Lateral Thinking XX *NEW PROBLEM*

[Holmes]

Here´s a real Lateral Thinking that will need a logical as the ones before this one, as many people know, like the one of the humans and dogs, the toughest ones: Lateral Thinking VI and VII, the one of the YES, NO, RANDOM people.
It is not THAAAT difficult but it has it little things.

THERE IS NO TIRCK, at this moment I´ll have to say this because of Lateral Thinking XIX, where there was a "trap" and no logical mind needed, no offence c-square.

Well, no more presentation needed:

Five competitors -- A, B, C, D, and E -- enter a swimming race that awards gold, silver, and bronze medals to the first three to complete it. Each of the following compound statements about the race is false, although one of the two clauses in each may be true.

â—¦A didn't win the gold, and B didn't win the silver.
â—¦D didn't win the silver, and E didn't win the bronze.
â—¦C won a medal, and D didn't.
â—¦A won a medal, and C didn't.
â—¦D and E both won medals.
Who won each of the medals?


Till Lateral Thinking XXI! And a new murder case ...there are five i think, not very popular ...

[c-square]

THERE IS NO TIRCK, at this moment I´ll have to say this because of Lateral Thinking XIX, where there was a "trap" and no logical mind needed, no offence c-square.

No offense taken.    I tend to think of these sorts of questions as logical thinking, whereas lateral thinking involves thinking outside the implied boundaries of the situation.  Just a difference in personal definitions, IMHO.

As for this question, isn't it the same as the one in Lateral Thinking XVIII?

[S.H.]

Isn't that the one on LTXVIII? 

[miakakiri]

Didn't we have this one already?

[Holmes]

Damm it! Too many Lateral Thinking problems and I forgot that I had already put it, here´s Lateral Thinking XX,

A card-shuffling machine always rearranges cards in the same way relative to the original order of the cards. All of the hearts, arranged in order from ace to king, were put into the machine. The cards were shuffled and then put into the machine again. After this second shuffling, the cards were in the following order: 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7. What order were the cards in after the first shuffle?

[c-square]

A card-shuffling machine always rearranges cards in the same way relative to the original order of the cards. All of the hearts, arranged in order from ace to king, were put into the machine. The cards were shuffled and then put into the machine again. After this second shuffling, the cards were in the following order: 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7. What order were the cards in after the first shuffle?

My solution:

Spoiler:

There are 13 positions for the cards, below numbered 1 through 13.  We know that after two shuffles, the cards get moved as follows:

1 ->  a  ->  8
2 ->  b  -> 12
3 ->  c  ->  6
4 ->  d  ->  7
5 ->  e  ->  9
6 ->  f  -> 11
7 ->  g  -> 13
8 ->  h  ->  4
9 ->  i  ->  2
10 ->  j  ->  1
11 ->  k  -> 10
12 ->  l  ->  3
13 ->  m  ->  5

where the letters a through m represent the cards' positions after the first shuffle.

Using these, we can make a series by just following the arrows and matching the numbers:

1 ->  a ->  8 ->  h ->  4 ->  d ->  7 ->  g -> 13 ->  m ->  5 ->  e ->  9 ->  i ->  2 ->  b -> 12 ->  l ->  3 ->  c ->  6 ->  f -> 11 ->  k -> 10 ->  j -> 1

Now, in the above series, each number shows up twice, once in numeral format (1 - 13) and once represented by one of the letters (a - m). 

So, let's start by figuring out which letter represents the number 1.  I'll guess the letter 'i' represents the number 1.  If that's the case, then we can rewrite the series as:

1 ->  a ->  8 ->  h ->  4 ->  d ->  7 ->  g -> 13 ->  m ->  5 ->  e ->  9 ->  i
i ->  2 ->  b -> 12 ->  l ->  3 ->  c ->  6 ->  f -> 11 ->  k -> 10 ->  j -> 1

Note, that once I've chosen the letter 'i' to be the number 1, all the other letters get paired up with a number.  'a' is paired with 2, 'b' with 8 and so on.  Replacing all the letters with numbers, we get:

1 -> 2 -> 8 -> 12 -> 4 -> 3 -> 7 -> 6 -> 13 -> 11 -> 5 -> 10 -> 9 -> 1

This is one method that the shuffler could use to go from a sorted set of 13 cards to the mixed set given in the puzzle in two shuffles.

So, by applying this only once to the sorted set, we can figure out what the order of the cards were after the first shuffle:

A -> 2nd spot
2 -> 8th spot
8 -> 12th spot, etc..

This results in an order of

9,  A,  4,  Q,  J,  7,  3,  2, 10,  5,  K,  8,  6

after the first shuffle.

Now, note, I've just shown one method.  Could there be more?

Luckily, the answer is no.  If we go back to the original series:

1 ->  a ->  8 ->  h ->  4 ->  d ->  7 ->  g -> 13 ->  m ->  5 ->  e ->  9 ->  i ->  2 ->  b -> 12 ->  l ->  3 ->  c ->  6 ->  f -> 11 ->  k -> 10 ->  j -> 1

by substituting any letter other than 'i' for the number 1, you end up with contradictions.  For example, say 'b' was the number 1.  That would mean our series would look like:

1 ->  a ->  8 ->  h ->  4 ->  d ->  7 ->  g -> 13 ->  m ->  5 ->  e ->  9 ->  i ->  2 ->  b
b -> 12 ->  l ->  3 ->  c ->  6 ->  f -> 11 ->  k -> 10 ->  j -> 1

If this were the case, 1 would be represented by 'b', but it would ALSO be represented by 'e' because of the paring:

m ->  5 ->  e
10 ->  j -> 1

Also, the 9, i, and 2 have nothing to pair to. 

Using any other letter than 'i' to represent the number 1 would give you the same bad result.

Therefore, the ONLY letter that can be used to represent the number 1 is 'i', meaning the ONLY shuffling algorithm possible is:

1 -> 2 -> 8 -> 12 -> 4 -> 3 -> 7 -> 6 -> 13 -> 11 -> 5 -> 10 -> 9 -> 1

and the ONLY result for the cards after one shuffle is:

9,  A,  4,  Q,  J,  7,  3,  2, 10,  5,  K,  8,  6

[Holmes]

Nice one! Correct!