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Though I somehow doubt this is a winner, I think it might qualify for the "bonus marks". You'll note that no division (or modulus) is used, so I consider it to be using the sieve with no special cases or wasted operations (though it probably has plenty of wasted comparisons O-:). Though it might be more efficient to break out of the grep when a non-prime is found (or, if we had things in a sorted order, to stop once \$_*\$_ passes \$n) and compensate for that by incrementing \$p{\$_} in an extra loop (rather than incrementing it, at most, one time per prime candidate), but I'm not even sure if that would be a net win (without resorting to sorting).

```sub sieve {
my%p=(2,2);for my\$n(3..pop){grep\$n==(\$p{\$_}+=\$_*(\$p{\$_}<\$n)),keys%p
or\$p{\$n}=\$n}keys%p
}

for( @ARGV ) {
print "\$_: ",join(" ",sort{\$a<=>\$b}sieve(\$_)),\$/;
}

I count 86 characters for the part that counts. And, no, I don't think it'd be fair to add 13 characters for sorting since the original challange didn't say anything about returning the list in sorted order. q-:

Sample output:

```1: 2
2: 2
3: 2 3
4: 2 3
5: 2 3 5
6: 2 3 5
50: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47
1000: 2 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
479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593
599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691
701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821
823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937
941 947 953 967 971 977 983 991 997

Update: BTW, this is strict compliant.

- tye (but my friends call me "Tye")

In reply to (tye)Re: (Golf): Sieve of Eratosthenes by tye
in thread (Golf): Sieve of Eratosthenes by tilly

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