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Topic: More True-or-False Statements (Read 1418 times) |
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ThudnBlunder
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More True-or-False Statements
« on: Jan 20th, 2005, 4:28am » |
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Below are 12 statements that are either true or false.
Can you find which are which?
1. This is a numbered list of 12 true-or-false statements.
2. Exactly 3 of statements 7 to 12 inclusive are true.
3. Exactly 2 of the even numbered statements are true.
4. If statement 5 is true, then statements 6 & 7 are both true.
5. Statements 2 to 4 are all false.
6. Exactly 4 of the odd numbered statements are true.
7. Either statement 2 or 3 is true, but not both of them.
8. If statement 7 is true, then statements 5 & 6 are both true.
9. Exactly 3 of statements 1 to 6 inclusive are true.
10. The following two statements are both true.
11. Exactly 1 of statements 7, 8 & 9 is true.
12. Exactly 4 of the other statements are true.
(By the way, here is an interesting article about Ed Witten.)
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| « Last Edit: Jan 20th, 2005, 6:52am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Three Hands
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Re: More True-or-False Statements
« Reply #1 on: Jan 20th, 2005, 4:49am » |
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Some initial thoughts:
Statements 1, 5, 8 and 11 are true, the rest are false, seems to work.
Statement 1 is true, fairly trivially. Assuming statement 5 is true makes 2, 3 and 4 are false. 4 being false means 6 or 7 is false, and 7 is false by definition. This makes 8 true (If... then are only false when the "if" part is true and the "then" false) For 9 to be true, 6 must be true, which then means that 3 is only false if at least one of 10 and 12 are true, and 6 will only be true if 11 is true. This would then mean both of 10 and 12 would be true, which leaves far too many true statements for 12 to be true, and 11 would be false. Therefore, both 6 and 9 are false. 11 is therefore true, but 12 cannot be true without 10 being true, which would make 12 false, so both 10 and 12 are false.
Are you still there?? 
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ThudnBlunder
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Re: More True-or-False Statements
« Reply #2 on: Jan 20th, 2005, 1:33pm » |
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Three Hands, your conclusion of 4 true statements implies that 12 (and therefore also 10) is correct, too.
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Grimbal
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Re: More True-or-False Statements
« Reply #3 on: Jan 20th, 2005, 2:52pm » |
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It depends whether statements refers to statements other than 7,8,9 or other than 12.
In the 1st case, TH's solution is OK, or we can take 10 and 12 to be true also.
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ThudnBlunder
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Re: More True-or-False Statements
« Reply #4 on: Jan 20th, 2005, 3:37pm » |
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on Jan 20th, 2005, 4:49am, Three Hands wrote:
Statement 1 is true, fairly trivially. |
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Do you think the first sentence of the puzzle is redundant then? 
on Jan 20th, 2005, 2:52pm, Grimbal wrote:| It depends whether statements refers to statements other than 7,8,9 or other than 12. |
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As usual with this type of puzzle, each statement stands alone and has no syntactical connection with any other. Therefore, 12 refers to all the preceding 11 statements.
on Jan 20th, 2005, 2:52pm, Grimbal wrote:| ...or we can take 10 and 12 to be true also. |
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10 and 12 True => 11 True
Hence 1, 5, 8, 10, 11, 12 are True
Therefore 12 is False.
I get that assuming 5 to be True leads to a general contradiction
(but don't have time to analyse further at the moment).
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| « Last Edit: Jan 21st, 2005, 9:17pm by ThudnBlunder » |
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Grimbal
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Re: More True-or-False Statements
« Reply #5 on: Jan 21st, 2005, 4:51am » |
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on Jan 20th, 2005, 3:37pm, THUDandBLUNDER wrote:10 and 12 True => 11 True
Hence 1, 5, 8, 10, 11, 12 are True
Therefore 12 is False. |
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Uh... right. What was on my mind?
But about the "other than 12" version:
:: True: 1 3 4 6 7 11
False: 2 5 8 9 10 12 ::
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Three Hands
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Re: More True-or-False Statements
« Reply #6 on: Jan 21st, 2005, 5:50am » |
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Bother - missed that one. Thought it looked too simple...
For the first statement, though, unless you come up with an incredibly good argument about the nature of "true-or-false statements", or what constitutes a "numbered list", or deny that there are exactly 12 members in the list which fit the criteria given, or debate what the term "this" refers to, then you would have to accept the validity of the statement. Probably the most successful attempt to prove the statement false, therefore, would be to suggest that "this" actually refers to either a) just the statement itself, or b) the entire post - neither of which is a numbered list of 12 true-or-false statements (although the latter contains it)
Perhaps it would be easier to say that statement 1's truth value is ambiguous
EDIT: Of course, denying the validity of statement 1 would then lead to a fairly obvious modification of my solution being true  - too bad I don't think the puzzle setters would agree with the reasoning for statement 1...
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| « Last Edit: Jan 21st, 2005, 5:53am by Three Hands » |
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Grimbal
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Re: More True-or-False Statements
« Reply #7 on: Jan 21st, 2005, 8:13am » |
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Hehe... and it tells about "the other statements", why should it be limited to these 12 statements?
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fatball
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Re: More True-or-False Statements
« Reply #8 on: Jan 21st, 2005, 11:59am » |
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ThreeHand, you only assumed that statement 5 is true to start with, but is it really true and what should follow if the statement is false? I really need soemone to help me think outside the loop...........
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ThudnBlunder
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Re: More True-or-False Statements
« Reply #9 on: Jan 21st, 2005, 10:42pm » |
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on Jan 21st, 2005, 4:51am, Grimbal wrote:
But about the "other than 12" version: |
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Grimbal, your solution checks out.
But were you able to logically deduce it?
on Jan 21st, 2005, 5:50am, Three Hands wrote:
For the first statement, though, unless you come up with an incredibly good argument about the nature of "true-or-false statements"... |
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However, when dealing with self-referential statements such as these, a third possibility is 'logically undecidable'.
Which is why I explicitly stated in the first sentence of the puzzle that all the statements are either true or false.
That is, if we cannot assume #1 to be True then probably there is not enough information to solve it.
Are you still there??
For a simple example, see Logical Signs I.
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| « Last Edit: Jan 22nd, 2005, 12:38pm by ThudnBlunder » |
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Three Hands
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Re: More True-or-False Statements
« Reply #10 on: Jan 24th, 2005, 4:06am » |
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Indeed. As has been shown, by Grimbal's solution and then my "attempted" version, if you deny the veracity of statement 1, then there is more than just one unique solution.
Fatball, I think I covered all possible lines of enquiry for assuming 5 to be true in my method, which results in no correct answers. Basically: 1 is true (for the sake of the puzzle, ignore the arguments about 1 potentially not being true...)
If 5 is true, then 2,3 and 4 must be false. This then prevents having 3 of the statements 7-12 inclusive being true, exactly 2 of the even-numbered statements being true, and both statements 6 and 7 being true. Due to neither statement 2 or 3 being true, 7 cannot be true, and, due to the nature of "if-then" statements only being false if the "if" is true, and the "then" is false, statement 8 is true without any direct consequence. 6,9,10,11 and 12 were all covered in my initial post, and the result therefore shows that 5 must be false...
Assuming 5 to be false leads to: Statement 1 is true. Statement 4 must be true, since the "if" for it's condition is not. If statement 7 is true, then statement 8 is false, and either statement 2 or 3 is true. Then, either statement 9 is true, and statements 6 and 11 are false, or statements 6 and 11 are true, and statement 9 is false. If 8 is true, then either 11 is true, or 9 and 6 are true. For 6 to be true, you need 4 odd-numbered statements true. In the case of 8 being true, we have 1 and 9, so 6 would be false, making 9 false and 11 true. This then gives us 1, 4, 8 and 11 true, which then requires either 3 to be true, and hence 7, which then leads to contradictions, or 12 to be true, which makes 10 true, and makes 12 false, etc. In the case of 7 is true, then we need to have either 2 or 3 true. for 2 to be true, 7, 9 and some other statement would be true, but none of 8, 10, 11 or 12 would work. Hence 3 must be true, and so one other even numbered statement must be true. This denies the truth of 10. This gives us 1, 3, 4, 7 as true, 2, 5, 8 and 10 as false, and 6, 9, 11 and 12 uncertain. For 12 to be true, no other statement would be true, but 6 and 11 would be true. Hence, 12 is false. If 9 is true, no other statement would be true, but 3 would not be satisfied. Hence, 1, 3, 4, 6, 7 and 11 is the only solution.
Sorry that's a bit garbled, I was in a bit of a rush...
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Barukh
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Re: More True-or-False Statements
« Reply #11 on: Jan 25th, 2005, 8:57am » |
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on Jan 21st, 2005, 10:42pm, THUDandBLUNDER wrote:
Grimbal, your solution checks out.
But were you able to logically deduce it? |
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There is a purely mechanical (although awful) method to solve such problems using Boolean algebra. I will give a sketch for the solution of this one (don’t try it at home!).
Let S1, …, S12 be Boolean variables denoting the truth/falsehood of the corresponding statement. Because of the nature of the given statements, we can write Boolean formulas for every one in terms of others. Some formulas will be extremely large (but finite ). For instance (~ is NOT, | is OR, & is AND):
S1 := 1
S4 := ~S5 | (S6 & S7)
S5 := ~(S2 & S4)
S7 := (S2 & ~S3) | (~S2 & S3)
S8 := ~S7 | (S5 & S6)
S10 := S11 & S12
I deliberately omitted other formulas to save place . For instance, the formula for S12 will contain 11-choose-4 = 330 terms combined by OR-operations, each term being an AND of exactly 4 variables.
To find the solutions, we need to evaluate the formula S1 & S2 & … & S12. If it simplifies to logical 0, there is no solution; it is simplifies to a single combination of AND-s, there is a unique solution; otherwise, there are multiple solutions.
After what has been done here, I suspect the formula should simplify to S1 & ~S2 & S3 & S4 & ~S5 & S6 & S7 & ~S8 & ~S9 & ~S10 & S11 & ~S12.
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Eigenray
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Re: More True-or-False Statements
« Reply #12 on: Jan 25th, 2005, 2:21pm » |
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I don't think it's that simple. Which would you define first, S2 or S7? And how would you apply it to
1. Statement 2 is true.
2. Statement 1 is true.
It seems like you'd need 24 variables, for example,
A4 = ~B5 | (B6&B7)
A5 = ~(B2|B3|B4)
A7 = B2 ^ B3
etc
and then solve (A1<->B1)&...&(A12<->B12)
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towr
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Re: More True-or-False Statements
« Reply #13 on: Jan 25th, 2005, 3:03pm » |
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Why not just temporarily assume truthvalues for each statement, evaluate each statement, and see if the resulting truthvalue equals the once you assumed. 2^12 isn't that much to let a computer muddle through.
Quote:| To find the solutions, we need to evaluate the formula S1 & S2 & … & S12. |
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No, then you're assuming they're all true, and obviously they can't all be true.
I think you'd have to eliminate variables, just like you'd do with a system in algebra. So replace S1 with 1, in S2..12 etc
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SWF
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Re: More True-or-False Statements
« Reply #14 on: Jan 25th, 2005, 5:33pm » |
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The solution Grimbal gives is the only one, as can be shown by deduction. To help with those confusing 'if' statments, a statement of the form "If A then B" can only be false if A is true and B is false. So two new rules which must be true are:
If 8 is false, 7 is true.
If 4 is false, 5 is true.
The True/False value of each statement can be deduced in the following order:
1, 10, 12, 11, 9, 4, 5, 8, 7, 6, 3, 2
It turns out that 2 can be any false statement.
You could, of course, run through all the values as towr suggests, but I think this is one of those cases where brute force defeats the purpose of the riddle. However, that is undeniably the easiest way to solve this.
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