## The Tower

I hope this **5th and last logic puzzle ****exceeds all your expectations**:

Two wise men were captured by a tyrant king. They were locked up in two separate cells in a tower in solitary confinement.

One of the cells pointed Eastwards and from there you could see half of the kingdom. The other pointed Westwards and you could see the other half. So, from their cells, the two wise men could see all the cities of the kingdom, but none of them was visible to both of them.

To prove their intelligence the tyrant told them the cities in his Kingdom were either 10 or 13. If any of them was able to guess correctly how many there were he would free them. A wrong answer would mean instant execution. Every evening a jailer would visit them, provide them with food and they could tell him the answer.

On the fifth evening both men were freed.

What logic reasoning did they follow to solve the problem? How many cities are there and how many did each man see?

Margot| Profile November 11th, 2009 - 12:28 amAn excellent puzzle! I have a few questions for clarification, if I may:

1) Are we to understand that the correct number of cities in the whole kingdom is EITHER 10 or 13? So the wise men have to determine WHICH is the correct choice?

2) I assume that each of the wise men can see at least one city?

3) Just to clarify, none of the cities is on the border between the two views and visible to both men? In other words, each city is wholly in view of EITHER one man or the other but NOT both?

4) Did the men have any other conversation with the Jailer that may be significant to the puzzle? Or did they say nothing until telling him the correct answer?

5) The men have no way to communicate with each other?

Thanks again!

TonyTKL| Profile November 11th, 2009 - 1:39 am10. If each could only see half, and none of them is visible to both, then the number has to be even. Otherwise there would be a “half-city” visible to both of them.

Migrated| Profile November 11th, 2009 - 5:43 amThis probably has nothing to do with the solution of the problem, but could they see the sunrise and sunset?

ahsergio| Profile November 11th, 2009 - 6:43 amI dont believe that any of my answers are correct (specially because i found 2) but i have to try anyway:

1-If the expression “half of the kingdom” is literal, than each man can see only 5 cities, since no city is commom to their sight.

2- If one of them can see 12,11 or 10 cities, he can guarantee that there are 13 cities, and can tell his friend after he is released.

I’ll try to figure it out, but as mentioned, idk if any of the answers are correct, specially because they are too obvious to be figured only in the 5th evening

bilbao| Profile November 11th, 2009 - 7:49 amI will try to clarify further the puzzle (in response to some of your questions):

1) The correct number of cities in the whole kingdom is either 10 or 13 (one or the other) so the wise men have to determine WHICH is the correct choice between those two.

2) Each of the wise men doesn’t necessarily see at least one city. (0,10) and (0,13) should be contemplated.

3) None of the cities is on the border between the two views and visible to both men. In other words, each city is wholly in view of EITHER one man or the other but NOT both.

4) No other conversations are mantained with the jailer until telling him the correct answer. No conversations take place between the wise men since they are in solitary confinement. Giving the correct answer to the jailer by ANY of the wise men means freedom to BOTH of the wise men.

5) The men have no way to communicate with each other.

6) It has nothing to do with the solution of the problem, but one of the wise men can see the sunrise (eastwards) and the other can see the sunset (westwards)

7) The expression ‘half of the kingdom’ is literal, and refers to the area/expanse of land. It doesn´t refer to ‘half of the cities’.

aaronlau| Profile November 11th, 2009 - 10:40 amMy first suspicion was the “fifth evening”. haven’t thought the logic thru but it should be something like that:

1st evening, if either saw >11 cities on their side means its 13. so both are looking 0 city.

3rd evening, if one saw 10 cities means it is 13. nothing happened, each looking 1 city.

5th evening, one of them was looking at 2 cities. So he immediately knew the other was looking at 8 (11 is eliminated on the first evening). And the answer is 10 cities.

Mashplum| PUZZLE MASTER | Profile November 11th, 2009 - 10:58 amIf one of the prisoners could see 13 cities, and the other zero (13,0) then the one who saw 13 would guess correctly and they would be freed the first evening. If they were seeing (10,0) the first man would wonder if the second man saw zero or three, and he would not guess. The second man would then know the answer was 10 by the fact that they were not freed, and he would guess correctly on the second night. If (10, 3) the first man would know by the 3rd evening that the answer was 13. If (7,3) the second man would know by the 4th night that it was 10. If (7,6) the first man would know it was 13 by the 5th evening, so that must be the answer.

Mashplum| PUZZLE MASTER | Profile November 11th, 2009 - 11:21 amFrom the diary of the first wise man…

Day 1 – I see seven cities. Does the other prisoner see six or three? If he sees six then he is wondering if I see seven or four. If he sees three then he is wondering if I see seven or 10. He won’t just guess, he will wait to see what I do.

Day 2 – He didn’t guess. And by now he knows I didn’t guess. Is he hoping I see 10? If I could see 10 then this would be simple.

Day 3 – Nothing happened again. Good. My fellow prisoner is clever. If he is looking at three cities, he is trying to find out if I see seven or 10. He hopes it is 10 because that would mean our freedom tonight. If I could see 10 cities I would guess 13 tonight. Since by now I would know that he sees three. I would know that he could not see zero because he would have guessed correctly by now. But I do not see 10, I see seven. Pity.

Day 4 – My counterpart was no doubt disappointed that I didn’t free us last evening. However, if he sees three cities, the knowledge that I do NOT see 10 will be all he needs to free us tonight.

Day 5 – Still here. I was wrong about my fellow captive. He does not see three. (Wishful thinking on my part, I suppose.) But finally our ordeal comes to a close. If he does not see three, then he must see six. With my seven, the answer has to be 13. I will free us both tonight.

Mashplum| PUZZLE MASTER | Profile November 11th, 2009 - 11:47 amFrom the diary of the second wise man…

Day 1 – Lock me up will he? And now he expects me to play games for my freedom! How do I even know there is another man in this tower? And who’s to say how “wise” he is – the king? That vacuous toffee-nosed malodorous pervert?! I refuse to cooperate. When that jailer comes for my guess I will not give him a number, but I will show him my longest digit!

Day 2 – head hurts! jailer hit me? bad man!

Day 3 – hed sill herts!!! hate bad jeller!!!!

Day 4 – mY 34r2 r 8l33d1n’! OMG WTF lolz

Day 5 – BARACK HUSSEIN OBAMA IS A RACIST KENYAN MUSLIM SOCIALIST COMMUNIST FACIST LIBERAL TERRORIST WHO WANT TO TAKE AWAY OUR FREEEEEEEEEEDOOOOOOM!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Day 6 – I’m out and feeling much better. The other prisoner guessed correctly. The fool. Sure it was 50:50, but those are miserable odds when my life hangs in the balance. He is probably to stupid to know that.

suineg| PUZZLE MASTER | Profile November 11th, 2009 - 12:29 pmFirst of all Congratulations Bilbao for such a nice puzzle, this puzzle purely represent what the title of your series is all about “LOGIC”, this one is the puzzle that I enjoy the most solving it, it was very cool and the right one to finish this wonderful series so here it goes my solution:

Reasoning:

The possible combinations are(Only the reference to 1 person the others are for symmetry):

13: (0,13)-(1,12)-(2,11)-(3,10)-(4,9)-(5,8)-(6,7)

10: (0,10)-(1,9)-(2,8)-(3,7)-(4,6)-(5,5)

both mens have this info so:

DAY 1:

Any intelligent men that sees 13, 12 ,11 already would answer, if not this combinations are eliminated: (0,13)-(1,12)-(2,11)

DAY 2:

A man that sees 0, 1 or 2 already know the combinations that were eliminated day 1 so he would have answered if he had seen that number of cities therefore this combinations are eliminated: (0,10)-(1,9)-(2,8)

DAY 3:

Now a man that sees 8 , 9 or 10 cities would know the number of cities, because he would know the combinations already eliminated so you can eliminated the following combinations: (4,9)-(5,8)-(3,10)

DAY 4

OK following the same logic the man that sees 3, 4 or 5 cities would have known the total number so you can now eliminate the following combinations4,6)-(5,5)-(3,7)

DAY 5

Now the only combination remaining is 6-7 so any of the man could shout 13!!!!

very cool.

Oneiric| Profile November 11th, 2009 - 3:40 pmWell, the only way I can conceive it is if one of the wise men could see 13 cities, meaning he is assured of seeing all the cities (as opposed to 10, where the other wise man could see 3 cities or none).

It is the only way they can be a 100% certain about how many cities there are in the empire.

But geez, it sounds almost too easy to be that simple.

Bobo The Bear| PUZZLE MASTER | Profile November 11th, 2009 - 3:51 pmOn the first evening, both prisoners deduce “My partner cannot have seen 11 or more cities, or he certainly would have declared the answer 13. The number of cites he sees is between 0 and 10.”

On the second evening, both prisoners deduce “Knowing that I saw 10 or fewer cities, my partner cannot have seen 2 or fewer cities, or he certainly would have eliminated 13 and declared the answer 10. The number of cites he sees is between 3 and 10.”

On the third evening, both prisoners deduce “Knowing that I saw 3 or more cities, my partner cannot have seen 8 or more cities, or he certainly would have eliminated 10 and declared the answer 13. The number of cites he sees is between 3 and 7.”

On the fourth evening, both prisoners deduce “Knowing that I saw 7 or fewer cities, my partner cannot have seen 5 or fewer cities, or he certainly would have eliminated 13 and declared the answer 10. The number of cites he sees is between 6 and 7.”

On the fifth evening, both prisoners declare that there are 13 cities, and that one sees 6 of them and the other sees 7 of them.

hex| PUZZLE MASTER | Profile November 11th, 2009 - 6:05 pmThis is an easy one. We have to assume the 2 guys are very smart (wise?).

The possibilities are the following, noting that the 1st column is what A sees, and the 2nd and 3rd columns are what B sees if there are 13 or 10 cities respectively:

0 13 10

1 12 9

2 11 8

3 10 7

4 9 6

5 8 5

6 7 4

7 6 3

8 5 2

9 4 1

10 3 0

11 2 x

12 1 x

13 0 x

At the end of the 1st day, we can rule out that neither A nor B sees 11,12 or 13 (they would have guessed that there were 13 cities)

Situation becomes:

0 x 10

1 x 9

2 x 8

3 10 7

4 9 6

5 8 5

6 7 4

7 6 3

8 5 2

9 4 1

10 3 0

x x x

x x x

x x x

At the end of the 2nd day, we can rule out that neither A nor B sees 0, 1, or 2 as neither has guessed

Situation becomes:

x x x

x x x

x x x

3 10 7

4 9 6

5 8 5

6 7 4

7 6 3

8 5 x

9 4 x

10 3 x

x x x

x x x

x x x

At the end of the 3rd day, we can rule out that neither A nor B sees 8, 9, or 10 as neither has guessed

Situation becomes:

x x x

x x x

x x x

3 x 7

4 x 6

5 x 5

6 7 4

7 6 3

x x x

x x x

x x x

x x x

x x x

x x x

At the end of the 4th day, we can rule out that neither A nor B sees 3, 4, or 5 as neither has guessed

Situation becomes:

x x x

x x x

x x x

x x x

x x x

x x x

6 7 x

7 6 x

x x x

x x x

x x x

x x x

x x x

x x x

At the end of the 5th day, both are bound to guess that there are 13 cities, as each sees either 6 or 7.

Jimmy Anders| PUZZLE MASTER | Profile November 11th, 2009 - 11:25 pmThere are 13 cities total, 6 cities on one side and 7 cities on the other.

Consider all combinations of the form (a, b), where one side sees “a” cities, and the other “b.” Naturally, the following combinations would be solved on the first day, since only one possibility corresponds to each “b” value:

(0, 13), (1, 12), (2, 11)

[For brevity, I list the possibilities only with the smallest first even though the following argument does not use that convention]

For each “a” value, there are 2 possible “b” values and vice-versa (since a+b is either 10 or 13). Therefore, each combination has an “a-partner” and a “b-partner”, meaning the combo that has the same “a” or “b”, except for these first-day combos, which have no b-partner. Therefore on day two, day one’s a-partners would be solved; and then on day three, day two’s b-partners and so on. Therefore the days on which each combo could be solved are:

Day two:

(0, 10), (1, 9), (2, 8)

Day three:

(3, 10), (4, 9), (5, 8)

Day four:

(3, 7), (4, 6), (5, 5)

Day five:

(6, 7)

Since it took five days, (6, 7) must have been the way the cities split.

Incidentally, the two men could not have been imprisoned longer than 5 days (assuming that they both use this reasoning), and the 5th day is the only one where we could answer for sure which split they saw.

Jimmy Anders| PUZZLE MASTER | Profile November 11th, 2009 - 11:26 pmUgh, I accidentally typed the code for some kind of face up there, so those smilies are really 8′s…

TexHex| Profile November 12th, 2009 - 2:52 amOne of the wise men(a) will see 10 cities and the other(b) will see 0

The wise man(a) who can’t see any cities will tell the jailer there has to be 10 cities because the other wise man(b) would have given the answer already if there were 13 cities.

Wise man(a) knows he can not see any cities viewing half the kingdom, so the other wise man(b) will be able to see all of the cities.

So he(a) waits a while to see if the other wise man(b) gives an answer. If that wise man(b) could see 13 cities he would have given the answer as it would be a give away. He(b) would know it can’t be 10 because he can see 13 cities.

RogHyde| Profile November 12th, 2009 - 5:41 amThe answer is 13 cities, divided 6, 7

Excellent puzzle – I thought it looked impossible but then realised the solution but found it difficult to explain. The secret is that there are only two possible combinations containing each number of cities and when neither prisoner declares the total, this eliminates three possible combinations. I hope this is clear:

There are 13 possible combinations:

13-0 12-1 11-2 10-3 9-4 8-5 7-6

10-0 9-1 8-2 7-3 6-4 5-5

On day 1: The prisdoner seeing 13, 12 or 11 cities (13-0, 12-1 and 11-2) would declare 13 and be freed.

On day 2 the remaining prisoner seeing 0, 1 and 2 (10-0, 9-1 and 8-2) would declare 10 and be freed.

On day 3 the remaining prisoner seeing 10, 9 and 8 (10-3, 9-4 and 8-5) would declare 13 and be freed.

On day 4 the remaining prisoner seeing 3, 4 and 5 (7-3, 6-4 and 5-5) would declare 10 and be freed.

This only leaves 6-7 on the fifth day so both prisoners declare 13 and are freed.

Roger

KPXXX| Profile November 12th, 2009 - 9:26 amHi! One more question to clarify, are each of the cities more or less the same size?

Wavey Davey| Profile November 12th, 2009 - 7:41 pmThere are 13 cities, one man can see 6 the other can see 7.

Consider the options (M1 = Man1, M2 – Man2)

M1 0 M2 13 – M2 states 13 straight away and both are released

M1 1 M2 12 – M2 states 13 straight away and both are released

M1 2 M2 11 – M2 states 13 straight away and both are released

M1 11 M2 2 – M1 states 13 straight away and both are released

M1 12 M2 1 – M1 states 13 straight away and both are released

M1 13 M2 0 – M1 states 13 straight away and both are released

M1 0 M2 10 – On day 2, M1 wonders why M2 hasn’t said 13. So he says 10 and both are released

M1 1 M2 9 – On day 2, M1 wonders why M2 hasn’t said 13. So he says 10 and both are released

M1 2 M2 8 – On day 2, M1 wonders why M2 hasn’t said 13. So he says 10 and both are released

M1 8 M2 2 – On day 2, M2 wonders why M1 hasn’t said 13. So he says 10 and both are released

M1 9 M2 1 – On day 2, M2 wonders why M1 hasn’t said 13. So he says 10 and both are released

M1 10 M2 0 – On day 2, M2 wonders why M1 hasn’t said 13. So he says 10 and both are released

M1 3 M2 10 – On day 3, M1 wonders why M2 hasn’t said 10. So he says 13 and both are released

M1 4 M2 9 – On day 3, M1 wonders why M2 hasn’t said 10. So he says 13 and both are released

M1 5 M2 8 – On day 3, M1 wonders why M2 hasn’t said 10. So he says 13 and both are released

M1 8 M2 5 – On day 3, M2 wonders why M1 hasn’t said 10. So he says 13 and both are released

M1 9 M2 4 – On day 3, M2 wonders why M1 hasn’t said 10. So he says 13 and both are released

M1 10 M2 3 – On day 3, M2 wonders why M1 hasn’t said 10. So he says 13 and both are released

M1 3 M2 7 – On day 4, M1 wonders why M2 hasn’t said 13. So he says 10 and both are released

M1 4 M2 6 – On day 4, M1 wonders why M2 hasn’t said 13. So he says 10 and both are released

M1 5 M2 5 – On day 4, both wonder why the other hasn’t said 13. So they both say 10 and both are released

M1 6 M2 4 – On day 4, M2 wonders why M1 hasn’t said 13. So he says 10 and both are released

M1 7 M2 3 – On day 4, M2 wonders why M1 hasn’t said 13. So he says 10 and both are released

M1 6 M2 7 – On day 5, both men know there are 13, so they are released

M1 7 M2 6 – On day 5, both men know there are 13, so they are released

There’s no way of knowing which of the men could see 6 and which could see seven ;-)

engjs1960| Profile November 12th, 2009 - 7:44 pmInitially, both men know that the other can see between 0 and 13 cities. After the first night, both men know that the other must see at most 10 cities, otherwise there would have to be 13 cities. After the second night, both men know that each must see at least 3 cities, otherwise there would have to be 10 cities. After the third night, both men know that each must see at most 7 cities, otherwise there would have to be 13 cities. After the fourth night, both men know that each must see at least 6 cities, otherwise there would have to be 10 cities. Two sixes are 12, so there must be 13 cities.

bizarette18| PUZZLE MASTER | Profile November 12th, 2009 - 9:24 pm7/6

jmpeax| Profile November 12th, 2009 - 11:36 pmThe two wise men are M1 and M2. We look from the perspective of M1. These are all of the possible combinations of cities we can see and deduce for M2:

day released) M1 cities seen / M2 cities seen

2) 0 / 10,13

2) 1 / 9,12

2) 2 / 8,11

4) 3 / 7,10

4) 4 / 6,9

4) 5 / 5,8

5) 6 / 4,7

5) 7 / 3,6

3) 8 / 2,5

3) 9 / 1,4

3) 10 / 0,3

1) 11 / 2

1) 12 / 1

1) 13 / 0

Day 1: If we see 11,12,13 then we tell the guard on day 1 that there are 13 cities.

Day 2: If we see 0,1,2 then M2 must have 10,9,8 and we can say there are 10 cities and leave, since M1 would’ve released us on day 1 if he had instead seen 13,12,11 cities.

Day 3: If we see 8,9,10 then M2 must have 5,4,3 and we can say there are 13 cities, since M2 would’ve released us on day 2 if he had instead seen 2,1,0 cities.

Day 4: If we see 3,4,5 then M2 must have 7,6,5 and we can say there are 10 cities, since M2 would’ve released us on day 3 if he had instead 10,9,8.

Day 5: We must see 6 or 7 cities, either way M2 sees 7 or 6 cities and there must be 13 cities.

Solution: Since they were released on the fifth day, then there are 13 cities, one sees 6 the other sees 7.

jmpeax| Profile November 12th, 2009 - 11:41 pmMade one small type in previous post. Day 2, I said that “since M1 would’ve”, it should be “since M2 would’ve”. Sorry about that.

bilbao| Profile November 13th, 2009 - 1:34 amHi kpxxx, it is not relevant to the puzzle. Consider them the size you want. When I tried this puzzle I had in mind a sort of fairytale landscape (very green, sunny,… and scattered villages (instead of cities).

FallenPhoenix| Profile November 13th, 2009 - 1:36 pmFirst time i’ve attempted one of these, but hopefully I got it right

The answer is 13 cities, 7 and 6 respectively. The reasoning is at the beginning we have 13 different possibilities (taking into account that 10 and 3 cities is the same as 3 and 10 cities). Because it took 5 days for the men to escape, we know that initially that neither man saw 13, 12 or 11 cities as theyd instantly know there were 13 cities. Because there was no escape on day two we can also discount that neither man saw 0,1 or 2 cities because using the information from the day before if either had these cities, theyd know there were 10 cities. On day 3 we know neither man saw 8, 9 or 10 cities due to neither escaping again. On day 4 we know neither man could see 5,4 or 3

On day 1 if either escaped we would instantly know that one man saw 13, 12 or 11 as these are the only combinations its possible to know there are 13 cities. Because neither escaped, we can discount these number. Thinking logically if one man could see 13, 12 or 11 and had escaped, the other must’ve seen either 0, 1 or 2 cities. If they hadnt escaped on day one the other man who could see 0, 1 or 2 cities would know that on day 2 there were 10 cities altogether, allowing him to escape. Alternating this theory means that on day three the other man would be able to escape if he could see 8, 9 or 10 cities (knowing there were 13 cities) and on day 4 the opposite man again would be able to escape if they could see 5, 4 or 3 cities. The important part of this is it took 5 days. After 5 days the only option either man would have left would be 6 or 7 cities which adds to 13 cities altogether.

Very long winded I know, but I did it on a piece of paper by knocking out each oppisite sets of cities for each number on alternate days (1st day 13 cities, 2nd day 10 cites, 3rd day 13 cities, 4th day 10 cites, 5 day 13 cities) till I was left with 6 and 7.

venkatg| Profile November 13th, 2009 - 1:55 pmI found that one of them sees 6 cities and the other 7 cities.

jamesbuc| Profile November 13th, 2009 - 5:09 pmHi Everyone, this is my first post. Lets hope it a good one…

I make it that there 13 cities. One prisoner sees 7, the other 6:

Let prisoner A be the prisoner seeing the (hypothetical) greater number of cities, let prisoner B be the prisoner seeing the fewer number.

On Day 1 prisoner A can go free if he can see 13, 12 or 11 cities (he would know there must be 13 cities in total).

On Day 2 prisoner B can go free if he can see 0, 1 or 2 cities (he would know there must be 10 cities in total or else he would have been freed by virtue of prisoner A giving the correct answer on day 1).

The following combinations remain:

Prisoner A sees 10 cities, Prisoner B sees 3 cities

A sees 9, B sees 4

8 5

7 6

7 3

6 4

5 5

Therefore on Day 3 any prisoner seeing 8, 9 or 10 cities can work out the answer since there is only one combination where each value occurs. Since the prisoners aren’t freed on day 3 the set reduces to the following for Day 4:

7 6

7 3

6 4

5 5

By the same logic on Day 4 any prisoner seeing 3, 4 or 5 cities can win both their freedom. Only one set remains for Day 5:

7 6

Therefore there are 13 villages, one prisoner seeing 7 the other 6.

BlindCupid| Profile November 13th, 2009 - 6:01 pmMy best guess would be if one man saw more than ten cities, theremore the answer must be 13.

Though I don’t think the tyrant king would give a question to something so simple..

Does the fact that they escaped on the 5th day have anything to do with the answer?

MarkS| Profile November 14th, 2009 - 10:15 amI’m not particularly good at these things, but here goes-

No one sees 11, 12 or 13 cities since then the solution (there are 13 cities) would be given the first night.

No one sees 0 cities since this means the other sees 13 or 10

cities but since there was no solution the first night the one

seeing 0 cities now knows that there must be 10 cities and gives this answer the second night. Likewise, no one sees

1 or 2 cities, else the solution would have been given the second night.

No one sees 10 cities since that means the other sees 0 or 3 cities but we know that 0 cities is not possible, the answer of 10+3 cities would be given the third night. Likewise no one sees 9 or 8 since 1 and 2 have been ruled out so the answer 9+4 or 8+5 would be given the third night.

No one sees 3 cities since the 10+3 answer was not

given the third night and so 3+7 would have been given on the

fourth night. The same reasoning shows that no one sees 4 or 5 cities.

The only possiblity left for the fifth night is 7+6.

kitchen| Profile November 14th, 2009 - 2:15 pmThere are only 13 possible combinations, (0,13)-(6,7) and (0,10)-(5,5).

If any man sees a # of cities that only occurs in one combination, he knows how many cities there are and both are freed. As each night goes by that they’re _not_ freed, both know that the other doesn’t see any of the distinct numbers, eliminating possibilities… it’s a lot like the island and eye color puzzle.

Any combination with 11-13 is eliminated after the first night, 0-2 2nd night, 8-10 3rd night, 3-5 4th night, and on the fifth night, (6,7) is the only remaining possibility.

Karbon| Profile November 14th, 2009 - 2:50 pmDo any of the wise men have a way of telling when the other one was released?

venkatg| Profile November 14th, 2009 - 3:05 pmBtw, the reasoning is based on induction. In fact this combination of cities is the only one that takes maximum 5 days . Other combination such as 10,3 would yield 3rd day.

bilbao| Profile November 16th, 2009 - 2:07 amHi BlindCupid, the answer to your question is yes, the fact that they escaped on the 5th day and not earlier is important.

Hi Karbon, when any of the wise men tells the jailer the correct solution, both wise men are released simultaneously, no matter who gave the correct solution. So they both know when they are released.

bilbao| Profile November 16th, 2009 - 3:39 amMany of you got the correct answer: 7/6 for a total of 13. Some good explanations above.

It is a pleasure to welcome some new visitors. It encourages me to post ever better puzzles…

bizarette18| PUZZLE MASTER | Profile November 17th, 2009 - 10:45 amHey, Mashplum, I like your scenario. I had Saruman doing the numerics in mine, and Gandalf patiently waiting to be rescued by eagles, but I like yours better.

Mashplum| PUZZLE MASTER | Profile November 17th, 2009 - 8:46 pmThanks bizarette. I had fun writing it.

Tsopi| Profile January 19th, 2010 - 2:30 pmI suppose it is as easy as that.As you said each man could see the half of the kingdom witch mean that there were 10 cities ’cause 13 cannot be devided by 2!Lol is that it?

bilbao| Profile January 20th, 2010 - 1:35 amHi Tsopi, not that easy :-)

The expression ‘half of the kingdom’ is literal, and refers to the area/expanse of land. It doesn´t refer to ‘half of the cities’.

Besides, none of the cities is on the border between the two views and visible to both men.

hex| PUZZLE MASTER | Profile January 20th, 2010 - 4:05 pmBut what if the area of the kingdom is not an even number? :mrgreen:

bambi_76258| Profile February 26th, 2010 - 10:19 pmI am not as smart as some of the people who answered this but I am going to give it a try. I think they saw 5 cities each and the way they got out was the dinners brought to them. Each day a dinner came from a different city, from the others vantage point. When the meals were brought in they were from a city that could not be seen by that prisioner. Knowing that there was not 12 cities and each could see half.

He knew the number was 10. Each knew none of the meals were from cities they could see. Bambi

bambi_76258| Profile February 26th, 2010 - 10:41 pmI think no matter how you try to rationalize the number of cities, it comes down to the meals brought to them each night. Well, in my opinion anyway. Bambi

bryan_always| Profile July 1st, 2010 - 12:22 amIndeed one of the best logic puzzles I have come across on the net. I wonder if it will be remembered as one of the ‘classic’ puzzles, years later. Just thought it worth mentioning, that though its best to look at 13-0 and the 10-0 as possible options, it might seem a bit absurd that a whole East ‘half’ of the kingdom or a whole West ‘half’ should have ZERO cities… and yet exist.

Settlers of Catan « thasentinel| Guest April 22nd, 2011 - 1:10 am[...] Yay! And then delved into philosophy, eventually leading me to one of my favorite riddles, which you can see here. While my friends pondered that one, my philosophy buddy pointed me to the blue-eyed islander [...]

CurlyDreadz| Profile March 31st, 2012 - 3:52 pmI don’t have a clue

y3llowfl4sh| Profile June 2nd, 2013 - 7:09 amOnly one jailer visit them to provide them with food. the first prisoner didn’t take his food at the fifth day which is the number of cities he sees from his cell. the second will notice that and sum both the cities he sees plus the 5 which corresponds to the number of evening it took for the other to hunger strike. since each look at half of the cities so the total number would be 10