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The sieve of Sundaram: Difference between revisions

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The sieve of Sundaram (view source)

Revision as of 19:22, 9 March 2024

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→‎{{header|EasyLang}}
m (→‎{{header|J}}: oops - bad copy/paste)
 
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The millionth odd prime number: 15485867
124.8262 ms</pre>P.S. for those (possibly faithless) who wish to have a conventional prime number generator, one can uncomment the <code>yield return 2</code> line at the top of the function.
 
=={{header|C++}}==
<syntaxhighlight lang="c++">
#include <cmath>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <vector>
 
std::vector<uint32_t> sieve_of_sundaram(const uint32_t& limit) {
std::vector<uint32_t> primes = {};
if ( limit < 3 ) {
return primes;
}
 
const uint32_t k = ( limit - 3 ) / 2 + 1;
std::vector<bool> marked(k, true);
for ( uint32_t i = 0; i < ( std::sqrt(limit) - 3 ) / 2 + 1; ++i ) {
uint32_t p = 2 * i + 3;
uint32_t s = ( p * p - 3 ) / 2;
for ( uint32_t j = s; j < k; j += p ) {
marked[j] = false;
}
}
 
for ( uint32_t i = 0; i < k; ++i ) {
if ( marked[i] ) {
primes.emplace_back(2 * i + 3);
}
}
return primes;
}
 
int main() {
std::vector<uint32_t> primes = sieve_of_sundaram(16'000'000);
std::cout << "The first 100 odd primes generated by the Sieve of Sundaram:" << std::endl;
for ( uint32_t i = 0; i < 100; ++i ) {
std::cout << std::setw(3) << primes[i] << ( i % 10 == 9 ? "\n" :" " );
}
std::cout << "\n" << "The 1_000_000th Sundaram prime is " << primes[1'000'000 - 1] << std::endl;
}
</syntaxhighlight>
{{ out }}
<pre>
The first 100 odd primes generated by the Sieve of Sundaram:
3 5 7 11 13 17 19 23 29 31
37 41 43 47 53 59 61 67 71 73
79 83 89 97 101 103 107 109 113 127
131 137 139 149 151 157 163 167 173 179
181 191 193 197 199 211 223 227 229 233
239 241 251 257 263 269 271 277 281 283
293 307 311 313 317 331 337 347 349 353
359 367 373 379 383 389 397 401 409 419
421 431 433 439 443 449 457 461 463 467
479 487 491 499 503 509 521 523 541 547
 
The 1_000_000th Sundaram prime is 15485867
</pre>
 
=={{header|EasyLang}}==
<syntaxhighlight>
func log n .
return log10 n / log10 2.71828182845904523
.
proc sundaram np . primes[] .
nmax = floor (np * (log np + log log np) - 0.9385) + 1
k = (nmax - 2) / 2
len marked[] k
for i to k
h = 2 * i + 2 * i * i
while h <= k
marked[h] = 1
h += 2 * i + 1
.
.
i = 1
primes[] = [ ]
while np > 0
if marked[i] = 0
np -= 1
primes[] &= 2 * i + 1
.
i += 1
.
.
sundaram 100 primes[]
print primes[]
sundaram 1000000 primes[]
print primes[len primes[]]
</syntaxhighlight>
 
{{out}}
<pre>
[ 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 ]
15485867
</pre>
 
=={{header|F_Sharp|F#}}==
Line 613 ⟶ 709:
1 millionth Prime Found: 15485867
</pre>
 
=={{header|FreeBASIC}}==
<syntaxhighlight lang="vbnet">Function sieve_of_Sundaram(n As Uinteger) As Uinteger Ptr
If n < 3 Then Return 0
Dim As Uinteger r = Cint(Sqr(n))
Dim As Uinteger k = Cint((n - 3) / 2) + 1
Dim As Uinteger l = Cint((r - 3) / 2) + 1
Dim As Uinteger Ptr primes = Callocate(k, Sizeof(Uinteger))
Dim As Boolean Ptr marked = Callocate(k, Sizeof(Boolean))
For i As Uinteger = 1 To l
Dim As Uinteger p = 2 * i + 1
Dim As Uinteger s = Cint((p * p - 1) / 2)
For j As Uinteger = s To k Step p
marked[j] = True
Next j
Next i
Dim As Uinteger count = 0
For i As Uinteger = 1 To k
If Not marked[i] Then
primes[count] = 2 * i + 1
count += 1
End If
Next i
Return primes
End Function
 
Const limit As Uinteger = 16e6
Dim As Uinteger Ptr s = sieve_of_Sundaram(limit)
Print "First 100 odd primes generated by the Sieve of Sundaram:"
For i As Uinteger = 0 To 99
Print Using "#####"; s[i];
If (i + 1) Mod 10 = 0 Then Print
Next i
Print !"\nSundaram primes start with 3."
Print !"\nThe 100th Sundaram prime is: "; s[99]
Print !"\nThe 1000000th Sundaram prime is: "; s[999999]
 
Sleep</syntaxhighlight>
{{out}}
<pre>First 100 odd primes generated by the Sieve of Sundaram:
3 5 7 11 13 17 19 23 29 31
37 41 43 47 53 59 61 67 71 73
79 83 89 97 101 103 107 109 113 127
131 137 139 149 151 157 163 167 173 179
181 191 193 197 199 211 223 227 229 233
239 241 251 257 263 269 271 277 281 283
293 307 311 313 317 331 337 347 349 353
359 367 373 379 383 389 397 401 409 419
421 431 433 439 443 449 457 461 463 467
479 487 491 499 503 509 521 523 541 547
 
Sundaram primes start with 3.
 
The 100th Sundaram prime is: 547
 
The 1000000th Sundaram prime is: 15485867</pre>
 
=={{header|Go}}==
Line 799 ⟶ 951:
<syntaxhighlight lang=J>
P=: sundaram 1e6
100{.10 10$P
3 5 7 11 13 17 19 23 29 31
3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547
37 41 43 47 53 59 61 67 71 73
79 83 89 97 101 103 107 109 113 127
131 137 139 149 151 157 163 167 173 179
181 191 193 197 199 211 223 227 229 233
239 241 251 257 263 269 271 277 281 283
293 307 311 313 317 331 337 347 349 353
359 367 373 379 383 389 397 401 409 419
421 431 433 439 443 449 457 461 463 467
479 487 491 499 503 509 521 523 541 547
{:P
15485867
</syntaxhighlight>
 
=={{header|Java}}==
<syntaxhighlight lang="java>
import java.util.ArrayList;
import java.util.List;
 
public final class TheSieveOfSundaram {
 
public static void main(String[] args) {
List<Integer> primes = sieveOfSundaram(16_000_000);
System.out.println("The first 100 odd primes generated by the Sieve of Sundaram:");
for ( int i = 0; i < 100; i++ ) {
System.out.print(String.format("%3d%s", primes.get(i), ( i % 10 == 9 ? "\n" :" " )));
}
System.out.println();
System.out.println("The 1_000_000th Sundaram prime is " + primes.get(1_000_000 - 1));
}
private static List<Integer> sieveOfSundaram(int limit) {
List<Integer> primes = new ArrayList<Integer>();
if ( limit < 3 ) {
return primes;
}
final int k = ( limit - 3 ) / 2 + 1;
boolean[] marked = new boolean[k];
for ( int i = 0; i < ( (int) Math.sqrt(limit) - 3 ) / 2 + 1; i++ ) {
int p = 2 * i + 3;
int s = ( p * p - 3 ) / 2;
for ( int j = s; j < k; j += p ) {
marked[j] = true;
}
}
for ( int i = 0; i < k; i++ ) {
if ( ! marked[i] ) {
primes.add(2 * i + 3);
}
}
return primes;
}
 
}
</syntaxhighlight>
{{ out }}
<pre>
The first 100 odd primes generated by the Sieve of Sundaram:
3 5 7 11 13 17 19 23 29 31
37 41 43 47 53 59 61 67 71 73
79 83 89 97 101 103 107 109 113 127
131 137 139 149 151 157 163 167 173 179
181 191 193 197 199 211 223 227 229 233
239 241 251 257 263 269 271 277 281 283
293 307 311 313 317 331 337 347 349 353
359 367 373 379 383 389 397 401 409 419
421 431 433 439 443 449 457 461 463 467
479 487 491 499 503 509 521 523 541 547
 
The 1_000_000th Sundaram prime is 15485867
</pre>
 
=={{header|JavaScript}}==
Line 1,451 ⟶ 1,672:
@sieve[$k] = 0;
 
hyperrace for (1 .. $k).batch(1000) -> \@i {
for @i my-> int $j = i; {
while (my int $lj = i + $j + 2 * i * $j) < $k {;
while (my int $l = $i + $j + 2 * $i * $j++) < $k {
@sieve[$l] = 1;
$j = $j + 1;
}
}
}
 
Line 1,571 ⟶ 1,793:
=={{header|Wren}}==
{{libheader|Wren-fmt}}
{{libheader|Wren-seq}}
I've worked here from the second (optimized) Python example in the Wikipedia article for SOS which allows an easy transition to an 'odds only' SOE for comparison.
<syntaxhighlight lang="ecmascriptwren">import "./fmt" for Fmt
import "/seq" for Lst
 
var sos = Fn.new { |n|
Line 1,629 ⟶ 1,849:
Fmt.print("Using the Sieve of Sundaram generated primes up to $,d in $,d ms.\n", limit, elapsed)
System.print("First 100 odd primes generated by the Sieve of Sundaram:")
for Fmt.tprint(chunk"$3d", in Lst.chunks(primes[0..99], 10)) Fmt.print("$3d", chunk)
Fmt.print("\nThe $,d Sundaram prime is $,d", 1e6, primes[1e6-1])
 
Chkas
1,978

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