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#2
February 13th, 2003, 11:55 PM
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RE: New scoring question
ABA=1+3+1=5
BAB=3+1+3=7 ABAB=1+3+1+3=8 BABA=3+1+3+1=8 ABABA=1+3+1+3+1=9 BABAB=3+1+3+1+3=11 ABABAB=1+3+1+3+1+3=12 5*5+7*7+8*8+8*8+9*9+11*11+12*12= 25+49+64+64+81+121+144= 548
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#3
February 14th, 2003, 03:26 AM
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RE: New scoring question
Wrong!
the words ABA, BAB and ABAB exist 2 times each in the word ABABAB. So the scor should be 548 + 25 + 49 + 64 = 686!
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#5
February 14th, 2003, 01:45 PM
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RE: New scoring question
so the output of the program for the above example would be:
<html> <body> 686 10 ABABAB BABAB ABABA BABA ABAB ABAB BAB BAB ABA ABA </body> </html> ? or is it: <html> <body> 686 10 ABABAB </body> </html> ?
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#7
February 15th, 2003, 02:03 AM
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RE: New scoring question
I know it's too late for anything to be changed in the rules, but it would make much more sense to first print the words you found and then the number of them and the scor. Now we have to temporarily save the words and display them after the scor.
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