Nuts and bolts - Programming contest
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No, you test the whole pile and each becomes two piles and a match. But since that means each pile is smaller than the source pile you end up with considerably less comparisons in total. If I remember Big O notation correctly - and it's been 40 years since I last had to - it's something like O(n2) vs O(n * log(n))
"I have no idea what I did, but I'm taking full credit for it." - ThisOldTony "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt AntiTwitter: @DalekDave is now a follower!
O(n * log(n)) would be an average, if you consistently select the wrong pivot you might end up with O(n2) :) It isn't just about the number of comparisons though, the number of swaps is also important
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
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42
But, that is the answer to everything!
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How about a little programming puzzle for the Holiday? I found this puzzle in a text by G. J. E. Rawlins. "You have a mixed pile of N nuts and N bolts and need to quickly find the corresponding pairs of nuts and bolts. Each nut matches exactly one bolt, and each bolt matches exactly one nut. By fitting a nut and bolt together, you can see which is bigger. But it is not possible to directly compare two nuts or two bolts." Selecting the winner will be heavily influenced by upvotes and Reactions™ :)
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
I'm going to be bold and say you're a nut :D
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Are nuts metric?
If you can keep your head while those about you are losing theirs, perhaps you don't understand the situation.
I believe that the official scale is neither metric nor imperial. Say you found yourself in a urologists' office with your pants down being told you had a tumor on old lefty and that it has to come out pronto. If this happened to you in Australia about 10 years ago the doctor would likely tell you that the government has a program offering free prosthetics and that he can whack in a replacement at the same time since he'll already have his hand in your scrotum. If you accepted this generous offer he would pull out something which looks a lot like a circle template a draftsman would of used many years ago to determine the appropriate size replacement. In a pretty dark day being told that you're a medium large in bollocks is a real highlight.
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Are nuts metric?
If you can keep your head while those about you are losing theirs, perhaps you don't understand the situation.
I believe that the official scale is neither metric nor imperial. Say you found yourself in a urologists' office with your pants down being told you had a tumor on old lefty and that it has to come out pronto. If this happened to you in Australia about 10 years ago the doctor would likely tell you that the government has a program offering free prosthetics and that he can whack in a replacement at the same time since he'll already have his hand in your scrotum. If you accepted this generous offer he would pull out something which looks a lot like a circle template a draftsman would of used many years ago to determine the appropriate size replacement. In a pretty dark day being told that you're a medium large in bollocks is a real highlight.
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The only way around that is to have manually pre-sorted lists, which I see as cheating. The requirements, as stated, would not survive the first sprint planning meeting. I'm the only person that has presented code, so I guess I win the contest. And here's a version that doesn't sort (but it won't be included in the final product because I'm the project lead dev and the customer does not determine technique used in the code):
foreach(Part nut in nuts) { foreach (Part bolt in bolts) { if (nut.Diameter == bolt.Diameter && nut.Pitch == bolt.Pitch) { Console.WriteLine("Pair: \[{0}\] - \[{1}\]", nut, bolt); } } }".45 ACP - because shooting twice is just silly" - JSOP, 2010
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You can never have too much ammo - unless you're swimming, or on fire. - JSOP, 2010
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When you pry the gun from my cold dead hands, be careful - the barrel will be very hot. - JSOP, 2013 -
How about a little programming puzzle for the Holiday? I found this puzzle in a text by G. J. E. Rawlins. "You have a mixed pile of N nuts and N bolts and need to quickly find the corresponding pairs of nuts and bolts. Each nut matches exactly one bolt, and each bolt matches exactly one nut. By fitting a nut and bolt together, you can see which is bigger. But it is not possible to directly compare two nuts or two bolts." Selecting the winner will be heavily influenced by upvotes and Reactions™ :)
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
Something like this, using Visual Prolog:
class predicates
match : (integer_list NutSizes, integer_list BoltSizes, integer_list CurrentMatches) -> integer_list SizesMatched.
clauses
match([], [], RevMM) = list::reverse(RevMM) :-
!.
match(NN, BB, CurrMM) = MM :-
N in NN,
B in BB,
N = B,
!,
UnatchedNN = list::remove(NN, N),
UnatchedBB = list::remove(BB, N),
write("\nUnmatched Nuts: ", NN),
write("\nUnmatched Bolts: ", BB),
write("\nCurrent matches: ", [N | CurrMM]),
MM = match(UnatchedNN, UnatchedBB, [N | CurrMM]).
match(_, _, _) = [] :-
exception::raise_error("Input lists contain an unmatched item or list lengths are unequal."). -
I'll post mine later. After viewing it with fresh eyes this morning. P.S. I have an idea for a change to mine, so maybe I'll have it ready tonight.
BTW, my sort version doesn't directly compare two nuts or two bolts. It compares a property in those objects in order to facilitate the sort process (as opposed to determining whether a given nut goes with a given bolt), so *technically*, I'm following the rules. Furthermore, I'm not matching a nut to a bolt via any kind of comparison. I'm simply iterating a list, and presenting the data in the order it exists in the lists.
".45 ACP - because shooting twice is just silly" - JSOP, 2010
-----
You can never have too much ammo - unless you're swimming, or on fire. - JSOP, 2010
-----
When you pry the gun from my cold dead hands, be careful - the barrel will be very hot. - JSOP, 2013 -
Something like this, using Visual Prolog:
class predicates
match : (integer_list NutSizes, integer_list BoltSizes, integer_list CurrentMatches) -> integer_list SizesMatched.
clauses
match([], [], RevMM) = list::reverse(RevMM) :-
!.
match(NN, BB, CurrMM) = MM :-
N in NN,
B in BB,
N = B,
!,
UnatchedNN = list::remove(NN, N),
UnatchedBB = list::remove(BB, N),
write("\nUnmatched Nuts: ", NN),
write("\nUnmatched Bolts: ", BB),
write("\nCurrent matches: ", [N | CurrMM]),
MM = match(UnatchedNN, UnatchedBB, [N | CurrMM]).
match(_, _, _) = [] :-
exception::raise_error("Input lists contain an unmatched item or list lengths are unequal.").And if you want to elaborate and analyze the Nuts and Bolts as structures:
domains
itemDOM = item(integer ID, integer Size).class predicates
matchItems : (itemDOM* Nuts, itemDOM* Bolts, tuple{itemDOM Nut, itemDOM Bolt}*) -> tuple{itemDom Nut, itemDOM BoltID}*.
clauses
matchItems([], [], RevMM) = list::reverse(RevMM) :-
!.
matchItems(NN, BB, CurrItems) = MM :-
item(IdN, SizeN) in NN,
item(IdB, SizeB) in BB,
SizeN = SizeB,
!,
UnmatchedNN = list::remove(NN, item(IdN, SizeN)),
UnmatchedBB = list::remove(BB, item(IdB, SizeB)),
MM = matchItems(UnmatchedNN, UnmatchedBB, [tuple(item(IdN, SizeN), item(IdB, SizeB)) | CurrItems]).
matchItems(_, _, _) = [] :-
exception::raise_error("Input lists contain an unmatched item or list lengths are unequal.").And invoke the matching like this:
clauses
run() :-
write("\n\n", "Testing", "\n\n"),
IDs = mkList(5, []), % create a list of 5 random integers
NutItems = [ item(ID, ID * 100) || ID in IDS ], % set sizes to be 5 X the ID
BoltItems = [ item(ID + 300, Size) || item(ID, Size) in NutItems ], % set Bolt IDs to 300 greater than Nut IDs
MM = matchItems(NutItems, BoltItems, []),
write(MM),
write("\nPress [Enter] to exit"),
_ = readLine(),
!. -
BTW, my sort version doesn't directly compare two nuts or two bolts. It compares a property in those objects in order to facilitate the sort process (as opposed to determining whether a given nut goes with a given bolt), so *technically*, I'm following the rules. Furthermore, I'm not matching a nut to a bolt via any kind of comparison. I'm simply iterating a list, and presenting the data in the order it exists in the lists.
".45 ACP - because shooting twice is just silly" - JSOP, 2010
-----
You can never have too much ammo - unless you're swimming, or on fire. - JSOP, 2010
-----
When you pry the gun from my cold dead hands, be careful - the barrel will be very hot. - JSOP, 2013 -
How about a little programming puzzle for the Holiday? I found this puzzle in a text by G. J. E. Rawlins. "You have a mixed pile of N nuts and N bolts and need to quickly find the corresponding pairs of nuts and bolts. Each nut matches exactly one bolt, and each bolt matches exactly one nut. By fitting a nut and bolt together, you can see which is bigger. But it is not possible to directly compare two nuts or two bolts." Selecting the winner will be heavily influenced by upvotes and Reactions™ :)
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
It's a binary tree sort: take the first nut and bolt and put them at the top of the tree, then continue to pick nuts and bolts at random, placing them recursively on left or right branches depending on whether they're larger or smaller than whatever's at any given node. Something like that. Matching nuts and bolts stay together. By the time you've tried all of them, they'll all have a match.
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O(n * log(n)) would be an average, if you consistently select the wrong pivot you might end up with O(n2) :) It isn't just about the number of comparisons though, the number of swaps is also important
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
Why would there be any swaps?
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How about a little programming puzzle for the Holiday? I found this puzzle in a text by G. J. E. Rawlins. "You have a mixed pile of N nuts and N bolts and need to quickly find the corresponding pairs of nuts and bolts. Each nut matches exactly one bolt, and each bolt matches exactly one nut. By fitting a nut and bolt together, you can see which is bigger. But it is not possible to directly compare two nuts or two bolts." Selecting the winner will be heavily influenced by upvotes and Reactions™ :)
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
So, if I have this right. I can tell if a Bolt+NUT is Bigger OR FITS or "not" meaning it is clearly smaller. The simplest algorithm is a bubble sort type loop/loop (O(n^2)).
// Z = Number of nuts/bolts, N[1..Z], B[1..Z] hold the nuts/bots
For X = 1 to Z
For Y = 1 to Z
If Fits(B[X],N[Y]) then F[X]=Y : Break;
Next Y
Next XFor X = 1 to Z
Print "Bolt " & X & " Fits with Nut " & F[X]
Next X// Accepting that if the output is all that is needed, do the print in the main loop, before the break;
for 0 or something to skip it if assigned// No optimizations here. But short, clear, concise.
// Simplest optimization is to process the second loop in reverse, deleting an element as it is matched// The list of Nuts will shrink by one with each pass, cutting the comparisons in half.
// Unfortunately requires a modifiable array.For X = 1 to Z
For Y = Length(N) to 1 Step -1
If Fits(B[X],N[Y]) then F[X]=Y : N.delete(Y) : Break;
Next Y
Next X -
Why would there be any swaps?
If you're sorting something you need to swap elements in the collection, right?
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
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If you're sorting something you need to swap elements in the collection, right?
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
No. And it's not sorting either. More like inserting into a sorted list -- at least that's what I'm doing.
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How about a little programming puzzle for the Holiday? I found this puzzle in a text by G. J. E. Rawlins. "You have a mixed pile of N nuts and N bolts and need to quickly find the corresponding pairs of nuts and bolts. Each nut matches exactly one bolt, and each bolt matches exactly one nut. By fitting a nut and bolt together, you can see which is bigger. But it is not possible to directly compare two nuts or two bolts." Selecting the winner will be heavily influenced by upvotes and Reactions™ :)
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
Assuming a mixed pile of nuts and bolts where 1. Each nut matches exactly one bolt. 2. Must fit a nut and bolt together to compare “which is bigger” (size) 3. Not possible to (cannot) directly compare two nuts or bolts Staying within the parameters: * From #1, we know that each bolt has a corresponding matching nut. Then the matching bolt and nut are eliminated from further searches. * From #2, ‘bigger’ assumes size. Not head configuration, pitch, left or right-handed, or composition. Also, since we need to fit the nut onto the bolt, we are only concerned about the width/diameter (of the bolt and nut) and not the length. * From #3, we cannot compare the nuts or the bolts therefore, we are not allowed to sort them. (This includes measuring the size of each bolt To go into the algorithm with this set of parameters, we are not to sort the list of bolts and nuts and we are only comparing the diameters. So, we know that we have a set of N nuts to match up with a similar number of bolts. Here is a simple algorithm:
int boltDiameters[n] // Resizable array
int nutDiameters[n] // Resizable arraydo {
boltsize = boltDiameters[0]
for (int i = 0; i < nutDiameters.Length; i++) {
if (nutDiameters[i] == boltsize) {
println(“Found nut-bolt with size of: “ + boltsize)
nutDiameters.RemoveAt(i)
boltDiameters.RemoveAt(0)
}
}
} Until (boltDiameters.Length == 0)Each iteration through the loop, we reduce the arrays by one element. So, the average number of comparisons are N/2 + (N-1)/2 + (N-2)/2 + … + (N-(N-1))/2 + 1 = (N + N-1 + N-2 + … + N-(N-1))/2 + 1, which by my guess is (N^2)/4
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No. And it's not sorting either. More like inserting into a sorted list -- at least that's what I'm doing.
Ah, but I was referring to Griffs original comment:
Griff wrote:
it's kinda using QuickSort to match 'em up.
That would most probably use swapping of elements, and that's where quicksort is excelling by doing fewer of them than most sorting algorithms.
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
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Ah, but I was referring to Griffs original comment:
Griff wrote:
it's kinda using QuickSort to match 'em up.
That would most probably use swapping of elements, and that's where quicksort is excelling by doing fewer of them than most sorting algorithms.
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
True, but this particular challenge doesn't require sorting or swapping. And at one point I thought he was talking about inserting to a binary tree.
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True, but this particular challenge doesn't require sorting or swapping. And at one point I thought he was talking about inserting to a binary tree.
I'm looking forward to see your solution
Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger
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Yeah, I think you're cheating. :-D The nuts and bolts should be presented in random order.
That wasn't stated as a requirement, however I did create a random collection of each. My example also assumes that there will only be one occurrence of each diameter and pitch, but changing that will only affect the sort comparison method, which everyone seems to think is not valid.
".45 ACP - because shooting twice is just silly" - JSOP, 2010
-----
You can never have too much ammo - unless you're swimming, or on fire. - JSOP, 2010
-----
When you pry the gun from my cold dead hands, be careful - the barrel will be very hot. - JSOP, 2013
