# Logic Puzzles

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## Re: Logic Puzzles

I have officially retired from logic puzzles after my attempts to explain the eye colour problem seemingly confused everyone and gave me high blood pressure.

However...

Since "Alex" can choose which card of the five is the hidden card he can express the suit by making the top card he hands me the same suit as the one I need to guess (Four suits, five cards, there will always be a duplicate).

That leaves three cards with which to express the value. The cards can be ranked from low to high by their value and a pre-agreed hierarchy of the suits). This means there are six combinations available. So the trick becomes how do we express a number between 1 and 13 with only 6 different values.

If we think of the range of numbers as a continuous cycle: (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A, 2, 3, 4...), since there are 13 distinct values, every value is within 6 of another by either counting forwards or backwards. For example, counting from 2 forwards to 10 is 8, counting from 2 backwards to 10 is 5. So now the only problem is how do we express whether we should count forwards or backwards. The simple answer is, we don't.

Remember, Alex can choose which card to hide and which to use to represent the suit, he can therefore ensure we only have to count forwards. Using our example of 2 and 10, counting from 2 to 10 is 8, however, counting from 10 forwards to 2 is just 5 (J, Q, K, A, 2). So by returning the 10 on top of the cards, and using the remaining three cards to represent the number 5, he can tell us exactly which card he has hidden from us.

Right, back into retirement.

However...

Since "Alex" can choose which card of the five is the hidden card he can express the suit by making the top card he hands me the same suit as the one I need to guess (Four suits, five cards, there will always be a duplicate).

That leaves three cards with which to express the value. The cards can be ranked from low to high by their value and a pre-agreed hierarchy of the suits). This means there are six combinations available. So the trick becomes how do we express a number between 1 and 13 with only 6 different values.

If we think of the range of numbers as a continuous cycle: (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A, 2, 3, 4...), since there are 13 distinct values, every value is within 6 of another by either counting forwards or backwards. For example, counting from 2 forwards to 10 is 8, counting from 2 backwards to 10 is 5. So now the only problem is how do we express whether we should count forwards or backwards. The simple answer is, we don't.

Remember, Alex can choose which card to hide and which to use to represent the suit, he can therefore ensure we only have to count forwards. Using our example of 2 and 10, counting from 2 to 10 is 8, however, counting from 10 forwards to 2 is just 5 (J, Q, K, A, 2). So by returning the 10 on top of the cards, and using the remaining three cards to represent the number 5, he can tell us exactly which card he has hidden from us.

Right, back into retirement.

**PaulC**- Starting player token
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## Re: Logic Puzzles

fantastic! Very well done Paul!

**Jamie**- Count of Carcassonne
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## Re: Logic Puzzles

I didn't follow your calculation, Jamie (too early for maths?). Instinctively I feel like the maximum range would be more like 10?

- Spoiler:
- You have to have at least two cards on the same suit, so the first card might represent the suit. Now we're down to 12 cards. Now the maximum range should be 6. Pick the "lower" card ie. the card you can get to the other one by adding, in the case of a king and a two king is the lower one because the values cycle). Then you name the remaining three cards low, middle, high (suits have different values), and the tell Peter what to add based on a code:

Low Middle High = 1

Low High Middle = 2

...

High Middle Low = 6

Good?

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

You're still wrong about Blue Eyes, though.

Ha, you posted that ages ago, I was a whole page behind. I'll feel smart about it anyway. Though it was Jamie's idea about the range that helped me out.

Ha, you posted that ages ago, I was a whole page behind. I'll feel smart about it anyway. Though it was Jamie's idea about the range that helped me out.

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

You still got it, counts as a win in my book.Sam wrote:Ha, you posted that ages ago, I was a whole page behind. I'll feel smart about it anyway. Though it was Jamie's idea about the range that helped me out.

**PaulC**- Starting player token
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## Re: Logic Puzzles

Well done Sam and Paul!

I wasn't clicking at all with that, even though I had all the elements of the solution ready to put together. I'd thought of using the first card to indicate suit but abandoned it when I couldn't get the other cards to indicate the number. Then when Jamie suggested that the card order can indicate an offset rather than a particular card I should have got it but for some reason I just didn't make the connection.

I wasn't clicking at all with that, even though I had all the elements of the solution ready to put together. I'd thought of using the first card to indicate suit but abandoned it when I couldn't get the other cards to indicate the number. Then when Jamie suggested that the card order can indicate an offset rather than a particular card I should have got it but for some reason I just didn't make the connection.

**BeardyTom**- Dominant Species
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## Re: Logic Puzzles

No, I'm right... I've just failed to explain why!Sam wrote:You're still wrong about Blue Eyes, though.

**PaulC**- Starting player token
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## Re: Logic Puzzles

You guys did all the groundwork that let me put the last piece in place. Standing on the shoulders of giants and all that.BeardyTom wrote:Well done Sam and Paul!

I wasn't clicking at all with that, even though I had all the elements of the solution ready to put together. I'd thought of using the first card to indicate suit but abandoned it when I couldn't get the other cards to indicate the number. Then when Jamie suggested that the card order can indicate an offset rather than a particular card I should have got it but for some reason I just didn't make the connection.

**PaulC**- Starting player token
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## Re: Logic Puzzles

Team win I think, with Paul finally hitting the ball in the back of net!

I was thinking of the relationship between 24 (possible arrangements of 4 cards) and the 48 other cards. We only needed one more (binary) bit of information to double the 24 to 48, and Paul got it; it was the 2 possible ways you could arrange the 2 same suited cards (you give Alex one of 2 cards).

I was thinking of the relationship between 24 (possible arrangements of 4 cards) and the 48 other cards. We only needed one more (binary) bit of information to double the 24 to 48, and Paul got it; it was the 2 possible ways you could arrange the 2 same suited cards (you give Alex one of 2 cards).

**Jamie**- Count of Carcassonne
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## Re: Logic Puzzles

It continues:

First step: work out what is actually being asked!

The Trainee Technician

A 120 wire cable has been laid firmly underground between two telephone exchanges located 10km apart.

Unfortunately after the cable was laid it was discovered to be the wrong type, the problem is the individual wires are not labeled. There is no visual way of knowing which wire is which and thus connections at either end is not immediately possible.

You are a trainee technician and your boss has asked you to identify and label the wires at both ends without ripping it all up. You have no transport and only a battery and light bulb to test continuity. You do have tape and pen for labeling the wires.

What is the shortest distance in kilometers you will need to walk to correctly identify and label each wire?

First step: work out what is actually being asked!

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

Already peeked at the answer on this one, so I'm out. Have fun with it though! *mwhahaha*

**Jamie**- Count of Carcassonne
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## Re: Logic Puzzles

Oh, I see. If this poor trainee -- probably unpaid, or on an apprentice wage -- doesn't find a better solution, he will have to connect a battery to one wire, walk 10km to the other end of the cable, test up to 120 wires until he finds the wire connected to the battery, label this wire at both ends, then do it again 119 times!

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

Can we tie the wires into bundles and test them in groups?

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

Sam wrote:Can we tie the wires into bundles and test them in groups?

Yes Sam. You absolutely can twist the wires into bundles to test them in groups.

**PaulC**- Starting player token
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## Re: Logic Puzzles

Sam wrote:Can we tie the wires into bundles and test them in groups?

I'm not an electrical engineer but for the purposes of logic I'm going to say yes we can. Presumably you could test half the wires at once and narrow things down that way. E.g. at exchange A, label all the wires in binary with 8 digits, from 00000001 to 11111011, then bundle all the ones with a one in the first position together and connect them to the battery, pop off to exchange B and find those which light up the bulb, adding a 1 to their label and a 0 on the labels of the rest, then go back to A and repeat for those with a 1 in the 2nd position, etc. You'd have to do that 8 times and I think that would end up with you walking 150km - 8 times walking A to B and 7 times walking back. Presumably after that, you've finished your job and any walking you do after that is on your own time. Maybe you get a bus home.

The trouble with this problem compared to the last one is it's an optimisation problem, so unless I can prove that this is the best way of doing it I can't be sure I've actually solved it and I'm not sure how to go about doing that. Are we allowed to know what the distance we're aiming for is, Jamie, or would that be cheating?

**BeardyTom**- Dominant Species
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## Re: Logic Puzzles

I'm pretty sure there's a better method than that one though. I think the trainee is missing an opportunity to fiddle with the wires at exchange B in some way in addition to just testing them.

**BeardyTom**- Dominant Species
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## Re: Logic Puzzles

That is what I came up with. Well, logically very similar but in practice I was testing half the wires and labeling them afterwards, going back and testing half of each group made each time, ending up with a binary code only at the end -- what can I say, he wasn't a forward thinking trainee. At any rate I believe you can do it with only 7 bits, so 130 km, but it isn't the right answer. You can test your answer if you go to the website and type it in. All it will tell you is whether you were correct or not.

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

I've got it.

- Hint:
- 1+2+3+4+...+14+15 = 120

**PaulC**- Starting player token
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## Re: Logic Puzzles

- My solution:

Take 15 of the wires, label them A and twist their ends together. Then take 14 different wires, label them B and twist their ends together. Then 13 wires, then 12, 11, etc until we have one wire left which we will label O. So we now have 15 groups of wires ranging from a single wire to 15 wires.

We must now walk 10km to the other site where we will be able to test each wire to determine which group it belongs to by choosing a wire and counting the number of other wires that make a connection. If 14 other wires make a connection, it's an A (and should be labelled as such); 13, it's a B, etc, etc until there is one wire left that doesn't make a connection - this is the O wire.

Number the A wires 1 to 15 (A1, A2, A3, ..., A15), the B wires 1 to 14, etc.

Now, connect all 15 wires that have a 1 together (A1, B1, C1, etc), connect all 14 wires that have a 2, etc, etc.

Now walk 10KM back to the original site, separate the connected wires and test again in exactly as we did at the other end (choose a wire, if 14 other wires make a connection, it's a 1; 13, it's a 2, etc, etc)

We now have every wire individually labelled and have walked 20KM.

Strictly speaking, we should walk another 10KM back to disconnect the wires we connected, but the puzzle only asked us to identify the wires, not leave them in a functional state.

**PaulC**- Starting player token
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## Re: Logic Puzzles

LA LA LA not looking! I'm so close to this one it's distracting!

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

Sorry!Sam wrote:LA LA LA not looking! I'm so close to this one it's distracting!

**PaulC**- Starting player token
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## Re: Logic Puzzles

Nah, wasn't as close as I thought I was That one was nasty! Good job Paul

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

I don't think I'd ever have got that one. Well done Paul.

**BeardyTom**- Dominant Species
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## Re: Logic Puzzles

My brain needs to recover from that last puzzle. Here's a plain old "Difficult":

The Most Intelligent Prince

A king wants his daughter to marry the smartest of 3 extremely intelligent young princes, and so the king's wise men devised an intelligence test.

The princes are gathered into a room and seated, facing one another, and are shown 2 black hats and 3 white hats. They are blindfolded, and 1 hat is placed on each of their heads, with the remaining hats hidden in a different room.

The king tells them that the first prince to deduce the color of his hat without removing it or looking at it will marry his daughter. A wrong guess will mean death. The blindfolds are then removed.

You are one of the princes. You see 2 white hats on the other prince's heads. After some time you realize that the other prince's are unable to deduce the color of their hat, or are unwilling to guess. What color is your hat?

Note: You know that your competitors are very intelligent and want nothing more than to marry the princess. You also know that the king is a man of his word, and he has said that the test is a fair test of intelligence and bravery.

**SamVS**- Count of Carcassonne
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## Re: Logic Puzzles

- Guess:
- Okay, I think this is an easier version of Blue Eyes. We know that if any prince sees two black hats he will know he has a white hat. I, the first prince, know neither of these princes see two black hats, but if I assume I have a black hat a second prince could also assume he has a black hat and is waiting for the third prince to speak. When the third prince doesn't speak then the second prince will speak. Neither speak, so I must assume I don't have a black hat after all. Of course, I expect the other two princes will realise this at the same time as me and we will call out in unison -- or I'm just smarter than them.

**SamVS**- Count of Carcassonne
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