making a number from zero by increments by 1 or doubling I am trying to solve one problem with the help of its available solution .Here is the strategy to make number 467 from number 0.However there are few claims in solution in reference to this approach that i am not understanding .
000000000 =0
000000001 =1
000000010 =2
000000011 =3
000000110 =6
000000111 =7
000001110 =14
000011100 =28
000011101 =29
000111010 =58
001110100 =116
011101000 =232
011101001 =233
111010010 =466
111010011 =467

For a number  Y of n bits, we first find the minimum number of moves needed to create the left-most n-1 bits. Then we need 1 doubling operation to shift those bits left, and we add 1 if the number y is odd.
WHAT does the above claim by the editorial means ?can anyone explain this more clearly ? 
Is the number of doubling operations equals to the number of bits in a number minus 1?
It will be more helpful if answers to the questions are given along with examples or required explanations ?
 A: Doubling a non-negative integer is the same as appending a $0$ to its binary representation. Since this does not change the other bits, clearly you can at each step $k$ double the number and then increment by 1 if and only if the $k$-th bit in the desired number is a $1$. This procedure will then produce the desired number.
Now there are clearly many ways one can get the desired number. For example one can use only incrementing. You didn't ask what is the fastest way. It turns out that following the above method except without doubling at step $1$ is the fastest, but the proof is not that trivial.
The easiest way is to prove by recursion that the least number of steps to get a positive integer $n$ is the number of bits plus the sum of the bits in the binary representation minus $1$. If $n$ is zero, no steps are needed. If $n$ is odd, the last step must be an incrementing, so the least number of steps is one more than that needed to get $n-1$, and hence the claim holds by recursion. If $n$ is even and positive, then $n$ has at most one bit more than $n-1$, and has a bit-sum at most that of $n-1$, so using a doubling for the last step is at least as good as using an incrementing, and hence the claim holds by recursion. (Note that a tie occurs when $n=2$, and indeed one can get to $2$ from $1$ by either doubling or incrementing.)
A: Claim. If $n$ is a natural number that has a binary representation of $k$ bits (i.e., $2^{k-1}\le n<2^k$) among which there are $m$ 1-bits, then it is possible to reach $n$ from $0$ in $k+m-1$ steps of doubling or incrementing, but not in less than $k+m-1$ steps.
Proof. By induction on $n$.
Clearly, $n=1$ (with $k=m=1$) can be reached in a single step and not in less than one step.
Assume $n>1$ and that the claim is true for all naturals $<n$.
If $n$ is odd, then $n-1$ has the same number of bits, but one 1-bit less. By induction hypothesis, it can be reached within $k+(m-1)-1$ steps. Appending one increment step allows us to reach $n$ in $k+m-1$ steps.
On the other hand, in any step sequence leading to $n$, the last step must be an increment. Hence if we reach $n$ in less than $k+m-1$ steps, we must have reached $n-1$ in less than $k+m-2$ steps, which is not possible, per induction hypothesis.
If $n$ is even, then $n/2$ has one bit less, but the same number of 1-bits. By induction hypothesis, it can be reached witin $(k-1)+m-1$ steps. Appending a doubling step allows us to reach $n$ in $k+m-1$ steps.
On the other hand, any step sequence that reached $n$ via $n/2$ must use at least $(k-1)+m$ steps. So maybe there exists a shorter sequence leading to $n$ that uses an increment as last step? For even $n$, the number of 1-bits in $n-1$ is at least as big as the number of 1-bits in $n$ (why?). On the other hand, the number of bits for $n-1$ is at most one less than the number of bits in $n$. Hence any step sequence reaching $n$ via $n-1$ and a final increment is again at least $k+m-1$ steps long. $\square$
