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书中有一个猜想,每次总是使用”刚刚得到“的那个数。 这个猜想可以尝试用反证法证明,不知道是否正确。 假设目标数X最少需要d次乘除法求出,那么在解答树上,X位于深度d,则根据猜想假设,X一定由X(d-1)求出: X = X(d-1) +/- Xk,0 <= k <= d-1。 现在给出反例,X无法由Xd-1得到,但可以由另外路径中两个结点Xi, Xj得到,即X = Xi +/- Xj,0 <= i <= j < d-1,那么在另一条路径,0--i--j,X = X = Xi +/- Xj,那X可以在第j+1层就被求出,即只需要j+1 次乘除法就可以求出,与假设最少需要d(>j+1)次乘除法相悖,因此反例不存在。
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书中有一个猜想,每次总是使用”刚刚得到“的那个数。
这个猜想可以尝试用反证法证明,不知道是否正确。
假设目标数X最少需要d次乘除法求出,那么在解答树上,X位于深度d,则根据猜想假设,X一定由X(d-1)求出: X = X(d-1) +/- Xk,0 <= k <= d-1。
现在给出反例,X无法由Xd-1得到,但可以由另外路径中两个结点Xi, Xj得到,即X = Xi +/- Xj,0 <= i <= j < d-1,那么在另一条路径,0--i--j,X = X = Xi +/- Xj,那X可以在第j+1层就被求出,即只需要j+1 次乘除法就可以求出,与假设最少需要d(>j+1)次乘除法相悖,因此反例不存在。
The text was updated successfully, but these errors were encountered: