A Big Bag of Chips

Assuming that the magician and his assistance both had a pencil and some paper (for notes), were proficient at adding and subtracting, and had practiced, the trick would take about an hour using my simple algorithm (the algorithm could probably be improved). A couple of programmable calculators would speed the trick up significantly. The puzzle is to figure out how someone can encode the information that specifies the sixth card using the five cards.

Carl G.

Hi Carl, Moshe,

Having considered what I believed might be the 'most difficult case' I felt I was merely a 'twist of an idea' from the solution. That it would come to me Monday evening as I watched Inspector Morse on 'BBC in America' (or some other very similar channel name). Having read over this exchange I believe I'll give 100% concentration to Morse. I'll look forward to reading the answer, however, but only if it is 5 pages or less.

Best wishes, Jim

The magician uses mod(6) to add up the 6 numbers. He uses the result to determine which card to take away. If the sum mod 6 is 0, he takes away the card with the lowest number etc.

Now the assistant has to do some calculations: He first assumes that the sum mod(6) of all cards equals 0. This means, that the first card was discarded. So he looks for all numbers lower than the lowest of the five numbers he sees, that result in 0 when added (mod 6) to the known numbers. Then he assumes that the sum mod(6) is 1. This means, that the second card was discarded. So he looks for all numbers between the lowest and second lowest of the numbers he sees that result in 1 when added (mod 6) to the known numbers.

etc.

Now he has 120 numbers left.

The left-over numbers can put in 120 combinations. 12345 (1 is lowest number, 5 highest) is combination 1 12345 is combination 2 etc.

If he has combination X, the sought after number is the X-lowest number of the numbers he has left.

I'm sure you mean 12354 for the second combo.

Very nice, and clearly generalizable to other numbers of chips, playing cards, &c. I was thinking about something along those lines. In fact, I thought of exactly that method for choosing which chip to exclude. But I couldn't convince myself that it would be uniquely decodable.

After seeing your description, I decided to work through an example. And having done so I figure I may as well post it here for other people. I still need to thoroughly convince myself by proving there will always be 120 possible solutions, but it seems like that would be true.

Suppose the original numbers are

130-259-527-555-630-631

The mod-6-sum is 2, so the magician takes away the third lowest chip

Cool problem by the way. I see a similar post about the 27 card variant of this over in alt.math.recreational.

Bob H

: highest). I may have been generous suggesting that the trick would take an : hour. A look-up chart with the 120 sorted combinations (12345 to 54321) : would make things easier for the assistant.

Why the assistant? Things are tougher for the magician, no?

I think most people (well, most people who read this group) could train themselves to do the magician part in less than a minute, and the assistant part in half a minute, without paper.

It is pretty easy to convert from combinations to the sort index simply by mimicking base conversion, but with a rolling base.

To take the example of 88, which occured above.

88 div 2 is 44, remainder 0, so the last digit in the rolling base version is 0

44 div 3 is 14 r 2

14 div 4 is 3 r 2.

So our rolling representation is 3220.

To convert to a permutation, start from the left and choose the i-th remaining digit, where 0 stands for the smallest remaining digit.

Thus 3220 goes to 43512. It helps to write 12345, and cross out digits as you choose them. Making the permutation, and then remembering it as you choose from the 'real' numbers, is a bit tricky to do in your head, I think. The base 6 calculations are also potentially confusing.

Going backwards is even easier. Start from the right, but with the last _choice_ that was made, the 1. Rank each digit with respect to those to its right only (how many is it bigger than). This retrieves the 3220. Then build the number by multiplying each digit by the appropriate rolling base and then adding the next digit.

3 x4+2 = 14 x3+2 = 44 x2+0 = 88

There should also be a simple way to jump from the number to the index (and back) by dividing by 6 and making some correct adjustment, although my brain is too fuddled now to get the adjustment right.

I really think you could train yourself to do this trick, but base 5 is sufficiently easier that I am thinking of training myself to do the simpler, 5-number version instead.

Jonathan Dushoff dushoff at eno.princeton.edu

You can shorten google links down to this form:
http://groups.google.com/groups?selm=2pj5i7%248jf% 40louhi.to.icl.fi

(You lost a '%' during the formatting ...)

Nis Jorgensen schrieb:

... and drop off the http:// if need be.

Have fun shortly

Well, how would you (or your newsreader) know the protocol then?

Nis Jorgensen schrieb:

I do, since I know google groups. My browser assumes http:// when I copy&paste this into the 'Location' box. Since I need to copy&paste anyhow if the line is broken, this is no disadvantage. My newsreader doesn't know, but then, I'm using it for news:, not for http:

ObPuzzle: Wouldn't 'nntp:' be more appropriate than 'news:'?

Have fun with the web

Assume HTTP unless otherwise stated.