mpsn Posted February 27, 2012 Share Posted February 27, 2012 Hi, for my assignment, I need to figure out how many possible paths through a bubblesort, Let n be the size of the array so my results are: if n = 0, 2^0 paths if n = 1, 2^1 paths if n = 2, 2^3 paths if n = 3, 2^6 paths if n = 4, 2^10 paths so I need to figure out a summation/product notation formula for the exponent, for a given size n so I mean 2^(some formula involving summation or multiplication notation) Any help appreciated Quote Link to comment https://forums.phpfreaks.com/topic/257850-finding-general-formula/ Share on other sites More sharing options...
silkfire Posted February 27, 2012 Share Posted February 27, 2012 The formula for adding series starting from 1 to n is: n(n+1)/2. So for 4 it will be 4 * (4 + 1) / 2 = 20 / 2 = 10. Did that answer your question? Quote Link to comment https://forums.phpfreaks.com/topic/257850-finding-general-formula/#findComment-1321646 Share on other sites More sharing options...
mpsn Posted February 27, 2012 Author Share Posted February 27, 2012 Thanks, appreciate it Quote Link to comment https://forums.phpfreaks.com/topic/257850-finding-general-formula/#findComment-1321676 Share on other sites More sharing options...
mpsn Posted February 27, 2012 Author Share Posted February 27, 2012 Wait, actually I need to have a summation and/or product notation used, b/c yours is simply: 2^[n(n+1)/2], but I need 2^(summation from start to n for some formula) for instance. Quote Link to comment https://forums.phpfreaks.com/topic/257850-finding-general-formula/#findComment-1321686 Share on other sites More sharing options...
Alex Posted March 2, 2012 Share Posted March 2, 2012 Wait, actually I need to have a summation and/or product notation used, b/c yours is simply: 2^[n(n+1)/2], but I need 2^(summation from start to n for some formula) for instance. The n(n+1)/2 is the value for the sum of numbers from 1 to n, so you're just looking for: Quote Link to comment https://forums.phpfreaks.com/topic/257850-finding-general-formula/#findComment-1323182 Share on other sites More sharing options...
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