User blog:Ynought/Part 1 of 2 The & System

                                                                                                 Here S(k&(#))=k+k-1&(#-1) and S(k)=2k                                   # - or + a is the array with every entry + or -  a                         # is the array in k&(#)                                                                 and the system takes the form of k&(#)    (# is solved from right to left)

k is always the k from k&(#)

linear system :
 * 1) if the first entry is 0 then remove that entry and the comma right to it and S(k&(#))
 * 2) if any of the following entries is a 0 then replace it by the entry next to it + k and decrement the next entry by one  and S(k&(#))
 * 3) if none is 0 then add the # with the last entry decremented by one and every other increased by k to every entry in the original # decrement the right most entry by one and S(k&(#))
 * 4) any tailing zeros are removed and every other entry is + the new # with every entry encremented by k and S(k&(#))
 * 5) when # is reduced to one entry then k&(n) gets reduced to k+n

extended & system:

lets define k&(a[b]c[d]...e)

a comma is a shorthand for [0]

rules: then lets define 0=[k] and k + S(the new array)
 * 1) any tailing 0 can be removed but the value of the seperator to the left of the 0 gets added to the next left seperator and S(every element(even the seperators) in the array where the leftmost element is decremented by one)
 * 2) if there is a 0 somewhere else then replace it by k and decrement the leftmost element by one and for every element S(the expression with the element that is being replaced by this decreased by one)
 * 3) if there is no 0 then S(the expression with the element that is being replaced by this decreased by one) and k+S(k&(#)) where # is the system and add the new k to every single entry except the last one with the last element decreased by one 
 * 4) with seperators in the seperators they get solved the same way

                        a=a-1[[a-1...k times...a-1a-1]]...]] and k + S(the new array)

with more brackets they also gets solved the same

part two will follow soon

and if you have any tips comment them please