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An entry in Fortnightly Topic Challenge #47: "Wacky Sudokus"


(h/t to Beastly Gerbil, whose 1,2,3... Trinary!!! puzzle provided inspiration for this puzzle)

This puzzle is a normal Sudoku, with each digit from 1-9 appearing once per row and column, and once in each square. The variation is in the clues; each clue is a consecutive substring of the Roman numeral expansion of the digit in the cell. There's no funny business: the Roman numeral for 4 is only IV, not IIII, and the substrings are given in order. As an example the clue "VI" cannot represent 4, but could represent 6, 7, or 8. I hope you enjoy!

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TEXT VERSION

═════════════════════════════════════
║   |I I|   ║ V | V |   ║I V|   |I I║
-------------------------------------
║III|   |I I║V I|   |   ║I I|I I|   ║
-------------------------------------
║V I| V |   ║   |I I|   ║   |III| V ║
═════════════════════════════════════
║ X |   | V ║   |   | V ║   |I I|VII║
-------------------------------------
║   |   |V I║   |I I| V ║   | X |   ║
-------------------------------------
║I I|   |   ║ X |   | V ║ V |   |I I║
═════════════════════════════════════
║III|   |   ║ I | I |I I║ V |   |   ║
-------------------------------------
║ V |   |VII║ I |   | X ║ V |   | V ║
-------------------------------------
║VII|   | V ║III|I I|III║ V | V |   ║
═════════════════════════════════════
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  • 3
    $\begingroup$ I won't have time to solve this, but for any who can here's a Penpa link. $\endgroup$
    – bobble
    Jan 17 at 23:58
  • $\begingroup$ Also, does "consecutive substring" mean a substring at any point in the numeral? So can both "V" and "I", as well as "IV", be in a cell that has a 4? $\endgroup$
    – bobble
    Jan 18 at 0:00
  • $\begingroup$ @bobble: Yes, it is a substring that may begin and end at any point of the numeral. So the digit 4 could be in a square labelled "I", "V", or "IV". $\endgroup$ Jan 18 at 0:01
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Solution (click to see large version):

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The first thing is to produce the following list:

I:    1234 6789
II:    23   78
III:    3    8
IV:      4
V:       45678
VI:        678
VII:        78
X:            9

From that, standard Sudoku techniques apply. I'll just list some middle steps below.

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2
  • $\begingroup$ That is the correct answer! $\endgroup$ Jan 18 at 0:52
  • $\begingroup$ I found a very useful tip is to add to the roman numerals when possible. For example if you have deduced that a certain cell can only be either a 7 or an 8, make the roman numeral in the corner into a VII. $\endgroup$ Jan 19 at 3:57

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