algorithmuscanorj

Double proof about permutations.

Jan 9th, 2013
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  1. >>>> Proof by Mathematical induction <<<<<
  2.  
  3. About this evident fact:
  4. According the standard convention for reading
  5. numbers in every positional number system: Decomposing
  6. the numbers as linear combinations of powers for a given
  7. radix, the concatenation in ascending order for the first
  8. digits in the given radix always assigns the smallest
  9. possible value.
  10.  
  11. Convention: When reading numbers from left to right,
  12. each digit have associated a power of a given radix.
  13. The exponents for those powers are distributed in
  14. descending order from left to right. So a number
  15. written in base r whose representation is written
  16. for example as "123bca" in the base r=16 (hexadecimal)
  17. will have assigned the following value in decimal (base-10):
  18.  
  19. "23bca"
  20. ---> ( 2 )*r^4 + ( 3 )*r^3 + ( b )*r^2 + ( c )*r^1 + ( a )*r^0
  21. ---> 146378 (decimal)
  22.  
  23. Now it will be observed in the further that for each
  24. case that the smallest possible k-digits permutation
  25. ( k < r ) for the first k digits in base r is the
  26. concatenation in ascending order of those digits.
  27.  
  28. k=1.
  29. Trivial since there is 1 and only 1 one-digit permutation: {"1"}
  30.  
  31. k=2.
  32. The two permutations are: {"12","21"}
  33.  
  34. The possible assignations are also two:
  35.  
  36. "12" <---> ( 1 )*r^1 + ( 2 )*r^0
  37.  
  38. "21" <---> ( 2 )*r^1 + ( 1 )*r^0
  39.  
  40. Subtracting these assignations:
  41.  
  42. "21 -12"  <--> (2-1)*r^1 + (1-2)*r^0 = 1*r^1 - 1*r^0 = (r-1)
  43.  
  44. "21" is greater than "12" by a difference of (r-1),
  45. then for two-digits permutations, "12" or the concatenation
  46. of the first two digits have assigned always the smallest
  47. possible value regardless the radix.
  48.  
  49. k=3. There are six (3!) possible assignations, but we only
  50. are interested in compare the concatenation of the first 3
  51. digits with the other five remaining permutations so there
  52. will be 5 differences to be checked:
  53.  
  54. Assignations:
  55.  
  56. "123" <---> ( 1 )*r^2 + ( 2 )*r^1 + ( 3 )*r^0
  57. "132" <---> ( 1 )*r^2 + ( 3 )*r^1 + ( 2 )*r^0
  58. "213" <---> ( 2 )*r^2 + ( 1 )*r^1 + ( 3 )*r^0
  59. "231" <---> ( 2 )*r^2 + ( 3 )*r^1 + ( 1 )*r^0
  60. "312" <---> ( 3 )*r^2 + ( 1 )*r^1 + ( 2 )*r^0
  61. "321" <---> ( 3 )*r^2 + ( 2 )*r^1 + ( 1 )*r^0
  62.  
  63. Differences:
  64.  
  65. "132" <---> ( 1 )*r^2 + ( 3 )*r^1 + ( 2 )*r^0
  66. "123" <---> ( 1 )*r^2 + ( 2 )*r^1 + ( 3 )*r^0
  67. ----------------------------------------------
  68. ( 0 )*r^2 + ( 1 )*r^1 + ( -1 )*r^0
  69.  
  70. Or: (r-1)
  71.  
  72. "213" <---> ( 2 )*r^2 + ( 1 )*r^1 + ( 3 )*r^0
  73. "123" <---> ( 1 )*r^2 + ( 2 )*r^1 + ( 3 )*r^0
  74. ----------------------------------------------
  75. ( 1 )*r^2 + ( -1 )*r^1 + ( 0 )*r^0
  76.  
  77. Or: r^2-r = r*(r-1)
  78.  
  79. Also: 10*(r-1) if we remember that 10 represents
  80. the base.
  81.  
  82. "231" <---> ( 2 )*r^2 + ( 3 )*r^1 + ( 1 )*r^0
  83. "123" <---> ( 1 )*r^2 + ( 2 )*r^1 + ( 3 )*r^0
  84. ----------------------------------------------
  85. ( 1 )*r^2 + ( 1 )*r^1 + ( -2 )*r^0
  86.  
  87. Or: (r+1)*(r-1)+(r-1) = (r+2)*(r-1)
  88.  
  89. "312" <---> ( 3 )*r^2 + ( 1 )*r^1 + ( 2 )*r^0
  90. "123" <---> ( 1 )*r^2 + ( 2 )*r^1 + ( 3 )*r^0
  91. ----------------------------------------------
  92. ( 2 )*r^2 + ( -1 )*r^1 + ( -1 )*r^0
  93.  
  94. Or: (2*r+1)*(r-1)
  95.  
  96. "321" <---> ( 3 )*r^2 + ( 2 )*r^1 + ( 1 )*r^0
  97. "123" <---> ( 1 )*r^2 + ( 2 )*r^1 + ( 3 )*r^0
  98. ----------------------------------------------
  99. ( 2 )*r^2 + ( 0 )*r^1 + ( -2 )*r^0
  100.  
  101. Or: 2*(r+1)*(r-1)
  102.  
  103. In general for k>3:
  104.  
  105. Let be c_i and c_(i+1) two digits inside the
  106. concatenation in ascending order for the first N
  107. digits of base-r (N <= r, 0<=i<=(N-1) ),
  108. then both c_i and c_(i+1) ranges in 0..(N-1),
  109. and if "i" is the corresponding count index
  110. from left to right, then clearly: c_i < c_(i+1).
  111.  
  112. According to the previous description the
  113. first digit at the left c_0 have associated
  114. the power r^(N-1) and the last digit most at
  115. right c_(N-1) have associated the power r^0.
  116. This additional comment is unnecessary in the
  117. present proof but it illustrates that while
  118. the coefficients increases by 1 from left to
  119. right, the exponents does exactly the opposite,
  120. decreases by 1 from left to right. So in the
  121. ascending order concatenation, two adjacent
  122. digits are represented in general by:
  123.  
  124. s0= c_i*r^j + c_(i+1)*r^(j-1)
  125.  
  126. Relative to the (base-10) value assigned to a permutation.
  127.  
  128. Now if the coefficients are commuted, we will have:
  129.  
  130. s1= c_(i+1)*r^j + c_i*r^(j-1)
  131.  
  132. And subtracting (s1-s0):
  133.  
  134. (s1-s0)= c_(i+1)*r^j + c_i*r^(j-1) +(-1)*( c_i*r^j + c_(i+1)*r^(j-1) )
  135.    = ( c_(i+1) - c_i )*r^j + ( c_i - c_(i+1) )*r^(j-1)
  136.    = ( c_(i+1) - c_i )*r^j -- ( c_(i+1) - c_i )*r^(j-1)
  137.    = ( c_(i+1) - c_i )*r^(j-1+1) -- ( c_(i+1) - c_i )*r^(j-1)
  138.    = ( c_(i+1) - c_i )*r^(j-1)*r -- ( c_(i+1) - c_i )*r^(j-1)
  139.    = ( ( c_(i+1) - c_i )*r^(j-1) )*(r-1)
  140.  
  141. (**Note: Some minus signs were duplicated for readability)
  142.  
  143. There is obtained a positive multiple of (r-1)
  144. like in the particular cases studied before.
  145.  
  146. So, continuing this process for k=4, k=5, ...
  147. and so on "ad infinitum", if it were provided
  148. that (k<r), then the mentioned concatenation
  149. in ascending order for the first digits in
  150. base-r will be found to be always the smallest
  151. possible permutation without repetition built
  152. using k digits.
  153.  
  154. Also notice the following interesting property
  155. about each one of the studied differences:
  156. All of them are multiples of (r-1).
  157.  
  158. (END)
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