Haybaler POW
From ThePlaz.com
IAG 1H POW # 8: The Haybaler Problem
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Contents |
Problem Statement
Not necessary to do.
Process
I started this POW with a lot of work which I didn’t need to do. I wrote down all the combos in the form of 1+2, 2+3, etc. I then guess some numbers to fill in. I tried lost of combos, only switching 4 and 5. I couldn’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’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:
80 | 1+2 |
82 | |
83 | |
84 | |
85 | |
86 | |
87 | |
88 | |
90 | |
91 | 4+5 |
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’t be 40 because you told us, and when I add 1+3 and 2+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+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."
1+2 | 2+3 | 3+4 | 4+5 |
1+3 | 2+4 | 3+5 | |
1+4 | 2+5 | ||
1+5 |
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:
80 | 1+2(3) |
82 | 4 |
83 | 5 |
84 | 5 |
85 | 6 |
86 | 6 |
87 | 7 |
88 | 7 |
90 | 8 |
91 | 4+5(9) |
1 | 39 |
2 | 41 |
3 | ? |
4 | ? |
5 | ? |
Because 90 is the only combo with a "weigh" of 8, 3+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+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:
80 | 1+2(3) |
82 | 1+3(4) |
83 | 5 |
84 | 5 |
85 | 6 |
86 | 6 |
87 | 7 |
88 | 7 |
90 | 3+5(8) |
91 | 4+5(9) |
1 | 39 |
2 | 41 |
3 | 43 |
4 | ? |
5 | ? |
I can now find 5. It must be 90(3+5)-43(#3)=47. So, we now have 5. We can also get combinations now. 83, with a "weight" of 5, must be 1+4 or 2+3. I will try 2+3 first. 41+43=88, which is not 83. This means that 83 must be 1+4. However we don’t know bale 4, but we do know that it is 83-39(combined #-1 bale known), which equals 44. Ta-da. Bale 4=44.
80 | 1+2(3) |
82 | 1+3(4) |
83 | 1+4(5) |
84 | 5 |
85 | 6 |
86 | 6 |
87 | 7 |
88 | 7 |
90 | 3+5(8) |
91 | 4+5(9) |
1 | 39 |
2 | 41 |
3 | 43 |
4 | 44 |
5 | ? |
Next up is 84. This must be 2+3, because our other 5 "weight" is already solved at 83. It works when I check it. 41+43=84.
Next up, we do the same finding our 6 and 7 "weights." 85 can be 1+5 or 2+4. Lets check, 2+4, 41+44=85 Check. Next up is 86, which must be 1+5. Wait we don’t know 5, but we know that it is 86-39+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+5 or 3+4. 41+47=88, so 88 is 2+4. This leaves 87 to be 3+4 or 43+44=87. There we go, POW solved.
Work
Solution
80 | 1+2(3) |
82 | 1+3(4) |
83 | 1+4(5) |
84 | 2+3(5) |
85 | 2+4(6) |
86 | 3+4(6) |
87 | 2+5(7) |
88 | 2+5(7) |
90 | 3+5(8) |
91 | 4+5(9) |
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+2 is 39+41+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’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’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’t think there are any more whol numbers that could fit. I also did lots of combos by guess and check where I couldn’t find an answer.
Evaluation
Not necessary to do.