Solve the no connection puzzle: Difference between revisions

=={{header|Nim}}==

{{trans|C++}}

I choose to use one-based indexing for the array of pegs. It seems more logical here and Nim allows to choose any starting index for static arrays.

<lang Nim>import strformat

const Connections = [(1, 3), (1, 4), (1, 5), # A to C, D, E

(2, 4), (2, 5), (2, 6), # B to D, E, F

(7, 3), (7, 4), (7, 5), # G to C, D, E

(8, 4), (8, 5), (8, 6), # H to D, E, F

(3, 4), (4, 5), (5, 6)] # C-D, D-E, E-F

type

Peg = 1..8

Pegs = array[1..8, Peg]

func valid(pegs: Pegs): bool =

for (src, dst) in Connections:

if abs(pegs[src] - pegs[dst]) == 1:

return false

result = true

proc print(pegs: Pegs; num: Positive) =

echo &"----- {num} -----"

echo &" {pegs[1]} {pegs[2]}"

echo &"{pegs[3]} {pegs[4]} {pegs[5]} {pegs[6]}"

echo &" {pegs[7]} {pegs[8]}"

echo()

proc findSolution(pegs: var Pegs; left, right: Natural; solCount = 0): Natural =

var solCount = solCount

if left == right:

if pegs.valid():

inc solCount

pegs.print(solCount)

else:

for i in left..right:

swap pegs[left], pegs[i]

solCount = pegs.findSolution(left + 1, right, solCount)

swap pegs[left], pegs[i]

result = solCount

when isMainModule:

var pegs = [Peg 1, 2, 3, 4, 5, 6, 7, 8]

discard pegs.findSolution(1, 8)</lang>

{{out}}

<pre>----- 1 -----

3 4

7 1 8 2

5 6

----- 2 -----

3 5

7 1 8 2

4 6

----- 3 -----

3 6

7 1 8 2

4 5

----- 4 -----

3 6

7 1 8 2

5 4

----- 5 -----

4 3

2 8 1 7

6 5

----- 6 -----

4 5

2 8 1 7

6 3

----- 7 -----

4 5

7 1 8 2

3 6

----- 8 -----

4 6

7 1 8 2

3 5

----- 9 -----

5 3

2 8 1 7

6 4

----- 10 -----

5 4

2 8 1 7

6 3

----- 11 -----

5 4

7 1 8 2

3 6

----- 12 -----

5 6

7 1 8 2

3 4

----- 13 -----

6 3

2 8 1 7

5 4

----- 14 -----

6 3

2 8 1 7

4 5

----- 15 -----

6 4

2 8 1 7

5 3

----- 16 -----

6 5

2 8 1 7

4 3</pre>