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Old 2021-08-08, 20:06   #3
Jan 2021

3×7×23 Posts

Originally Posted by Uncwilly View Post
I can choose to subtract n-2 from n and go right to the final in a single step. Or I can divide n by n (no cost division) and then add 1, thus a single step.
No - as stated, the moves that cost one are replace n with n+1 or n-1, not n-x. The no cost move is to replace n with n/d where d <= square root(n).

For the example given to demonstrate the process the factoring strategy works (but you'll have to write the divisions in reverse) because it's smooth and has 4 as a factor. You can't go from 106 to 2 because 53^2 > 106.

All powers of 2 are trivially 0 move numbers, and any 0 move number n multiplied by a number m where m < n results in another 0 move number.

Last fiddled with by slandrum on 2021-08-08 at 20:18
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