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352 lines
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Groff
352 lines
11 KiB
Groff
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.\" ========================================================================
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.\"
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.IX Title "RPNTUTORIAL 1"
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.TH RPNTUTORIAL 1 "2013-05-23" "1.4.8" "rrdtool"
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.\" For nroff, turn off justification. Always turn off hyphenation; it makes
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.\" way too many mistakes in technical documents.
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.if n .ad l
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.nh
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.SH "NAME"
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rpntutorial \- Reading RRDtool RPN Expressions by Steve Rader
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.SH "DESCRIPTION"
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.IX Header "DESCRIPTION"
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This tutorial should help you get to grips with RRDtool \s-1RPN\s0 expressions
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as seen in \s-1CDEF\s0 arguments of RRDtool graph.
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.SH "Reading Comparison Operators"
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.IX Header "Reading Comparison Operators"
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The \s-1LT\s0, \s-1LE\s0, \s-1GT\s0, \s-1GE\s0 and \s-1EQ\s0 \s-1RPN\s0 logic operators are not as tricky as
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they appear. These operators act on the two values on the stack
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preceding them (to the left). Read these two values on the stack
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from left to right inserting the operator in the middle. If the
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resulting statement is true, then replace the three values from the
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stack with \*(L"1\*(R". If the statement if false, replace the three values
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with \*(L"0\*(R".
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.PP
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For example, think about \*(L"2,1,GT\*(R". This \s-1RPN\s0 expression could be
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read as \*(L"is two greater than one?\*(R" The answer to that question is
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\&\*(L"true\*(R". So the three values should be replaced with \*(L"1\*(R". Thus the
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\&\s-1RPN\s0 expression 2,1,GT evaluates to 1.
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.PP
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Now consider \*(L"2,1,LE\*(R". This \s-1RPN\s0 expression could be read as \*(L"is
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two less than or equal to one?\*(R". The natural response is \*(L"no\*(R"
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and thus the \s-1RPN\s0 expression 2,1,LE evaluates to 0.
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.SH "Reading the IF Operator"
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.IX Header "Reading the IF Operator"
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The \s-1IF\s0 \s-1RPN\s0 logic operator can be straightforward also. The key
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to reading \s-1IF\s0 operators is to understand that the condition part
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of the traditional \*(L"if X than Y else Z\*(R" notation has *already*
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been evaluated. So the \s-1IF\s0 operator acts on only one value on the
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stack: the third value to the left of the \s-1IF\s0 value. The second
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value to the left of the \s-1IF\s0 corresponds to the true (\*(L"Y\*(R") branch.
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And the first value to the left of the \s-1IF\s0 corresponds to the false
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(\*(L"Z\*(R") branch. Read the \s-1RPN\s0 expression \*(L"X,Y,Z,IF\*(R" from left to
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right like so: \*(L"if X then Y else Z\*(R".
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.PP
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For example, consider \*(L"1,10,100,IF\*(R". It looks bizarre to me.
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But when I read \*(L"if 1 then 10 else 100\*(R" it's crystal clear: 1 is true
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so the answer is 10. Note that only zero is false; all other values
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are true. \*(L"2,20,200,IF\*(R" (\*(L"if 2 then 20 else 200\*(R") evaluates to 20.
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And \*(L"0,1,2,IF\*(R" ("if 0 then 1 else 2) evaluates to 2.
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.PP
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Notice that none of the above examples really simulate the whole
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\&\*(L"if X then Y else Z\*(R" statement. This is because computer programmers
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read this statement as \*(L"if Some Condition then Y else Z\*(R". So it's
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important to be able to read \s-1IF\s0 operators along with the \s-1LT\s0, \s-1LE\s0,
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\&\s-1GT\s0, \s-1GE\s0 and \s-1EQ\s0 operators.
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.SH "Some Examples"
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.IX Header "Some Examples"
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While compound expressions can look overly complex, they can be
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considered elegantly simple. To quickly comprehend \s-1RPN\s0 expressions,
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you must know the algorithm for evaluating \s-1RPN\s0 expressions:
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iterate searches from the left to the right looking for an operator.
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When it's found, apply that operator by popping the operator and some
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number of values (and by definition, not operators) off the stack.
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.PP
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For example, the stack \*(L"1,2,3,+,+\*(R" gets \*(L"2,3,+\*(R" evaluated (as \*(L"2+3\*(R")
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during the first iteration and is replaced by 5. This results in
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the stack \*(L"1,5,+\*(R". Finally, \*(L"1,5,+\*(R" is evaluated resulting in the
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answer 6. For convenience, it's useful to write this set of
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operations as:
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.PP
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.Vb 3
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\& 1) 1,2,3,+,+ eval is 2,3,+ = 5 result is 1,5,+
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\& 2) 1,5,+ eval is 1,5,+ = 6 result is 6
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\& 3) 6
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.Ve
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.PP
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Let's use that notation to conveniently solve some complex \s-1RPN\s0 expressions
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with multiple logic operators:
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.PP
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.Vb 1
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\& 1) 20,10,GT,10,20,IF eval is 20,10,GT = 1 result is 1,10,20,IF
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.Ve
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.PP
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read the eval as pop \*(L"20 is greater than 10\*(R" so push 1
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.PP
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.Vb 1
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\& 2) 1,10,20,IF eval is 1,10,20,IF = 10 result is 10
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.Ve
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.PP
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read pop \*(L"if 1 then 10 else 20\*(R" so push 10. Only 10 is left so
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10 is the answer.
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.PP
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Let's read a complex \s-1RPN\s0 expression that also has the traditional
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multiplication operator:
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.PP
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.Vb 4
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\& 1) 128,8,*,7000,GT,7000,128,8,*,IF eval 128,8,* result is 1024
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\& 2) 1024 ,7000,GT,7000,128,8,*,IF eval 1024,7000,GT result is 0
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\& 3) 0, 7000,128,8,*,IF eval 128,8,* result is 1024
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\& 4) 0, 7000,1024, IF result is 1024
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.Ve
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.PP
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Now let's go back to the first example of multiple logic operators,
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but replace the value 20 with the variable \*(L"input\*(R":
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.PP
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.Vb 1
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\& 1) input,10,GT,10,input,IF eval is input,10,GT ( lets call this A )
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.Ve
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.PP
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Read eval as \*(L"if input > 10 then true\*(R" and replace \*(L"input,10,GT\*(R"
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with \*(L"A\*(R":
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.PP
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.Vb 1
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\& 2) A,10,input,IF eval is A,10,input,IF
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.Ve
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.PP
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read \*(L"if A then 10 else input\*(R". Now replace A with it's verbose
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description again and\*(--voila!\-\-you have an easily readable description
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of the expression:
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.PP
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.Vb 1
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\& if input > 10 then 10 else input
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.Ve
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.PP
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Finally, let's go back to the first most complex example and replace
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the value 128 with \*(L"input\*(R":
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.PP
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.Vb 1
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\& 1) input,8,*,7000,GT,7000,input,8,*,IF eval input,8,* result is A
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.Ve
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.PP
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where A is \*(L"input * 8\*(R"
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.PP
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.Vb 1
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\& 2) A,7000,GT,7000,input,8,*,IF eval is A,7000,GT result is B
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.Ve
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.PP
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where B is \*(L"if ((input * 8) > 7000) then true\*(R"
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.PP
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.Vb 1
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\& 3) B,7000,input,8,*,IF eval is input,8,* result is C
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.Ve
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.PP
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where C is \*(L"input * 8\*(R"
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.PP
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.Vb 1
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\& 4) B,7000,C,IF
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.Ve
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.PP
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At last we have a readable decoding of the complex \s-1RPN\s0 expression with
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a variable:
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.PP
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.Vb 1
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\& if ((input * 8) > 7000) then 7000 else (input * 8)
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.Ve
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.SH "Exercises"
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.IX Header "Exercises"
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Exercise 1:
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.PP
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Compute \*(L"3,2,*,1,+ and \*(R"3,2,1,+,*" by hand. Rewrite them in
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traditional notation. Explain why they have different answers.
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.PP
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Answer 1:
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.PP
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.Vb 3
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\& 3*2+1 = 7 and 3*(2+1) = 9. These expressions have
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\& different answers because the altering of the plus and
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\& times operators alter the order of their evaluation.
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.Ve
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.PP
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Exercise 2:
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.PP
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One may be tempted to shorten the expression
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.PP
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.Vb 1
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\& input,8,*,56000,GT,56000,input,*,8,IF
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.Ve
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.PP
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by removing the redundant use of \*(L"input,8,*\*(R" like so:
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.PP
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.Vb 1
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\& input,56000,GT,56000,input,IF,8,*
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.Ve
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.PP
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Use traditional notation to show these expressions are not the same.
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Write an expression that's equivalent to the first expression, but
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uses the \s-1LE\s0 and \s-1DIV\s0 operators.
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.PP
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Answer 2:
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.PP
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.Vb 2
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\& if (input <= 56000/8 ) { input*8 } else { 56000 }
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\& input,56000,8,DIV,LE,input,8,*,56000,IF
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.Ve
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.PP
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Exercise 3:
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.PP
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Briefly explain why traditional mathematic notation requires the
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use of parentheses. Explain why \s-1RPN\s0 notation does not require
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the use of parentheses.
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.PP
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Answer 3:
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.PP
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.Vb 6
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\& Traditional mathematic expressions are evaluated by
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\& doing multiplication and division first, then addition and
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\& subtraction. Parentheses are used to force the evaluation of
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\& addition before multiplication (etc). RPN does not require
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\& parentheses because the ordering of objects on the stack
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\& can force the evaluation of addition before multiplication.
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.Ve
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.PP
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Exercise 4:
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.PP
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Explain why it was desirable for the RRDtool developers to implement
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\&\s-1RPN\s0 notation instead of traditional mathematical notation.
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.PP
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Answer 4:
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.PP
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.Vb 5
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\& The algorithm that implements traditional mathematical
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\& notation is more complex then algorithm used for RPN.
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\& So implementing RPN allowed Tobias Oetiker to write less
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\& code! (The code is also less complex and therefore less
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\& likely to have bugs.)
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.Ve
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.SH "AUTHOR"
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.IX Header "AUTHOR"
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Steve Rader <rader@wiscnet.net>
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