I was trying to write a short MIPS32 code for computing the median of three registers.

The rules:

  • Assume that some values are pre-loaded into $t0, $t1, $t2.
  • The median of the three values should be placed in $t0.
  • No branches are allowed, but otherwise any instruction is okay.
  • You may arbitrarily modify the values of all registers.

I was able to optimize it down to 15 instructions. Can you do better?

  • \$\begingroup\$ Welcome to Code Golf! I added the code-golf tag, but other than that this challenge looks good. (You might still want to check out the sandbox for future challenges, it can be pretty useful sometimes) \$\endgroup\$ Commented Nov 2, 2021 at 16:59
  • \$\begingroup\$ @RedwolfProgrammed - thanks! \$\endgroup\$
    – M A
    Commented Nov 2, 2021 at 17:00
  • 1
    \$\begingroup\$ I've retagged this as atomic-code-golf since the count seems to be the number of instructions, although this could be made more clear in the body of the challenge. \$\endgroup\$
    – Wheat Wizard
    Commented Nov 2, 2021 at 20:08
  • \$\begingroup\$ May I modify some specific addresses in memory? May I use jump instructions j/jal/jr? May I modify the code during execution? \$\endgroup\$
    – Bubbler
    Commented Nov 3, 2021 at 0:45
  • \$\begingroup\$ @Bubbler - anything is allowed except branches. \$\endgroup\$
    – M A
    Commented Nov 4, 2021 at 9:49

2 Answers 2


7 instructions

slt $t3, $t0, $t1
slt $t4, $t2, $t1
slt $t5, $t2, $t0

Compare all three pairs of values.

xor $t3, $t3, $t4
movn $t0, $t1, $t3

$t1 is the median if it is greater than exactly one of the other values.

xor $t5, $t5, $t4
movn $t0, $t2, $t5

$t2 is the median if it is less than exactly one of the other values.

(Otherwise, $t0 is already the median.)

  • \$\begingroup\$ Very nice. I followed the same logic, but my implementation required more than twice the number of instructions :). \$\endgroup\$
    – M A
    Commented Nov 4, 2021 at 8:40

8 instructions

slt $t4, $t2, $t1
movn $t3, $t1, $t4
movn $t1, $t2, $t4
movn $t2, $t3, $t4

Rearrange $t1 and $t2 so that $t1$t2.

slt $t3, $t2, $t0
slt $t4, $t0, $t1
movn $t0, $t2, $t3
movn $t0, $t1, $t4

If $t0 falls between $t1 and $t2, it is already the median. If it's outside that range, the one it lies beyond is the median.


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