From: nma@12000.org   
      
   On 10/16/2017 9:04 PM, Albert Rich wrote:   
   > On Monday, October 16, 2017 at 5:57:45 AM UTC-10, Nasser M. Abbasi wrote:   
   >   
   >> Thanks. Yes, this is the normalized mean leaf size. I did this for   
   >> the FULL test suite.   
   >   
   > That's good, but it's not clear to me how you compute the normalized mean   
   leaf size. Do you add up all the result leaf sizes and all the optimal leaf   
   sizes, and then take their ratio. Or do you compute the ratio for each   
   problem, and then take the    
   average of those ratios?   
   >   
   > Albert   
   >   
      
      
   Lets say there are 14,000 integrals which are passed.   
      
   Then for each one of these integrals, I divide its leaf size by the   
   leaf size of the optimal result for it. The reasume is new   
   list of 14,000 numbers (each is ratio).   
      
   Then find the mean of this long list of numbers using the Mean command.   
      
   So I am doing the second method you mentioned above. Is this the   
   right way? I am not a statistic person and only managed to get   
   B in my probability and statics course with lots of struggle.   
      
   The code is this one line:   
      
      meanNormalized = Mean[  allPassed[[All,3]] / allPassed[[All,4]]  ];   
      
   The 3rd field is the leaf size of the result from CAS, and the 4th   
   field is the leaf size of the optimal from your input file.   
      
   allPassed is the array which contains all the integrals which passed   
   from all the 200 test files.   
      
   I can change it if it not the correct way.   
      
   --Nasser   
      
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