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Suppose I have a list of 17 objects, out of these 17 objects some have a certain property say P1. I separate them and say they are n in number where n < 17. Out of these n objects some have another property say P2. I separate them and say they are m in number where m < n. Out of these m objects some have another property say P3. I separate them and say they are k in number where k < m. I want to print these k objects only.
I was thinking of a long way that is I separate n, m and k objects all from 17 objects according to their respective property and then look for common index, the index that appear in all of three calculations.
Either I need to derive this common index or I do what I have written in the first paragraph that is to filter through and through according to the three properties.
Example:
list_1 = [17, 23, 15, 37, 43, 52, 57, 93, 55, 85, 11, 13, 7, 22, 24]
list_odd = [17, 23, 15, 37, 43, 57, 93, 55, 85, 11, 13, 7] #P1 is a number is odd
list_odd_div3 = [15, 57, 93] #P2 is a number divisible by 3
list_odd_div5 = [15, 55, 85] #P3 is a number divisible by 5
required_list = [15] #A number having P1, P2 and P3
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
I have just started my journey with Wolfram Mathematica and I want to implement a simple genetic algorithm. The construction of the data is given and I have to start with such rows/columns.
Here is what I have:
chromosome := RandomSample[CharacterRange["A", "G"], 7]
chromosomeList = Table[chromosome, 7] // MatrixForm
This gives me a matrix, where every row represents a chromosome:
yPos = Flatten[Position[chromosomeList, #], 1] & /# {"A", "B", "C",
"D", "E", "F", "G"};
yPos = yPos[[All, 3 ;; 21 ;; 3]] // Transpose
Now every column represents a letter (From A to G) and every row it's index in every chromosome:
Here is a given efficiency matrix, where very row represents different letter (From A to G) and every column gives the value that should be applied on the particular position:
efficiencyMatrix = {
{34, 31, 20, 27, 24, 24, 18},
{14, 14, 22, 34, 26, 19, 22},
{22, 16, 21, 27, 35, 25, 30},
{17, 21, 24, 16, 31, 22, 20},
{17, 29, 22, 31, 18, 19, 26},
{26, 29, 37, 34, 37, 20, 21},
{30, 28, 37, 28, 29, 23, 19}}
What I want to do is to create a matrix with values that correspond to the letter and it's position. I have done it like that:
values = Transpose[{ efficiencyMatrix[[1, yPos[[1]]]],
efficiencyMatrix[[2, yPos[[2]]]],
efficiencyMatrix[[3, yPos[[3]]]],
efficiencyMatrix[[4, yPos[[4]]]],
efficiencyMatrix[[5, yPos[[5]]]],
efficiencyMatrix[[6, yPos[[6]]]],
efficiencyMatrix[[7, yPos[[7]]]]}]
How can I write it in more elegant way?
You can apply a list of functions to some variable using the function Through, which is helpful when applying Position multiple times. Because Position[patt][expr] == Position[expr, patt], we can do
Through[ (Position /# CharacterRange["A","C"])[{"B", "C", "A"}] ]
to get {3, 1, 2}.
Position can also operate on lists, so we can simplify finding ypos by doing
Transpose#Map[Last, Through[(Position /# characters)[chromosomeList]], {2}]
where characters is the relevant output of CharacterRange.
We can also simplify dealing with ranges of integers by mapping over the Range function, so in total we end up with
characters = CharacterRange["A","G"]
efficiencies = ...
chromosomes = ...
ypos = Transpose#Map[Last, Through[(Position /# characters)[chromosomes]], {2}];
efficiencies[[#, ypos[[#]]]]& /# Range[Length[characters]] //Transpose ]
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've written a similar question which was closed I would like to ask not the code but an efficiency tip. I haven't coded but if I can't find any good hint in here I'll go and code straightforward. My question:
Suppose you have a function listNums that take a as lower bound and b as upper bound.
For example a=120 and b=400
I want to print numbers between these numbers with one rule. 120's permutations are 102,210,201 etc. Since I've got 120 I would like to skip printing 201 or 210.
Reason: The upper limit can go up to 1020 and reducing the permutations would help the running time.
Again just asking for efficiency tips.
I am not sure how you are handling 0s (eg: after outputting 1 do you skip 10, 100 etc since technically 1=01=001..).
The trick is to select a number such that all its digits are in increasing order (from left to right).
You can do it recursively. AT every recursion add a digit and make sure it is equal to or higher than the one you recently added.
EDIT: If the generated number is less than the lower limit then permute it in such a way that it is greater than or equal to the lower limit. If A1A2A3..Ak is your number and it is lower than limit), then incrementally check if any of A2A1A3...Ak, A3A1A2...Ak, ... , AkA1A2...Ak-1 are within limit. If need arises, repeat this step to with keeping Ak as first digit and finding a combination of A1A2..Ak-1.
Eg: Assume we are selecting 3 digits and lower limit is 99. If the combination is 012, then the lowest permutation that is higher than 99 is 102.
When the lower bound is 0, an answer is given by the set of numbers with non-decreasing digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99, 111, 112...) that fall in the requested range.
This sequence is easily formed by incrementing an integer, and when there is a carry, replicate the digit instead of carrying. Exemple: 73 is followed by 73+1 = 74 (no carry); 79 is followed by 79+1 = 80 (carry), so 88 instead; 22356999 is followed by 22356999+1 = 22357000, hence 22357777.
# Python code
A= 0 # CAUTION: this version only works for A == 0 !
B= 1000
N= A
while N < B:
# Detect zeroes at the end
S= str(N)
P= S.find('0')
if P > 0:
# Replicate the last nonzero digit
S= S[:P] + ((len(S) - P) * S[P-1])
N= eval(S)
# Next candidate
print N
N+= 1
Dealing with a nonzero lower bound is a lot more tricky.