Okay, just one conclusion. I need a math tutor. I cannot for the life of me calculate the worst case scenarios when either increasing or inserting an element into a hash table. I won't even cover the whole probability thing. If anyone lives like within 10 miles of UC-Berkeley, email me.
Try google.
dwk
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I was thinking about maybe trying LSD in an effort to bootstrap myself further. Yeah, google did help (sort of).
http://groups.google.com/group/comp....3b014a9abaceaf
Or Wikipedia. And FTLOG, read up on the other CS tutorials there. Free tutor.
If you don't want to learn, there is no shortcut. A greek mathematician once told a king: "There is no royal road to geometry." Meaning: start learning yourself the hard way like we all did, or don't bother at all. Nobody's gonna drive all the way to your house and help you pass your CS course for you.
Depends on your collision resolution policy. A good double hashing algorithm (eg. p-2 modular or quadratic search) typically leads to near-constant times.
Of course, the worst-case is only bounded by the size of your hash table. Yes, no matter how you build it. It would be much more worthwhile to measure the amortized complexities of hashing algorithm implementations.
Code:#include <stdio.h> void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){ puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9 /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i] ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][ t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}
