Author |
Topic: Solder a cube (Read 2173 times) |
|
anshil
Guest
|
Hi, I've a nice riddle I've been even asked in an interview.
Say you have some bendable wires (any number, any length)
------------------------------------------
MODIFICATION INSERTED BY WILLIAM WU:
What is the minimum number of solder connections needed to make a cube? Prove it.
Also, what is the minimum number of wires necessary? Prove it.
------------------------------------------
-
Yet that company had the wrong solution! BTW: Yes, I dind't job there. This is where riddles in interviews are really stupid, if the asker believes in a wrong solution, or didn't understand the problem himself.
|
| « Last Edit: Oct 23rd, 2003, 7:28pm by Icarus » |
 IP Logged |
|
|
|
william wu
wu::riddles Administrator
Gender: 
Posts: 1291
|
 |
Re: New puzzle: solder a cube
« Reply #1 on: Jul 29th, 2002, 1:16pm » |
 Quote  Modify
|
Some clarification. Can I have multiple wires overlapping onto each other at an edge? Or must every edge consist of exactly one wire?
Nice riddle BTW. I plan to add it to the archive, and I'd like to credit you with your real name; send an e-mail to wwu@ocf.berkeley.edu if you'd like this. Else, it will just be credited as [anshil].
Thanks!
|
| « Last Edit: Jul 29th, 2002, 1:17pm by william wu » |
IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
anshil
Guest
|
 |
Re: New puzzle: solder a cube
« Reply #2 on: Jul 29th, 2002, 9:22pm » |
Quote Modify
 Remove
|
Oh 1 write for every edge, or it would be to simple and you don't need to solder at all.
|
|
IP Logged |
|
|
|
yosenl
Guest
|
|
Re: New puzzle: solder a cube
« Reply #3 on: Jul 29th, 2002, 11:58pm » |
Quote Modify
Remove
|
This problem is similar to the euler tour problem for graphs, using corners as vertices and edges as edges. With 8 vertices of odd degrees, you'll need 4 wires to make the edges, and 1 solder at each of the vertices, for 8 total.
|
|
IP Logged |
|
|
|
Kozo Morimoto
Guest
|
|
Re: New puzzle: solder a cube
« Reply #4 on: Jul 30th, 2002, 1:59am » |
Quote Modify
Remove
|
Can you go along diagonals of the "surface" of the cube? Or diagonals of the 'inside' of the cube? This will still be "one wire per edge" rule...
|
|
IP Logged |
|
|
|
jmlyle
Newbie
Gender:
Posts: 31
|
|
Re: New puzzle: solder a cube
« Reply #5 on: Jul 30th, 2002, 8:42am » |
Quote Modify
|
Assuming that each edge can be only one wire, and that you can solder three ends together, you can do it in 7.
The pieces would be:
|_| |_| |_| |_ |
I definately lack the skills to inteligebly build this cube in ASCII, but it is yet another interesting problem for you to work on....
--jmlyle
|
|
IP Logged |
|
|
|
Your name here
Guest
|
|
Re: New puzzle: solder a cube
« Reply #6 on: Jul 30th, 2002, 4:17pm » |
Quote Modify
Remove
|
It seems to me that you must solder 8 times, because each corner of the cube must be soldered. There are three wires coming out from each corner, which can't be done with one wire, so each corner requires soldering.
I can understand how jmlyle makes the cube from three U's, an L and an I, but I can't see how 7 solders can join these pieces together. Perhaps I'm missing something.
|
|
IP Logged |
|
|
|
Nicodemus
Guest
|
|
Re: New puzzle: solder a cube
« Reply #7 on: Jul 30th, 2002, 6:19pm » |
Quote Modify
Remove
|
I agree with "Your Name Here". If you are only allowed one wire per edge and you cannot have any diagonals/internals/superfluous edges, then each corner of the cube has exactly three wires arriving at it. That means that each corner must be soldered. So the answer is 8.
Interestingly, the problem doesn't change if we eliminate the restriction against extra diagonals. As long as we assume that intersections must be soldered (e.g. two V-bends meeting at a cube corner need to be soldered to each other), then the answer is still 8. (Though we can do it in 1 wire.)
|
|
IP Logged |
|
|
|
jmlyle
Newbie
Gender:
Posts: 31
|
|
Re: New puzzle: solder a cube
« Reply #8 on: Jul 30th, 2002, 10:47pm » |
Quote Modify
|
Well, by god, I AM an idiot. I had spent the majotiy of the day working on another problem with everyone at my office, and just jumped into the cube problem for a little breather.
Low and behold, I must have sprained my brain somewhere along the way. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. Of course you can't do it in 7 solders. I had even double checked myself a number of times. I'm sure glad I wasn't doing anything important today, or I might have accidentally killed people....
On the other hand, IF you ARE allowed to use diagonals (which I think is bad form, frankly), then you can definately do it in less than 8. If you are allowed to have an edge consist of more than one wire, you can do it with 1, but I'm not sure if you could do it just with diagonals and one wire per edge. That's a 3D postman problem, I guess, which I can't remember how to solve off hand....
--jmlyle
|
|
IP Logged |
|
|
|
anshil
Newbie
Posts: 41
|
|
Re: New puzzle: solder a cube
« Reply #9 on: Aug 1st, 2002, 4:48am » |
Quote Modify
|
Well there most something be wrong with me. Am I the only one in the world who needs only 3 soldering points?
As I've understood the situation, you don't need to solder if the wires hold themselfs. For me it's a minimalists (least production costs) puzzle.
Oh yes a normal wire cube, meaning no wire going twice on an edge, and no diagonal wires.
|
|
IP Logged |
|
|
|
jmlyle
Newbie
Gender:
Posts: 31
|
|
Re: New puzzle: solder a cube
« Reply #10 on: Aug 1st, 2002, 7:26am » |
Quote Modify
|
Ah, I see what you're saying, anshil. If the wires are pretty rigid, then you can assume that it will hold its shape.
 or  or lots of other possiblities...
So 4 pieces (a,b,c, & d) and 3 solders (o).
|
| « Last Edit: Aug 1st, 2002, 7:27am by jmlyle » |
IP Logged |
|
|
|
Troy
Guest
|
|
Re: New puzzle: solder a cube
« Reply #11 on: Aug 7th, 2002, 4:43pm » |
Quote Modify
Remove
|
jmlyle, Interesting, but the logic just doesn't hold up in a real world application. Even if the wires hold their shape, they must still complete the cube.
Let's use a Wire hanger as an example. It can make a triangular shape and hold that shape without any wrapping around the "stem". Now try to hang something on it and suddenly the shape is not rigid enough to hold that item.
I mean, I suppose the puzzle didn't say that the frame must be rigid enough to handle this, but the assumption is that the end result will appear to be a painted wire cube where you can't see the connections (or where the connections can't bend away from the rest of the cube) in any way. In other words the corners of the cube must be solidly connected with no breaks or spaces between the wires.
So, with that in mind, the correct answer is 8 solders, and 4 wires.
The rest of us assumed the end result is a solid cube with every corner connected and not just "standing" there without it being solidly connected in the end result....
|
|
IP Logged |
|
|
|
jmlyle
Newbie
Gender:
Posts: 31
|
|
Re: New puzzle: solder a cube
« Reply #12 on: Aug 8th, 2002, 12:48pm » |
Quote Modify
|
I also initially assumed that it was a rigid, structurally sound cube. I was responding to anshil, who thought it could be done with 3 solders. I thought about it, and realized what he meant, so I tried to illustrate it with un-put together pieces. The cube would be visually indistinguishable from a rigid cube if the pieces are formed correctly and precisely, and then the 3 solders are made, and the wires are relatively rigid on their own (that's an assumption, but an assumption which allows the answer to lean towards "minimum number of solders and wires" as asked.)
It is the best answer to the question though, since our assumption about the resulting cube's rigidity is not a part of the problem. That assumption leads to a higher number of solders and wires, and is therefore not a good assumption. Thinking the cube is rigid is thinking "inside the box."
|
|
IP Logged |
|
|
|
alan
Guest
|
My instinct with this problem is to say 6 solder points. My first wire produces one side of the cube, the connection to the opposite side and then all 4 sides of that side too. I now have 3 edges missing, connecting my 2 complete faces and completing the cube. Each edge requiring 2 solder points, making 6.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Solder a cube
« Reply #14 on: May 26th, 2004, 6:28pm » |
Quote Modify
|
If you want a rigid cube, then you've forgotten the solders at each end of your first, long, wire. If you are after a less rigid cube, wherein the edges only need to be anchored at one end to keep them in place, then you only need to solder one end each of your three short wires.
So you still end up with 8 solders for total rigidity and 3 solders for the less rigid version. 6 solders is too much for one and too little for the other.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
The_Rookie_Riddler
Newbie
Gender:
Posts: 2
|
If a "perfect" cube is not a requirement (which by the way I interpreted the riddle is NOT), you can do it with a single long wire which you need to fold like shown in the picture.
As we all know, this will form a cube when folded properly.
You can solder it at 5 (or 6) points to increase rigidity.
|
|
IP Logged |
 |
|
|
rmsgrey
Uberpuzzler
    
Gender:
Posts: 2874
|
|
Re: Solder a cube
« Reply #16 on: Sep 15th, 2004, 8:01am » |
Quote Modify
|
And, as we also all know, the outside edges of the unfolded version will pair off so you have two lengths of wire per edge, which is ruled out by the clarification in the third post in the thread.
If you require precisely one "thickness" of wire on each edge, then doing it with only one piece of wire becomes much trickier.
|
|
IP Logged |
|
|
|
The_Rookie_Riddler
Newbie
Gender:
Posts: 2
|
|
Re: New puzzle: solder a cube
« Reply #17 on: Sep 15th, 2004, 11:31pm » |
Quote Modify
|
I apologize, my bad...  It is true that in my solution all but four of the edges will be double wired. Can I get away with my cube being rock solid?
And all the time I was wondering what the following sentence meant...
on Jul 29th, 2002, 9:22pm, anshil wrote:| Oh 1 write for every edge, or it would be to simple and you don't need to solder at all. |
|
But then again, what is the way to do it with a single wire IF we are not to use diagonals? Any suggestions?
|
|
IP Logged |
|
|
|
TenaliRaman
Uberpuzzler
I am no special. I am only passionately curious.
Gender:
Posts: 1001
|
|
Re: Solder a cube
« Reply #18 on: Sep 16th, 2004, 1:18am » |
Quote Modify
|
Single wire cannot do it ...
Consider the cube as a graph
if we are looking to do this with a single piece of wire then we are looking for a euler circuit on this graph but all the nodes on this have odd degree hence not possible ....
|
|
IP Logged |
Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Solder a cube
« Reply #19 on: Sep 17th, 2004, 6:00pm » |
Quote Modify
|
Or, in case you are not familiar with graph theory terminology, look at it this way:
Assume we could do the whole cube in a single wire without repeating any edges or going along diagonals. Our wire must start some where - and we might as well do it at a vertex. Look at the 2nd vertex: The vertex has 3 edges meeting. The wire can come in on one edge and leave on another. At some point, the wire must come back to the vertex along the third edge. Now it has no where to go. All three edges have already been used. The wire must end here. All well and good, except the same situation holds at all 8 vertexs of the cube. At each vertex the wire passes through on 2 of the edges, but must terminate along the third. This means our wire must have at least 8 ends. So one wire will not work. It requires at least 4. It is fairly easy to find paths that use exactly 4 wires, so 4 wires is the minimum.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Nigel_Parsons
Junior Member
Gender:
Posts: 63
|
|
Re: Solder a cube
« Reply #20 on: Sep 25th, 2004, 10:41am » |
Quote Modify
|
My first thoughts on reading the question were along the lines proposed by Alan:
One long wire foming one face,one side at right angles to it, then the opposite face. The three remaining edges then being added as unit pieces of wire. This would require 3 soldered joints.
As long as all the pieces touch at the corners I see no reason to require soldering as the figure will have the appearance of a cube.
Even if you insist on soldering all eight joints you do still not really have a cube, as a cube has 6 faces, and this has Spaces for Faces. If you can ignore the fact that the faces are immaterial, surely you can ignore the fact that the connections at corners are too.
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Solder a cube
« Reply #21 on: Sep 25th, 2004, 12:12pm » |
Quote Modify
|
What about cutting a tiny cube out of the thickness of the wire? That would be a real cube with faces. And no solder at all.
|
|
IP Logged |
|
|
|
Speaker
Uberpuzzler
Gender:
Posts: 1118
|
|
Re: Solder a cube
« Reply #22 on: Sep 26th, 2004, 6:39pm » |
Quote Modify
|
Grimbal, is that what they call thinking outside the box? 
How about bending a tiny hook/loop in the ends of the four wires. This way you would not need to solder them, just twist them around.
|
|
IP Logged |
They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. <Ben Franklin>
|
|
|
coolnfundu
Junior Member
Posts: 116
|
|
Re: Solder a cube
« Reply #23 on: Oct 21st, 2004, 5:00am » |
Quote Modify
|
can you make a cube with a bendable wire (as the question says) without the wire going through the same point more than once 
|
|
IP Logged |
::  Obvious is always wrong  ::
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2874
|
|
Re: Solder a cube
« Reply #24 on: Oct 21st, 2004, 5:21am » |
Quote Modify
|
If you can make a cube at all, the laws of physics suggest that you can do it without the wire passing through the same point more than once...
Of course, if you actually just want to make the skeleton of the cube (the 12 edges) then you can't do it for reasons most clearly posted by Icarus just a few responses ago...
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print