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\$\begingroup\$ You can save a bit by only dividing by 2 once. And you don't actually gain anything by assigning to variables: (a=v[0])a is longer than v[0]v[0]. \$\endgroup\$

Peter Taylor
– Peter Taylor

2013-03-27 10:43:15 +00:00
Commented Mar 27, 2013 at 10:43
\$\begingroup\$ If I divided by 2 only one time, like s=(v[0]+v[1]+v[2])/2 with a,b,c=3,4,5 would result in "345"/2=172.5" and not 6. Improved without a, b` and c though. \$\endgroup\$

tomsmeding
– tomsmeding

2013-03-27 16:32:43 +00:00
Commented Mar 27, 2013 at 16:32
\$\begingroup\$ Ah, JavaScript's wonderful type system. Ok, s=(-v[0]-v[1]-v[2])/2 and change the other - to +. It's an even number of terms, so it cancels out. \$\endgroup\$

Peter Taylor
– Peter Taylor

2013-03-27 16:39:28 +00:00
Commented Mar 27, 2013 at 16:39

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