In lambda calculus you can take a beta reduction as the step.<br>But Haskell is not normally implemented by lambda calculus so you have to pick something else.<br>There are measures of reduction that you can come up with but they will vary, 
e.g., by compiler, optimization level, etc.<br>I think time is a much more interesting measure, since that&#39;s what you really care about in the end.<br><br><div class="gmail_quote">On Jan 14, 2008 2:03 PM, Ben Franksen &lt;
<a href="mailto:ben.franksen@online.de">ben.franksen@online.de</a>&gt; wrote:<br><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"><div class="Ih2E3d">
Lennart Augustsson wrote:<br>&gt; What is a reduction anyway?<br><br></div>I am not an expert but I thought in lambda calculus one has primitive rules<br>for evaluation, e.g. beta reduction. So a reduction is a &#39;smallest step&#39; in
<br>reducing an expression to normal form, no?<br><br>Cheers<br><font color="#888888">Ben<br></font><div><div></div><div class="Wj3C7c"><br>_______________________________________________<br>Haskell-Cafe mailing list<br><a href="mailto:Haskell-Cafe@haskell.org">
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