Subscribe via feed.

Posted by luv2fap on February 18, 2011 – 2:06 pm

A well{known logical joke is a card saying on one side “The statement
on the other side is false” and on the other, “The statement on the
other side is true”. Clearly, it is impossible for either of the statements
to be true or false without leading to an inconsistency. However, if
both sides said “The statement on the other side is false” or “The
statement on the other side is true”, it is possible for these statements
to be logically consistent.
Now suppose you are given a stack of n cards, each of which says either
“The statement on the next card is false” or “The statement on the
next card is true”. Is it possible to arrange the cards into an order
which makes the whole pack logically consistent?

Tags:
This post is under “Tom” and has 8 respond so far.

### 8 Responds so far- Add one»

1. 1. Karl Sharman Said：

I would only say no as the last card will not have a following card to validate the statement.
If n however, is infinite, then yes.

2. 2. DP Said：

I’m going to guess you mean that the final card will refer you to this first (circular referencing).
In this case, as long as when you see a “false” card, you follow it with another “false” you should be ok.
could look something like this:
the statement below is true
the statement below is true
the statement below is true
the statement below is false
the statement below is false
the statement below is true
the statement below is false
the statement below is false
the first statement is true

3. 3. Chris Said：

Using the shorthand t and f denote the literal message printed on the cards, t => “the next is true”, f => “the next is false”, and T and F to represent the cumulative message that is effectively printed on the card.

Then Tf => TF => F, Tt => Tt => T, Ff => FT => T, Ft => FF => F
where the single card is the net effect of the two cards.

It is pretty obvious that T’s have no effect on the state of the effective messages on the subsequent cards, and that each f reverse the effective message on the subsequent cards.

To determine the effective message on a given card, count how many f cards there are preceding it. If #f is an even number, then the effective message = the actual message and if #f is an odd number, the effective message is the reverse of the actual message.

If the deck is not circular, any sequence of cards is valid, as no contradictions arise. If the deck is circular, then it only seems sensible that each card has a unique state i.e. it must be the same every time you go round the loop. So there must be an even number of f cards. That conveniently means that it doesn’t matter where you start in the circle provided you take the first card to be literally telling the truth.

It gets a bit silly if you assume the first card is preceded by an F for the circular case.

4. 4. Chris Said：

Rats. Third para, first line should have been t’s not T’s

5. 5. Chris Said：

Double rats. Second para, I wrote a Tt when I meant a TT.

In the circular case, if you start on a t (and assume it’s a T) then you’ll get one set of effective messages. But if you start on an f and assume it’s an F, all the effective messages will be the reverse of the t starting case. But as long as the number of f’s is even, there are no inconsistencies.

I think I’ve got that right; I’m too tired to think straight.

6. 6. luv2fap Said：

We assume that the last card in an arrangement refers to the rst
card. We are going to show that regardless of the order, the cards can
be logically consistent if and only if the number of cards which state
that the next card is false is even.
Suppose we have the deck of cards together with some consistent assignment
of truth value to each card. Label each card by T if it says
that the statement on the next card is true, and by F if it says that
the statement on the next card is false. Also label each card by t if it
is true, and by f if it is false. Thus each card has one of the labels Tt,
Tf, Ft, or Ff.
We rst show that any card labelled T can be removed from the deck
without altering the remaining labels. If it is labelled Tt, the preceding
card must be either Tt or Ff and the next card must be either Tt or
Ft. Thus removing the given Tt card does not change anything in the
rest. If it is labelled Tf, then the preceding card must be labelled Tf or
Ft and next card must be either Ft or Ff. Once again, removing the
given Tf card does not change the rest.
Hence we now assume that the remaining cards are all labelled Ft or
Ff. Next we show that removing a pair of adjacent F cards does not
alter the remaining labels. First note that a pair (Ft, Ft) cannot be
consistent, since Ft truthfully claims that the statement on the next
card is false. Similarly, (Ff, Ff) cannot be consistent, since Ff falsely
claims that the statement on the next card is false, so that statement
must be true. Hence any adjacent pair must be (Ft, Ff), in which case
the preceding card is Ff and the next card is Ft; or (Ff, Ft) in which
case the preceding card is Ft and the next card is Ff. In both cases,
the pair can be removed.

7. 7. luv2fap Said：

*Thus we finally reach a pair (Ft, Ff) or (Ff, Ft) which are consistent, or
a single Ft or Ff which corresponds to `The statement on the other side
is false’ which is inconsistent, a contradiction. Hence the original deck
was consistent if there are an even number of F cards, and inconsistent
if there are an odd number.

8. 8. Sean Said：

No, the N+1 card will never exist and therefore a statement about it is implicitly false.

PHP Warning: PHP Startup: Unable to load dynamic library 'C:\Program Files (x86)\Parallels\Plesk\Additional\PleskPHP5\ext\php_mssql.dll' - The specified module could not be found. in Unknown on line 0 PHP Warning: PHP Startup: Unable to load dynamic library 'C:\Program Files (x86)\Parallels\Plesk\Additional\PleskPHP5\ext\php_pdo_mssql.dll' - The specified module could not be found. in Unknown on line 0