Nuts and bolts - Programming contest

[Lost User]

I think the "cannot compare" rule applies. Assume you have mittens on. You can't tell if the nut / bolt is bigger or smaller; only that it does or does not match.

It was only in wine that he laid down no limit for himself, but he did not allow himself to be confused by it. ― Confucian Analects: Rules of Confucius about his food

[OriginalGriff]

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!

[realJSOP]

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
-----
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

[Kornfeld Eliyahu Peter]

Can I bribe you by sending a box of matching nuts and bolts of your choice?

"The only place where Success comes before Work is in the dictionary." Vidal Sassoon, 1928 - 2012

[Jorgen Andersson]

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

[Rusty Bullet]

But, that is the answer to everything!

[Sander Rossel]

I'm going to be bold and say you're a nut :D

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[Josh Gray2]

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.

[Josh Gray2]

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.

[PIEBALDconsult]

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.

[Harrison Pratt]

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.").

[realJSOP]

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

[Harrison Pratt]

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(),
!.

[PIEBALDconsult]

Yeah, I think you're cheating. :-D The nuts and bolts should be presented in random order.

[Member_5893260]

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.

[PIEBALDconsult]

Why would there be any swaps?

[Kirk 10389821]

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 X

For 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

[Jorgen Andersson]

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

[PIEBALDconsult]

No. And it's not sorting either. More like inserting into a sorted list -- at least that's what I'm doing.

[James Lonero]

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 array

do {
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