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Topic: Strip Teez (Read 2847 times) 

rloginunix
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Strip Teez
« on: Aug 5^{th}, 2016, 9:24am » 
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Strip Teez A strip of paper 2^{n} sequentially numbered unit squares long is placed on a table along the OX axis in such a way that its leftmost square, numbered as 1, is fixed at origin in the XOY plane. The strip then undergoes n like transformations as follows:  the strip is folded in half upward in such a way that its current rightmost square is rotated counterclockwise 180 degrees in the ZOX plane about the current strip's middle until it is placed on top of the strip's leftmost square After the nth fold is complete, the strip is one squarelong along the OX axis with 2^{n} squares piled one on top of the other along the OZ axis*:  if n = 1 fold is performed then the initial length of the strip is 2^{1} and at the end of the transformation the distribution of numbers in the vertical direction is: 1 2  if n = 2 folds are performed then the initial length of the strip is 2^{2} and at the end of the transformation the distribution of numbers in the vertical direction is: 1 4 3 2  for n = 3 folds: 1 8 5 4 3 6 7 2  for n = 4 folds: 1 16 9 8 5 12 13 4 3 14 11 6 7 10 15 2  for n = 5 folds: 1 32 17 ... and so on.  what type(s) of toy model emulating the actual folding process can be thought of?  given a square in the original strip numbered with the current year, what will its position be in the folded strip after 30 folds are performed? The actual answer to the last question is, of course, irrelevant  the idea here is to devise a way to compute the new position of an arbitrary number after an arbitrary number of foldings very fast  without modelling the actual folding process. * A way to obtain the new sequence of numbers: holding the folded strip by its 'book' end, cut all the forward facing folds, thumb through the 'pages' to reveal their numbers. PS Conversely, the problem may be stated as follows: if the squares in the folded strip are numbered sequentially, from 1, what will the sequence of these numbers be when the strip is unfolded or, conversely, how the squares in the folded strip should be numbered so that after unfolding the strip they are sequential, etc.


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Grimbal
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Re: Strip Teez
« Reply #1 on: Aug 19^{th}, 2016, 9:32am » 
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Use 0based numbering. Take number X, write it in binary, reverse the order of the n digits, leftshift, if X is odd reverse all digits (0's become 1 and 1's become 0), convert to decimal. For example: 1 16 9 8 5 12 13 4 3 14 11 6 7 10 15 2 12th number: 0based it is 11 in binary: 1011 reversed: 1101 leftshifted: 1010 not'd (11 is odd): 0101 That is 5 (0based) or 6 (1based). The 12th number is 6, the 6th number is 12. [edit] made meaning of 'reversed' less ambiguous [/edit]

« Last Edit: Aug 21^{st}, 2016, 8:01am by Grimbal » 
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rloginunix
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Re: Strip Teez
« Reply #2 on: Aug 20^{th}, 2016, 8:34am » 
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That's right. Purely for terminology sake: reversing the digits = reversing the order (of digits) = geometrically rotating the binary digits about the (vertical) axis of the bitpattern's symmetry, abcd > dcba not'ting the digits = flipping the binary value of each digit to its binary counterpart = applying the C's one's complement operator ~ (my order was: left shift, ~ if odd, reverse, six of one or half a dozen)


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Grimbal
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Re: Strip Teez
« Reply #3 on: Aug 21^{st}, 2016, 8:06am » 
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I forgot to mention the surprising thing that that to reverse the transformation, just apply the process again. Yet another meaning of reverse.

« Last Edit: Aug 21^{st}, 2016, 8:07am by Grimbal » 
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towr
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Re: Strip Teez
« Reply #4 on: Aug 21^{st}, 2016, 1:06pm » 
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It's not really surprising if you swap the x and y axis.


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Grimbal
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Re: Strip Teez
« Reply #5 on: Aug 22^{nd}, 2016, 1:44am » 
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Indeed, that explains it. (It is X and Z actually). But now all the magic has gone. Thanks, towr!


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towr
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Re: Strip Teez
« Reply #6 on: Aug 22^{nd}, 2016, 12:57pm » 
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I wonder if it would work well as a sort of obfuscation/encryption like rot13. The necessary padding is a bit of a downside, though.


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rloginunix
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Re: Strip Teez
« Reply #7 on: Aug 23^{rd}, 2016, 8:09am » 
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Should work  upper casetrailing six characters are all printable: left square bracket [, backslash \, right square bracket ], caret ^, underscore _ and back tick `. Another interesting thing I noticed about this transformation: stationary points. For example, for towrsuggested 5fold obfuscator: 1based: 8, 12, 13, 19, 22, 26, 31 6fold: 9, 21, 29, 35, 43, 55, 63 7fold: 16, 24, 25, 37, 44, 52, 61, 67, 78, 86, 91, 103, 106, 114, 127 etc. Feels like some weird cosine is waving through.


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rloginunix
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Re: Strip Teez
« Reply #8 on: Aug 23^{rd}, 2016, 9:16am » 
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HILL] []^LP KTEF ^KLIS


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Grimbal
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Re: Strip Teez
« Reply #9 on: Aug 24^{th}, 2016, 8:34am » 
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I thought of scrambling the letters (anagramlike). That would require a length of a power of 2. Or padding to such a length. Instead of padding, you could cut the text in parts that are powers of 2 long. (How this can be achieved is left as an exercise to the reader). rloginunix proposed instead the encoding of each letter. I.e. use the scrambling not on the text, but on the alphabet to create a substitution table. For that case, you could also use the 64 characters of base 64 encoding: AZ, az, 09, +, /.


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rloginunix
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Re: Strip Teez
« Reply #10 on: Aug 24^{th}, 2016, 9:25am » 
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on Aug 24^{th}, 2016, 8:34am, Grimbal wrote:Instead of padding, you could cut the text in parts that are powers of 2 long 
 Beautiful: any (natural) number is representable as a sum of powers of 2; then it can be done lefttoright or vice versa (how do we encode a 0length string though?)


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