Since you're asking for the complexity of the given code,
bool s_a (int a)
{
while (a % 2 == 0)
a = a / 2;
while (a % 3 == 0)
a = a / 3;
while (a % 5 == 0)
a = a / 5;
if (a == 1)
return true;
else
return false;
}
void s_b (int n)
{
unsigned int k = 1,nc = 1;
while ( k <= n)
{
if(s_a(nc) == true)
{
cout << nc << " ";
k++;
}
nc++;
}
}
” would also appreciate an answer on its complexity
… it seems clear that this needs a more in-depth answer than just “use a sieve”.
The code computes numbers of the form 2a×3b×5c, where a, b and c are non-negative integers.
These numbers are called Hamming numbers in computer science, and more generally they're called Regular numbers.
In passing, the code above can be made more clear by
using self-describing names,
using a boolean expression directly in a return statement, rather than an if-else construction, and saying just if(expression) rather than if(expression==true), and
not doing i/o in basic computational functions (consider: such a function can't be used in a GUI), but rather returning the results.
Then modulo formatting and basic syntax – I use my preferences here – the code can look like this:
auto is_hamming_number( int a )
-> bool
{
while( a % 2 == 0 ) { a = a / 2; }
while( a % 3 == 0 ) { a = a / 3; }
while( a % 5 == 0 ) { a = a / 5; }
return (a == 1);
}
auto hamming_numbers( int const n )
-> vector<int>
{
vector<int> result;
for( int i = 1; n != n_items( result ); ++i )
{
if( is_hamming_number( i ) )
{
result.push_back( i );
}
}
return result;
}
… where n_items is defined as e.g.
template< class Container >
auto n_items( Container const& c )
-> ptrdiff_t
{ return c.size(); }
… in order to avoid problems with signed/unsigned comparisons, and associated compilation warnings.
In the first short version of this answer, which I posted to get an editable answer in place before the question was possibly closed, I wrote that
” The short answer to the complexity is that the above code scans all numbers up to and including n up to and including apparently nk where k is rather small, e.g. k = 1.36, and for each number checks it by removing all factors of 2, 3 and 5, which I believe on average is a linear logarithmic process, hence (under these assumptions) its complexity is O(n1.36logn). A mathematical analysis must surely be available on the net, but I didn't find it.
I now have better data, from coding up an O(n log n) time sequence generator:
n max Ham
1 1
2 2
4 4
8 9
16 25
32 90
64 450
128 3375
256 43200
512 1049760
1024 60466176
This indicates that the nth Hamming number increases faster than the square of n. Intuitively I'd expect exponential behavior. But I'm no mathematician, and simple googling did not lead me to any description or mathematical analysis of the behavior.
The complexity of the simple code is necessarily at least as high as the behavior of the nth Hamming number, since it checks all numbers up that. Then multiply with the time needed to remove 2, 3 and 5 factors from each number, which I believe is logarithmic in the size of the number. So we're talking at least O( n2×log(n) ), but I'm not sure about how bad it is.
Researching (googling) the behavior of the nth Hamming number, I found a partial duplicate of this question, namely (Finding Hamming Numbers - not code or distance). The second code example in M Oehm's answer there is almost exactly what you ask for, modulo the linear complexity requirement. However, as you clarified in a comment on this answer,
” It doesn't do exactly what I want in the sense that it shows the hamming numbers smaller than n and not the first n hamming numbers. Also the code uses some things that I did not learn yet like the sieve,therefore it's a bit hard to understand
Instead of a sieve the following code uses the main idea of the second of Oehm's examples, namely to directly generate all products of (2,3,5) factor combinations that are smaller than some limit, and then sort them.
In order to generate at least n such products the code below makes a guess about the nth Hamming number. If that guess turns out to be too low – because fewer than n Hamming numbers were generated – the guess is doubled and the Hamming numbers are generated anew. The doubling means that the complexity of the final generation overwhelms the complexity of any earlier generations, so that they can be ignored wrt. complexity.
#include <algorithm> // std::sort
#include <assert.h> // assert
#include <iostream>
#include <limits.h> // INT_MAX
#include <math.h> // pow
#include <stddef.h> // ptrdiff_t
#include <unordered_set> // std::unordered_set
#include <vector> // std::vector
using namespace std;
using Size = ptrdiff_t; // Signed integer corresponding to size_t.
template< class Container >
auto n_items( Container const& c )
-> Size
{ return c.size(); }
class Hamming_sequence
{
private:
int n_;
vector<int> numbers_;
public:
auto computed_numbers() const
-> vector<int> const&
{ return numbers_; }
auto size() const -> Size { return n_; }
auto begin() const { return numbers_.begin(); }
auto end() const { return begin() + size(); }
auto operator[]( Size const i ) const
-> int
{ return numbers_[i]; }
Hamming_sequence( int const n )
: n_( n )
{
double const fp_max_number = pow( n, 2.5 ); // Reasonable
assert( fp_max_number <= INT_MAX );
int max_number = static_cast<int>( fp_max_number );
while( n_items( numbers_ ) < n )
{
numbers_.clear();
clog << ": trying with max number " << max_number << endl;
for( int two = 1; two <= max_number/2; two *= 2 )
{
for( int three = 1; three <= max_number/two/3; three *= 3 )
{
for( int five = 1; five <= max_number/two/three/5; five *= 5 )
{
int const number = two*three*five;
numbers_.push_back( number );
}
}
}
assert( max_number <= INT_MAX/2 );
max_number *= 2;
}
sort( numbers_.begin(), numbers_.end() ); // O( n*log(n) )
}
};
#include <iomanip> // setw
auto main() -> int
{
for( int const number : Hamming_sequence{ 720 } )
{
cout << number << " ";
}
cout << "\n";
cout << "\n";
Hamming_sequence const h{ 1024 };
for( int i = 1; i <= n_items( h ); i *= 2 )
{
cout << setw( 5 ) << i << setw( 15 ) << h[i-1] << endl;
}
}
The asserts are not assertions about what I believe to be true at those points in the execution, but are rather a device to make the program crash if the numbers get too large to be handled. So this is not “production code”. For more robust code replace with e.g. exception throwing.
Regarding complexity, the final guess about the max Hamming number is necessarily within a factor of 2 of the true value, for not too small n. And because doubling the value of the Hamming number quite evidently (see the behavior table above) does not double the sequence length, so the number of computed Hamming numbers is within 2n. With the added sorting this gives O(n log n) in the number of Hamming numbers.
The Wikipedia article describes Edsger W. Dijkstra's linear time algorithm, which in short is to merge the three number sequences 2×H, 3×H and 5×H, where H is the basic Hamming number sequence. This works because every Hamming number h is of at least one of the forms 2×h, 3×h and 5×h, where h is some smaller Hamming number. And so one can just work forward:
#include <algorithm> // std::min_element
#include <array> // std::array
#include <assert.h> // assert
#include <iostream>
#include <limits.h> // INT_MAX
#include <math.h> // pow
#include <stddef.h> // ptrdiff_t
#include <unordered_set> // std::unordered_set
#include <vector> // std::vector
using namespace std;
using Size = ptrdiff_t; // Signed integer corresponding to size_t.
template< class Container >
auto n_items( Container const& c )
-> Size
{ return c.size(); }
class Hamming_sequence
{
private:
int n_;
vector<int> numbers_;
template< int factor >
struct H_
{
vector<int>& h_;
int i_ = 0;
auto operator*()
-> int
{
assert( h_[i_] <= INT_MAX/factor );
return factor*h_[i_];
}
void operator++() { ++i_; }
H_( vector<int>& numbers ): h_( numbers ) {}
};
public:
auto computed_numbers() const
-> vector<int> const&
{ return numbers_; }
auto size() const -> Size { return n_; }
auto begin() const { return numbers_.begin(); }
auto end() const { return begin() + size(); }
auto operator[]( Size const i ) const
-> int
{ return numbers_[i]; }
Hamming_sequence( int const n )
: n_( n )
{
H_<2> h2{ numbers_ };
H_<3> h3{ numbers_ };
H_<5> h5{ numbers_ };
numbers_ = {1};
array<int, 3> h = {*h2, *h3, *h5};
int previous = 1;
while( n_items( numbers_ ) < n_ )
{
int const i = min_element( h.begin(), h.end() ) - h.begin();
if( h[i] != previous )
{
numbers_.push_back( h[i] );
previous = h[i];
}
switch( i )
{
case 0: ++h2; h[0] = *h2; break;
case 1: ++h3; h[1] = *h3; break;
case 2: ++h5; h[2] = *h5; break;
}
}
}
};
#include <iomanip> // setw
auto main() -> int
{
for( int const number : Hamming_sequence{ 720 } )
{
cout << number << " ";
}
cout << "\n";
cout << "\n";
Hamming_sequence const h{ 1024 };
for( int i = 1; i <= n_items( h ); i *= 2 )
{
cout << setw( 5 ) << i << setw( 15 ) << h[i-1] << endl;
}
cout << "\n";
cout << "The 1500'th Hamming number is " << Hamming_sequence( 1'500 )[1'499] << "." << endl;
}
Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 45 48 50 54 60 64 72 75 80 81 90 96 100 108 120 125 128 135 144 150 160 162 180 192 200 216 225 240 243 250 256 270 288 300 320 324 360 375 384 400 405 432 450 480 486 500 512 540 576 600 625 640 648 675 720 729 750 768 800 810 864 900 960 972 1000 1024 1080 1125 1152 1200 1215 1250 1280 1296 1350 1440 1458 1500 1536 1600 1620 1728 1800 1875 1920 1944 2000 2025 2048 2160 2187 2250 2304 2400 2430 2500 2560 2592 2700 2880 2916 3000 3072 3125 3200 3240 3375 3456 3600 3645 3750 3840 3888 4000 4050 4096 4320 4374 4500 4608 4800 4860 5000 5120 5184 5400 5625 5760 5832 6000 6075 6144 6250 6400 6480 6561 6750 6912 7200 7290 7500 7680 7776 8000 8100 8192 8640 8748 9000 9216 9375 9600 9720 10000 10125 10240 10368 10800 10935 11250 11520 11664 12000 12150 12288 12500 12800 12960 13122 13500 13824 14400 14580 15000 15360 15552 15625 16000 16200 16384 16875 17280 17496 18000 18225 18432 18750 19200 19440 19683 20000 20250 20480 20736 21600 21870 22500 23040 23328 24000 24300 24576 25000 25600 25920 26244 27000 27648 28125 28800 29160 30000 30375 30720 31104 31250 32000 32400 32768 32805 33750 34560 34992 36000 36450 36864 37500 38400 38880 39366 40000 40500 40960 41472 43200 43740 45000 46080 46656 46875 48000 48600 49152 50000 50625 51200 51840 52488 54000 54675 55296 56250 57600 58320 59049 60000 60750 61440 62208 62500 64000 64800 65536 65610 67500 69120 69984 72000 72900 73728 75000 76800 77760 78125 78732 80000 81000 81920 82944 84375 86400 87480 90000 91125 92160 93312 93750 96000 97200 98304 98415 100000 101250 102400 103680 104976 108000 109350 110592 112500 115200 116640 118098 120000 121500 122880 124416 125000 128000 129600 131072 131220 135000 138240 139968 140625 144000 145800 147456 150000 151875 153600 155520 156250 157464 160000 162000 163840 164025 165888 168750 172800 174960 177147 180000 182250 184320 186624 187500 192000 194400 196608 196830 200000 202500 204800 207360 209952 216000 218700 221184 225000 230400 233280 234375 236196 240000 243000 245760 248832 250000 253125 256000 259200 262144 262440 270000 273375 276480 279936 281250 288000 291600 294912 295245 300000 303750 307200 311040 312500 314928 320000 324000 327680 328050 331776 337500 345600 349920 354294 360000 364500 368640 373248 375000 384000 388800 390625 393216 393660 400000 405000 409600 414720 419904 421875 432000 437400 442368 450000 455625 460800 466560 468750 472392 480000 486000 491520 492075 497664 500000 506250 512000 518400 524288 524880 531441 540000 546750 552960 559872 562500 576000 583200 589824 590490 600000 607500 614400 622080 625000 629856 640000 648000 655360 656100 663552 675000 691200 699840 703125 708588 720000 729000 737280 746496 750000 759375 768000 777600 781250 786432 787320 800000 810000 819200 820125 829440 839808 843750 864000 874800 884736 885735 900000 911250 921600 933120 937500 944784 960000 972000 983040 984150 995328 1000000 1012500 1024000 1036800 1048576 1049760 1062882 1080000 1093500 1105920 1119744 1125000 1152000 1166400 1171875 1179648 1180980 1200000 1215000 1228800 1244160 1250000 1259712 1265625 1280000 1296000 1310720 1312200 1327104 1350000 1366875 1382400 1399680 1406250 1417176 1440000 1458000 1474560 1476225 1492992 1500000 1518750 1536000 1555200 1562500 1572864 1574640 1594323 1600000 1620000 1638400 1640250 1658880 1679616 1687500 1728000 1749600 1769472 1771470 1800000 1822500 1843200 1866240 1875000 1889568 1920000 1944000 1953125 1966080 1968300 1990656 2000000 2025000 2048000 2073600 2097152 2099520 2109375 2125764 2160000 2187000 2211840 2239488 2250000 2278125 2304000 2332800 2343750 2359296 2361960 2400000 2430000 2457600 2460375 2488320 2500000 2519424 2531250 2560000 2592000 2621440 2624400 2654208 2657205 2700000 2733750 2764800 2799360 2812500 2834352 2880000 2916000 2949120 2952450 2985984 3000000 3037500 3072000 3110400 3125000 3145728 3149280 3188646 3200000 3240000 3276800 3280500 3317760 3359232 3375000 3456000 3499200 3515625 3538944 3542940 3600000 3645000 3686400 3732480 3750000 3779136 3796875 3840000 3888000 3906250 3932160 3936600 3981312 4000000 4050000 4096000 4100625 4147200 4194304 4199040 4218750 4251528 4320000 4374000 4423680 4428675 4478976 4500000 4556250 4608000 4665600 4687500 4718592 4723920 4782969 4800000 4860000 4915200 4920750 4976640 5000000 5038848 5062500 5120000 5184000 5242880 5248800 5308416 5314410 5400000 5467500 5529600 5598720 5625000 5668704 5760000 5832000 5859375 5898240 5904900 5971968 6000000 6075000 6144000 6220800 6250000 6291456 6298560 6328125 6377292 6400000 6480000 6553600 6561000 6635520 6718464 6750000 6834375 6912000
1 1
2 2
4 4
8 9
16 25
32 90
64 450
128 3375
256 43200
512 1049760
1024 60466176
The 1500'th Hamming number is 859963392.
