**Paradox Papers**

Robin Robertson

Martin Gardner's "Paradox Papers" was first published in the July 1971 issue of the magic magazine,
*The Pallbearer's Review*. It presented a new topological principle in magic using nothing more than a square
sheet of paper. I'll summarize the principle in its most elementary form, then present a way to force a number
using this principle.

Start with a square piece of paper. A good size is to take a normal 8-1/2" X 11" sheet and cut it to
8-1/2" square. Fold it in half twice in each direction, forming 16 squares. Make all the folds several times
in each direction, so they fold easily along any crease.

Mark large X's and O's in alternating squares like this:

X |
O |
X |
O |

O |
X |
O |
X |

X |
O |
X |
O |

O |
X |
O |
X |

Then turn the paper over--it doesn't matter if you turn it over sideways, or end-over-end--and mark X'S and O'S on this side the same way. You'll find that an X on this side is backed by an O on the other side, and vice versa.

Now fold the square IN ANY WAY you like to make a packet that is only one square high and wide. Just as one of
many possibilities, you might fold the left column over to the right, making a figure 3 columns wide and 4 high.
Then fold the top row down under, and so on. Any way will do. When you're finished, take a pair of scissors and
cut all four edges. Make sure you cut away all folds, as it's easy to miss some if you're not careful.

If you examine the little squares of paper, you'll find that all the X's are face-up and all the O's face-down
(or vice versa). That's the basic principle and Gardner explained several possibilities using the principle. Here's
another.

**Paradox Squares Force**

Assume that instead of X's and O's, you're going to fill the 32 possible squares (i.e, 16 on each side) with unique numbers from 1 to 32. Their sum then will be 32/2 X (32 +1) = 16 X 33 = 528. But, more importantly, if you fill in the sides carefully, when folded and cut, either side will total 16/2 X 33 = 8 X 33 = 264.

In order to do that, simply make sure that if any number "N" goes in an "X" square, "33-N"
also goes in an "X" square, on either side. Similarly, if "M" goes in an "O" square,
"33-M" goes in another "O" square. Here's an example of a filled-in square:

11 | 31 | 8 | 21 |

5 | 27 | 18 | 20 |

1 | 14 | 4 | 28 |

9 | 25 | 16 | 13 |

Side A

32 | 26 | 6 | 17 |

12 | 10 | 2 | 23 |

3 | 7 | 22 | 24 |

19 | 30 | 15 | 29 |

Side B

Presented as a magic trick, you write a prediction of "264" in advance, seal it in an envelope and ask someone to hold it. Then bring out the square and allow another person to fold it any way they like, then cut off the edges. Have someone read off the numbers on top of the 16 squares while another person adds them using a calculator. They'll come up with "264". You then have the envelope opened and the prediction read. Sure enough, you predicted "264" even though you had no way of knowing how the paper would be folded.

With a 16 cell square, you can force any number of the form 264 = 32 n . Any number > 264 can be forced if you're
willing to allow duplicate numbers.

A 9-cell square on each side adds up to either 85 or 86. Just remember that if the front side starts with an X
in the upper left corner, the back side starts with an O (since this is an odd numbered square). Then make sure
all numbers pair up with complements of 19 on the same type of square, as with the 16-cell square discussed above.