Multiple choice

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.

boothead• 11 years ago

E 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?

scoot• 11 years ago

Spoiler alert. View the question before reading answers posted here.

peteretep2• 11 years ago

How does knowing the answer help you solve the problem?

eCa• 11 years ago

It doesn't. Knowing the answer means you can't solve it without being influenced of other people's answers/thoughts.

peteretep2• 11 years ago

Simple 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• 11 years ago

F is false because D is true.

F is false because E is true.

jtheory• 11 years ago

That'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• 11 years ago

This 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• 11 years ago

Yup, 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

... 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/

[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• 11 years ago

I 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• 11 years ago

Absolutely - thanks for sharing; you were far more concise than how I was going to explain my solution. :)

jethroalias97• 11 years ago

Figured I'd dust off my old constraint programming language (https://github.com/dbunker/SABR/blob/master/test/Real/Multip...)

with result (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.

Multiple choice

Related Stories