>Both satisfy the conditions of the puzzle
The winning condition is to have an answer set that implies a unique solution. Both of your suggested answer sets allow for multiple pin numbers:
Sbe nafjre frg (0,0,1,0,1) lbhe fhttrfgrq fbyhgvba 5039 vf abg gur bayl bar: r.t. 3139 naq 3138 jbhyq nyfb or inyvq.Fnzr sbe (0,1,0,0,1): 5219 vf abg gur bayl fbyhgvba, r.t. 3217, 3218 naq 3219 jbhyq nyfb jbex.
By the same logic, the same is true of the original puzzle!
Gur rkcrpgrq nafjref (0,1,1,0,1) qb abg ryvzvangr 3139 naq 3138 nal zber guna (0,0,1,0,1) be (0,1,0,0,1) qb. 3 vf abg 1 naq 3 vf abg bqq, fb dhrfgvbaf 1 naq 4 qb abg ryzvangr vg. Jung qrgrezvarf gur havdhr fbyhgvba vf gur fnzr va nyy pnfrf: gung lbh bayl *trg* n havdhr fbyhgvba vs gur fhz bs gur svefg guerr qvtvgf vf 8. Guvf vf jung sbeprf gur svefg qvtvg gb or 3 sbe gur rkcrpgrq nafjre frg naq 5 sbe gur bguref.
To put it another way, va lbhe bevtvany cbfg lbh rkcyvpvgyl nffhzr
gung gur nafjref gb dhrfgvbaf 2 naq 3 ner obgu "lrf", ohg guvf vf whfg
na nffhzcgvba, naq gur snpg gung vg yrnqf gb *n* fbyhgvba qbrfa'g zrna
vg yrnqf gb gur *bayl* fbyhgvba. Vg pbhyq pbagvahr vafgrnq:
Yrg'f svk gur frpbaq naq guveq qvtvg naq nffhzr gur nafjre gb dhrfgvba 2 vf "ab" naq dhrfgvba 3 vf "lrf".
Vs
gur nafjre gb uvag 5 jnf "ab", jr pbhyqa'g qrgrezvar gur sbhegu qvtvg
ng nyy. Gur bayl jnl jr pna or fher bs gur sbhegu qvtvg vf vs gur nafjre
gb uvag 5 vf "lrf", naq gur bayl jnl guvf qrgrezvarf n fvatyr qvtvg vf
vs gur fhz bs gur svefg guerr qvtvgf vf 8.
Gur svefg qvtvg zhfg
or bqq (be dhrfgvba 1 jbhyq or erqhaqnag) naq abg 1 (be dhrfgvba 4 jbhyq
or erqhaqnag), fb vg'f rvgure 3, 5 be 7. 7 vf gbb ovt - gur fhz jbhyq
or ng yrnfg 10 - fb bayl 3 be 5 ner cbffvoyr. Ohg vs gur svefg qvtvg vf
3, gur frpbaq zhfg or 2, juvpu pbagenqvpgf bhe "ab" nafjre sbe uvag 2.
Fb gur svefg qvtvg zhfg or 5 naq gur frpbaq 0.
Fb abj jr'ir svkrq na nafjre frg (ab, ab, lrf, ab, lrf) gung yrnirf hf jvgu rknpgyl bar cva ahzore: svir mreb guerr avar.
I'm not sure I know what you're getting at because:
>But if the first digit is 3, the second must be 2
Why?
>So now we've fixed an answer set (no, no, yes, no, yes) that leaves us with exactly one pin number: five zero three nine.
This is not correct, because with (no, no, yes, no, yes) the pin 3039 would also be valid, so five zero three nine is not a unique solution.
Re: "Why?", I think I can only repeat part of what I said above: Vs gur nafjre gb uvag 5 jnf "ab", jr pbhyqa'g qrgrezvar gur sbhegu qvtvg ng nyy. Gur bayl jnl jr pna or fher bs gur sbhegu qvtvg vf vs gur nafjre gb uvag 5 vf "lrf", naq gur bayl jnl guvf qrgrezvarf n fvatyr qvtvg vf vs gur fhz bs gur svefg guerr qvtvgf vf 8.
That's also why 3039 is not valid in this case: vs gur svefg qvtvg vf guerr, jr pna'g fbyir gur ceboyrz. Jr'er gbyq jr pna fbyir gur ceboyrz, fb gur svefg qvtvg zhfg or 5.
I think the clearest way I can make this point is: Whfg sebz pbafvqrevat uvag 5, jr xabj gung gur fhz bs gur svefg guerr qvtvgf zhfg or 8: nal bgure inyhr yrnirf rvgure ab fbyhgvba be zhygvcyr fbyhgvbaf. Xabjvat gung, jr pna nafjre uvagf 2 naq 3 nal jnl rkprcg "ab, ab" naq trg n fvatyr fbyhgvba rnpu jnl.
>Just from considering hint 5, we know that the sum of the first three digits must be 8: any other value leaves either no solution or multiple solutions. Knowing that, we can answer hints 2 and 3 any way except "no, no" and get a single solution each way.
I think this is where both our reasonings are "the other way round": With this being a Constraint Satisfaction Problem without an a priori fixed set of constraints, you're instead treating this more like an iterative deduction based problem?
Also sorry if you've already stated that elsewhere and I didn't catch that, but what exactly is your actual problem with this puzzle? I mean it clearly has a single unique solution (ie. exactly 1 set of answers that results in exactly 1 pin) which you can proof by just bruteforcing all permutations of pin numbers and answers.
I think the puzzle is very clever! I just don't think the accepted answer actually fits the parameters, and I think two other answers do. The point of the puzzle is to deduce the values of the digits from the hints, and you can deduce from hint 5 (together with the promise that the puzzle is solvable) what the fourth digit is and what the other three sum to. With that information, hints 1 and 4 are redundant, and two answers are possible. The accepted answer is only unique and correct (with no redundant hints) if you ignore some of the information hint 5 gives you.
Just to see if I'm clear on the disagreement here: it's about whether or not it's valid to use "there is a unique solution" as part of the constraints?
Because I had started by assuming that since hint 5 is the only thing that constrains digit four, in order to have a unique solution, the first three digits must sum to either eight or zero. And that still seems obviously true to me.
But your position is that "there is a unique solution" is sort of an external check and may not be used as an actual constraint? I can see that if you're not allowed to say "there is a unique solution, therefore the digits must sum to 8," then you're left with "IF the OTHER hints fail to force the digits to sum to 8, THEN you don't have a unique solution." And that makes hints 1 and 4 are NOT redundant in the yes/yes/no case for hints 2/3/5.
Do I have that right?
It still does seem very strange to me that unique solution would not be an allowable constraint when non-redundant hints is. So yeah, seems like a "both reasonings are the other way around" problem, and therefore a poorly-designed puzzle. But hey, game jam.
Ah, dustbunnies. Now I see the error in my logic. It's true that hint 5 says that ifthere is a unique solution, it must have this property. But that doesn't help you answer the question of whether a particular set of hint answers rules out all of the non-unique answers that may or may not have that property. Gah. OK.
