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DLosc
DLosc
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BitCycle -u, 90 bytes

 ~  ~!
?v C/v
v<   <
A\\ B^
>/\/C =v
  Cvv  <
  v~v/
  >   ^
  v =
>> >>^
\~~~
 ~v~^
^ + ~

Try it online! Or, watch it in action herehere.

Algorithm

The main loop starts with the current number \$n\$ in unary in the A collector. We divide the number by 2, splitting off two bits at a time; one of the halves, \$\lfloor \frac n 2 \rfloor\$, goes into the uppermost C collector; the other half goes into the middle C collector; and the remainder, \$n\text{ mod }2\$, goes into the bottom C collector.

Once the number is completely divided up in this way, the C collectors open.

  • The top C collector sends a 1 to the sink at the top right, adding 1 to the output, and sends all of its bits back into A.
  • If the bottom C collector is empty (i.e. \$n\text{ mod }2 = 0\$, i.e. \$n\$ was even), the first bit from the middle collector hits the bottommost switch = and activates it pointing right, which discards the bits. This leaves A with just the \$\lfloor\frac n 2\rfloor\$ bits it got from the top C collector: \$n\text{ even}\to n/2\$.
  • If the bottom C collector contains a 1 bit (i.e. \$n\text{ mod }2 = 1\$, i.e. \$n\$ was odd), a negated copy of it hits the bottommost switch and activates it pointing left. This sends the bits from the bottom and middle C collectors into the big collection of dupnegs ~ at the bottom, which makes five copies of its input and discards one bit. All the copies are then sent back into A: \$n\text{ odd}\to \lfloor\frac n 2\rfloor + 5\left( \lfloor\frac n 2\rfloor + 1\right) - 1 = 6\lfloor\frac n 2\rfloor + 4 = 3n + 1\$\$n\text{ odd}\to \lfloor\frac n 2\rfloor + 5\left( \lfloor\frac n 2\rfloor + 1\right) - 1 = 6\lfloor\frac n 2\rfloor + 3 + 1 = 3n + 1\$.

This whole process repeats until \$n=1\$, at which point the two halves are 0; this means the only C collector with data is the bottommost one that holds the remainder. The remainder bit is directed up to the uppermost switch =. Normally, this switch would have been activated by the bits from the middle C collector already, and the remainder bit would follow them into the 5-times circuitry. But since the middle C collector is empty, the remainder bit passes through, activating the switch, and continues northward off the playfield. Since there are no bits remaining on the playfield, the program halts and displays the number of steps taken.

BitCycle -u, 90 bytes

 ~  ~!
?v C/v
v<   <
A\\ B^
>/\/C =v
  Cvv  <
  v~v/
  >   ^
  v =
>> >>^
\~~~
 ~v~^
^ + ~

Try it online! Or, watch it in action here.

Algorithm

The main loop starts with the current number \$n\$ in unary in the A collector. We divide the number by 2, splitting off two bits at a time; one of the halves, \$\lfloor \frac n 2 \rfloor\$, goes into the uppermost C collector; the other half goes into the middle C collector; and the remainder, \$n\text{ mod }2\$, goes into the bottom C collector.

Once the number is completely divided up in this way, the C collectors open.

  • The top C collector sends a 1 to the sink at the top right, adding 1 to the output, and sends all of its bits back into A.
  • If the bottom C collector is empty (i.e. \$n\text{ mod }2 = 0\$, i.e. \$n\$ was even), the first bit from the middle collector hits the bottommost switch = and activates it pointing right, which discards the bits. This leaves A with just the \$\lfloor\frac n 2\rfloor\$ bits it got from the top C collector: \$n\text{ even}\to n/2\$.
  • If the bottom C collector contains a 1 bit (i.e. \$n\text{ mod }2 = 1\$, i.e. \$n\$ was odd), a negated copy of it hits the bottommost switch and activates it pointing left. This sends the bits from the bottom and middle C collectors into the big collection of dupnegs ~ at the bottom, which makes five copies of its input and discards one bit. All the copies are then sent back into A: \$n\text{ odd}\to \lfloor\frac n 2\rfloor + 5\left( \lfloor\frac n 2\rfloor + 1\right) - 1 = 6\lfloor\frac n 2\rfloor + 4 = 3n + 1\$.

This whole process repeats until \$n=1\$, at which point the two halves are 0; this means the only C collector with data is the bottommost one that holds the remainder. The remainder bit is directed up to the uppermost switch =. Normally, this switch would have been activated by the bits from the middle C collector already, and the remainder bit would follow them into the 5-times circuitry. But since the middle C collector is empty, the remainder bit passes through, activating the switch, and continues northward off the playfield. Since there are no bits remaining on the playfield, the program halts and displays the number of steps taken.

BitCycle -u, 90 bytes

 ~  ~!
?v C/v
v<   <
A\\ B^
>/\/C =v
  Cvv  <
  v~v/
  >   ^
  v =
>> >>^
\~~~
 ~v~^
^ + ~

Try it online! Or, watch it in action here.

Algorithm

The main loop starts with the current number \$n\$ in unary in the A collector. We divide the number by 2, splitting off two bits at a time; one of the halves, \$\lfloor \frac n 2 \rfloor\$, goes into the uppermost C collector; the other half goes into the middle C collector; and the remainder, \$n\text{ mod }2\$, goes into the bottom C collector.

Once the number is completely divided up in this way, the C collectors open.

  • The top C collector sends a 1 to the sink at the top right, adding 1 to the output, and sends all of its bits back into A.
  • If the bottom C collector is empty (i.e. \$n\text{ mod }2 = 0\$, i.e. \$n\$ was even), the first bit from the middle collector hits the bottommost switch = and activates it pointing right, which discards the bits. This leaves A with just the \$\lfloor\frac n 2\rfloor\$ bits it got from the top C collector: \$n\text{ even}\to n/2\$.
  • If the bottom C collector contains a 1 bit (i.e. \$n\text{ mod }2 = 1\$, i.e. \$n\$ was odd), a negated copy of it hits the bottommost switch and activates it pointing left. This sends the bits from the bottom and middle C collectors into the big collection of dupnegs ~ at the bottom, which makes five copies of its input and discards one bit. All the copies are then sent back into A: \$n\text{ odd}\to \lfloor\frac n 2\rfloor + 5\left( \lfloor\frac n 2\rfloor + 1\right) - 1 = 6\lfloor\frac n 2\rfloor + 3 + 1 = 3n + 1\$.

This whole process repeats until \$n=1\$, at which point the two halves are 0; this means the only C collector with data is the bottommost one that holds the remainder. The remainder bit is directed up to the uppermost switch =. Normally, this switch would have been activated by the bits from the middle C collector already, and the remainder bit would follow them into the 5-times circuitry. But since the middle C collector is empty, the remainder bit passes through the switch and continues northward off the playfield. Since there are no bits remaining on the playfield, the program halts and displays the number of steps taken.

Source Link
DLosc
DLosc
  • 39.2k
  • 5
  • 83
  • 141

BitCycle -u, 90 bytes

 ~  ~!
?v C/v
v<   <
A\\ B^
>/\/C =v
  Cvv  <
  v~v/
  >   ^
  v =
>> >>^
\~~~
 ~v~^
^ + ~

Try it online! Or, watch it in action here.

Algorithm

The main loop starts with the current number \$n\$ in unary in the A collector. We divide the number by 2, splitting off two bits at a time; one of the halves, \$\lfloor \frac n 2 \rfloor\$, goes into the uppermost C collector; the other half goes into the middle C collector; and the remainder, \$n\text{ mod }2\$, goes into the bottom C collector.

Once the number is completely divided up in this way, the C collectors open.

  • The top C collector sends a 1 to the sink at the top right, adding 1 to the output, and sends all of its bits back into A.
  • If the bottom C collector is empty (i.e. \$n\text{ mod }2 = 0\$, i.e. \$n\$ was even), the first bit from the middle collector hits the bottommost switch = and activates it pointing right, which discards the bits. This leaves A with just the \$\lfloor\frac n 2\rfloor\$ bits it got from the top C collector: \$n\text{ even}\to n/2\$.
  • If the bottom C collector contains a 1 bit (i.e. \$n\text{ mod }2 = 1\$, i.e. \$n\$ was odd), a negated copy of it hits the bottommost switch and activates it pointing left. This sends the bits from the bottom and middle C collectors into the big collection of dupnegs ~ at the bottom, which makes five copies of its input and discards one bit. All the copies are then sent back into A: \$n\text{ odd}\to \lfloor\frac n 2\rfloor + 5\left( \lfloor\frac n 2\rfloor + 1\right) - 1 = 6\lfloor\frac n 2\rfloor + 4 = 3n + 1\$.

This whole process repeats until \$n=1\$, at which point the two halves are 0; this means the only C collector with data is the bottommost one that holds the remainder. The remainder bit is directed up to the uppermost switch =. Normally, this switch would have been activated by the bits from the middle C collector already, and the remainder bit would follow them into the 5-times circuitry. But since the middle C collector is empty, the remainder bit passes through, activating the switch, and continues northward off the playfield. Since there are no bits remaining on the playfield, the program halts and displays the number of steps taken.