Showing posts with label logic puzzles. Show all posts

Saturday, May 23, 2015

Sailor's Delight

Games Reviewer 8:11 AM , , , , , , , , No comments :
10 pirates are ranked in order, first to last. After finding a treasure chest of 100 gold coins, they are discussing how to divide up the booty. They allow the lowest ranked sailor to divide up the coins and then vote on his idea. If the number of pirates who like the division is equal to or greater than the others who don't like it, then the boss will say, "Make it So." (The proposer of the idea also has a vote.)

Otherwise... well, being pirates their simple solution is to dump the unfortunate sailor into the deep blue sea and let the next pirate in line decide how to divide up the spoils.

pirates and coins game

How many pirates will be thrown into the sea?

 

Notes:
  • Pirates are smart, want money, and love life, especially their own!
  • This one is harder than average. If you are stuck, think of fewer pirates...
  • Why would #1 ever vote for any schemes?
  • Why would #2 ever vote for any schemes?
  • ... hmmmm

Sailor's Delight Puzzle Solution

We asked you readers to send us a solution to this puzzle, and we kept an open mind about it.
The first person with a simple, elegant, and, in our opinion, valid solution, is Saurabh Gupta, a faithful reader and avid puzzler. Here is Saurabh's solution.

How many pirates will be thrown into the sea? None. And the correct distribution is:
Pirate
with rank		 Number
of coins
1 0
2 1
3 0
4 1
5 0
6 1
7 0
8 1
9 0
10 96

Pirate 10 divided the coins. He will get the votes of pirates 2, 4, 6, 8, and himself. This is taking into consideration what each of the pirates will get if this plan is not passed.

Starting with a situation when there is only pirate 1. He keeps all the 100 coins for himself and live happily by passing the division with his only vote.

In case that there are two pirates, pirate 2 divides and he keeps 100 coins for himself while giving none to pirate 1. He still gets the division passed with his vote and live happily ever after.
In case there are three pirates, pirate 3 divides and gives pirate 1 a single coin and keeps the other 99 coins for himself. Pirate 1 would now vote in his favor because if he votes against, then pirate 2 would get a chance to divide and would keep all the loot for himself.

If four pirates are present, pirate 4 divides and now gives pirate 2 a single coin to gain his vote (who otherwise gets nothing if pirate 3 has a chance to divide the coins). In this case, pirates 1 and 3 get nothing.

Therefore, in a similar manner, the distribution when there is an extra pirate is achieved as follows:

With 5 pirates,
pirate 5 divides:
Pirate
with rank		 Number
of coins
1 1
2 0
3 1
4 0
5 98
With 6 pirates,
pirate 6 divides:
Pirate
with rank		 Number
of coins
1 0
2 1
3 0
4 1
5 0
6 98
With 7 pirates,
pirate 7 divides:
Pirate
with rank		 Number
of coins
1 1
2 0
3 1
4 0
5 1
6 0
7 97
With 8 pirates,
pirate 8 divides:
Pirate
with rank		 Number
of coins
1 0
2 1
3 0
4 1
5 0
6 1
7 0
8 97
With 9 pirates,
pirate 9 divides:
Pirate
with rank		 Number
of coins
1 1
2 0
3 1
4 0
5 1
6 0
7 1
8 0
9 96
With 10 pirates,
pirate 10 divides:
Pirate
with rank		 Number
of coins
1 0
2 1
3 0
4 1
5 0
6 1
7 0
8 1
9 0
10 96
This solution holds because all pirates are rational and think through the situation. They understand that they don't get anything if the next ranked pirate gets a chance to make the division.

There is, however, still some doubt about this solution. After all, it was said that pirates are smart. With this solution, based on a method of dividing the booty supposedly approved by all, it is shown that only one (the last one) is very smart, while 4 of them get only a coin, and 5 of them get nothing. This seems to be in conflict with the statement that pirates are smart. Even if the method was approved only by half of them (the ones who get at least a coin), 4 of these voters don't seem very smart if they get only one coin. Certainly, it couldn't have been a dictatorship decision taken by the leader, as he ends up empty-handed.

An alternative, to make the pirates look brighter, is that the dividing pirate actually shares the loot evenly between the pirates likely to give him a positive vote. In that case, the pirates who previously got only one coin, end up getting just as much as the lowest ranked, while the others get nothing again. This would entail the following solution:
Pirate
with rank		 Number
of coins
1 0
2 20
3 0
4 20
5 0
6 20
7 0
8 20
9 0
10 20

This alternative solution, however, defies a bit the mechanism of the division logic and process explained in the main solution. Perhaps, it would be fair to make a re-wording of the puzzle, stating that, although pirates are smart, the method of dividing the booty is forced on them by some external entity, and no pirate, not even the leader, can oppose to it. However, a rewording of the puzzle will NOT take place, as this is how it was presented on Brent's page, and we are going to preserve it.

Also, it might be possible that this is not the only acceptable solution, and another alternative solution might exist, that doesn't raise any doubts with the puzzle premises.

So, we are still asking you: send us your solution!

Monday, February 2, 2015

Bread And Water

Games Reviewer 7:05 AM , , , , , , , , No comments :
bread and water riddle
The unforgiving heat of the desert sun was unbearable. Nearing total exhaustion, Alek the Polish traveller stumbled ever onwards through the endless expanse of sand. His camel had fled, his water reserves were long exausted, and there was not even a lizard to catch for sustenance -- not that he would have had the energy to catch it in any case. All he had left with him were the clothes he wore, eight golden coins, and his need for food and water.

He collapsed, looked up and thought that the sun had had the better of even his eyes; he saw two bedouins walking towards him. Mirages, he thought. He shook his head and rubbed his eyes to clear it, but the bedouins were still there, getting closer. When they reached him, Alek weakly asked for water and some food, and promised he would repay them generously.

They introduced themselves as Azad and Mohammed. "Water," one of them said, "is free." As far as food was concerned, they would share with him, which consisted of bread only. Azad had three slices, and Mohammed had five. They put the slices together, split them in three equal parts, and each of them ate his portion quietly. When they finished their meal, Alek pulled out his eight golden coins, and set them before the bedouins, telling them to share them fairly. He thanked them for saving his life, promised to call them sometime, and with renewed energy continued his journey.
When the traveller was gone, the two bedouins looked at the eight golden coins for a little while, and then Mohammed moved his hand to grab five coins.

"Hold it there!" said Azad. "We will share them as good friends; four coins each!"

Mohammed was convinced he deserved five coins, but Azad would not agree, and the argument grew louder. Before the first punch was thrown, the Great Sage happened to be passing by on his camel. He enquired about the matter, which was quickly explained by the bedouins.

"Neither 5 - 3 nor 4 - 4 are fair," stated the Great Sage, before sharing his wisdom.

The bedouins got their fair share of the coins, and the Great Sage went on his way, satisfied that yet another problem had been resolved.

How much did each of the bedouins get?

Bread And Water Puzzle Solution

Each slice of bread was divided into 3 equal pieces, making a total of 24 pieces. These were then divided between the three men, 8 pieces each.

Mohammed had 5 slices, and so contributed 15 pieces. He ate 8 pieces himself, so 7 were eaten by Alek.

Azad had 3 slices, which contributed 9 pieces. Azad ate 8, leaving 1 piece for Alek.

Therefore, Mohammed gave 7 pieces of bread away, and Azad only gave 1 piece. So Mohammed deserves 7 gold coins and Azad only 1. If he were smarter or less greedy, Azad should have accepted Mohammed's initial offer of 5 - 3.

Tuesday, January 13, 2015

Visit At The Kremlin

Games Reviewer 6:50 AM , , , , , , , , No comments :
"This," explained Colonel Nevskij to the Druggar of Bongo Ghango - chief of a large country by River Ghango - "is the Buttons Room."

"I've seen a room like this in Washington," replied the big Chief, smiling with satisfaction, "there too, you couldn't see a single button. The only ones I can see here are, uh.. hehe, the ones on your uniform, Colonel!"

"Ahah, comrade Druggar likes joking. But the buttons are here," replied the Soviet, pointing at a large panel at the end of the room, "they are behind that panel."

"A very large panel," observed the Chief, "much larger than the one at the White House: I presume there are more buttons here."

"Of course, comrade Druggar: in America, only a bunch of opportunistic capitalists has a saying in the big decisions, while here, uh, here is different: the entire Soviet community, through its representatives, takes part in the decision process of the Union."

"Any citizen could then come here and press the buttons?"

"Err, no, not exactly. If I tried to do it, the room would fill up with narcotic gas, an alarm would set off, and... Well, no need to talk about that. For each button there is a slot, into which a magnetic card must be inserted, to activate the corresponding button. Therefore, no card, no button. To launch the missiles, every button must be activated and pressed, and only a handful of comrades holds the magnetic cards, which of course, each of them has a different code from the others. The personalities holding the cards are the Secretary of the Communist Party, the President of the Praesidium, the Chief of the KGB, and five comrades, Heroes of the Soviet Union. The distribution - and here is the originality of our system - is such that the Secretary of the Party holds the complete set of codes, and so he can launch the missiles by himself; if the Secretary is not available, the missiles can be launched by the President of the Praesidium together with the Chief of the KGB, or by anyone of these two, together with any two of the five Heroes of the Union. If - Marx forbid - the Secretary, the President, and the Chief have all been victimised by an imperialistic attack, our nuclear response can be initiated by any four of the five Heroes of the Soviet Union; any four of them would be sufficient to have the entire set of magnetic cards to activate the buttons."

"So, how many buttons are there?" asked the Chief.

What is the minimum number of buttons, and how are they distributed?

Visit At The Kremlin Puzzle Solution

There are 20 buttons. This is because there are 10 combinations of Heroes in pairs of 2, and this is multiplied by 2 because these combinations have to be mapped to two different persons (the President of the Praesidium or the Chief of KGB). The magnetic cards, as held by the eight persons, and marked with an X, are distributed as follows:
Person Buttons
Secretary of Party X X X X X X X X X X X X X X X X X X X X
President of Praesidium X X X X X X X X X X X X X X X - - - - -
Chief of KGB - - - - - X X X X X X X X X X X X X X X
1st Hero X X X X - - - - - - - X X X X X X X X -
2nd Hero X X X - X X X X - - - - - - X X X X - X
3rd Hero X X - X X - - X - X X - - X - X X - X X
4th Hero X - X X X - X - X - X - X - - X - X X X
5th Hero - X X X X X - - X X - X - - - - X X X X

Wednesday, October 15, 2014

Cheers To Statistics

Games Reviewer 5:15 AM , , , , , , , No comments :
Two and a half artists spend two and a half hours painting two and a half models on two and a half canvases.
How many artists are necessary to paint twenty-four models on twenty-four canvases in twenty hours?

Cheers To Statistics Puzzle Solution


Three artist would do the trick. This is because twenty-four artists would paint twenty-four models in two and a half hours. Since the available time increases eight-fold (2.5 * 8 = 20), it is possible to reduce the number of painters by the same number of times (24 / 8 = 3).

Sunday, October 5, 2014

Faulty Batches

Games Reviewer 5:09 AM , , , , , , , , No comments :
"This time," said the Treasury Minister, "I ditched those dodgy Europeans, and I have assigned the manufacture of our gold coins to five American companies. Look, they are all shining and beautiful, and they are all exactly the same!"

The secretary looked at the coins, weighed some of them, and cleared his throat. "Ahem, Sir. I would like to point out that here we have at least three different kinds of coin; they all look the same, but their weight is different. Would you please come close to the scale? This coin weights 10 grams, as it should, but this other one is 11 grams, while this one is only 9 grams. Obviously two of our manufacturing companies haven't done a good job."

Sad as he could have been, for having been tricked agin by other dodgy companies, the Minister managed to raise his head. "Well.. it's just a matter of finding the fauly ones using the trick that you've showed me, by using the scale only once..."

"Sir. Actually, this is a different problem altogether, we need to find two sources of errors, rather than just one. One batch is heavier, another is lighter. The method I used before will not be sufficient this time. But we can nevertheless find the two offending batches by using the scale once."

How did they manage to use the scale only once?



Notes:
  • You may assume that each batch is made of a large amount of coins (thousands, millions, up to you! :)
  • All coins of the same batch weight the same amount.
  • The storyline in this puzzle follows from the story in Faulty Batch. It is however NOT necessary to have previously read/solved that puzzle in order to solve this one, even though it may be preferable.

Faulty Batches Puzzle Solution

They had to weigh 1 coin from the 1st batch, 2 from the 2nd, 4 from the 3rd, 8 from the 4th, and 16 from the 5th one.

If all coins weighed 10 grams as they should, the scale would display 310 grams ((1 + 2 + 4 + 8 + 16) * 10). However, since one batch has 9 grams coins, and another 11 grams coins, then the total weight of this combination of coins will be:
Total Weight Number of
9g coins		 Number of
11g coins
311 1 2
313 1 4
317 1 8
325 1 16
312 2 4
316 2 8
324 2 16
314 4 8
322 4 16
318 8 16
309 2 1
307 4 1
303 8 1
295 16 1
308 4 2
304 8 2
296 16 2
306 8 4
298 16 4
302 16 8

After seeing the solution to this puzzle, it is clear that it would be a lot easier to simply use the scales up to 5 times rather than go through all this, but where is the fun in that?
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