In article <3D51FEBC.906@clotho.com>, Chris Dolan <chris@clotho.com> writes: > Eugene van der Pijll wrote: >> >> Unfortunately, his solution doesn't seem to work here. >> >> $ cat factorial.pl >> #!perl -l >> $_*=$`%9e9,//for~4=~/0*$/..pop;print$`%10 >> $ perl factorial.pl 15 >> 0 >> $ perl --version >> >> This is perl, v5.6.1 built for alpha-dec_osf >> >> Can someone explain, please? >> > > It's a 32- vs 64- bit thing. On a 32-bit box, ~4 == 4294967291 (which > ends in "1"). On a 64 bit alpha, it is different. > Ok, the rules says the solution has to work on all architectures, so it should in fact work on 64-bit machines. So his solution is invalid. However, we as referees should have caught this. Juho originally had +1 there, and only changed it to ~4 when hunting the best tie break. So I'm going to apply referees discretion and behave as if his top entry is #!perl -l $_*=$`%9e9,??for+1=~?0*$?..pop;print$`%10 which is certainly what he would have submitted if the refs had caught ~4. This gives factorial.pl: 44.1212 strokes (ok) postorder.pl: 51.0831 strokes (ok) ------- total: 95.2044 strokes which is still ahead of Eugene, so i think it's fair to still award Juho first place. Hopefully everybody will see this as acceptable. (and BoB falls back to #!perl -l $_*=$`%9e9,??for??=~?0*$?..pop;print$`%10 which scores 44.1175)Thread Previous | Thread Next