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Colourful Knight's Tour (Posted on 2008-01-24) Difficulty: 3 of 5
A knight enters the grid at number 1 and then moves, as if he were on a chessboard, to number 2. He visits every square in numerical order, before exiting at number 64.

KEY:
Blue = cube numbers (1, 8, 27 and 64)
Green = squares which are not cubes (4, 9, 16, 25, 36 and 49)
Red = prime numbers (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59 and 61)
Yellow = multiples of ten (10, 20, 30, 40, 50 and 60)
Purple = multiples of 11 which are not prime (22, 33, 44 and 55)

One number has been added to get you started.

35

No Solution Yet Submitted by Josie Faulkner    
Rating: 4.9091 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Step by step Walkthrough | Comment 9 of 19 |


I don't know if this is the cleanest way you can move through the puzzle, but it's something.

AKMA = a knight's move away
TIOOECA = there is only one empty cell available
Columns will be called A-H from left to right, rows will be called 1-8 from top to bottom.

A helpful piece of information: For a moment, imagine the board colored like a checkerboard (CB).  Let's say 35 (which is odd) happens to be on a black CB square.  Then all the odd numbers will be on a black CB square, and all the even numbers will be on a red CB square.  Why?  Because a knight's move will make you travel 3 units squares, it will always alternate landing on red-black-red-black and so on.  And since the knight will be visiting numbers in numerical order, he will always alternate landing on even-odd-even-odd and so on.  Since the board ISN'T colored like a CB, I will refer to this as "Diagonal Cardinality" or DC.  What I mean is, all the cells along any diagonal will be the same cardinality (odd or even).  So if I am considering a certain cell, and I know another cell in the same diagonal is odd, then the cell I am considering will also be odd.  And two diagonals that intersect at a cell (not a corner) will have to be the same cardinality since they will share a common cell.  Moving onů

36, being a square, must be green and AKMA from 35.  There is only one green cell that is AKMA from D8, so B7 is 36.
37, being a prime, must be red and AKMA from 36.  There is only one red cell that is AKMA from B7, so C5 is 37.
34, not being a special number here, must be white and AKMA from 35.  There is only one white cell that is AKMA from D8, so C6 is 34.
33, being a multiple of 11, must be purple and AKMA from 34.  There is only one purple cell that is AKMA from C6, so A5 is 33.

Oooooo!  There is only one prime number that is also even.  So 2 must be in a red cell that will never be diagonal from an odd cell due to DC.  So C2 is 2.
1, being a cube, must be blue and AKMA from 2.  There is only one blue cell that is AKMA from C2, so A1 is 1.

Of the remaining cubes, only one is odd.  So 27 must be in a blue cell that is diagonal from other odd cells due to DC.  So B8 is 27.
25, being a square, must be green and two knight's moves away from 27.  There is only one green cell that is two knight's moves away from B8, so F8 is 25.
26 must be AKMA from both 27 and 25.  There is only one cell that is AKMA from both B8 and F8, so D7 is 26.
28 must be AKMA from 27 and TIOOECA.  So A6 is 28.
29, being a prime, must be red and AKMA from 28.  There is only one red cell that is AKMA from A6, so B4 is 29.
30, being a multiple of 10, must be yellow and AKMA from 29.  There is only one yellow cell that is AKMA from B4, so A2 is 30.

Let's look at A8.  It is yellow, so it is a multiple of 10.  There are only two cells that are AKMA from it: B6 and C7.  Since A8 isn't blue, it won't be 1 or 64, so both A8-1 and A8+1 have to be AKMA from it.  Therefore A8-1 and A8+1 must be in B6 and C7 (but we don't know which goes where yet).
Well, B6 is green, so it must be a square.  And it must be odd (we know this from DC and the fact that it will be one more or less than a multiple of 10, which must be even).  The odd squares are 9, 25 and 49.  25 is not one more or one less than a multiple of 10, so B6 cannot be 25.  If B6 were 9, then A8 would be 10, and this would force C7 to be 11.  But 11 is a prime and must be red.  C7 is not red.  So B6 cannot be 9.  So B6 is 49.

Going back to what we learned about A8, we now know A8 is 50.
51 must be AKMA from 50 and TIOOECA.  So C7 is 51.

Let's look at H8.  It is red, so it is a prime.  There are only two cells that are AKMA from it: G6 and F7.  Since H8 isn't blue, it won't be 1 or 64, so both H8-1 and H8+1 have to be AKMA from it.  Therefore A8-1 and A8+1 must be in G6 and F7 (but we don't know which goes where yet).
Well, F7 is blue, so it must be a cube.  It can only be 8 or 64 since the other two have been placed.  64 is not one more or one less than a prime, so F7 cannot be 64.  So F7 is 8.
Going back to what we learned about H8, we now know that H8 is 7.
6 must be AKMA from 7 and TIOOECA.  So G6 is 6.
9, being a square, must be green and AKMA from 8.  There is only one green cell that is AKMA from F7, so H6 is 9.
10, being a multiple of 10 (imagine that), must be yellow and AKMA from 9.  There is only one yellow cell that is AKMA from H6, so G8 is 10.
11, being a prime, must be red and AKMA from 10.  There is only one red cell that is AKMA from G8, so E7 is 11.

24, not being a special number here, must be white and AKMA from 25.  There is only one white cell that is AKMA from F8, so H7 is 24.

23, being a prime, must be red and AKMA from 24.  There is only one red cell that is AKMA from H7, so G5 is 23.

64 is the only remaining cube, so it has to go in the only remaining blue cell.  So H1 is 64.
63 must be AKMA from 64.  63 is not prime so it cannot be in a red cell.  So F2 is 63.

Of the remaining multiples of 11, only one is odd.  So 55 must be in a purple cell that is diagonal from other odd cells due to DC.  So F6 is 55.

4, being a square, must be green.  There are only two green cells available: B1 and G2.  If B1 were 4, then 3 would have to be in A3 in order to be AKMA from both 4 and 2.  But 3 is a prime and A3 is not red.  So G2 is 4.
5, being a prime, must be red and AKMA from both 4 and 6.  There is only one red cell that is AKMA from both G2 and G6, so H4 is 5.

16, being a square, must be green.  There is only one green cell left, so B1 is 16.
15, not being a special number here, must be white and AKMA from 16.  There is only one white cell that is AKMA from B1, so A3 is 15.

32, not being a special number here, must be white and AKMA from 33.  There are only two such cells available: B3 and C4.
31, being a prime, must be red and AKMA from 32.  If 32 went in C4, then 31 could only go in D2, E3 or E5.  But 31 also has to be AKMA from 30 and none of D2, E3 or E5 are AKMA from A2.
So 32 cannot go in C4.  So B3 is 32.
31 must be AKMA from both 32 and 30.  There is only one cell that is AKMA from both A2 and B3.  So C1 is 31.  And just to confirm, it is red which it should be.

44, being a multiple of 11, must be purple.  There are only two purple cells left, so 44 is either H3 or E6.
43, being a prime, must be red and AKMA from 44.  There are no red cells available that are AKMA from H3.
So E6 is 44.
22, being a multiple of 11, must be purple.  There is only one purple cell left, so H3 is 22.

45, not being a special number here, must be white and AKMA from 44.  There is only one white cell available that is AKMA from E6, so F4 is 45.
21, not being a special number here, must be white and AKMA from 22.  There is only one white cell available that is AKMA from H3, so G1 is 21.
20, being a multiple of 10, must be yellow and AKMA from 21.  There is only one yellow cell that is AKMA from G1, so F3 is 20.

Let's look at H5, which is yellow.  The only remaining multiples of 10 are 40 and 60.  If H5 were 40, then 39 would have to be AKMA from H5.  But the only cells that are available are red, and 39 is not prime.  So H5 is 60.

61 must be AKMA from 60.  There are only two cells available that are AKMA from H5: G3 and G7.  If G7 were 61 there would be no way to place 62 so that it is AKMA from both 61 and 63.  So G3 is 61.
62 must be AKMA from both 61 and 63.  There is only one cell available that is AKMA from both G3 and F2, so E4 is 62.
59 must be AKMA from 60.  There is only one cell available that is AKMA from H5, so G7 is 59.  And just to confirm, it is red which it should be.

43, being a prime, must be red and AKMA from 44.  There is only one red cell available that is AKMA from E6, so D4 is 43.

Let's look at F1, which is white.  All four cells that are AKMA from it are red.  That means F1 must be a number that is one less than one prime number and one more than another prime number.  The remaining primes are 3, 13, 17, 19, 41, 47, 53.  There is only one pair of primes that are 1 away from the same number: 17 and 19.  So F1 is 18.

17 must be AKMA from both 16 and 18.  There is only one cell that is AKMA from both B1 and F1, so D2 is 17.
19 must be AKMA from both 18 and 20.  There is only one cell available that is AKMA from both F1 and F3, so H2 is 19.

40, being a multiple of 10, must be yellow.  There is only one yellow cell left, so D1 is 40.
39, not being a special number here, must be white and AKMA from 40.  There is only one white cell available that is AKMA from D1, so B2 is 39.

Let's look at E1.  There are four cells that are AKMA from it: C2, D3, F3, G2 which are currently 2, empty, 20 and 4.  E1 must link up with exactly two of these cells.  It can't link up with 20 because 19 and 21 have already been placed.  So it has to link up with at least one of either 2 or 4.  Well 1 and 5 are already placed, so the only way it E1 can link up with either 2 or 4 is if it is 3.  So E1 is 3.

Let's look at G4.  There are 6 cells that are AKMA from it, but three of them contain fully linked numbers (61, 19 and 9), so we'll disregard those.  Instead we will focus on E3, E5 (both of which are empty and red) and F6 (which is 55).  Since G4 must be linked to two of these cells.  So G4 is either between two prime numbers, or between one prime number and 55.  Of the remaining primes, there are no pairs that are 1 away from the same number.  So G4 must be between one prime number and 55.  Well, 57 is not prime, but 53 is.  So G4 is 54.

53 can't be E5 because there would not be a cell that is AKMA from both 53 and 51 (which 52 would have to be).  So E3 is 53.
52 must be AKMA from both 53 and 51.  There is only one cell that is AKMA from both E3 and C7, so D5 is 52.
56 must be AKMA from 55 and TIOOECA, so E8 is 56.
57 must be AKMA from 56 and TIOOECA, so D6 is 57.
58 must be AKMA from both 57 and 59 and TIOOECA, so F5 is 58.

12 must be AKMA from 11 and TIOOECA, so C8 is 12.
13 must be AKMA from 12 and TIOOECA, so A7 is 13.  And it is red, so that checks out.
14 must be AKMA from 13 and TIOOECA, so B5 is 14.  And that is AKMA from 15, so that checks out.

41 must be AKMA from 40 and TIOOECA, so C3 is 41.  And it is red, so that checks out.

42 must be AKMA from both 41 and 43 and TIOOECA, so E2 is 42.

47, being a prime, must be red.  There is only one red cell left, so E5 is 47.
46 must be AKMA from both 45 and 47 and TIOOECA, so D3 is 46.
48 must be AKMA from both 47 and 49 and TIOOECA, so C4 is 48.

38 must be AKMA from both 37 and 39 and TIOOECA, so A4 is 38.

Whew! That was lengthy! =) Later!

Edited on January 28, 2008, 10:57 am
  Posted by nikki on 2008-01-24 23:37:12

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