The Silver Bar Puzzle - 01/02/2013

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The Silver Bar Puzzle - 01/02/2013

Postby showmyiq » Thu Jan 31, 2013 11:48 am

Even silver traders somehow distracted by their plans could be trapped in no-money situations. Like our main character, who was not able to pay his rent to the housekeeper for March,2013. As we know the month has exactly 31 days, so he proposed to his housekeeper this strange advanced-payment scheme.

I should add the name of our main character – Silvestar, which compared to the housekeeper’s name Scrooge, is instantly making him our good character. On the other hand Uncle Scrooge the Housekeeper would put nothing in front him as important as the money themselves. Anyway, that’s another story and unfortunately very common one.

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Oh yes, now about the advanced-payment scheme they did. Fortunately Silvestar possessed a silver bar exactly 31 inches long. He proposed to give to Uncle Scrooge exactly 1 inch of his silver bar everyday through the March. Because the silver bar is exactly 31 inches long and March has exactly 31 days, the bar will be sufficient to fulfill this arrangement. At the end of the March Silvestar will gather the money, pay to Uncle Scrooge the rent and take back his silver bar (unfortunately at pieces; silver is not counted, but weighted anyway). If of course the noble silver trader fails to complete his part of the deal, the housekeeper will keep the silver in his possession (maybe greedily laughing on a full-moon night, trying to reassemble the pieces).

What Silvestar thought is - it will be a really hardworking dividing the silver bar to 31 pieces. He pour a hot tea in his silver cup (fortunately we are not involved in dividing the cup between them) and give it some thought. Nothing. Then try harder. Something! What if he gave to Uncle Scrooge 1 inch the first day of March, then again 1 inch of the silver bar the second day of March and instead giving him again 1 inch on the third day of March – he could just take the 2 pieces of 1 inch back from Scrooge and give the old man a whole 3 inches piece. He just invented a way to fulfill his part of the deal with lesser cutting and lesser dividing of the silver bar. Which on the other hand will not only save his efforts in dividing the silver bar on 31 pieces, but defacing his shiny treasure so brutally will be avoided too.

After countless sleepless nights Silvestar invented the best approach to this puzzle. He was astonished by the elegant solution he just discovered. Your task is to find this solution – what is the optimal/minimum dividing we need in order to complete the deal? How many pieces we need? How many inches long will be each piece? Take your shovel and start digging.

And maybe you are eager to read what happen between Silvestar and Scrooge?
Well,
Silvestar tried to explain.
Scrooge tried to understand.
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Re: The Silver Bar Puzzle - 01/02/2013

Postby bjars » Thu Jan 31, 2013 5:56 pm

1, 2, 4, 8, and 16 (5 pieces)

Day 1: give 1 inch bar
Day 2: take back 1 inch and give 2 inch
Day 3: give 1 inch and let him keep 2 inch
Day 4: take back 1 inch and 2 inch, and give 4 inch
Day 5: give 1 inch and let him keep 4 inch
Day 6: take back 1 inch, give 2 inch, and let him keep 4 inch
Day 7: give 1 inch and let him keep 2 inch and 4 inch
Day 8: take back 1 inch, 2 inch, and 4 inch and give 8 inch
Day 9: give 1 inch and let him keep 8 inch
Day 10: take back 1 inch, give 2 inch, and let him keep 8 inch

Day 11 - 31: repeat pattern until Day 31 when you give him the whole bar, then pay him and take it back in 5 pieces
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Re: The Silver Bar Puzzle - 01/02/2013

Postby showmyiq » Thu Jan 31, 2013 6:06 pm

Correct Answer!

This is the optimal solution available! Can we prove that such construction is always optimal solution? Imagine in Puzzle – Land there are months with X days each, for some very large integer X, and in the same time the silver bar is X inches large? If X is some integer equal to 2^Y – 1, for some random integer Y, that is ok. But what if X is for example 100? Or 1000?

Imagine we have X days in one month and a silver bar with length of X inches. Let’s say Y is the largest possible integer which fulfilled the statement that 2^Y – 1 <= X.

On the other hand we can be sure that X=<2^(Y+1) -1 .

Can we state that, if X is equal to the left point of this interval (2^Y -1) then the optimal pieces will be exactly Y, and if X is situated on any other point of this interval the pieces will be exactly Y+1?


It’s out of the topic, because you already solved the puzzle, just I am curious and wanted to share that with you.

Puzzle is closed anyway! The optimal count of the pieces is exactly 5, as you described it.
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Re: The Silver Bar Puzzle - 01/02/2013

Postby showmyiq » Fri Feb 01, 2013 9:35 am

The Puzzle is Archived here: The Silver Bar Puzzle
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Re: The Silver Bar Puzzle - 01/02/2013

Postby gerince » Fri Feb 01, 2013 12:19 pm

It was a really interesting one with a beautiful solution - 1,2,4,8,16. I've solved it a few minutes ago because I didn't have time earlier, but I see that bjars have already written the solution. I'll go to check the new one! :) Have a nice day!
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