11
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You are a knight, far from your familiar checkered board. You have 9 moves to escape, but there are some rules.

  • Each region must contain all letters matching the number move you are making. So, the first region must contain the letters O, N, and E
  • The space you land on must be the first letter of the word
  • With the exception of the letter E, there are to be no duplicate letters in any row/column
  • Every row/column must contain at least one E
  • The region for 7 cannot have two Es in the same row/column

puzzle

And a bad attempt at a text version with === as start and ~~~ as exit:

+---+===+---+---+---+---+  
| N   O |           | U |  
+   +---+   +---+---+   +  
| E |   |   | W |       |  
+---+   +---+   +   +---+  
| X     |       |   |   |  
+---+---+---+---+---+   +  
|       |       |     E |  
+   +   +   +---+   +   +  
|     E | E |   |       |  
+   +---+   +   +---+---+  
|   |       |           |  
+---+---+---+---+~~~+---+
$\endgroup$
11
  • 1
    $\begingroup$ Do we start on O? $\endgroup$ Jan 9, 2017 at 18:06
  • $\begingroup$ Are we assuming that the knight can jump, or that it moves in an L? $\endgroup$ Jan 9, 2017 at 18:11
  • $\begingroup$ Is the "one exception" for one occurrence of duplication? Or for one letter entirely? It appears to be the latter. $\endgroup$
    – Rubio
    Jan 9, 2017 at 18:15
  • 1
    $\begingroup$ I think there is a mistake. E has to go in the bottom part of the U bit due to the row rule, the u bt which can only be for four and there is no E in four $\endgroup$ Jan 9, 2017 at 18:15
  • 1
    $\begingroup$ Brilliant puzzle. I'm curious how you came up with it? $\endgroup$
    – undefined
    Jan 10, 2017 at 6:15

3 Answers 3

10
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Full solution with explanation

Of the numbers from one to nine:

  • ONE, TWO, and SIX have three letters; we already know which region is ONE, and TWO and SIX are then also given by the W and X already in place.
  • FOUR, FIVE, and NINE have four letters; FOUR is given by the U already in place, NINE must be the bottom right region, so FIVE must be the top middle one.
  • THREE, SEVEN, and EIGHT have five letters.

Since "the space you land on must be the first letter of the word", we can immediately fill in the rest of the TWO and the first N of NINE.

Now the F of FOUR is reachable in two moves from the T of TWO, so it must be either below the U or the lowest cell of the FOUR region. The latter is impossible since the intermediate step must be in one of the five-cell regions. This also tells us where the T of THREE must be, so THREE is the rightmost five-cell region:

first pass

Then we know exactly where the F of FIVE must be, and we can fill in the whole of SIX. Since the third row must contain at least one E, it must be the second E of THREE, which gives us the "one exception" to letter duplication in a single row/column. (Clearly this "one exception" means the entire letter E, rather than the single pair of E's we can already see adjacent in the initial setup.)

second pass

Now that no-duplication rule tells us that the O of FOUR can't be in the third row and the R of THREE can't be in the fifth column, enabling us to finish filling in those regions.

The E of EIGHT (the only E in the region) must be two moves away from the S of SIX. This tells us which way round the SEVEN and EIGHT regions are, and where the S of SEVEN must be.

Also, the no-duplication rule tells us the N of SEVEN can't be in the first column, and then the fourth (SEVEN-specific) rule enables us to complete this region.

The no-duplication rule also tells us where the T of EIGHT and the second N of NINE must be:

third pass

Now we have the I,V,E of FIVE, the I,G,H of EIGHT, and the I,E of NINE to fill in. The fourth and fifth columns must each contain an E, so we can fill in the two last E's. The no-duplication rule tells us where the I of EIGHT must be, and the rest is free. So the final, most general answer is:

final answer

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3
  • $\begingroup$ Your solution has two e's in the column furthest to the right. Pretty sure we can't obey all the rules set $\endgroup$ Jan 9, 2017 at 18:30
  • $\begingroup$ @Beastly You're misinterpreting the 3rd restriction, I think ... $\endgroup$ Jan 9, 2017 at 18:48
  • $\begingroup$ @Beastly See clarifying edit to OP. $\endgroup$ Jan 9, 2017 at 21:03
4
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I am assuming the "one exception" means we can ignore the no-duplication rule for one entire LETTER, as no solution is possible otherwise.

 +---+===+---+---+---+---+
 | N   O | V*  F   E | U |
 +   +---+   +---+---+   +
 | E | S | I*| W | O   F |
 +---+   +---+   +   +---+
 | X   I | T   O | R | E |
 +---+---+---+---+---+   +
 | S   N | G*  I | T   E |
 +   +   +   +---+   +   +
 | V   E | E | N | H   R |
 +   +---+   +   +---+---+
 | E | T   H*| E   N   I |
 +---+---+---+---+~~~+---+

The starred cells allow some freedom - V/I can be swapped, as can G/H.
The two Es in SEVEN cannot be moved, nor the Ns in NINE, so it is not possible to prevent a same-row duplication of E in the bottom row; nor can we avoid a same-column duplication of (at least) one E in THREE.

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4
  • $\begingroup$ you have duplicate E's in Three (right column), and several from Seven to other letters $\endgroup$
    – Sconibulus
    Jan 9, 2017 at 18:25
  • $\begingroup$ See my note at the top. There has to be some odd play with the Es. It's not clear from the rules exactly what "one exception" means. $\endgroup$
    – Rubio
    Jan 9, 2017 at 18:29
  • $\begingroup$ @Sconibulus, Rubio: see clarifying edit to OP. $\endgroup$ Jan 9, 2017 at 21:03
  • 1
    $\begingroup$ I'd already solved it using that assumption, over half an hour before he provided it :) $\endgroup$
    – Rubio
    Jan 9, 2017 at 21:05
2
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The one exception to the duplicate rule is obviously E. There are far too many Es in these number names to make them not collide.

+---+===+---+---+---+---+  
| N   O | V   F   E | U |  
+   +---+   +---+---+   +  
| E | S | I | W | O   F |  
+---+   +---+   +   +---+  
| X   I | T   O | R | E |  
+---+---+---+---+---+   +  
| S   N | G   I | T   E |  
+   +   +   +---+   +   +  
| V   E | E | N | H   R |  
+   +---+   +   +---+---+  
| E | T   H | E   N   I |  
+---+---+---+---+~~~+---+
$\endgroup$
3
  • $\begingroup$ Ah I see now that Rubio beat me to it. $\endgroup$ Jan 9, 2017 at 18:39
  • $\begingroup$ I like that, despite some freedom in some of the letter positions, you and I have the identical solution. Great minds think alike! $\endgroup$
    – Rubio
    Jan 9, 2017 at 18:41
  • $\begingroup$ I filled in all the "first letters" and then just ended up randomly filling in things and swapping them around until they worked. Haha $\endgroup$ Jan 9, 2017 at 18:57

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