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A Plumbing Puzzle: Solution

 0 Comments- Add comment Written on 17-Apr-2010 by asqui

Faucet AeratorLast week I posed an obscure plumbing problem where the bathroom basin was behaving in a rather peculiar manner. When you wash your hands with warm water then switch to maximum cold, the water coming out of the faucet is hot for some time.

After significant head scratching I managed to establish the root cause of the problem: The “faucet aerator” was to blame.

The faucet aerator is a thing screwed on to the end of the spout; it mixes some air into the stream of water in order to make it all soft and fluffy. The particular faucet aerator fitted had a “water saving” feature, which intentionally limited the flow of water passing through. (Blue mesh in the picture)

This combined with the pressure differential between the hot water and the cold water, meant that when using the mixer to create warm water what was actually happening was this:

  1. Hot and cold water rush in to the mixing chamber together.
  2. The warm water can’t exit fast enough because the spout is blocked by the “water saver”.
  3. The hot water pressure is significantly higher, which overpowers the cold water.
  4. Hot water dominates and begins to flow directly into the cold water pipe!
  5. As the pipes between the boiler and the basin warm up, the hot water becomes progressively hotter.

Then you desperately attempt to rinse the soap from your hands before they catch fire, but you can’t do it quick enough, at which point you desperately swing the mixer over to maximum cold only to have the scalding hot water that has just backed up into the cold pipe dump out on you, followed by some warm water for a time, and eventually cold water (by which time your hands are already burnt).

Needless to say I quickly did away with the extra “water saving” part of the faucet aerator, and that faucet has been fine ever since!

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A Plumbing Puzzle: Hint

 2 Comments- Add comment Written on 11-Apr-2010 by asqui

Here is a slight hint for the “Plumbing Puzzle”:

As ‘DIY’ correctly noted in this comment, the root cause of the observed effect is that the hot water is at a higher pressure than the cold. The reason for this is that the cold water is fed from a water tank in the loft, whilst the hot water comes from the combi-boiler (which is fed directly from the water mains).

Aside:

If the mains water pressure is higher than that of the water coming from the water tank then what’s the point of having a tank? I’m not sure; I’ve yet to consult a plumber for the answer to this.

It could be a historic artefact. Maybe once upon a time the mains water pressure was not that great. Or maybe the water mains pressure is good with only one or two faucets on, but would be unable to supply all four flats in the property if everyone happened to turn on their faucets and flush the toilet at the same time? Or maybe it’s for isolation, so that someone in another flat flushing the toilet won’t make your shower go hot. (But that doesn’t really work out, because it would make your shower go cold instead, since the hot water from the combi-boiler is still fed from the cold water mains!) I don’t know.

What I do know, however, is that this isn’t the full solution to the puzzle. Even with a pressure differential, why would the hot water push back into the cold pipe rather than coming out of the faucet? Surely it's easier to come out from the faucet than to push back against the cold water?

So the question remains: How does the hot water end up in the cold water pipe?

Give up? Click here for the solution.

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A Plumbing Puzzle

 3 Comments- Add comment Written on 05-Apr-2010 by asqui

Mixer tap Here’s a puzzle based on the true story of a slight plumbing problem we had when we first moved in to our current home.

Consider the following observations regarding the behaviour of the mixer tap in the bathroom basin:

  • The cold water flow rate is fairly paltry.
  • The hot water flow rate is more reasonable.
  • After washing hands with warm water, turning the mixer to maximum cold yields warm water for a time.
  • When trying to investigate this phenomenon, feeling the water pipes under the basin shows they are both warm.

From this information it should be possible to conclude where the problem lies. However, it is probably not immediately obvious what is going on. (It certainly wasn’t to me, at the time!)

Can you work out where the problem lies?

Click here for a hint.

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Harry Eng's Impossible Bottles

 6 Comments- Add comment Written on 28-Jan-2009 by asqui

Harry Eng BottleHarry Eng's instructions for producing this Impossible Bottle on the right:

"Find a piece of wood from the High Chaparral (Manginita wood). Drill Deck. Put case in bottle. Put cards in case. Put rope through deck. Tie knot. Put nut, bolt, and lock parts into bottle. Hold bolt with a magnet - screw nut on with dental floss. Assemble and lock padlock. Finally sign the pack of cards."

Although this one seems rather labour intensive to assemble — much like a ship-in-a-bottle — some of his others are downright impossible.

Harry Eng Bottle

 

For example, this block of wood and the metal lock are both too large to fit through the neck of the bottle. And although the lock could conceivably be taken apart and re-assembled inside, the block of wood most certainly cannot.

As a kicker, the wood block is engraved with his "Think" logo.

I give up... do you have any idea how this could be achieved?

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GEB Chapter 1: The MU-Puzzle

 2 Comments- Add comment Written on 21-May-2008 by asqui

The real first chapter of Gödel, Escher, Bach is an introduction to Formal Systems[?] (what I previously thought was Chapter 1 was actually an introduction to the entire book).

Here's a puzzle from that chapter:

Given the starting string "MI", what is the quickest way to turn it into the string "MU" using only the following rules: (where x and y represent any string)

  1. Any string ending in I may have U appended to the end:
    xI → xIU (e.g. MI → MIU)
  2. Any string beginning with M can have the following letters doubled:
    Mx → Mxx (e.g. MIU → MIUIU)
  3. Any occurrence of III may be replaced with U:
    xIIIyxUy (e.g. MIUIII → MIUU)
  4. Any occurrence of UU may be removed:
    xUUyxy (e.g. MIUU → MI)
Answers on the back of a postcard; Prizes for the correct solution.
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When is a Triangular Number a Square Number?

 0 Comments- Add comment Written on 14-May-2008 by asqui

Whilst watching a Swan Lake ballet at The Barbican Centre a couple of months ago (and confirming my suspicions that ballet is not very captivating to me) I was distracted by the geometrical formations that the ballerinas took on — in one sequence they were neatly arranged in a triangular layout, and then quickly moved into a square layout.

That was my cue to set about trying to deduce which triangular numbers also happen to be square numbers (and therefore work out how many ballerinas were on stage, without actually counting them) with the additional challenge of doing it all in my head, whilst continuing to watch the performance. (I figured that whipping out a notebook might be considered somewhat unsavoury, and more importantly, I had omitted to bring one with me!)

I managed to conclude, during the course of the performance, that the first integer greater than one to be a triangular number as well as a square number is 36. However, the number of ballerinas present in the curtain call were far fewer. I must have made a mistake in my observations somewhere — perhaps the square and/or triangle were incomplete? Perhaps there were 15 in the triangle and then an extra one sneaked in to make 16 for the square?

Later in the week I had a crack at coming up with a general solution.

Triangular and Square Numbers

As you can see, I didn't get very far (perhaps my mathematics is getting a little rusty through disuse) but a quick brute-force in Excel came up with the following results:

Numbers which are both square and triangular:

1
36
1225
41616
1413721
48024900
1631432881
55420693056

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Mr P and Mr Q Solution

 2 Comments- Add comment Written on 04-Apr-2008 by asqui

Here is my solution to the Mr P and Mr Q Logic Puzzle I posted recently:

Five Of DiamondsAssumptions:

  1. "number" refers to one of {2, 3, 4, 5, 6, 7, 8, J, Q, K, A} (ie. 'Picture cards' have a "number")
  2. "suit" refers to one of {H, S, C, D}
  3. Let U be the set of all (number,suit) card tuples.
    U = {
    (A,H), (Q,H), (4,H),
    (J,S),
    (8,C), (4,C), (2,C), (7,C), (3,C)
    (A,D), (5,D)
    }
  4. Mr P knows the number.
  5. Mr Q knows the suit.
Phase 1:
  • P cannot ascertain what card it is.
  • Therefore it must be a card which is not uniquely identifiable by its number (which P already knows).
Let A be the set of cards in U which share a number.
A = {
(A,H),(Q,H),(4,H)
(A,D),(5,D),
}
  • Q now knows that the soultion is in the set A.

Phase 2:
  • Q knew that P wouldn't be able to uniquely identify the card, based on knowing it's suit.
  • Therefore the suit of the card must be one for which all numbers of that suit are duplicated (if a card is not one whose number is duplicated then P would have known what card it was, given that he already knows the number)
  • So, the suit of the card must be a suit which appears in set A. Namely, hearts or diamonds.
Phase 3:
  • Given Q's statement, P makes the deduction in Phase 2.
  • P knows the number of the card, and knows that it is in set A, and from this he concludes the solution.
  • Therefore the number of the card must be one which uinquely identifies a card from set A, namely, the number must be one of {Q, 4, 5}.
Let C be the set of cards in A for which this holds true.
C = {
(Q,H),(4,H)
(5,D),
}

Phase 4:
  • Given P's statement, Q makes the deduction in Phase 3.
  • Q knows the suit of the card, and now knows that it is in set C, and this allows him to discover the solution.
  • So the solution is a card in set C which is uniquely identifiable by its number, namely (5,D).
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Mr P and Mr Q Play Cards

 2 Comments- Add comment Written on 02-Apr-2008 by asqui
Playing Cards

One day, Mr P and Mr Q were having a test on logic by Professor White.

On the table in front of them there were 16 cards and both of them knew what they were:

Hearts: A, Q, 4
Spades: J, 8, 4, 2, 7, 3
Clubs: K, Q, 5, 4, 6
Diamonds: A, 5

Then all the 16 cards were turned upside down and reshuffled.

Professor White picked a card randomly and told Mr P the number of the card had and told Mr Q the suit of the card.

Professor White then asked if they could find out what the card was.

Mr P said, "No I can't."
When Mr Q said, "I knew you couldn't."
Having heard this, Mr P said, "ah, now I know."
Then Mr Q said, "oh, so do I now."

Can you tell what card it was by logical deduction?

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Lateral Fishermen (solution)

 1 Comment- Add comment Written on 21-Mar-2008 by asqui

Here is the solution to the lateral-thinking puzzle I posted recently. If you'd like to have another attempt at solving it before you give up, this is your final chance. The question is how did the fishermen manage to decrease their prices but increase their profit at the same time?

Dr. Robert Jensen at Brown University conducted a study of Indian fishermen who started using mobile phones to find the best coastal marketplaces for their catch. While the fishermen saw profits increase 8 percent, consumer prices actually dropped by 4 percent...

...because less fish was being wasted.

If you give up, highlight the text in the red box to see the sentence ending, which I omitted previously.

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Lateral Fishermen

 3 Comments- Add comment Written on 20-Mar-2008 by asqui

Fish on iceHere's an inadvertent lateral-thinking puzzle I pulled out from the current issue of The Architecture Journal:

Dr. Robert Jensen at Brown University conducted a study of Indian fishermen who started using mobile phones to find the best coastal marketplaces for their catch. While the fishermen saw profits increase 8 percent, consumer prices actually dropped by 4 percent...

So they dropped their prices yet managed to increase their profits at the same time? How curious.

How did they manage that? Can you guess?

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