I’m not sure if you can do it without being able to turn all the cards.

]]>Let me think a bit more.

]]>In Blaine’s puzzle all of the ‘cards’ have to be the same way up, rather than having to all be face up. They could be all face down.

This makes flipping all of the cards unnecessary.

My solution is the same as his with extra steps to flip all of the cards.

]]>I’m afraid I can’t make the spoiler tag work properly. I’ve tried several times including copying from another comment.

So don’t read this if you don’t want to know my solution.

[spoiler] Suppose the starting states of the four cards is a, b, c, d. I always show the cards in the starting orientation regardless of how the table is rotated.

1) Starting position

a b

c d

2) Flip all cards. We now have covered all scenarios where zero of four cards are flipped.

-a -b

-c -d

3) Flip two cards down a diagonal. As we don’t know the orientation of the table we don’t know which diagonal will be flipped.

a –b OR -a b

-c d c -d

4) Flip all cards. This gives the same state as if we had flipped the other diagonal in step 3. We have now covered all states where a diagonal is flipped.

-a b OR a -b

c -d -c d

5) Flip two cards along a side. Again we don’t know which side is flipped, nor which of the two states we are in at step 4. However we do know that we will have a state with one side flipped relative to the starting position at step 1)

a b

-c -d

Steps 6-8 repeat steps 2-3. This gives all four states with this property (one side flipped). To see this you need to see that each of these states can be got from any other by flipping a diagonal or flipping all the cards.

I only show one possibility. The order will depend on the table rotations.

6) Flip all cards, as step 2.

-a -b

c d

7) Flip a diagonal, as step 3. In this case a and d are flipped.

a -b

c -d

8) Flip all cards, as step 4.

-a b

-c d

Now we have covered all eight possibilities with an even number of cards flipped relative to the starting position.

There are a further eight possibilities with an odd number of cards flipped. To get these we flip one card and then repeat the previous steps.

Flipping one card will put into some state with an odd number of cards flipped. As the further steps lead to an even number of cards being flipped and are distinct this will give us all eight possibilities with an odd number flipped.

I show one possibility.

9) Flip one card. In this case a is flipped.

a b

-c d

10) Flip all cards, as step 2.

-a -b

c -d

11) Flip two cards down a diagonal, as step 3. In this case a and d are flipped.

a –b

c d

12) Flip all cards, as step 4.

-a b

-c -d

13) Flip two cards along a side, as step 5. In this case we flip a and c.

a b

c -d

14) Flip all cards, as step 6.

-a -b

-c d

15) Flip a diagonal, as step 7. In this case we flip a and d.

a -b

-c -d

16) Flip all cards, as step 8.

-a b

c d

So there we are. All sixteen possibilities accounted for. At some point they must all be face up! [/spoiler]

That said, I’m still doubtful that these problems work out to be the same. Your problem on your site calls for the switches to be either all on or all off. All face down isn’t a solution set here.

I’ll look in more detail in a couple of days. Unfortunately I’m packing right now for a trip and I will probably have little time over the next four days or so.

]]>http://puzzles.blainesville.com/2009/05/friday-fun-rapidly-rotating-electronic.html

[spoiler]My puzzle assumed that you couldn’t start with all cards(switches) facing the same way. My solution required no more than 7 steps. With the additional possibility of all cards up or down, you would need one extra step where you flip any cards you like. So the total steps would be 8 in this case.

Summary:

Step 1: Flip any cards you like (1, 2 or 3)

Step 2: Flip two opposite cards.

Step 3: Flip two adjacent cards.

Step 4: Flip two opposite cards.

Step 5: Flip one card (or equivalently, three cards)

Step 6: Flip two opposite cards.

Step 7: Flip two adjacent cards.

Step 8: Flip two opposite cards.[/spoiler]