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lib/auto/TT2/Play/Area/public/codemirror/mode/lua/lua.js  view on Meta::CPAN

    return new RegExp("^(?:" + words.join("|") + ")$", "i");
  }
  var specials = wordRE(parserConfig.specials || []);

  // long list of standard functions from lua manual
  var builtins = wordRE([
    "_G","_VERSION","assert","collectgarbage","dofile","error","getfenv","getmetatable","ipairs","load",
    "loadfile","loadstring","module","next","pairs","pcall","print","rawequal","rawget","rawset","require",
    "select","setfenv","setmetatable","tonumber","tostring","type","unpack","xpcall",

    "coroutine.create","coroutine.resume","coroutine.running","coroutine.status","coroutine.wrap","coroutine.yield",

    "debug.debug","debug.getfenv","debug.gethook","debug.getinfo","debug.getlocal","debug.getmetatable",
    "debug.getregistry","debug.getupvalue","debug.setfenv","debug.sethook","debug.setlocal","debug.setmetatable",
    "debug.setupvalue","debug.traceback",

    "close","flush","lines","read","seek","setvbuf","write",

    "io.close","io.flush","io.input","io.lines","io.open","io.output","io.popen","io.read","io.stderr","io.stdin",
    "io.stdout","io.tmpfile","io.type","io.write",

lib/auto/TT2/Play/Area/public/codemirror/mode/stex/index.html  view on Meta::CPAN

           As $G$ is \termref[cd=balanced-binary-trees,name=balanced-binary-tree]{balanced} and $\gdepth{G}=n>i$, $\livar{V}{i-1}$ cannot contain
            leaves.
          \end{spfstep}
          \begin{spfstep}[type=conclusion]
           Thus $\eq{\card{\livar{V}i},{\atimes[cdot]{2,\card{\livar{V}{i-1}}}},{\atimes[cdot]{2,\power2{i-1}}},\power2i}$.
          \end{spfstep}
        \end{spfcase}
      \end{spfcases}
    \end{sproof}
  \item 
    \begin{assertion}[id=fbbt,type=corollary]	
      A fully balanced tree of depth $d$ has $\power2{d+1}-1$ nodes.
    \end{assertion}
  \item
      \begin{sproof}[for=fbbt,id=fbbt-pf]{}
        \begin{spfstep}
          Let $\defeq{G}{\tup{V,E}}$ be a fully balanced tree
        \end{spfstep}
        \begin{spfstep}
          Then $\card{V}=\Sumfromto{i}1d{\power2i}= \power2{d+1}-1$.
        \end{spfstep}