Is it known for which pairs of integers Ackermann 's function is commutative?

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I was recently writing test for checking the termination of programs and I wanted to write some difficult condition to verify so that the solver didn't use the if-condition to simplify the goal to proof.



It turned that writing in the if, ackermann(x,y) == 1 (for the definition of Ackermann see Wikipedia) was sufficient to confuse the solver, but I also tried ackermann(x,y) == ackermann(y,x) and of course the solver didn't go better on this instance.



I'm thinking of other programs that imply difficult problems in mathematics, for instance the so-called Collatz problem (also Syracuse, Kakutani, Hasse or Ulam's problem):



while n > 1 do
 if(n mod 2) != 0 then n = 3n + 1 
 else n = n div 2


So my question is, is there any problem with Ackermann's commutativity. Is it known what pairs (if any) of numbers make it commutative?




discrete-mathematics ackermann-function




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    I was recently writing test for checking the termination of programs and I wanted to write some difficult condition to verify so that the solver didn't use the if-condition to simplify the goal to proof.



    It turned that writing in the if, ackermann(x,y) == 1 (for the definition of Ackermann see Wikipedia) was sufficient to confuse the solver, but I also tried ackermann(x,y) == ackermann(y,x) and of course the solver didn't go better on this instance.



    I'm thinking of other programs that imply difficult problems in mathematics, for instance the so-called Collatz problem (also Syracuse, Kakutani, Hasse or Ulam's problem):



    while n > 1 do
     if(n mod 2) != 0 then n = 3n + 1 
     else n = n div 2


    So my question is, is there any problem with Ackermann's commutativity. Is it known what pairs (if any) of numbers make it commutative?




    discrete-mathematics ackermann-function




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I was recently writing test for checking the termination of programs and I wanted to write some difficult condition to verify so that the solver didn't use the if-condition to simplify the goal to proof.



      It turned that writing in the if, ackermann(x,y) == 1 (for the definition of Ackermann see Wikipedia) was sufficient to confuse the solver, but I also tried ackermann(x,y) == ackermann(y,x) and of course the solver didn't go better on this instance.



      I'm thinking of other programs that imply difficult problems in mathematics, for instance the so-called Collatz problem (also Syracuse, Kakutani, Hasse or Ulam's problem):



      while n > 1 do
       if(n mod 2) != 0 then n = 3n + 1 
       else n = n div 2


      So my question is, is there any problem with Ackermann's commutativity. Is it known what pairs (if any) of numbers make it commutative?




      discrete-mathematics ackermann-function




      share|cite|improve this question













      I was recently writing test for checking the termination of programs and I wanted to write some difficult condition to verify so that the solver didn't use the if-condition to simplify the goal to proof.



      It turned that writing in the if, ackermann(x,y) == 1 (for the definition of Ackermann see Wikipedia) was sufficient to confuse the solver, but I also tried ackermann(x,y) == ackermann(y,x) and of course the solver didn't go better on this instance.



      I'm thinking of other programs that imply difficult problems in mathematics, for instance the so-called Collatz problem (also Syracuse, Kakutani, Hasse or Ulam's problem):



      while n > 1 do
       if(n mod 2) != 0 then n = 3n + 1 
       else n = n div 2


      So my question is, is there any problem with Ackermann's commutativity. Is it known what pairs (if any) of numbers make it commutative?





      discrete-mathematics ackermann-function





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      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago
























      asked 2 days ago









      Javier

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          2 Answers
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          Aside from $m=n$ the only cases of commutativity are $A(1,0)=A(0,1)$ and $A(2,0)=A(0,2)$. In all other cases $m gt n implies A(m,n) gt A(n,m)$. I don't have a nice proof of that, but the height of the tower is much more important than the numbers that make up the tower. You can find cases where different pairs of arguments give equal values, like $A(0,12)=A(1,11)=A(2,5)=A(3,1)=A(4,0)=13$ and $A(0,65534)=A(1,65533)=A(2,32765)=A(3,13)=A(4,1)=65533$

          Looking at the definition, given $A(m,n)$ you can also express it as $A(p,q)$ for every $p lt m$.






          share|cite|improve this answer




























            up vote
            3
            down vote













            To finish up Ross Millikan's answer:



            Note that if $m>n>1$ we have



            beginalignA(m,n)&=A(m-1,A(m,n-1))\&ge A(n,A(m,n-1))\&ge A(n,m+n-1)\&>A(n,m)endalign



            using the fact that the Ackermann function is increasing in all arguments.






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            • very helpful addition indeed!
              – Javier
              yesterday










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            2 Answers
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            Aside from $m=n$ the only cases of commutativity are $A(1,0)=A(0,1)$ and $A(2,0)=A(0,2)$. In all other cases $m gt n implies A(m,n) gt A(n,m)$. I don't have a nice proof of that, but the height of the tower is much more important than the numbers that make up the tower. You can find cases where different pairs of arguments give equal values, like $A(0,12)=A(1,11)=A(2,5)=A(3,1)=A(4,0)=13$ and $A(0,65534)=A(1,65533)=A(2,32765)=A(3,13)=A(4,1)=65533$

            Looking at the definition, given $A(m,n)$ you can also express it as $A(p,q)$ for every $p lt m$.






            share|cite|improve this answer

























              up vote
              1
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              Aside from $m=n$ the only cases of commutativity are $A(1,0)=A(0,1)$ and $A(2,0)=A(0,2)$. In all other cases $m gt n implies A(m,n) gt A(n,m)$. I don't have a nice proof of that, but the height of the tower is much more important than the numbers that make up the tower. You can find cases where different pairs of arguments give equal values, like $A(0,12)=A(1,11)=A(2,5)=A(3,1)=A(4,0)=13$ and $A(0,65534)=A(1,65533)=A(2,32765)=A(3,13)=A(4,1)=65533$

              Looking at the definition, given $A(m,n)$ you can also express it as $A(p,q)$ for every $p lt m$.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Aside from $m=n$ the only cases of commutativity are $A(1,0)=A(0,1)$ and $A(2,0)=A(0,2)$. In all other cases $m gt n implies A(m,n) gt A(n,m)$. I don't have a nice proof of that, but the height of the tower is much more important than the numbers that make up the tower. You can find cases where different pairs of arguments give equal values, like $A(0,12)=A(1,11)=A(2,5)=A(3,1)=A(4,0)=13$ and $A(0,65534)=A(1,65533)=A(2,32765)=A(3,13)=A(4,1)=65533$

                Looking at the definition, given $A(m,n)$ you can also express it as $A(p,q)$ for every $p lt m$.






                share|cite|improve this answer













                Aside from $m=n$ the only cases of commutativity are $A(1,0)=A(0,1)$ and $A(2,0)=A(0,2)$. In all other cases $m gt n implies A(m,n) gt A(n,m)$. I don't have a nice proof of that, but the height of the tower is much more important than the numbers that make up the tower. You can find cases where different pairs of arguments give equal values, like $A(0,12)=A(1,11)=A(2,5)=A(3,1)=A(4,0)=13$ and $A(0,65534)=A(1,65533)=A(2,32765)=A(3,13)=A(4,1)=65533$

                Looking at the definition, given $A(m,n)$ you can also express it as $A(p,q)$ for every $p lt m$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered 2 days ago









                Ross Millikan

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                    up vote
                    3
                    down vote













                    To finish up Ross Millikan's answer:



                    Note that if $m>n>1$ we have



                    beginalignA(m,n)&=A(m-1,A(m,n-1))\&ge A(n,A(m,n-1))\&ge A(n,m+n-1)\&>A(n,m)endalign



                    using the fact that the Ackermann function is increasing in all arguments.






                    share|cite|improve this answer





















                    • very helpful addition indeed!
                      – Javier
                      yesterday














                    up vote
                    3
                    down vote













                    To finish up Ross Millikan's answer:



                    Note that if $m>n>1$ we have



                    beginalignA(m,n)&=A(m-1,A(m,n-1))\&ge A(n,A(m,n-1))\&ge A(n,m+n-1)\&>A(n,m)endalign



                    using the fact that the Ackermann function is increasing in all arguments.






                    share|cite|improve this answer





















                    • very helpful addition indeed!
                      – Javier
                      yesterday












                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    To finish up Ross Millikan's answer:



                    Note that if $m>n>1$ we have



                    beginalignA(m,n)&=A(m-1,A(m,n-1))\&ge A(n,A(m,n-1))\&ge A(n,m+n-1)\&>A(n,m)endalign



                    using the fact that the Ackermann function is increasing in all arguments.






                    share|cite|improve this answer













                    To finish up Ross Millikan's answer:



                    Note that if $m>n>1$ we have



                    beginalignA(m,n)&=A(m-1,A(m,n-1))\&ge A(n,A(m,n-1))\&ge A(n,m+n-1)\&>A(n,m)endalign



                    using the fact that the Ackermann function is increasing in all arguments.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered 2 days ago









                    Simply Beautiful Art

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                    • very helpful addition indeed!
                      – Javier
                      yesterday
















                    • very helpful addition indeed!
                      – Javier
                      yesterday















                    very helpful addition indeed!
                    – Javier
                    yesterday




                    very helpful addition indeed!
                    – Javier
                    yesterday












                     

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