The New Order Oracle Coding Challenge 2

2 08 2011

August 2, 2011

(Back to the Previous Post in the Series) (Forward to the Next Post in the Series)

In the previous article in this series we tried to develop different methods for identifying special numbers between 1 and 1,000,000, such that the number formed by reversing the order of the digits will evenly divide into the original number.  I initially predicted that there would be at least three unique solutions provided before the undocumented REVERSE function was mentioned.  A fellow Oak Table member quickly provided a solution using the REVERSE function, which happens to be the fastest of the various approaches.  Another reader noticed a pattern that helps find those special numbers – if that pattern identifies all of the special numbers, it is easily the most efficient approach to finding these special numbers.

Today’s blog article also is on the topic of finding special numbers, determined as special after changing the order of the number’s digits.  I used my fancy, on-point introduction for this article in the earlier article (letters on the dice, trying to form words from those letters), so I thought that I would jump the little pony over the edge of the cliff in search of another introduction (there are probably five readers with an odd image stuck in their head at this moment).   ;-)

Have you ever noticed the disparity between the Environment Protection Agency’s (EPA or the equivalent in other countries) estimates for fuel efficiency and the actual fuel efficiency calculated with the hand held calculator?  Take for instance a car that I own – that car has never achieved the EPA’s estimated fuel efficiency rating.  A 412 horsepower (at the flywheel as claimed by the manufacturer, other tests have concluded that it might be higher) V8 achieving 25 miles per U.S. gallon (MPG) of fuel?

Time for the Oracle Database test case.  On a frigid 32 degree day (in Celsius, that is a hot 90 degrees Fahrenheit), I lowered the windows, topped off the fuel tank with premium grade fuel, reset the car’s fuel efficiency gauge (the MPG gauge… oddly, this gauge always reads 6% to 8% lower than the actual fuel efficiency), and then set out on a roughly 117 mile (188 KM) drive down the highway with the radar detector on the dash board.  While I have heard that the car’s speed limit is set at the factory to roughly 160 miles per hour (258 KM/H), the posted speed limit was typically 55 MPH (miles per hour) except in the roughly half-dozen or so cities along the way.  The car never did hit the EPA’s estimated fuel efficiency rating.  Somebody was playing with the numbers a bit, I think. :-)

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Along the route selected for the 117 mile test loop, I captured cell phone pictures showing the car’s fuel efficiency gauge (MPG gauge).  Below are the cell phone pictures captured at 15 miles into the drive, roughly half way through the drive, and near the end of the 117 mile drive. 

So, how does Oracle Database play a part in the introduction to this blog article?  Let’s take a guess at the car’s actual fuel efficiency using a SQL statement:

SELECT
  29.5 CAR,
  ROUND(29.5 * 1.06, 2) LOW_ACTUAL,
  ROUND(29.5 * 1.08, 2) HIGH_ACTUAL
FROM
  DUAL;

       CAR LOW_ACTUAL HIGH_ACTUAL
---------- ---------- -----------
      29.5      31.27       31.86

So much for EPA estimates – between 25.08% and 27.44% lower than the actual fuel efficiency.  Does this test case then suggest that if we do not use all of the features that are available in a tool, that we are then more efficient than expected?  I think that I need another test case to decide.  ;-)

Side note: For the tank of fuel when the car rolled over the 1,000 mile mark, the car achieved 28.75 miles per gallon while touring some of the sites around Michigan, specifically the hills and curves along the north-west shoreline of the lower peninsula (several of those hills were steep – for one hill in particular the car downshifted three gears to maintain speed going down the hill).  The car’s overall fuel efficiency since it left the factory is 25.58 MPG (excluding the 117 mile test case), so I guess that one could argue that if you TRUNC the fuel efficiency number, at the root the EPA is right.  Make like a tree and leave (that was a movie quote from “Back to the Future”).

After returning from the test case I set out on my other 412 HP vehicle… cough, that would be a 4 cylinder 12 horsepower 1946 Farmall A tractor.  Time to mow the lawn in second gear.  Cutting a 6 foot (1.83 meter) wide path through grass that is up to 12 inches (30.48 cm) high, and capable of mowing down 2 inch diameter trees as necessary.  Oh, the power of the horse…

OK, now that we have drifted far from the topic of this article and have had a chance to play with numbers in the long introduction, on to the challenge.  Consider the four digit integer numbers between 1,000 and 9,999.  There are 4! (1 * 2 * 3 * 4 = 24) possible combinations of the four digits of each of these numbers.  For example, consider the number 8712 – the 24 possible combinations are:

1278
1287
1728
1782
1827
1872
2178
2187
2718
2781
2817
2871
7128
7182
7218
7281
7812
7821
8127
8172
8217
8271
8712 8721

The objective is to find all combinations of the four digit numbers between 1,000 and 9,999 that evenly divide into the original number.  For the above example with the original number 8712, the combination 2178 evenly divides into 8712.  Thus, the output would include:

ORIGINAL COMBINATION
-------- -----------
    8712        2178

The restrictions:

  • The four digit original numbers between 1,000 and 9,999 should not be included in the output if the last digit of those numbers is 0 – the combination number may end with 0 if the original number included a 0 digit.  Thus, do not include the original numbers 1000, 1010, 1020, 1030, 1040, etc.
  • The original value cannot be equal to the combination value (in the above example, it is not valid to output the combination value 8712).
  • Each original number and combination pair should be output at most one time.
  • You may not reuse a digit position twice; all digit positions must be used once.  For example, the number 7141 includes the digit 1 twice, therefore, all generated combinations of that number must include two number 1 digits.

This challenge is a bit more difficult than was the challenge in part 1 of this series.  The greatest difficulty likely involves creating the 23 valid digit combinations (there are 24, but one of those combinations will always be invalid) of the four digit numbers. 

The problem is solvable – will your solution be the same as mine?  I will give the readers a couple of days to ponder the thoughts of why a horse entered a race between a car and a tractor before sharing my approach to solving the problem.








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