, 34, 6, 38, 14, 32, 16, 22, 28, 18, 24, 10 });
I'm trying to implement simplified DES for learning purposes in python, but I am having trouble figuring out how to do the permutations based on a "schedule." Essentially, I have a tuple with the appropriate permutations, and I need to bit shift to the correct location.
For example, using a key:
K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001
Would move the 57st bit to the first bit spot, 49th bit to the second bit spot, etc...
K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111
Current code:
def keyGen(key):
PC1table = (57, 49, 41, 33, 25, 17, 9,
1, 58, 50, 42, 34, 26, 18,
10, 2, 59, 51, 43, 35, 27,
19, 11, 3, 60, 52, 44, 36,
63, 55, 47, 39, 31, 23, 15,
7, 62, 54, 46, 38, 30, 22,
14, 6, 61, 53, 45, 37, 29,
21, 13, 5, 28, 20, 12, 4)
keyBinary = bin(int(key, 16))[2:].zfill(64)
print keyBinary
permute(PC1table, keyBinary)
def permute(permutation, permuteInput):
elements = list(enumerate(permutation))
for bit in permuteInput:
***magic bitshifting goes here***
keyGen("133457799BBCDFF1")
The logic I thought would work was to enumerate the tuple of permutations, and for each bit of my old key, look in the enumeration to find the index corresponding the the bit, and bit shift the appropriate number of times, but I just can't figure out how to go about doing this. It may be that I am approaching the problem from the wrong angle, but any guidance would be greatly appreciated!
Ok, I ended up figuring a way to make this work, although this probably isn't the most efficient way...
prior to calling the function, turn the binary number into a list:
keyBinary = bin(int(key, 16))[2:].zfill(64)
keyBinary = [int(i) for i in keyBinary]
Kplus = permute(PC1table, keyBinary)
def permute(mapping, permuteInput):
permuteOutput = []
for i in range(len(mapping)):
permuteOutput.append(permuteInput[mapping[i % 56] - 1])
return permuteOutput
if anyone has a better way of tackling this, I'd love to see your solutions!
I have a question which can be divided into two subquestions.
I have created a table the code of which is given below.
Problem 1.
xstep = 1;
xmaximum = 6;
numberofxnodes = 6;
numberofynodes = 3;
numberofzlayers = 3;
maximumgridnodes = numberofxnodes*numberofynodes
mnodes = numberofxnodes*numberofynodes*numberofzlayers;
orginaltable =
Table[{i,
node2 = i + xstep, node3 = node2 + xmaximum,
node4 = node3 - xstep,node5 = i + maximumgridnodes,
node6 = node5 + xstep,node7 = node6 + xmaximum,
node8 = node7 - xstep},
{i, 1, mnodes}]
If I run this I will get my original table. Basically I want to remove the sixth element and multiples of the sixth element from my original table. I am able to do this by using this code below.
modifiedtable = Drop[orginaltable, {6, mnodes, 6}]
Now I get the modified table where every sixth element and multiples of sixth element of my original table is removed. This solves my Problem 1.
Now my Problem 2:
** MAJOR EDITED VERSION**:(ALL THE CODES GIVEN ABOVE IS CORRECT)
Thanks a lot for the answers, but I wanted something else and I made a mistake
while explaining it initially so I'm making another try.
Below is my modified table: I want the elements in between
"/** and **/" deleted and remaining there.
{{1, 2, 8, 7, 19, 20, 26, 25}, {2, 3, 9, 8, 20, 21, 27, 26}, {3, 4,10, 9, 21, 22, 28, 27}, {4, 5, 11, 10, 22, 23, 29, 28}, {5, 6, 12, 11, 23, 24, 30, 29}, {7, 8, 14, 13, 25, 26, 32, 31}, {8, 9, 15, 14, 26, 27, 33, 32}, {9, 10, 16, 15, 27, 28, 34, 33}, {10, 11, 17, 16, 28, 29, 35, 34}, {11, 12, 18, 17, 29, 30, 36, 35}, /**{13, 14, 20, 19, 31, 32, 38, 37}, {14, 15, 21, 20, 32, 33, 39, 38}, {15, 16, 22, 21, 33, 34, 40, 39}, {16, 17, 23, 22, 34, 35, 41, 40}, {17, 18, 24, 23, 35, 36, 42, 41},**/ {19, 20, 26, 25, 37, 38, 44, 43}, {20, 21, 27, 26, 38, 39, 45, 44}, {21, 22, 28, 27, 39, 40, 46, 45}, {22, 23, 29, 28, 40, 41, 47, 46}, {23, 24, 30, 29, 41, 42, 48, 47}, {25, 26, 32, 31,43, 44, 50, 49}, {26, 27, 33, 32, 44, 45, 51, 50}, {27, 28, 34, 33, 45, 46, 52, 51}, {28, 29, 35, 34, 46, 47, 53, 52}, {29, 30, 36, 35, 47, 48, 54, 53}, /**{31, 32, 38, 37, 49, 50, 56, 55}, {32, 33, 39, 38,50, 51, 57, 56}, {33, 34, 40, 39, 51, 52, 58, 57}, {34, 35, 41, 40, 52, 53, 59, 58}, {35, 36, 42, 41, 53, 54, 60, 59},**/ {37, 38, 44, 43,55, 56, 62, 61}, {38, 39, 45, 44, 56, 57, 63, 62}, {39, 40, 46, 45, 57, 58, 64, 63}, {40, 41, 47, 46, 58, 59, 65, 64}, {41, 42, 48, 47,59, 60, 66, 65}, {43, 44, 50, 49, 61, 62, 68, 67}, {44, 45, 51, 50, 62, 63, 69, 68}, {45, 46, 52, 51, 63, 64, 70, 69}, {46, 47, 53, 52, 64, 65, 71, 70}, {47, 48, 54, 53, 65, 66, 72, 71}, /**{49, 50, 56, 55, 67, 68, 74, 73}, {50, 51, 57, 56, 68, 69, 75, 74},{51,52, 58, 57, 69, 70, 76, 75}, {52, 53, 59, 58, 70, 71, 77, 76}, {53, 54, 60, 59, 71, 72, 78, 77}}**/
Now, if you observe, I wanted the first ten elements
(1st to 10th element of modifiedtable) to be there in my final table
( DoubleModifiedTable ). the the next five (11th to 15th elements of modifiedtable) deleted.
Then the next ten elements ( 16th to 25th elements of modifiedtable)
to be present in my final table ( DoubleModifiedTable )
then the next five deleted (26th to 30th elements of modifiedtable) and so on for the whole table.
Let say we solve this problem and we name the final table DoubleModifiedTable.
I am basically interested in getting the DoubleModifiedTable. I decided to subdivide the problem as it easy to explain.
I want this to happen automatically through the table since as this is just an example table but in reality I have huge table. If I can understand how I can solve this problem for this table, then I can solve it for my large table too.
Perhaps simpler:
DoubleModifiedTable =
Module[{copy = modifiedtable},
copy[[Flatten[# + Range[5] & /# Range[10, Length[copy], 10]]]] = Sequence[];
copy]
EDIT
Even simpler:
DoubleModifiedTable =
Delete[modifiedtable,
Transpose[{Flatten[# + Range[5] & /# Range[10, Length[modifiedtable], 10]]}]]
EDIT 2
Per OP's request: one only has to change a single number (10 to 15) in any of my solutions to get the answer to a modified problem:
DoubleModifiedTable =
Delete[modifiedtable,
Transpose[{Flatten[# + Range[5] & /# Range[10, Length[modifiedtable], 15]]}]]
Another way is to do something like
DoubleModifiedTable = With[{n = 10, m = 5},
Flatten[{{modifiedtable[[;; m]]},
Partition[modifiedtable, n - m, n, {n - m + 1, 1}, {}]}, 2]]
Edit
The edited version of Problem 2 is actually slightly simpler to solve than the original version. You could for example do something like
DoubleModifiedTable =
With[{n = 10, m = 5}, Flatten[Partition[modifiedtable, n, n + m, 1, {}], 1]]
Edit 2
What my second version does is to split the original list modifiedtable into sublists using Partition and then to flatten these sublists to form the final list. If you look at the Documentation for Partition you can see that I'm using the 6th form of Partition which means that the length of the sublists is n and the offset (the distance be is n+m. The gap between the sublists is therefore n+m-n==m.
The next argument, 1, is actually equivalent to {1,1} which tells Mathematica that the first element of modifiedtable should appear at position 1 in the first sublist and the last element of modifiedtable should appear on or after position 1 of the last sublist.
The last argument, {} is to indicate that no padding should be used for sublists with length <=n.
In summary, if you want to delete the first 10 elements and keep the next 5 you want sublists of length n=5 with gap m=10. Since you want the first sublist to start with the (m+1)-th element of modifiedtable, you could replace the fourth argument in Partition with something of the form {k,1} for some value of k but it's probably easier to just drop the first m elements of modifiedtable beforehand, i.e.
DoubleModifiedTable =
With[{n = 5, m = 10},
Flatten[Partition[Drop[modifiedtable, m], n, n + m, 1, {}], 1]]
DoubleModifiedTable=
modifiedtable[[
Complement[
Range[Length[modifiedtable]],
Flatten#Table[10 i + j, {i, Floor[Length[modifiedtable]/10]}, {j, 5}]
]
]]
or, slightly shorter
DoubleModifiedTable=
#[[
Complement[
Range[Length[#]],
Flatten#Table[10 i + j, {i, Floor[Length[#]/10]}, {j, 5}]
]
]] & # modifiedtable