Using this code implement backtracking in the code to print the solution:

from itertools import permutations as perm

def dict$\_$perms$($ category$\_$list $)$:

for p in perm$($range$(1,6))$: yield $\{$a:b for $($a$,$b$)$ in zip$($category$\_$list,p$)\}$

def key$($d$,$ v$)$: # find a key in dictionary d with value v

return $[$name for name, val in d$.$items$()$ if val $==$ v$][0]$

colours $=[$ colour for colour in dict$\_$perms$([\text{'}$Red$\text{'},$'Green','Indigo','Yellow','Blue'$])$

if colour$[\text{'}$Green$\text{'}]==$ colour$[\text{'}$Indigo$\text{'}]+1]$

colour$\_$nation $=[($colour$,$nation$)$ for colour in colours

for nation in dict$\_$perms$([\text{'}$Englishman$\text{'},$'Spaniard','Ukranian','Norweigan','Japanese'$])$

if nation$[\text{'}$Norweigan$\text{'}]==1$

and colour$[\text{'}$Red$\text{'}]==$ nation$[\text{'}$Englishman$\text{'}]$

and abs$($nation$[\text{'}$Norweigan$\text{'}]-$ colour$[\text{'}$Blue$\text{'}])==1]$

colour$\_$nation$\_$drink $=[($colour$,$ nation, drink$)$ for colour,nation in colour$\_$nation

for drink in dict$\_$perms$([\text{'}$Coffee$\text{'},$'Milk','Orange Juice','Water','Tea'$])$

if drink$[\text{'}$Coffee$\text{'}]==$ colour$[\text{'}$Green$\text{'}]$

and nation$[\text{'}$Ukranian$\text{'}]==$ drink$[\text{'}$Tea$\text{'}]$

and drink$[\text{'}$Milk$\text{'}]==3]$

colour$\_$nation$\_$drink$\_$smoke $=[($colour$,$ nation, drink, smoke$)$

for colour, nation, drink in colour$\_$nation$\_$drink

for smoke in dict$\_$perms$([\text{'}$Old Gold','Kools','Chesterfield','Lucky Strike','Parliament'$])$

if smoke$[\text{'}$Kools$\text{'}]==$ colour$[\text{'}$Yellow$\text{'}]$

and smoke$[\text{'}$Lucky Strike'$]==$ drink$[\text{'}$Orange Juice'$]$

and nation$[\text{'}$Japanese$\text{'}]==$ smoke$[\text{'}$Parliament$\text{'}]]$

solutions $=[($colour$,$ nation, drink, smoke, pet$)$

for colour, nation, drink, smoke in colour$\_$nation$\_$drink$\_$smoke

for pet in dict$\_$perms$([\text{'}$Dog$\text{'},$'Snails','Fox','Horse','Zebra'$])$

if smoke$[\text{'}$Old Gold'$]==$ pet$[\text{'}$Snails$\text{'}]$

and abs$($smoke$[\text{'}$Chesterfield$\text{'}]-$ pet$[\text{'}$Fox$\text{'}])==1$

and abs$($smoke$[\text{'}$Kools$\text{'}]-$ pet$[\text{'}$Horse$\text{'}])==1$

and nation$[\text{'}$Spaniard$\text{'}]==$ pet$[\text{'}$Dog$\text{'}]]$

if len$($solutions$)>1$:

print "multiple solutions"

else:

$($colour$,$nation,drink,smoke,pet$)=$ solutions$[0]$

for pos in range$(1,6)$:

print pos, $"$:$",$ key$($colour$,$pos$),",",$key$($nation$,$pos$),",",$key$($drink$,$pos$),",",$key$($smoke$,$pos$),",",$key$($pet$,$pos$)$