ThePlaz: imp takedown

2008-12-11T16:21:39Z

imp takedown

ThePlaz: post

2007-06-22T02:42:22Z

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<big>'''[[IAG 1H]] [[POW]] <nowiki>#</nowiki> 8: The Haybaler Problem'''</big>

{{Word|Image:Haybaler_Problem.doc}}
{{PDF|Image:Haybaler_Problem.pdf}}

==Problem Statement==
Not necessary to do.

==Process==
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.

: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:

{|border="2" cellspacing="0" cellpadding="4"
|80
|1<nowiki>+</nowiki>2
|-
|82
| 
|-
|83
| 
|-
|84
| 
|-
|85
| 
|-
|86
| 
|-
|87
| 
|-
|88
| 
|-
|90
| 
|-
|91
|4<nowiki>+</nowiki>5
|-
|}

{|border="2" cellspacing="0" cellpadding="4"
|1
|?
|-
|2
|?
|-
|3
|?
|-
|4
|?
|-
|5
|?
|-
|}

Throughout the problem, I updated this chart when I found more info.

: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.

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."

{|border="2" cellspacing="0" cellpadding="4"
|1<nowiki>+</nowiki>2
|2<nowiki>+</nowiki>3
|3<nowiki>+</nowiki>4
|4<nowiki>+</nowiki>5
|-
|1<nowiki>+</nowiki>3
|2<nowiki>+</nowiki>4
|3<nowiki>+</nowiki>5
| 
|-
|1<nowiki>+</nowiki>4
|2<nowiki>+</nowiki>5
| 
| 
|-
|1<nowiki>+</nowiki>5
| 
| 
| 
|-
|}

{|border="2" cellspacing="0" cellpadding="4"
|3
|5
|7
|9
|-
|4
|6
|8
| 
|-
|5
|7
| 
| 
|-
|6
| 
| 
| 
|-
|}

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:

{|border="2" cellspacing="0" cellpadding="4"
|80
|1<nowiki>+</nowiki>2(3)
|-
|82
|4
|-
|83
|5
|-
|84
|5
|-
|85
|6
|-
|86
|6
|-
|87
|7
|-
|88
|7
|-
|90
|8
|-
|91
|4<nowiki>+</nowiki>5(9)
|-
|}

{|border="2" cellspacing="0" cellpadding="4"
|1
|39
|-
|2
|41
|-
|3
|?
|-
|4
|?
|-
|5
|?
|-
|}

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:

{|border="2" cellspacing="0" cellpadding="4"
|80
|1<nowiki>+</nowiki>2(3)
|-
|82
|1<nowiki>+</nowiki>3(4)
|-
|83
|5
|-
|84
|5
|-
|85
|6
|-
|86
|6
|-
|87
|7
|-
|88
|7
|-
|90
|3<nowiki>+</nowiki>5(8)
|-
|91
|4<nowiki>+</nowiki>5(9)
|-
|}

{|border="2" cellspacing="0" cellpadding="4"
|1
|39
|-
|2
|41
|-
|3
|43
|-
|4
|?
|-
|5
|?
|-
|}

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.

{|border="2" cellspacing="0" cellpadding="4"
|80
|1<nowiki>+</nowiki>2(3)
|-
|82
|1<nowiki>+</nowiki>3(4)
|-
|83
|1<nowiki>+</nowiki>4(5)
|-
|84
|5
|-
|85
|6
|-
|86
|6
|-
|87
|7
|-
|88
|7
|-
|90
|3<nowiki>+</nowiki>5(8)
|-
|91
|4<nowiki>+</nowiki>5(9)
|-
|}

{|border="2" cellspacing="0" cellpadding="4"
|1
|39
|-
|2
|41
|-
|3
|43
|-
|4
|44
|-
|5
|?
|-
|}

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.

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.

===Work===
*[[:Image:Haybaler Solution Pg. 1 Not Helped.jpg|Page 1 - Did Not Help]]
*[[:Image:Haybaler Solution Pg. 2.jpg|Page 2]]

==Solution==
{|border="2" cellspacing="0" cellpadding="4"
|80
|1<nowiki>+</nowiki>2(3)
|-
|82
|1<nowiki>+</nowiki>3(4)
|-
|83
|1<nowiki>+</nowiki>4(5)
|-
|84
|2<nowiki>+</nowiki>3(5)
|-
|85
|2<nowiki>+</nowiki>4(6)
|-
|86
|3<nowiki>+</nowiki>4(6)
|-
|87
|2<nowiki>+</nowiki>5(7)
|-
|88
|2<nowiki>+</nowiki>5(7)
|-
|90
|3<nowiki>+</nowiki>5(8)
|-
|91
|4<nowiki>+</nowiki>5(9)
|-
|}

{|border="2" cellspacing="0" cellpadding="4"
|1
|39
|-
|2
|41
|-
|3
|43
|-
|4
|44
|-
|5
|47
|-
|}

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.

==Extension==
#Are there more weights?
#Can it be done faster?

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.

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.

==Evaluation==
Not necessary to do.

[[Category:IAG 1H]]
[[Category:POW]]

Haybaler POW - Revision history

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