“ As a side challenge for this quest of the week series, I propose something I call "LotR LCG Golf". Given a quest, what is the bare minimum number of cards required to beat it, in any universe? This means that every element of randomness is under your control. You can define the exact ordering of any deck at any time as it suits you.

To be fair, we will count heroes and contracts as well. So for JatA here, I used 5 cards. Is it doable with less?”

I can get as low as 2 (two different— though related— ways). I’ll hold off for a few days on how in case anyone else wants to take a crack at it.

@Some Sort Nicely done! Do your two related ways happen to be (https://rot13.com/) GRbjla + Tevzorbea/Obebzve?

The nice thing about this golf game is that even when someone gives the answer it can still be fun to go and try it out to see if you can figure out how they did it. I'm sure there are solutions in the >5 card range with less optimal cards that might actually pose more of a puzzle!

@Seastan Nope! Is it possible with either of those pairs? That's neat, if so-- that would give us four different two-card pairs that are theoretically capable of beating Anduin.

Lbh'er gelvat gb fbyir vg snfg, V bcgrq gb gnxr vg fybj: Gunyva + Netnynq/Gvyob. Erirny gur Uvyy Gebyy be n gernpurel qhevat frghc, unir Gunyva dhrfg rirel ebhaq, naq yrg rirel rapbhagre erirny sbe gur erfg bs gur tnzr or na Rnfgrea Pebj (juvpu vf xvyyrq ol Gunyva naq fuhssyrq onpx, ernql sbe arkg ebhaq'f erirny) hagvy lbh uvg Fgntr 3, juvpu tbrf Ybpngvba / Ybpngvba.

Rvgure Netnynq be Ovyob pna xvyy gur gebyy va avar ebhaqf, jryy orsber lbhe guerng uvgf vgf ratntrzrag guerfubyq, Gunyva puhtf nybat cynpvat 1 cbvag bs cebterff hagvy vg'f qrnq naq gura 2 (jvgu Ovyob) be 3 (jvgu Netnynq) nsgre, naq gur jubyr nssnve gheaf vagb n avpr, erynkrq, 15-17 ebhaq fgebyy qbja n fpravp evire.

@Some Sort

Very nice!

Lrf, V pbasvezrq vg'f qbnoyr jvgu gur cnvef V yvfgrq, va nobhg 6 ebhaqf V guvax. Lbhe nccebnpu vf irel pyrire, V qvqa'g pbafvqre qverpg qnzntr gb gnxr qbja gur gebyy. Jvgu gung va zvaq, lbh'ir urycrq zr vqragvsl 2-pneq nabgure fgengrtl, juvpu vf Netnynq + nal 4 jvyycbjre ureb. Znlor lbh pna svther bhg ubj vg jbexf :)

Gur Onaxf bs gur Naqhva fbsgybpx. Trg n qhq frghc, erirny Onaxf bs gur Naqhva va Ebhaq 1, geniry gb vg. Erirny Onaxf bs gur Naqhva va Ebhaq 2, oybj hc gur svefg Onaxf, geniry gb gur frpbaq. Erirny gur svefg Onaxf ntnva va Ebhaq 3, oybj hc gur frpbaq, geniry gb gur svefg ntnva. Gung’yy trg lbh guebhtu Fgntr 1 naq gur gebyy jvgubhg nal vffhrf.

Hasbeghangryl Onaxf pna’g znvagnva n shyy fbsgybpx ba Fgntr 2 jura lbh’er erirnyvat gjb pneqf (ng yrnfg abg jvgubhg fbzr Abegurea Genpxref be Eubinavba Bhgevqref nebhaq gb oybj vg hc va fgntvat), ohg gurer ner rabhtu qhq gernpurevrf gb znxr vg guebhtu gur 8+ ebhaqf— Qrfcnve, Qevira ol Funqbj, Chefhrq ol Funqbj, Rivy Fgbez, gernpurebhf sbt.

I think there is also a solution with 2 cards using only 1 hero. (Solution in code below if anybody wants to have a crack first).

Gn Ovyob+BUNHU Tnaqnys

11 pneqf unir ab erny rssrpg.

Dhrfg jvgu Ovyob nyjnlf, gnetrggvat Uvyy Gebyy

Gheaf 1-3 Erirny 3*Rivy Fgbez Ghea 4 Erirny Onaxf bs gur Naqhva, geniry gurer Ghea 5 Cynl Tnaqnys, Erirny Onaxf, Rkcyber Onaxf, frggvat hc gur Ybbc Ghae 7 abg gevttrevat Onaxf qhr gb rkcybevat, geniry gb onaxf va FN Ghea 8 Erirny Qevira ol Funqbj, rkcyber 2aq Onaxf, ab Erfcbafr ntnva Ghea 9 Erirny Guernpurebhf Sbt, xvyy Gebyy (9gu qnzntr), cynpr 6 cebterff ba fgntr 2.

Ghea 10: Erirny 2 Chefhrq ol Funqbj, 11 cebterff ba 2 Ghea 11: Erirny 1Anpebznapref Ernpu naq 1*Gernpurebhf Sbt, 16 cebterff ba 2, erirny 2 ybpngvbaf sbe fgntr 3, jva

Guerng: 9 (Ovyob)+12 (Tnaqnys)+10 (abezny ebhaqf)=31

Nice challenge!

@Some Sort Yep!

@Mormegil Nice solution! I feel like that's the trickiest one to figure out so far if you just tell people the 2 cards.