Kvant Math Problem 1086

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m51s
Source on kvant.digital

Problem

You are allowed to perform two operations on a number: "double" and "increase by 1". What is the minimum number of operations required to obtain from 0 the following numbers:

1. 100?
2. $n$?

M. V. Sapir

Exploration

Consider the problem of reaching a target number from $0$ using only two operations: doubling the current number or adding $1$. Small examples help clarify the dynamics. To reach $1$ from $0$, one must increment once. To reach $2$, one can increment to $1$ and then double, requiring two operations. To reach $3$, increment to $1$, double to $2$, increment to $3$, also three operations. For $4$, increment to $1$, double to $2$, double to $4$, three operations. Observing these examples suggests that sometimes doubling is more efficient than incrementing repeatedly, and sometimes the last step must be an increment.

It is natural to consider the reverse process: starting from the target number and reasoning backward to $0$ using the inverse operations: halve the number if it is even, or subtract $1$ if it is odd. Testing this strategy on small numbers reproduces the minimal number of steps correctly: for $3$, $3-1=2$, $2/2=1$, $1-1=0$, three steps. This indicates that the reverse approach may provide a systematic, rigorous method for any $n$.

The potential subtlety lies in ensuring that no sequence of operations can produce a smaller number of steps than this backward strategy. One must justify that greedily halving when even and subtracting $1$ when odd indeed minimizes the total number of steps.

For $100$, the binary representation is $1100100_2$. Converting from $0$ using operations aligns with building this binary expansion from left to right, each doubling corresponding to a left shift and each increment corresponding to a $1$ in the binary expansion. The number of operations appears closely related to the number of binary digits plus the number of $1$ bits minus one.

Problem Understanding

The task is to compute the minimal number of operations required to transform $0$ into a given integer $n$ using only the operations "double" and "increment by $1$". This is a Type C problem: one seeks to find the minimal number of operations and to justify that no sequence can be shorter. The core difficulty is proving that a greedy strategy—working backward by halving when even and subtracting $1$ when odd—produces a minimal sequence for any $n$. The answer for a specific $n$ is intuitively the sum of the number of bits in the binary representation of $n$ minus one for the leftmost bit plus the number of $1$ bits, corresponding to the sequence of operations.

Proof Architecture

Lemma 1: For any positive integer $n$, in a minimal sequence of operations from $0$ to $n$, the last operation is either an increment (if $n$ is odd) or a doubling (if $n$ is even). This is true because doubling an integer always produces an even number and adding $1$ always produces an odd number.

Lemma 2: If $n$ is even, halving it and finding the minimal sequence to the quotient produces a minimal sequence to $n$ by appending a doubling. This holds by induction on the number of operations: if a shorter sequence existed, removing the last doubling would give a shorter sequence to the half, contradicting minimality.

Lemma 3: If $n$ is odd, subtracting $1$ and finding the minimal sequence to $n-1$ produces a minimal sequence to $n$ by appending an increment. The proof mirrors Lemma 2.

Hardest step: verifying that alternating halving and subtraction exactly minimizes operations, especially when multiple consecutive doublings or increments could be rearranged.

Solution

Let $f(n)$ denote the minimal number of operations required to reach $n$ from $0$. By definition, $f(0)=0$. For $n>0$, define

$$f(n) = \begin{cases} f\left(\frac{n}{2}\right)+1,& \text{if $n$ is even},\ f(n-1)+1,& \text{if $n$ is odd}. \end{cases}$$

Lemma 1 ensures that the last operation in a minimal sequence is uniquely determined by the parity of $n$. Lemmas 2 and 3 ensure that the recursive construction of $f(n)$ using these last steps produces a minimal sequence.

Applying this to $n=100$, we compute recursively. $100$ is even, so $f(100) = f(50)+1$. $50$ is even, $f(50)=f(25)+1$. $25$ is odd, $f(25)=f(24)+1$. $24$ is even, $f(24)=f(12)+1$. $12$ is even, $f(12)=f(6)+1$. $6$ is even, $f(6)=f(3)+1$. $3$ is odd, $f(3)=f(2)+1$. $2$ is even, $f(2)=f(1)+1$. $1$ is odd, $f(1)=f(0)+1=1$. Tracing back, $f(2)=2$, $f(3)=3$, $f(6)=4$, $f(12)=5$, $f(24)=6$, $f(25)=7$, $f(50)=8$, $f(100)=9$.

Thus the minimal number of operations to reach $100$ is $9$.

For general $n$, $f(n)$ can be computed recursively by repeatedly halving when even and subtracting $1$ when odd, counting one operation per step. This method produces the minimal number of operations by Lemmas 2 and 3.

Verification of Key Steps

The critical steps are Lemmas 2 and 3. Testing $n=7$, the backward method gives $7-1=6$, $6/2=3$, $3-1=2$, $2/2=1$, $1-1=0$, totaling $5$ operations. Any alternative sequence must begin from $0$ and combine doublings and increments. Direct checking shows that sequences with different orders, e.g., doubling first or incrementing multiple times before doubling, require at least $5$ operations. Similarly, testing $n=15$ confirms that halving odd numbers after subtracting $1$ indeed minimizes steps.

Alternative Approaches

One could represent $n$ in binary and interpret the sequence of operations as building the binary expansion from the most significant bit. Each bit corresponds to either a doubling (for left shifts) or an increment (for a $1$ bit). This approach leads to the same recursive formula, but the backward halving method is preferable because it directly constructs the minimal sequence, requires no preprocessing of the binary representation, and naturally aligns with induction, simplifying rigorous proof.

For $100$, binary expansion $1100100_2$ has seven digits and four ones. The minimal number of operations equals the number of digits minus one plus the number of ones: $6+3=9$, matching the recursive calculation. This confirms consistency and efficiency of the backward method.

$\boxed{9}$