Re: Re: Brace For This

You're very close, but you forgot or missed an important detail:

You are generating the character index from the number of sub-arrayrefs and elements counted recursively.

At the risk of being pedantic, not exactly :-)

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

Comment on Re: Re: Brace For This

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Re: Re: Re: Brace For This by jmcnamara (Monsignor) on Apr 29, 2001 at 18:47 UTC
At the risk of being pedantic, not exactly You are right. I should have been more explicit. The algorithm is a binary encoding of the ordinal values of the characters based on the accumulated level of recursion. For example the letter "J" is encoded as follows: `[ [[[[]]], [[]]], [[[]], []], [[]] ] Adding brackets to highlight the nesting of the arrayrefs: [ ([ ([[[ ]]]) , ([[ ]])) ]), ([ ([[ ]]) , ([ ]) ]), ([ ([ ] +) ]) ] 4 2 2 1 1 This gives: = 2(4+2) +2(2+1) +2(1) = 2(6) +2(3) +2(1) = 1001010b = 74 = ord('J');` [download] Clever and artistic. ;-) John. --	[reply] [d/l]

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Re: Re: Re: Brace For This
by jmcnamara (Monsignor) on Apr 29, 2001 at 18:47 UTC

At the risk of being pedantic, not exactly

    [ [[[[]]], [[]]], [[[]], []], [[]] ]

    Adding brackets to highlight the nesting of the arrayrefs:    
    [ ([ ([[[ ]]]) , ([[ ]])) ]),    ([ ([[ ]]) , ([ ]) ]),    ([ ([ ]
+) ])  ]
             4          2                  2        1               1
    
    This gives:
    
    = 2**(4+2) +2**(2+1) +2**(1)

    = 2**(6) +2**(3) +2**(1)
    
    = 1001010b
    
    = 74
    
    = ord('J');
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