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{{Short description|Algorithm for minimising register usage}}
When [[Code generation|generating code]] for arithmetic expressions, the [[compiler]] has to decide which is the best way to translate the expression in terms of number of instructions used as well as number of registers needed to evaluate a certain subtree (especially if free registers are scarce). The '''Sethi-Ullman algorithm''' (also known as '''Sethi-Ullman numbering''') fulfills the property of producing code which needs the least number of instructions possible as well as the least number of storage references (under the assumption that at the most [[commutativity]] and [[associativity]] apply to the operators used, but distributive laws i.e. <math>a * b + a * c = a * (b + c)</math> do not hold). Please note that the algorithm succeeds as well if neither [[commutativity]] nor [[associativity]] hold for the expressions used, and therefore arithmetic transformations can not be applied.▼
In [[computer science]], the '''Sethi–Ullman algorithm''' is an [[algorithm]] named after [[Ravi Sethi]] and [[Jeffrey D. Ullman]], its inventors, for translating [[abstract syntax tree]]s into [[machine code]] that uses as few [[Processor register|registers]] as possible.
==Overview==
==Simple Sethi-Ullman algorithm==▼
▲When [[
The '''simple Sethi-Ullman algorithm''' works as follows (for a [[RISC|load-store architecture]]):▼
▲The '''simple
# Traverse the [[abstract syntax tree]] in pre- or postorder
## For every leaf node, if it is a non-constant
## For every non-leaf node
# To emit code, if the subtrees need different numbers of registers, evaluate the subtree needing the most registers first (since the register needed to save the result of one subtree may make the other one [[Register spilling|spill]]), otherwise the order is irrelevant.
===Example===
For an arithmetic expression <math>a = (b + c
=
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+ +
/ \ / \
+ *
/ \ / \
To continue with the algorithm, we need only to examine the arithmetic expression <math>(b + c) * (d + 3)</math>, i.e. we only have to look at the right subtree of the assignment '=':▼
b c f g
▲To continue with the algorithm, we need only to examine the arithmetic expression <math>(b + c + f * g) * (d + 3)</math>, i.e. we only have to look at the right subtree of the assignment '=':
*
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+ +
/ \ / \
+ *
/ \ / \
Now we start traversing the tree (in preorder for now), assigning the number of registers needed to evaluate each subtree (note that the last summand in the expression <math>(b + c) * (d + 3)</math> is a constant):▼
b c f g
▲Now we start traversing the tree (in preorder for now), assigning the number of registers needed to evaluate each subtree (note that the last summand in the expression <math>(b + c + f * g) * (d + 3)</math> is a constant):
*<sub>'''2'''</sub>
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+<sub>'''2'''</sub> +<sub>'''1'''</sub>
/ \ / \
+<sub>'''1'''</sub> / \ / \
b<sub>'''1'''</sub> c<sub>'''0'''</sub>f<sub>'''1'''</sub> g<sub>'''0'''</sub>
From this tree it can be seen that we need 2 registers to compute the left subtree of the '*', but only 1 register to compute the right subtree. Nodes 'c' and 'g' do not need registers for the following reasons: If T is a tree leaf, then the number of registers to evaluate T is either 1 or 0 depending whether T is a left or a right subtree (since an operation such as add R1, A can handle the right component A directly without storing it into a register). Therefore we shall start to emit code for the left subtree first, because we might run into the situation that we only have 2 registers left to compute the whole expression. If we now computed the right subtree first (which needs only 1 register), we would then need a register to hold the result of the right subtree while computing the left subtree (which would still need 2 registers), therefore needing 3 registers concurrently. Computing the left subtree first needs 2 registers, but the result can be stored in 1, and since the right subtree needs only 1 register to compute, the evaluation of the expression can do with only 2 registers left.▼
▲From this tree it can be seen that we need 2 registers to compute the left subtree of the '*', but only 1 register to compute the right subtree. Therefore we shall start to emit code for the left subtree first, because we might run into the situation that we only have 2 registers left to compute the whole expression. If we now computed the right subtree first (which needs only 1 register), we would then need a register to hold the result of the right subtree while computing the left subtree (which would still need 2 registers), therefore needing 3 registers concurrently. Computing the left subtree first needs 2 registers, but the result can be stored in 1, and since the right subtree needs only 1 register to compute, the evaluation of the expression can do with only 2 registers left.
In an advanced version of the '''
==See also==
▲==Advanced Sethi-Ullman algorithm==
*[[Strahler number]], the minimum number of registers needed to evaluate an expression without any external storage
▲In an advanced version of the '''Sethi-Ullman algorithm''', the arithmetic expressions are first transformed, exploiting the algebraic properties of the operators used.
*[[Ershov Number]], basically the same concept as Strahler number
==References==
*
==External links==
▲*[http://portal.acm.org/citation.cfm?doid=321607.321620 The Generation of Optimal Code for Arithmetic Expressions] [[Ravi Sethi]], [[Jeffrey D. Ullman|J. D. Ullman]], Journal of the [[Association for Computing Machinery|ACM]], Vol. 17, No. 4, October 1970, pp. 715-728
▲*[http://lambda.uta.edu/cse5317/fall02/notes/node40.html Code Generation for Trees]
{{DEFAULTSORT:Sethi-Ullman algorithm}}
[[Category:Compiler
[[Category:Graph algorithms]]
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