Project Euler Problem Number 282

Ackerman Function And Problem Statement
        Start . A(1,0)   (Use rule 2)
        Next .. A(0,1)   (Use rule 1)
        Next .. 2        (1+1 is 2)
        .
        . Next
        .
        Start . A(2,2) ............................... Rule 3
        Then .. A(1, A(2,1)) ......................... Rule 3
        Next .. A(1, A(1, A(2,0))) ................... Rule 2
        Next .. A(1, A(1, A(1,1))) ................... Rule 3
        Next .. A(1, A(1, A(0, A(1,0)))) ............. Rule 3
        Next .. A(1, A(1, A(0, A(0,1)))) ............. Rule 2
        Next .. A(1, A(1, A(0, 2))) .................. Rule 1
        Next .. A(1, A(1, 3)) ........................ 2+1 is 3
        Next .. A(1, A(0, A(1,2))) ................... Rule 3
        Next .. A(1, A(0, A(0, A(1,1)))) ............. Rule 3
        Next .. A(1, A(0, A(0, A(0, A(1,0))))) ....... Rule 3
        Next .. A(1, A(0, A(0, A(0, A(0,1))))) ....... Rule 2
        Next .. A(1, A(0, A(0, A(0, 2)))) ............ Rule 1
        Next .. A(1, A(0, A(0, 3))) .................. 2+1 is 3
        Next .. A(1, A(0, 4)) ........................ 3+1 is 4
        Next .. A(1, 5) .............................. 4+1 is 5
        Next .. A(0, A(1,4)) ......................... Rule 3
        Next .. A(0, A(0, A(1,3))) ................... Rule 3
        Next .. A(0, A(0, A(0, A(1,2)))) ............. Rule 3
        Next .. A(0, A(0, A(0, A(0, A(1,1))))) ....... Rule 3
        Next .. A(0, A(0, A(0, A(0, A(0, A(1,0)))))) . Rule 3
        Next .. A(0, A(0, A(0, A(0, A(0, A(0,1)))))) . Rule 2
        Next .. A(0, A(0, A(0, A(0, A(0, 2))))) ...... Rule 1
        Next .. A(0, A(0, A(0, A(0, 3)))) ............ 2+1 is 3
        Next .. A(0, A(0, A(0, 4))) .................. 3+1 is 4
        Next .. A(0, A(0, 5)) ........................ 4+1 is 5
        Next .. A(0, 6) .............................. 5+1 is 6
        Next .. 7 .................................... 6+1 is 7
        

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This page last updated Saturday, 13 Mar, 2010 11:59:45 GMT