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Infix to Postfix
how do you convert infix to postfix in case of exponents?
is 2^3^2 in postfix
23^2^
or
232^^
quite confused on this. thanks.
is 2^3^2 in postfix
23^2^
or
232^^
quite confused on this. thanks.
ASKER CERTIFIED SOLUTION
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Matlab does infix calculatiosns. Here is a screen paste:
>> 2^3^2
ans =
64
My HP calculator does postfix
23^2^ = 64
The two exponent operators have equal precedence, so you work left to right.
>> 2^3^2
ans =
64
My HP calculator does postfix
23^2^ = 64
The two exponent operators have equal precedence, so you work left to right.
SOLUTION
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This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
when in doubt use ( )
calculators use two main types of arithmatic which will produce different results if ( ) are not used.
TI calculators (and most others) use standard algrbraic operations.
HP and a few others use reverse polish notation which does operations in a different order if ( ) are not used.
TI calculators (and most others) use standard algrbraic operations.
HP and a few others use reverse polish notation which does operations in a different order if ( ) are not used.
23^2^
or
232^^
On an HP calculator the last line would not work.
You use
2 enter
3 ^
2^
and the result is 64.
In your last line you have two operators in a row, which procedure does not work.
In your first line your notation is difficult in that you do not know whether you are dealing with 2 and 3 or 23. You could develop a computer program which would take care of this particular problem but you would still hve a problem with longer first numbers.
or
232^^
On an HP calculator the last line would not work.
You use
2 enter
3 ^
2^
and the result is 64.
In your last line you have two operators in a row, which procedure does not work.
In your first line your notation is difficult in that you do not know whether you are dealing with 2 and 3 or 23. You could develop a computer program which would take care of this particular problem but you would still hve a problem with longer first numbers.
There is a cute HP Calculator online: http://www.hpmuseum.org/simulate/hp35sim/hp35sim.htm
Two operators in a row is not a problem. You do have to use enter twice to terminate digit entry.
2
Enter
3
Enter
2
y^x Returns 8
Y^x Returns 64
Two operators in a row is not a problem. You do have to use enter twice to terminate digit entry.
2
Enter
3
Enter
2
y^x Returns 8
Y^x Returns 64
Ignore my last post, except for the HP link.
The answer is
23^2^
There are tons of converters on the web eg. http://webplaza.pt.lu/dostert/infixtopostfix.htm
23^2^
There are tons of converters on the web eg. http://webplaza.pt.lu/dostert/infixtopostfix.htm
Part of the reason this is confusing is that your example is symmetric:
2^3^5 = (2^3)^5 = 32768
2 3 ^ 5 ^ = 32768 on my HP-15C where the exponential function is y^x
5 3 2 ^ ^ = 32768 on the HP-35 where the exponential function is x^y
2^3^5 = (2^3)^5 = 32768
2 3 ^ 5 ^ = 32768 on my HP-15C where the exponential function is y^x
5 3 2 ^ ^ = 32768 on the HP-35 where the exponential function is x^y
> how do you convert infix to postfix in case of exponents?
> is 2^3^2 in postfix
> 23^2^
> or
> 232^^
It depends on whether you have defined your exponent operator as left-associative or right-associative. I don't think there's any authoritative answer on this. Without defining your exponent operator as left-associative or right associative, a^b^c results in an ambiguous expression.
Take a look at http://mathworld.wolfram.com/ExponentLaws.html . Notice that there are no expressions like x^y^m. However, there is an expression like (x^m)^n.
> is 2^3^2 in postfix
> 23^2^
> or
> 232^^
It depends on whether you have defined your exponent operator as left-associative or right-associative. I don't think there's any authoritative answer on this. Without defining your exponent operator as left-associative or right associative, a^b^c results in an ambiguous expression.
Take a look at http://mathworld.wolfram.com/ExponentLaws.html . Notice that there are no expressions like x^y^m. However, there is an expression like (x^m)^n.
The TI-92, which allows entry of large equations using infix, evaluates 2^3^2 as 512.
ASKER
i guess i would go for 232^^ = 512
i found these websites and a book saying that exponents are evaluated right to left.
http://www.reed.edu/~mcphailb/applets/calc/
http://www.qiksearch.com/javascripts/infix-postfix.htm
i found these websites and a book saying that exponents are evaluated right to left.
http://www.reed.edu/~mcphailb/applets/calc/
http://www.qiksearch.com/javascripts/infix-postfix.htm
Glad i could help
mlmcc
mlmcc
The first is correct
(23^)2^ = 82^ = 64
But 232^^ = 2(3^2)^ = 2 9^ = 512