5
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There are two* logical, discrete ways to solve this Futoshiki puzzle, each with its own unique solution. You can solve it using the letters given, or use the numbers 1-5, but if you map the letters to numbers, there must be a justification for how you map them (ie. you can't arbitrarily assign a letter to a number). The rules of Futoshiki are:

  1. Each row and column must contain one of each digit from 1-5 (or letter in "DUCKY").
  2. All inequality constraints between boxes must be satisfied (ie. 1 < 5; you'll have to figure out what the comparison between letters is depending on which route you take).

futoshiki board

Here is a penpa-edit (without the letters entered in case you want to use numbers instead), and here is a transcription:

[ ] [ ] [D] [ ] [ ]  
 v   v           ^   
[ ] [ ] [U]<[ ] [ ]  
                     
[D] [U] [C] [K] [Y]
 v                   
[ ] [ ] [K] [ ]<[ ]
                 v   
[ ]>[ ] [Y] [ ]<[ ]

A full answer of both solutions is preferable, but if you want to submit a partial answer of only one solution, I will accept the other solution (since it was left to last, it seems that it was harder), if that makes sense.

*Technically, there are three, but two of these methods are simply "inverses" of each other, so only one of them will count.

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1
  • $\begingroup$ Why do I have the feeling that the title is a spoonerism? $\endgroup$ – xhienne Feb 6 at 21:26
4
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First Solution

We can try the following

Assign the letters to numbers depending on the order in which they appear in the alphabet (C=1, D=2, K=3, U=4, Y=5)

Step 1

First we can fill in the Cs in a unique way.
enter image description here

Step 2

The inequalities in the bottom right force some letters which gives us some more of the rest of the grid.
enter image description here

Step 3

Noticing that the top of the fourth column must be a U we find that the rest of the grid solves quite quickly.
enter image description here

Second Solution

I have also noticed that

If we look at the Scrabble scores of D,U,C,K,Y we get D=2, U=1, C=3, K=5, Y=4, so maybe this will lead to a solution

Step 1

We can fill in all of the Us quickly
enter image description here

Step 2

The bottomright corner then resolves pretty quickly (the last two squares in the right column cannot contain K or Y so must contain C and D) and we can fill in a bit more of the grid using that.
enter image description here

Step 3

Now the bottom left is filled in in just one way and we can easily finish from here.
enter image description here

Hence, we have two valid and different solutions motivated by different aspects of the letters.

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  • $\begingroup$ I saw you were on the right track with the rot13(fpenooyr fpberf) but your original solution wasn't quite right. I was tempted to give you a nudge in the right direction in case someone else found the second solution from your original answer and then I'd have to wrestle with which one to accept, now I have nothing to worry about! :) $\endgroup$ – samm82 Feb 7 at 17:16
  • 1
    $\begingroup$ @samm82 Yeah, sorry about that, I first misread one of the inequalities and then convinced myself it didn't work. I think I was just very tired yesterday. $\endgroup$ – hexomino Feb 7 at 17:31

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