@Jonah How about this?

@Bubbler That's nice, though I have 1 additional requirement I didn't specify clearly enough. I need to be able to "tick up". That is, I either need to be able to give a number and get the odometer equivalent of that number, or else be able to continuously increment the odometer to its next highest value, as with my original example.

After some thoughts, mine converged to yours but with two less ~s.

@Bubbler, Nice. Thanks for looking.

And looks like J got a new cool set of primitives quite recently

Excellent. I need to read up on those.

⎕←{n←⌊3⍟1+⍵×2⋄P Q←w↓⍨¨¯1+0⍳⍨¨=/w↑⍨¨⌊/≢¨w←(n/¨3)⊤¨⍵+2÷⍨1-3*n⋄'S' 'NE' 'NW'[⍬{0=≢⍵:⍺⋄t←⌊/⍵∘{⊃⌽⍸⍺=⍵}¨⍳3⋄(⍺,t⌷⍵)∇⍵↑⍨t}P],'N' 'SW' 'SE'[Q]}299792458 1000

@Sherlock9
INDEX ERROR

Got it down to 132

Oh right

⋄⎕IO←0⋄⎕←{n←⌊3⍟1+⍵×2⋄P Q←w↓⍨¨¯1+0⍳⍨¨=/w↑⍨¨⌊/≢¨w←(n/¨3)⊤¨⍵+2÷⍨1-3*n⋄'S' 'NE' 'NW'[⍬{0=≢⍵:⍺⋄t←⌊/⍵∘{⊃⌽⍸⍺=⍵}¨⍳3⋄(⍺,t⌷⍵)∇⍵↑⍨t}P],'N' 'SW' 'SE'[Q]}299792458 1000

@Sherlock9
┌──┬─┬─┬──┬──┬──┬──┬──┐
│NE│S│S│SE│SW│SE│SW│SE│
└──┴─┴─┴──┴──┴──┴──┴──┘

⋄⎕IO←0⋄⎕←{n←⌊3⍟1+⍵×2⋄P Q←w↓⍨¨0⍳⍨¨=/w↑⍨¨⌊/≢¨w←(n/¨3)⊤¨⍵+2÷⍨1-3*n⋄'S' 'NE' 'NW'[⍬{0=≢⍵:⍺⋄t←⌊/⍵∘{⊃⌽⍸⍺=⍵}¨⍳3⋄(⍺,t⌷⍵)∇⍵↑⍨t}P],'N' 'SW' 'SE'[Q]}299792458 1000

@Sherlock9
┌──┬─┬─┬──┬──┬──┬──┬──┬─┐
│NE│S│S│SE│SW│SE│SW│SE│N│
└──┴─┴─┴──┴──┴──┴──┴──┴─┘

Bug fix saved bytes!

I forgot to finish converting to ⎕IO←0 :D

This bounty applies to this challenge. I've found a 9-byte Extended Dyalog solution.

Oh right I should claim a bounty on this answer

Also, if you have any golfing suggestions for that answer, I would very much appreciate it :D

@Sherlock9 Why only 100? Isn't it the shortest answer?

Oh whoops. I forgot about the bonuses XD

@Sherlock9 Willing to go Extended?

@Adám I think so, what do you have in mind?

⎕CY'dfns'
{P Q←w↓⍨¨⊃⍸≠⌿↑1+w←(⍳3)∘adic¨⍵⋄'S' 'NE' 'NW'[⍬{0=≢⍵:⍺⋄t←⌊/⍵∘{⊃⌽⍸⍺=⍵}¨∪⍵⋄(⍺,t⌷⍵)∇t↑⍵}P],('N' 'SW' 'SE')[Q]}
→
{P Q←w↓⍨¨⊃⍸≠⌿↑1+w←(⍳3)∘⌂adic¨⍵⋄'S' 'NE' 'NW'[⍬{=≢⍵:⍺⋄t←⌊/⊃∘⌽∘⍸¨⍵∘=¨∪⍵⋄(⍺,t⌷⍵)∇t↑⍵}P],Q⊇'N' 'SW' 'SE'}

@Adám why's that (⍳3) in parentheses?

@Ven ⍳3∘⌂adic¨⍵

oh, it's not a ⋄. I'll blame firefox for that one...

@Sherlock9 In any case, you don't need to parenthesise ('N' 'SW' 'SE')[Q]

Heck, I was sure I'd fixed that

I'll edit it in at the same time as changing to Extended

@Sherlock9 Full program at 99

@Adám I've edited it in! I should probably rearrange the post so that the Extended solution is the more prominent one but I don't get why =≢⍵ is equal to 0=≢⍵ here

@Sherlock9 Because in Extended, all monadic scalar comparisons imply the prototypical element as left argument, so numbers are compared to 0 and characters to space.

@Adám I've a 10-byter for non-extended

@J.Sallé That's good. Hurry up and cash in!

Yeah I'm just finishing testing to see if I didn't miss anything

@Adam Did you consider :47640307 any further?

@Ven What is ":47640307"?

I guess you can't reply after a while.

@Ven Which one of them? I did add ⍮ to Extended Dyalog APL.

@Ven pretty sure you can, you just can't have anything before the : and must have stuff after :num

@dzaima ah, maybe then!

@Adám Done

@J.Sallé ⍵*2 could just be |⍵, no? Anyway, I don't see it handling ¯5 5right; it needs to return 5: if the temperatures are -5 and 5, output 5.

@Adám |⍵ wouldn't handle that case, while ⍵*2 does

@Adám TIO

@J.Sallé both give 25, no? Try it online!

@Adám Yeah I hadn't tried the negative number before the positive

shucks

@Adám I think I've fixed it for an extra 2 bytes

Still testing to see if I missed anything though, but I think {⊃⍵[⌽⍋⍵×-⍵]} works

@J.Sallé That doesn't look right. Again, positive and negative are treated the same since ⍵×-⍵ is the same as -|⍵*2

I find it strange that ⍵[⍒⍵×-⍵] doesn't match ⍵[⌽⍋⍵×-⍵]

@J.Sallé Because of duplicates. Think about how you'd grade 1 1 2 where ⍋ gives 1 2 3 and ⍒ gives 3 1 2.

@Adám I was using the wrong test case >.< goddamn it

@Adám Yeah I assumed that was it.

Ah hell there's a bug in the code

@Sherlock9 I'll wait for a fix before bountying.

Sure thing

I'm a little distressed as a) I should have seen this earlier, and b) while I have written a correct answer in Ruby before, I'm not sure how to translate it into APL

The idea is that I can't just chop off everything before the point of mismatch. It's not efficient enough. I need to find a way to keep the two paths and the common ancestor and also search the first path and the common ancestor for the shortest backtracking step

And it's all so jargon-y in my head, I'm not sure where to start

Webinar right now!

I have an idea for a fix but I'm not sure how to implement it. In a case like 66 to 1, where the paths are P←1 3 3 3 and Q←1, the point of mismatch is index 2. ⍸P∘=¨∪P gets us (1)(2 3 4) and then ⊃⌽⍸P∘=¨∪P gets us 1 4. Now I need to get, not the minimum, but whichever index in that last list is closest to the point of mismatch, 2. Now how do I do that?

@Sherlock9 can P and Q contain 0s?

if not, maybe something like ⎕io++/=⌿↑P Q?

@ngn I can fairly easily change the 0 1 2 of the alphabet to 1 2 3 by just adding 1 to P and Q

Sorry I keep swapping between ⎕IO←0 and ⎕IO←1 and forget to specify

@ngn Okay so with ⎕IO←0, that's +/=⌿↑1+P Q

@Sherlock9 shame on apl for allowing that :)

:D

Okay I'm not sure how this solves my problem of needing to find from a list of ancestors, the one closest to the point of mismatch. So of ancestors 3 7 14 and point of mismatch 9, the answer should be 7

@Sherlock9 that just finds the index of the point of mismatch. if there's more, i don't quite understand what the rest of the task is.

Ahhh fair enough

"whichever index in that last list is closest to the point of mismatch" - isn't that always pointOfMismatch+1?

Basically once I've found the indices of the ancestors, say, indices 3 7 14. I need to find the index that's closest or equal to that point of mismatch

@Sherlock9 ok, let me try to re-state it in my own words: so you've got a list of numbers a and a number b and you want to find the element of a that is closest to b?

Exactly!

Sorry for not being clear! I'm having trouble understanding everything myself

a[⊃⍋|a-b]?

To misquote Tom Lehrer

Thank you very much. One moment while I write all this down

googling "Tom Lehrer" ... :)

@ngn I just realized the problem. I'm looking for the first point where P and Q mismatch. That gets me the number of matches between the two lists.

lol :) i knew that voice sounds familiar - it's the guy who turned the periodic table into a song

⎕←{M←⊃⍸≠⌿↑1+P Q←⍵{(⍵/3)⊤⍺-+/3*⍳⍵}¨⌊3⍟1+⍵×2⋄I←⍬{M≥≢⍵:⍺⋄t←{⍵[⊃⍋|⍵-M]}⍵∘{⊃⌽⍸⍺=⍵}¨∪⍵⋄(⍺,t)∇t↑⍵}P⋄'S' 'NE' 'NW'[P[I]],'N' 'SW' 'SE'[Q↓⍨⊃⌽I]}66 5

@Sherlock9
┌─┬──┐
│S│SW│
└─┴──┘

Oh I didn't specify ⎕IO

⋄⎕IO←0⋄⎕←{M←⊃⍸≠⌿↑1+P Q←⍵{(⍵/3)⊤⍺-+/3*⍳⍵}¨⌊3⍟1+⍵×2⋄I←⍬{M≥≢⍵:⍺⋄t←{⍵[⊃⍋|⍵-M]}⍵∘{⊃⌽⍸⍺=⍵}¨∪⍵⋄(⍺,t)∇t↑⍵}P⋄'S' 'NE' 'NW'[P[I]],'N' 'SW' 'SE'[Q↓⍨⊃⌽I]}66 5

@Sherlock9
┌─┬─┬──┐
│S│N│SW│
└─┴─┴──┘

@Sherlock9 in your answer you must specify ⎕fr←1287 too, otherwise it fails for 1162261466 1743392199

@ngn It looks fine to me with ⎕FR←645

I'll keep it in mind though

interesting...

it works on tio, so you're right

in my interpreter it fails, i'm using a model for ⍸: where←{(,⍵)/,⍳⍴⍵}

@Adám Should be fixed

I should add the reason for the algorithm being incorrect to the explanation at some point

I found and fixed another bug while writing up an explanation to the previous bug! XD

reminder: ≈1h left

I really liked this answer and I'd appreciate any golfs you guys might find. There's an awful lot of parenthesizing going on in there

@Sherlock9 115

for my bounty, @H.PWiz gets 350 rep for his 28-byte answer and @dzaima gets an honourable mention for (probably) having come up with a shorter answer but never publishing it.

:)

@Sherlock9 130 for the non-Extended.

@ngn my 23-byte solution: {⍉⍵⌹¯1*+⌿×/2⊥⍣¯1↑⍳⍴⍵}⍣2

@ngn yep, my part of 27 was ¯1*≠⌿∧/2⊥⍣¯1↑⍳⍴⍵, though as i have no idea what matrix multiplication or division is, i had no ability to golf that part

@ngn What was your 27? Like mine+dzaima's?

Also, cool solution!

@H.PWiz yeah, equivalent to those combined

when you said you had 28 i knew you'd figured out the trick with ⌹ :)

and when dzaima said he had -1, i was almost certain he had something equivalent to my way to generate the Hadamard matrix

proof that the 23 solution is correct, starting with H.PWiz's formula:

(⍵⌹h)+.×⌹h  ←→  ⍉(⍉⌹h)+.×⍉⍵⌹h  ←→  ⍉(⌹h)+.×⍉⍵⌹h  ←→  ⍉(⍉⍵⌹h)⌹h  ←→  (⍵(⍉⌹)h)(⍉⌹)h  ←→  h(⍉⌹⍨)h(⍉⌹⍨)⍵  ←→  h(⍉⌹⍨)⍣2⊢⍵

it uses the identities a+.×b ←→ ⍉(⍉b)+.×(⍉a), a⌹b ←→ (⌹b)+.×a, and ⍉h ←→ h (because the Hadamard matrix is symmetric)

@J.Sallé 44 and can probably cut another couple with Extended.