prize18-4  Exercise 1: Surrogate Pairs by Hand
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Method for U+1F600: 0x1F600 - 0x10000 = 0x0F600 = 0000 1111 0110 0000 0000 (20 bits).
Top 10 bits 0000111101 = 0x03D, so high surrogate = 0xD800 + 0x3D = 0xD83D.
Bottom 10 bits 1000000000 = 0x200, so low surrogate = 0xDC00 + 0x200 = 0xDE00.  Pair: D83D DE00.

ex1.py:

  def to_surrogates(cp):
      v = cp - 0x10000
      return 0xD800 + (v >> 10), 0xDC00 + (v & 0x3FF)

  for cp in [0x1F600, 0x24B62]:
      v = cp - 0x10000
      hi, lo = to_surrogates(cp)
      print(f"U+{cp:X}: minus 0x10000 = {v:#07x} = {v:020b}")
      print(f"   top 10 bits {v >> 10:010b}, low 10 bits {v & 0x3FF:010b}")
      print(f"   pair {hi:04X} {lo:04X}")
      print("   Python:", chr(cp).encode("utf-16-be").hex(" "))

Real output (Python 3.14):

  U+1F600: minus 0x10000 = 0x0f600 = 00001111011000000000
     top 10 bits 0000111101, low 10 bits 1000000000
     pair D83D DE00
     Python: d8 3d de 00
  U+24B62: minus 0x10000 = 0x14b62 = 00010100101101100010
     top 10 bits 0001010010, low 10 bits 1101100010
     pair D852 DF62
     Python: d8 52 df 62

Both match. Every high surrogate lies in D800-DBFF and every low one in DC00-DFFF, because 10 bits can add at most 0x3FF to the base.