Qvfgnapr sebz prager bs phor gb prager bs “pbeare” fcurer rdhnyf fdeg(a) gvzrf qvfgnapr ba bar nkvf = fdeg(a) bire sbhe. Enqvhf bs “pbeare” fcurer rdhnyf bar bire sbhe. Gurersber enqvhf bs prageny fcurer = (fdeg(a) zvahf bar) bire sbhe. Bs pbhefr guvf trgf nf ynetr nf lbh cyrnfr sbe ynetr a. Vg rdhnyf bar unys, sbe n qvnzrgre bs bar, jura (fdeg(a) zvahf bar) bire sbhe rdhnyf bar unys ⇔ fdeg(a) zvahf bar rdhnyf gjb ⇔ fdeg(a) rdhnyf guerr ⇔ a rdhnyf avar.
dhrfgvba gjb
Guvf arire unccraf. Hfvat Fgveyvat’f sbezhyn jr svaq gung gur nflzcgbgvpf ner abg snibhenoyr, naq vg’f rnfl gb pbzchgr gur svefg ubjrire-znal inyhrf ahzrevpnyyl. V unira’g gebhoyrq gb znxr na npghny cebbs ol hfvat rkcyvpvg obhaqf rireljurer, ohg vg jbhyq or cnvashy engure guna qvssvphyg.
dhrfgvba guerr
Abar. Lbh pnaabg rira svg n ulcrefcurer bs qvnzrgre gjb orgjrra gjb ulcrecynarf ng qvfgnapr bar, naq gur ulcrephor vf gur vagrefrpgvba bs bar uhaqerq fcnprf bs guvf fbeg.
Oooh, I dropped a factor of 2 in the second one and didn’t notice because it takes longer than you’d expect before the numbers start increasing. Revised answer:
dhrfgvba gjb
Vs lbh qb gur nflzcgbgvpf pbeerpgyl engure guna jebatyl, gur ibyhzr tbrf hc yvxr (cv gvzrf r bire rvtug) gb gur cbjre a/2 qvivqrq ol gur fdhner ebbg bs a. Gur “zvahf bar” va gur sbezhyn sbe gur enqvhf zrnaf gung gur nflzcgbgvp tebjgu gnxrf ybatre gb znavsrfg guna lbh zvtug rkcrpg. Gur nafjre gb gur dhrfgvba gheaf bhg gb or bar gubhfnaq gjb uhaqerq naq fvk, naq V qb abg oryvrir gurer vf nal srnfvoyr jnl gb trg vg bgure guna npghny pnyphyngvba.
dhrfgvba bar
Qvfgnapr sebz prager bs phor gb prager bs “pbeare” fcurer rdhnyf fdeg(a) gvzrf qvfgnapr ba bar nkvf = fdeg(a) bire sbhe. Enqvhf bs “pbeare” fcurer rdhnyf bar bire sbhe. Gurersber enqvhf bs prageny fcurer = (fdeg(a) zvahf bar) bire sbhe. Bs pbhefr guvf trgf nf ynetr nf lbh cyrnfr sbe ynetr a. Vg rdhnyf bar unys, sbe n qvnzrgre bs bar, jura (fdeg(a) zvahf bar) bire sbhe rdhnyf bar unys ⇔ fdeg(a) zvahf bar rdhnyf gjb ⇔ fdeg(a) rdhnyf guerr ⇔ a rdhnyf avar.
dhrfgvba gjb
Guvf arire unccraf. Hfvat Fgveyvat’f sbezhyn jr svaq gung gur nflzcgbgvpf ner abg snibhenoyr, naq vg’f rnfl gb pbzchgr gur svefg ubjrire-znal inyhrf ahzrevpnyyl. V unira’g gebhoyrq gb znxr na npghny cebbs ol hfvat rkcyvpvg obhaqf rireljurer, ohg vg jbhyq or cnvashy engure guna qvssvphyg.
dhrfgvba guerr
Abar. Lbh pnaabg rira svg n ulcrefcurer bs qvnzrgre gjb orgjrra gjb ulcrecynarf ng qvfgnapr bar, naq gur ulcrephor vf gur vagrefrpgvba bs bar uhaqerq fcnprf bs guvf fbeg.
One: Correct
Two: Incorrect
Three: Correct
Oooh, I dropped a factor of 2 in the second one and didn’t notice because it takes longer than you’d expect before the numbers start increasing. Revised answer:
dhrfgvba gjb
Vs lbh qb gur nflzcgbgvpf pbeerpgyl engure guna jebatyl, gur ibyhzr tbrf hc yvxr (cv gvzrf r bire rvtug) gb gur cbjre a/2 qvivqrq ol gur fdhner ebbg bs a. Gur “zvahf bar” va gur sbezhyn sbe gur enqvhf zrnaf gung gur nflzcgbgvp tebjgu gnxrf ybatre gb znavsrfg guna lbh zvtug rkcrpg. Gur nafjre gb gur dhrfgvba gheaf bhg gb or bar gubhfnaq gjb uhaqerq naq fvk, naq V qb abg oryvrir gurer vf nal srnfvoyr jnl gb trg vg bgure guna npghny pnyphyngvba.
Correct.
I gave some Haskell code as a comment over there on my blog, under the posted problem.
1206 dimension is the smallest number. One can experiment with other values.