S2T2

Jim Randell 10:10 am on 1 July 2021 Permalink Reply
Tags: by: C Higgins ( 36 ), by: H Bradley ( 36 )   

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  Jim Randell 10:11 am on 1 July 2021 Permalink | Reply

  This is a similar idea to Enigma 1732, except that this is for a 1 way journey.

  See: Jeep Problem [ @wikipedia ].

  This Python program runs in 54ms.

  Run: [ @replit ]

  from enigma import (Rational, irange, inf, printf)
  
  Q = Rational()  # choose an implementation of rationals
  
  T = 420  # total amount of fuel available
  R = 105  # range
  
  # consider amount of fuel required for a maximal distance in k trips
  f = d = 0
  for k in irange(1, inf):
    f += R
    d += R * Q(1, 2 * k - 1)
    printf("{k} trips, fuel = {f}, distance = {x:.2f} ({d})", x=float(d))
    # stop when the fuel required reaches (or exceeds) the available fuel
    if not (f < T): break
  

  Solution: The distance between the camps is 176 miles.

  The journey will require 4 trips:

  Trip 1:
  With a full load of fuel (105 miles), drive 15 miles (= 105 / 7).
  Make a fuel dump with 75 miles of fuel, and use the remaining 15 miles to drive back.
  There is 315 miles of fuel remaining at base camp.

  Trip 2:
  With a full load (105 miles), drive 15 miles, and fill up at dump 1 (leaving 60 miles at dump 1).
  Drive a further 21 miles (= 105 / 5), and make a fuel dump with 63 miles of fuel.
  Drive back to dump 1 using the remaining 21 miles of fuel, and refuel for the last 15 miles (leaving 45 miles at dump 1).
  There is 210 miles of fuel remaining at base camp.

  Trip 3:
  With a full load (105 miles), drive 15 miles to dump 1 and refuel (leaving 30 miles at dump 1).
  Drive 21 miles to dump 2 and refuel (leaving 42 miles of fuel at dump 2).
  Drive 35 miles (= 105 / 3) to dump 3 and leave 35 miles of fuel.
  Drive back to dump 2 using the remaining 35 miles of fuel, and refuel for the 21 miles to dump 1 (leaving 21 miles at dump 2).
  Drive back to dump 1 and refuel for the 15 miles to base camp (leaving 15 miles at dump 1).
  There is 105 miles of fuel remaining at base camp.

  Trip 4:
  Fill up with the remaining fuel (105 miles), drive 15 miles to dump 1 and refuel (exhausting dump 1).
  Drive 21 miles to dump 2 and refuel (exhausting dump 2).
  Drive 35 miles to dump 3 and refuel (exhausting dump 3).
  Drive 105 miles (= 105 / 1) and arrive at main camp.
  Total distance = 15 + 21 + 35 + 105 = 176 miles.

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  Frits 1:44 pm on 8 July 2021 Permalink | Reply

  Similar (next one R=315, T = 5*R)

  T = 420       # total amount of fuel available
  R = 105       # range
  
  k = T // R    # number of trips
  
  # number of times traject <n-1>th depot to the <n>th depot is travelled 
  V = lambda k, n : 2 * (k - n) + 1
  
  # T / R = k = 4 so 4 trips can be made setting up 3 depots.
  # The traject <n-1>th depot to the <n>th depot (n = 1..3) is travelled 
  # V(n) = 2 * (k - n) + 1 times (where 0th depot is the base camp) so we need
  # to split R into V(n) equal parts and store V(n) - 2 parts at the depot 
  # leaving 2 parts to reach and return from the <n>th depot
  # this means the <n>th depot is sum (i=1..n, R / V(n)) miles from base camp
   
  # consider amount of fuel required for a maximal distance in k trips
  print("answer:", R + sum(R // V(k, n) for n in range(1, k)), "miles")
  

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  • 

    Frits 3:39 pm on 8 July 2021 Permalink | Reply

    or

    from math import log
    
    def H(n):
        """Returns an approximate value of n-th harmonic number.
    
           http://en.wikipedia.org/wiki/Harmonic_number
        """
        # Euler-Mascheroni constant
        gamma = 0.57721566490153286060651209008240243104215933593992
        return gamma + log(n) + 0.5 / n - 1 / (12 * n**2) + 1 / (120 * n**4)
        
    R = 105    
    n = 4
    print("answer:", round((H(2 * n - 1) - 0.5 * H(n - 1)) * R), "miles")
    

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  Frits 1:43 pm on 17 July 2021 Permalink | Reply

  Different order:

  # make 3 return trips to 1st depot (15 miles from Base Camp) and 
  #      each time drop 75 miles of fuel at 1st depot
  # make 4th trip to 1st depot and fill up (210 miles of fuel left at 1st depot)
  # make 2 return trips from 1st to 2nd depot (21 miles from 1st depot) and
  #      each time drop 63 miles of fuel at 2nd depot
  # make 3th trip from 1st to 2nd depot and fill up 
  #    (0 / 105 miles of fuel left at 1st/2nd depot)
  # make a return trip from 2nd to 3rd depot (35 miles from 2nd depot) and
  #      drop 35 miles of fuel at 3rd depot
  # make 2nd trip from 2nd to 3rd depot and fill up
  #    (0 / 0 / 0 miles of fuel left at 1st/2nd/3rd depot)
  # drive 150 miles to Main Camp
  
  # Total miles: 15 + 21 + 35 + 150 = 176 miles
  

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