5 ms·
Multiple choice
- stuartmalcolm 11y agoe
- toabi 11y agowas my first guess too
- jamies888888 11y agob
- widdma 11y agoBut then "d" will be true, contradicting "b".
- sanjeetsuhag 11y agoThe answer is E.
- d_theorist 11y agoe
- fao_ 11y agoEither e or f, I can't think of a reason why it can't be f...
- keerthiko 11y agoBecause e == f, and if f is true then e is true, in which case f is false, hence a contradiction, hence it cannot be f.
- addandsubtract 11y ago>Because e == f, and if f is true then e is true, in which case f is false e != f, because they produce different results. You could say e() == f(), each having the previous answer choices as arguments and each requiring them to be false. But since e() evaluates to true, f() can't be true.
- contravariant 11y agoA better way to phrase it would be that, f implies e, but e doesn't imply f. In fact e implies 'not f'.
- fao_ 11y agoAaaaah, I understand now. Thanks :3
- boothead 11y agoE A very interesting mental brute force process to work it out though. How did other people work it out, is there a more elegant method?
- deleted 11y ago[deleted]
- scoot 11y agoSpoiler alert. View the question before reading answers posted here.
- peteretep 11y agoSimple solution: * A and B can't both be true, which means that C is false. * If C is false, A must be false. * B and D have a paradoxical relationship, given the others - if B were true, D would be true, which would make B false, so B must be false. * D is false, given A, B, C are false. * E is true, given A, B, C, D are false. * F is false because E is true (typo corrected as per comment)
- imron 11y agoF is false because D is true. F is false because E is true.
- jtheory 11y agoThat's a fun quick brainteaser! Reminds me of those old puzzles where you'd have a list of facts -- like there are 7 houses in a row, all different colors and with different species of pets in the yard, the blue house is next to the orange house, the green house has a pet with 2 legs, etc. etc., and they provide just enough info that you can logically chain together the order of colors and pets. They're a bit like the word-problem version of working out sudoku (vs. guessing at it, which is a different game).
- eutectic 11y agoThis is a perfect use case for a SAT solver. Here is a solution using the Python Z3 bindings: from z3 import * def all(vars): result = BoolVal(True) for var in vars: result = And(result, var) return result def any(vars): result = BoolVal(False) for var in vars: result = Or(result, var) return result def none(vars): return Not(any(vars)) def oneof(vars): result = any(vars) for i, a in enumerate(vars): for b in vars[i + 1:]: result = And(result, Not(And(a, b))) return result vars = [Bool(c) for c in "abcdef"] def above(i): return vars[:i] def below(i): return vars[i + 1:] s = Solver() s.add(vars[0] == all(below(0))) s.add(vars[1] == none(below(1))) s.add(vars[2] == all(above(2))) s.add(vars[3] == oneof(above(3))) s.add(vars[4] == none(above(4))) s.add(vars[5] == none(above(5))) s.check() print(s.model()) For what it's worth, the answer is E. :)
- shoo 11y agoYup, I agree! After figuring it out by hand I was having fun encoding it for sat via this horrible pipeline: step 1 : write it out by hand in PBL [alpha] step 2 : convert it to CNF format using [beta] step 3 : hand-reformat it tediously to produce [gamma] step 4 : plug machine readable CNF into [delta] step 5 : receive a feasible solution! [epsilon] [alpha] Main_Exp : ((a <=> (b & c & d & e & f)) & (b <=> (~c & ~d & ~e & ~f)) & (c <=> (a & b)) & (d <=> (a | b | c)) & (e <=> (~a & ~b & ~c & ~d)) & (f <=> (~a & ~b & ~c & ~d & ~e))) [beta] http://formal.cs.utah.edu:8080/pbl/PBL.php http://formal.cs.utah.edu:8080/pbl/PBL.php ... find replace, find replace, ... [gamma] c multiple choice p cnf 6 29 -1 2 0 -1 3 0 -1 4 0 -1 5 0 -1 6 0 -2 -3 -4 -5 -6 1 0 -2 -3 0 -2 -4 0 -2 -5 0 -2 -6 0 3 4 5 6 2 0 -3 1 0 -3 2 0 -1 -2 3 0 -4 1 2 3 0 -1 4 0 -2 4 0 -3 4 0 -5 -1 0 -5 -2 0 -5 -3 0 -5 -4 0 1 2 3 4 5 0 -6 -1 0 -6 -2 0 -6 -3 0 -6 -4 0 -6 -5 0 1 2 3 4 5 6 0 [delta] http://www.msoos.org/2013/09/minisat-in-your-browser/ http://www.msoos.org/2013/09/minisat-in-your-browser/ [epsilon] This is MiniSat 2.0 beta ============================[ Problem Statistics ]============================= | | | Number of variables: 6 | | Number of clauses: 29 | | Parsing time: 0.00 s | ============================[ Search Statistics ]============================== | Conflicts | ORIGINAL | LEARNT | Progress | | | Vars Clauses Literals | Limit Clauses Lit/Cl | | =============================================================================== | 0 | 6 29 75 | 9 0 nan | 0.000 % | =============================================================================== Verified 29 original clauses. restarts : 1 conflicts : 2 (2 /sec) decisions : 5 (0.00 % random) (5 /sec) propagations : 10 (10 /sec) conflict literals : 5 (28.57 % deleted) CPU time : 1 s SATISFIABLE v -1 -2 -3 -4 5 -6 0 5 is true, aka e. edit: I think I am someone who interprets "one of the above" as "at least one of the above but possibly more" not "exactly one of the above". not that the interpretation makes a difference for this problem.
- StavrosK 11y agoI just went down one by one and ruled them out, I'm surprised that people have harder ways to do it. My reasoning for ruling out each one: a) b contradicts it. b) cde encompass all the choices, so you can't have none of those. c) a contradicts b. d) abc have already been rejected. e) works. f) e already works.
- mikehawkins 11y agoAbsolutely - thanks for sharing; you were far more concise than how I was going to explain my solution. :)
- jethroalias97 11y agoFigured I'd dust off my old constraint programming language (https://github.com/dbunker/SABR/blob/master/test/Real/MultipleChoice/multi.tb https://github.com/dbunker/SABR/blob/master/test/Real/Multip...) with result (https://github.com/dbunker/SABR/blob/master/test/Real/MultipleChoice/multi-expected.txt https://github.com/dbunker/SABR/blob/master/test/Real/Multip...). Fifth position being E. A bit on the verbose side, but might be preferable to some since it explicitly enumerates all the possibilities for each rule.