Difference between revisions of "Haybaler POW"

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<big>'''[[IAG 1H]] [[POW]] <nowiki>#</nowiki> 8: The Haybaler Problem'''</big>
 
<big>'''[[IAG 1H]] [[POW]] <nowiki>#</nowiki> 8: The Haybaler Problem'''</big>
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{{Word|Image:Haybaler_Problem.doc}}
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{{PDF|Image:Haybaler_Problem.pdf}}
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==Problem Statement==
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Not necessary to do.
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==Process==
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I started this POW with a lot of work which I didn<nowiki>’</nowiki>t need to do.  I wrote down all the combos in the form of 1<nowiki>+</nowiki>2, 2<nowiki>+</nowiki>3, etc.  I then guess some numbers to fill in.  I tried lost of combos, only switching 4 and 5.  I couldn<nowiki>’</nowiki>t get anything to work.  I then started over with changing numbers.  When ever I would change a number, lots of combos would change, and I was not having fun.  Looking back, I was getting close at the end, but I didn<nowiki>’</nowiki>t know that.  I spent an hour with this strategy.  I put the problem aside for a few days.
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:I then received the hint sheet on Friday.  After getting the sheet, I solved the basic solution to the POW in 20 minutes.  I used the hints page that you passed out to us.  The first 2 hints were my best friends.  I first found what the heaviest and lightest bails weighed.  Well actually I did that the other way around.  I found that the lightest must be 1 and 2.  (Note: 1 is always the lightest and 5 is always the heaviest, and the other numbers are in order)  This is because when you combine the 2 lightest, they add up to be the lightest.  Likewise the heaviest, must be bails 4 and 5.  I then made this chart:  I also made a chart of what I thought each bail weighed:
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{|border="2" cellspacing="0" cellpadding="4"
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|80
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|1<nowiki>+</nowiki>2
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|82
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|83
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|84
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|85
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|86
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|87
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|88
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|90
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|91
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|4<nowiki>+</nowiki>5
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{|border="2" cellspacing="0" cellpadding="4"
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|1
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|2
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Throughout the problem, I updated this chart when I found more info.
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:Now I wanted to find 1 and 2.  I know that they add up to 80.  I also know that they both can<nowiki>’</nowiki>t be 40 because you told us, and when I add 1<nowiki>+</nowiki>3 and 2<nowiki>+</nowiki>3, they would but equal the same thing.  I decided to make 1 equal 39, and 2 =41.  It could not be the other way around (1=41, 2=39) because then Bale 1 would not be lighter.  However, it really makes no difference, but they are easier in order.
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I then made these charts of possible combos (without repeating):  I also "added" the possible solutions to see how "big" they would be. (ex. 1<nowiki>+</nowiki>2=3)  Numbers which are big, in this simple form, would be big when I added the hay bails.  I guess this would kind of be like variables.  This is what I call a "weight."
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{|border="2" cellspacing="0" cellpadding="4"
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|1<nowiki>+</nowiki>2
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|2<nowiki>+</nowiki>3
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|3<nowiki>+</nowiki>4
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|4<nowiki>+</nowiki>5
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|1<nowiki>+</nowiki>3
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|3<nowiki>+</nowiki>5
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|1<nowiki>+</nowiki>4
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|2<nowiki>+</nowiki>5
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|1<nowiki>+</nowiki>5
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To find these "weights," I played around.  These weights also go onto the combos chart like this:  They must go in order.  Here is also an updated answer chart:
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{|border="2" cellspacing="0" cellpadding="4"
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|80
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|1<nowiki>+</nowiki>2(3)
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|83
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|90
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|4<nowiki>+</nowiki>5(9)
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{|border="2" cellspacing="0" cellpadding="4"
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Because 90 is the only combo with a "weigh" of 8, 3<nowiki>+</nowiki>5 must be that combination, because it also is the only pair with a "weight" of 8.  This can also be done in reverse with 84, which has the exclusive "weight" of 4, so it must be 1<nowiki>+</nowiki>3.  I can now find bale 3.  It must be 82-39 (Which is Bale 1) =43.  Bail 3 weighs 43 kg.  Here is a recap so far:
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{|border="2" cellspacing="0" cellpadding="4"
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|80
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|1<nowiki>+</nowiki>2(3)
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|1<nowiki>+</nowiki>3(4)
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|83
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|3<nowiki>+</nowiki>5(8)
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|4<nowiki>+</nowiki>5(9)
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{|border="2" cellspacing="0" cellpadding="4"
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I can now find 5.  It must be 90(3<nowiki>+</nowiki>5)-43(<nowiki>#</nowiki>3)=47.  So, we now have 5.  We can also get combinations now.  83, with a "weight" of 5, must be 1<nowiki>+</nowiki>4 or 2<nowiki>+</nowiki>3.  I will try 2<nowiki>+</nowiki>3 first.  41<nowiki>+</nowiki>43=88, which is not 83.  This means that 83 must be 1<nowiki>+</nowiki>4.  However we don<nowiki>’</nowiki>t know bale 4, but we do know that it is 83-39(combined <nowiki>#</nowiki>-1 bale known), which equals 44.  Ta-da.  Bale 4=44.
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{|border="2" cellspacing="0" cellpadding="4"
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|80
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|1<nowiki>+</nowiki>2(3)
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|1<nowiki>+</nowiki>3(4)
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|83
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|1<nowiki>+</nowiki>4(5)
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Next up is 84.  This must be 2<nowiki>+</nowiki>3, because our other 5 "weight" is already solved at 83.  It works when I check it.  41<nowiki>+</nowiki>43=84.
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Next up, we do the same finding our 6 and 7 "weights."  85 can be 1<nowiki>+</nowiki>5 or 2<nowiki>+</nowiki>4.  Lets check, 2<nowiki>+</nowiki>4, 41<nowiki>+</nowiki>44=85 Check.  Next up is 86, which must be 1<nowiki>+</nowiki>5.  Wait we don<nowiki>’</nowiki>t know 5, but we know that it is 86-39<nowiki>+</nowiki>47.  Bada-Bing Bale 5=47.  We now know all the bails.  Lets just finish up, by doing 87 and 88.  They are either 2<nowiki>+</nowiki>5 or 3<nowiki>+</nowiki>4.  41<nowiki>+</nowiki>47=88, so 88 is 2<nowiki>+</nowiki>4.  This leaves 87 to be 3<nowiki>+</nowiki>4 or 43<nowiki>+</nowiki>44=87. There we go, POW solved.
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===Work===
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*[[:Image:Haybaler Solution Pg. 1 Not Helped.jpg|Page 1 - Did Not Help]]
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*[[:Image:Haybaler Solution Pg. 2.jpg|Page 2]]
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==Solution==
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{|border="2" cellspacing="0" cellpadding="4"
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|80
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I then checked all of the weights to make sure they can work.  (ex. 1<nowiki>+</nowiki>2 is 39<nowiki>+</nowiki>41<nowiki>+</nowiki>80 Check)  I know none can repeat because all of the combs are listed here, and all work with none left over.
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==Extension==
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#Are there more weights?
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#Can it be done faster?
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Let<nowiki>’</nowiki>s start with 2.  I think my way is fast to find answers.  It just takes a lot of writing.  Someone told me that you can do it with averages of all the numbers, but I know no details.  It might not be faster.
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1.  I don<nowiki>’</nowiki>t think I can find more answers.  This all works with a delicate balance of numbers, where if you change one, you change 5 of them.  There also is an acceptable range where nothing else would fit.  I don<nowiki>’</nowiki>t think there are any more whol numbers that could fit.  I also did lots of combos by guess and check where I couldn<nowiki>’</nowiki>t find an answer.
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==Evaluation==
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Not necessary to do.
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[[Category:IAG 1H]]
 
[[Category:IAG 1H]]
 
[[Category:POW]]
 
[[Category:POW]]

Latest revision as of 16:21, 11 December 2008

IAG 1H POW # 8: The Haybaler Problem

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I was asked to take down POW solutions. Remember it's best to think about how to solve the problems on your own.

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