Hourglass puzzle: Difference between revisions

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=={{header|Julia}}==
=={{header|Julia}}==
Implemented as a game solver rather than as a game with user input.
Implemented as a game solver rather than as a game with user input.
<lang julia>function findsumfromend(target, series)
<lang julia>function euclidean_hourglassflipper(hourglasses, target)
for startpoint in length(series):-1:1
if sum(series[startpoint:end]) == target
return startpoint
end
end
return nothing
end

function euclidean_hourglassflipper(hourglasses, target)
gcd(hourglasses) in hourglasses && throw("Hourglasses fail sanity test (not relatively prime enough")
gcd(hourglasses) in hourglasses && throw("Hourglasses fail sanity test (not relatively prime enough")
flippers, series = deepcopy(hourglasses), Int[]
flippers, series = deepcopy(hourglasses), Int[]

Revision as of 22:33, 29 December 2020

Hourglass puzzle is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Given two hourglass of 4 minutes and 7 minutes, the task is to measure 9 minutes.

Notes

Implemented as a 1-player game.

Julia

Implemented as a game solver rather than as a game with user input. <lang julia>function euclidean_hourglassflipper(hourglasses, target)

   gcd(hourglasses) in hourglasses && throw("Hourglasses fail sanity test (not relatively prime enough")
   flippers, series = deepcopy(hourglasses), Int[]
   for i in 1:typemax(Int32)
       n = minimum(flippers)
       push!(series, n)
       flippers .-= n
       for (i, n) in enumerate(flippers)
           if n == 0
               flippers[i] = hourglasses[i]
           end
       end
       for startpoint in length(series):-1:1
           if sum(series[startpoint:end]) == target
               println("Series: $series")
               return startpoint, length(series)
           end
       end
   end

end

println("Flip an hourglass every time it runs out of grains, and note the interval in time.") i, j = euclidean_hourglassflipper([4, 7], 9) println("Use hourglasses from step $i to step $j (inclusive) to sum 9")

</lang>

Output:

Series: [4, 3, 1, 4, 2, 2]

Flip an hourglass every time it runs out of grains, and note the interval in time.
Use hourglasses from step 3 to step 6 (inclusive) to sum 9

tested with FMSlogo <lang logo> to bb Make "small_capacity 4 Make "big_capacity 7 make "small 0 make "big 0 make "t 0 print "_____________decision_0_game_over print "_________decision_1_start_timing print "_______decision_2_flip_small print "____decision_3_flip_big print "__decision_4_flip_both print "_________any_other_number________________wait do.until [show list list :small :big :t print "your_decision_0_1_2_3_4 human_decision if :my_decision>1 [machine_computes] ] [:my_decision=0] print list :t "minutes_passed end

to human_decision make "my_decision readword if :my_decision=1 [print "reset_timer make "t 0] if :my_decision=2 [print "flip_small make "small :small_capacity-:small] if :my_decision=3 [print "flip_big make "big :big_capacity-:big] if :my_decision=4 [print "flip_both make "small :small_capacity-:small make "big :big_capacity-:big ] if :my_decision>4 [print "wait] end

to machine_computes ifelse :small>:big [make "my_selection :big] [make "my_selection :small] if :small=0 [make "my_selection :big] if :big=0 [make "my_selection :small] make "small :small-:my_selection make "big :big-:my_selection make "t :t+:my_selection if :small<0 [make "small 0] if :big<0 [make "big 0] end

to zzz

A. 7 minutes with 4- and 5-minute timers
B. 15 minutes with 7- and 11-minute timers
C. 14 minutes with 5- and 8-minute timers

ifelse YesNoBox [Welcome] [run / show me the code] [bb] [edall]

A is possible
Turn both the 5 and the 4. When the 4 runs out, flip it over.Now, when the 5 runs out, start timing. The 4 will run for three more minutes, after which, you can flip it over to reach 7.
B is possible
Turn both the 7 and the 11. When the 7 runs out, start timing. The 11 will run for 4 more minutes, after which it can be flipped to reach 15.
C is possible
Turn both the 5 and the 8. When the 5 runs out, flip it. The 8 will then run out after 3 minutes, leaving 2 minutes in the 5. Flip the 8 then. When the 5 runs out, start timing. There are now 6 minutes left in the 8, and flipping the 8 after those 6 minutes gives 6 + 8 = 14 minutes.

end

Make "big 0 Make "big_capacity 5 Make "my_decision " Make "my_selection 4 Make "small 0 Make "small_capacity 4 Make "startup [zzz] Make "t 0


</lang>

Python

There isn't much of a task description as I write this, but, here goes...

<lang python>def hourglass_puzzle():

   t4 = 0
   while t4 < 10_000:
       t7_left = 7 - t4 % 7
       if t7_left == 9 - 4:
           break
       t4 += 4
   else:
       print('Not found')
       return 
   print(f"""

Turn over both hour glasses at the same time and continue flipping them each when they individually run down until the 4 hour glass is flipped {t4//4} times, wherupon the 7 hour glass is immediately placed on its side with {t7_left} hours of sand in it. You can measure 9 hours by flipping the 4 hour glass once, then flipping the remaining sand in the 7 hour glass when the 4 hour glass ends.

""")

hourglass_puzzle()</lang>

Output:
Turn over both hour glasses at the same time and continue flipping them each
when they individually run down until the 4 hour glass is flipped 4 times,
wherupon the 7 hour glass is immediately placed on its side with 5 hours 
of sand in it.
You can measure 9 hours by flipping the 4 hour glass once, then
flipping the remaining sand in the 7 hour glass when the 4 hour glass ends.