### Chess 5 Piece Puzzle

Source: Mailed to me by Smriti Mittal (Final year student, IIT Bombay)

Problem:
Green numbers indicate how many pieces could move to that square on the next move. Blue squares show the possible locations of the following five different chess pieces: King Queen Rook Bishop Knight

How are the five pieces arranged? Lets mark the squares as (r,c) where r is the row number and c is the column number. So, the squares where the pieces are to be kept are: (1,1), (1,8), (5,8), (6,3), (8,2)

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2. (1,1)knight
(1,8)queen
(5,8)rook
(6,3)king
(8,2)bishop

3. (1,1)->Knight
(1,8)->Rook
(5,8)->Kinght
(6,3)->Bishop
(8,2)->Queen
OR
(1,1)->King
(1,8)->Queen
(5,8)->Rook
(6,3)->Knight
(8,2)->Bishop
OR
(1,1)->Rook
(1,8)->Queen
(5,8)->King
(6,3)->Knight
(8,2)->Bishop
These are the only possible solutions

4. (1,1)->Knight
(1,8)->Rook
(5,8)->Kinght
(6,3)->Bishop
(8,2)->Queen
OR
(1,1)->King
(1,8)->Queen
(5,8)->Rook
(6,3)->Knight
(8,2)->Bishop
OR
(1,1)->Rook
(1,8)->Queen
(5,8)->King
(6,3)->Knight
(8,2)->Bishop
These are the only possible solutions

1. I think only soln 1 is correct.

in soln 2,
if (1,1) is king and (1,8) is queen, then box (1,2) will display 2, instead of 1.

in soln 3,
if (1,1) is rook and (1,8) is queen, then box (1,2) will display 2, instead of 1. also box (1,6) will display 2, instead of 1.

Regards.

5. Eliminate at least 3 locations for the queen, place it at one of the remaining possible slots and the remaining pieces become obvious.

1,1: Knight
1,8: Queen
5,8: Rook
6,3: King
8,2: Bishop

6. I didn't use any algorithm to solve this problem (just did it by trial-and-error and on the basis of my knowledge of chess):
the only way a piece can move to position (4,6) is from (5,8) or (8,2) and it would make sense to put the knight at (5,8) because placing the knight anywhere else will not enable to reach a green square. Thus, place the knight at (5,8).
Placing the rook at (8,2) would enable it to move to (1,2) but not to (4,6) which is already covered by the knight. Thus, place the rook at (8,2).
Now if we place the bishop at (1,8), it will cover the green square at (4,5) but not anything else. Thus, place the rook at (1,8).
Also, in order for any other piece not to cover (4,5) again, we can place the king at (6,3).
Finally, place the queen at (1,1) to cover remaining green squares.

1. Queen at (1,1) wont work. see that queen can go diagonally , but that square at (3,3) has value 0

7. (1,1): knight
(1,8): Queen
(6,3): King
(8,2): Bishop
(5,8): Rook
Did you want to count the number of possibilities or just give a solution?

8. (1,1)- knight
(1,8)- queen
(5,8)- rook
(6,3)- king
(8,2)- bishop

9. knight, queen, rook, bishop and king respectively

10. Lets name these cells first.
Say cells containing green numbers are represented by a,b,c,d,e
a= (1,2) is 1
b= (1,6) is 1
c= (4,5) is 1
d= (4,6) is 1
e= (1,3) is 0

and all blue cells are represented by p,q,r,s,t
p= (1,1)
q= (1,8)
r= (5,8)
s= (6,3)
t= (8,2)

In my solution, by "a piece covering a cell" I mean that piece can move to that cell in its next move making its count as 1.

Now to make "b" count as one, "q" has to be a queen or a rook which will also cover count for "a". Therefore "a" should not be covered from any other piece.

Thus,
"t" cannot have Queen or Rook (to avoid covering "a" twice) ---- eq1
Also since "e" is 0, no one should cover it. Therefore,
"s" cannot have Queen or Rook ------ eq2

This implies "p","q","r" will have queen,rook and one of the remaining third --- eq3

Now to cover "c", either "q" has to be bishop or "q" has to be queen or "s" has to be bishop.

Take this cases one by one.
Say "q" is bishop
therefore "p" and "r" should have queen and rook (from eq2)
Now "d" cannot be covered because "r" cannot be knight (its either queen or rook) and "t" cannot be bishop as bishop is already been taken.

Lets say "s" has bishop,
to cover "d", "r" has to be knight
since "t" cannot have queen or rook (eq1), "t" is king
since "p" cannot be queen to avoid covering "e", "q" is queen and "p" is rook thus making count of "a" and "b" as two

Final option is "q" is queen
so "a","b" and "c" are covered by this queen
now "p" cannot have rook (to avoid make "a","b" as two) thus "r" is rook (from eq1 & e2).
To cover "d", "t" has to be bishop
"p" cannot have king to avoid making count of a as two. So "p" has knight and "s" has king.

Solutions is
p = (1,1) is Knight
q = (1,8) is Queen
r = (5,8) is Rook
s = (6,3) is King
t = (8,2) is Bishop

11. One of the possible solution is
(1,1) - Knight
(1,8) - Queen
(5,8) - Rook
(6,3) - King
(8,2) - Bishop

12. knight(1,1), Queen(1,8), Rook(5,8), King(6,3), bishop(8,2)

13. (1,1) N
(1,8) Q
(5,8) R
(6,3) K
(8,2) B

14. (1,1)KNIGHT, (1,8)ROOK, (5,8)KING, (6,3)BISHOP, (8,2)QUEEN

15. (1,1)KNIGHT, (1,8)QUEEN, (5,8)ROOK, (6,3)BISHOP, (8,2)KING