Problem C
ARMPIT Computations

The popular ARMPIT processor has a simple architecture that is built around a single register that is $12$ bits long. The most significant bit is viewed as being at the left end, and the least significant bit at the right end. The contents of the register are often represented as an unsigned integer written in base $10$. The processor supports the following operations:




modifies the register by performing a bitwise OR with <v>

LSL <amt>

bitwise left shifts the register by <amt> positions

ROR <amt>

bitwise right rotates the register by <amt> positions


increments the register by $1$


inverts (complements) each bit in the register


adds the register to a copy of itself that has been bitwise


left shifted by <amt> positions

In the above table:

  • <v> is an unsigned integer stored in $4$ bits; bitwise OR combines this with the $4$ least significant bits of the register

  • <amt> is an unsigned integer stored in $3$ bits

  • a left shift by <amt> positions moves <amt> $0$’s into the right end of the register, and discards the <amt> bits that “fall off” the left end

  • a bitwise rotation by $1$ position puts the least significant bit into the most significant bit position, and all other bits shift right by $1$ position; a bitwise rotation by <amt> positions is equivalent to <amt> successive rotate-by-$1$ operations

  • for operations ADDONE and ADDSHIFT, addition is performed mod $2^{12}$, i.e., with truncation on overflow

Note that with the exception of ADDSHIFT, similar operations are found in most CPU instruction sets. Here are examples of the supported operations, assuming that the register stores $6$ ($0000\, 0000\, 0110$) before each operation, <v> is $12$ ($1100$), and <amt> is $5$ ($101$). For convenience, leading (high-end) zeros in the register are often omitted.

  • ORWITH makes the register $14$, since the OR of binary $0110$ and $1100$ is $1110$.

  • LSL makes the register $192$, since left shifting $0110$ by $5$ positions gives binary $1100\, 0000$. If, on the other hand, the register initially stored $1111\, 1111\, 1111$, left shifting by $5$ positions would yield $4064$ (binary $1111\, 1110\, 0000$), because the result is truncated to $12$ bits.

  • ROR makes the register $768$, since the binary value $0110$ becomes $0011\, 0000\, 0000$ when rotated rightward by $5$ positions.

  • ADDONE makes the register store $7$. If the register had initially stored $4095$, it would contain $0$ after the ADDONE operation.

  • NOT makes the register store $4089$ (binary $1111\, 1111\, 1001$).

  • ADDSHIFT makes the register store $198$, because $6$ is added to $192$ (see the example for LSL to explain the $192$).

For each of a given list of $12$-bit target values, determine the length of the shortest sequence of ARMPIT instructions that puts that value into the register, under the assumption that the register is initially filled with zeros. For example, for a target value of $769$, the shortest sequence contains $3$ operations (one possiblity is: ORWITH $6$, ROR $5$, ADDONE).


The first line of the input contains a positive integer, $T$ ($T \leq 200$), representing the number of test cases. Each of the next $T$ lines contains a single integer between $0$ and $4095$, inclusive, the target value for that test case.


For each of the $T$ test cases, output a line containing a single integer, the minimum number of ARMPIT instructions that must be executed to put the target value in the register, assuming that the register is initially filled with zeros (for each test case).

Sample Input 1 Sample Output 1

Please log in to submit a solution to this problem

Log in