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What is the largest number of pieces you can get by cutting a pizza

  • Thread starter wubie
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wubie

Hello,

I am trying to follow an example/explanation, but with no luck. Here is the question:

What is the largest number of pieces you can get by cutting a pizza with n straight vertical cuts if the pieces are not to be moved between cuts?
Let n be the number of cuts and let an be the maximum number of pieces.

Now it is clear that a0 = 1 and a1 = 2. Since we want to maximize the number of pieces, we ensure that two cuts cross. Hence, a2 = 4.

(Now here is the part in the example where I get confused):

Similarly, a third cut should cross both of the previous two, but not where they cross each other. This yields a3 = 7.

How did they do that? I don't understand. Wait a minute, I think I do understand how they did it. But, the pieces couldn't be equal could they? Is it possible to make three cuts, get seven pieces, but have all seven pieces equal?

The example goes further:

Suppose n - 1 cuts have been made and an - 1 pieces are obtained. As before, the n-th cut should cross each of the others at distinct points. Hence there will be n - 1 points of intersection on it, dividing it into n segments. Each segment cuts an existing piece into two. Hence we get the recurrence relation

an = an - 1 + n, 1 <= n

a0 = 1.

Ok. I didn't understand this part. Could someone reword this for me if it isn't too much trouble? I can't see what they are saying as of yet.

Any help is appreciated. Thankyou.
 
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HallsofIvy

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Wait a minute, I think I do understand how they did it. But, the pieces couldn't be equal could they? Is it possible to make three cuts, get seven pieces, but have all seven pieces equal?
Don't go off on a tangent. Was there anything in the problem that said the pieces were SUPPOSED to be equal?

Suppose n - 1 cuts have been made and an - 1 pieces are obtained. As before, the n-th cut should cross each of the others at distinct points. Hence there will be n - 1 points of intersection on it, dividing it into n segments. Each segment cuts an existing piece into two. Hence we get the recurrence relation

an = an - 1 + n, 1 <= n

a0 = 1.
There are n-1 cuts, meaning the pizza is crossed by n-1 lines. Your new cut will be a straight line. One straight line can't cross another twice, but if you are very careful, your new cut can cross every cut already made once. Since your new line is crossing n-1 lines, that will make n-1 points on your new line where the other cross it. That actually divides the new line into n segments (think about that: 1 point on a line divides it into 2 parts. 2 points on a line divide it into 3 parts, etc.
Of course, the section between two of the "old" cuts was a piece of pizza. Your new cut divides that piece into two pieces. Suppose the pizza had already been cut into 7 pieces (thats what I got dividing a circle by 3 lines) and your new cut crosses 4 of those pieces (i.e. crosses all 3 old lines). One way of counting the new number of pieces is to say "3 of the pieces were not cut. 4 we divided in two (NOT of equal size- the problem says nothing about "equal") by the new cut- that's 8 pieces. We now have 3+ 8= 11 pieces. It's easier, however, to say that there are 4 "new" pieces, counting 1 of the 2 halves of the divided piece as the "old" piece. 7 "old" pieces plus 4 "new" pieces is 11 just as before. If there were n-1 cuts, making an pieces, adding a new cut that crosses each of the old ones divides n "old" pieces in two, effectively adding n "new" pieces. an= an-1+ n.

Look at this from the start: with 0 cuts, you have 1 piece. Adding 1 new cut, so that n=1, you divide the pizza into 2 pieces. A second cut crosses that first cut once, of course, meaning that your second cut crosses 2 pieces, adding 2 pieces to the previous total: a2= a1+ 2= 2+2= 4. The third line crosses both previous lines and, so, crosses 3 pieces, dividing them in two and creating 3 "new" pieces: a3= a2+ 3= 7 pieces.
Hope that helps.

Of course, with 0 cuts, the pizza is in one piece: a0= 1.
 

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