10079  Pizza Cutting
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10079 Pizza Cutting  I want to know the proof.
It's a very easy problem, and I solved it within 10 min.
In my intuition, I can see that answer makes following arithmetic progression
a(n) = 1 + (n(n+1))/2;
However, I want to know why pizza cutting makes this expression.
Anyone tell me the proof of this expression?
In my intuition, I can see that answer makes following arithmetic progression
a(n) = 1 + (n(n+1))/2;
However, I want to know why pizza cutting makes this expression.
Anyone tell me the proof of this expression?
I love Problem Solving!!!
I don't know if it really is a proof, but when I coded my first version of this problem I made an iterative solution, when at each step I just added the number of pieces for the best possible cut.
Drawing it on the paper, I saw that I could ever cut all the previous lines with a single line, thus creating the sequence
1+2+3+4+...
It ran for almost 25 seconds, so after that i figured out that it was actually an arithmetic progression, and then just used this formula.
Drawing it on the paper, I saw that I could ever cut all the previous lines with a single line, thus creating the sequence
1+2+3+4+...
It ran for almost 25 seconds, so after that i figured out that it was actually an arithmetic progression, and then just used this formula.
Thiago Sonego Goulart  UFMG/Brazil
Well, the way I approached it was with common differences and regression.
Obviously, if you have zero cuts in the pizza, you'll have 1 slice. If you have one slice you'll have two pieces. It's given that you'll have 7 pieces with 3 slices, and you should also know that two slices would equate to four pieces.
(0,1)
(1,2)
(2,4)
(3,7)
Therefore: Quadratic.
Using y = ax^2+bx+c (and knowing that 1 is your c value, as (0,1))
2 = a(1)^2 + b(1) + 1
1 = a + b
4 = a(2)^2+b(2) + 1
3 = 4a + 2b
Then eliminate:
2*2 = (a + b)*2
 3 = 4a + 2b
1 = 2a
a = 1/2
1 = a + b
1 = 1/2 + b
1/2 = b
Thusly:
y = ax^2+bx+c
y = (x^2)/2+(x/2)+1
Or
y = ((x^2+x)/2)+1
That's how I looked at it anyways.
Obviously, if you have zero cuts in the pizza, you'll have 1 slice. If you have one slice you'll have two pieces. It's given that you'll have 7 pieces with 3 slices, and you should also know that two slices would equate to four pieces.
(0,1)
(1,2)
(2,4)
(3,7)
Code: Select all
1  2  4  7
D1: 1 2 3
D2: 1 1
Using y = ax^2+bx+c (and knowing that 1 is your c value, as (0,1))
2 = a(1)^2 + b(1) + 1
1 = a + b
4 = a(2)^2+b(2) + 1
3 = 4a + 2b
Then eliminate:
2*2 = (a + b)*2
 3 = 4a + 2b
1 = 2a
a = 1/2
1 = a + b
1 = 1/2 + b
1/2 = b
Thusly:
y = ax^2+bx+c
y = (x^2)/2+(x/2)+1
Or
y = ((x^2+x)/2)+1
That's how I looked at it anyways.
 saman_saadi
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 Posts: 31
 Joined: Sun Feb 25, 2007 6:33 pm
 Location: Tehran
The proof is by mathematical induction:
f(n) = the maximum number of pieces with n cuts
Proof: suppose we know f(n  1) now consider f(n):
if we remove the nth cut we know the answer by the hypothesis. The question is how to add the nth cut and gain the maximum number of pieces. It's obvious that the nth cut not be parallel with any cuts and no three cuts intersect each other in one point (We say these cuts (lines) are in General Position). So the nth cut must intersect with all n  1 cuts (with above conditions) then n new pieces we have:
f(n) = f(n  1) + n
f(0) = 1
if we solve above recurrence we have f(n) = 1 + n * (n +1) / 2
Hope it's useful.
f(n) = the maximum number of pieces with n cuts
Proof: suppose we know f(n  1) now consider f(n):
if we remove the nth cut we know the answer by the hypothesis. The question is how to add the nth cut and gain the maximum number of pieces. It's obvious that the nth cut not be parallel with any cuts and no three cuts intersect each other in one point (We say these cuts (lines) are in General Position). So the nth cut must intersect with all n  1 cuts (with above conditions) then n new pieces we have:
f(n) = f(n  1) + n
f(0) = 1
if we solve above recurrence we have f(n) = 1 + n * (n +1) / 2
Hope it's useful.
Re: 10079
First I'm sorry for my bad english.. Please forgive me because I am a korean and i'm an Elementry School student.....Red Scorpion wrote:
Hi, I am trying to solve problem 10079, and got failed.
I use this method :
for n=an, MAX = bn
n=0 , MAX = 0
n=1, MAX = 2
n=2, MAX = 4 (2 + 2 )
n=3, MAX = 7 (4 + 3 )
n=4, MAX = 11 (7 + 4)
n=5, MAX=16 (11 + 5)
n=6, MAX=22 (b(n1) + an)
n=7, MAX=29
n=8, MAX=37
n=9, MAX=46
n=10, MAX=56
... and so forth (and the number will be very huge)
If Anyone have time, Please tell me what wrong with this method ?
Thanks.
eh?? I made it like you and I got AC in 1.4..... did you used for? that will make better
Re: 10079 Pizza Cutting  I want to know the proof.
I am sorry but I think your code is wrong.soyoja wrote:It's a very easy problem, and I solved it within 10 min.
In my intuition, I can see that answer makes following arithmetic progression
a(n) = 1 + (n(n+1))/2;
However, I want to know why pizza cutting makes this expression.
Anyone tell me the proof of this expression?
I did as same as you but I got WA.
I'm so sorry about my bad english. Thank you.
Re: 10079  Pizza Cutting
No, Soyoja algorithm is right. I've tried it and it's worked. It's just need some carefully arithmetic for doing so.mintae71 wrote:I am sorry but I think your code is wrong.soyoja wrote:It's a very easy problem, and I solved it within 10 min.
In my intuition, I can see that answer makes following arithmetic progression
a(n) = 1 + (n(n+1))/2;
However, I want to know why pizza cutting makes this expression.
Anyone tell me the proof of this expression?
I did as same as you but I got WA.
I'm so sorry about my bad english. Thank you.
But for the prove I can't give a thing why it's worked sorry.
Re: 10079  Pizza Cutting
Trying to sove the problem.
I tried five times.
But still got WA.
I tried five times.
But still got WA.
Re: 10079  Pizza Cutting
a
Last edited by dtopic on Sat Feb 26, 2011 7:32 pm, edited 1 time in total.
WA10079
it's gettung WA...???
Code: Select all
#include<stdio.h>
int main()
{
long long int n;
for( ; ; )
{
if(n<0)
{
break;
}
scanf("%lld",&n);
printf("%lld\n",(n*(n+1)/2)+1);
}
return 0;
}
Last edited by brianfry713 on Tue Jan 27, 2015 9:26 pm, edited 1 time in total.
Reason: Added code blocks
Reason: Added code blocks
Re: 10079  Pizza Cutting
WA....
Code: Select all
#include<stdio.h>
int main()
{
long long int n;
for( ; ; )
{
if(n<0)
{
break;
}
scanf("%lld",&n);
printf("%lld\n",(n*(n+1)/2)+1);
}
return 0;
}
Last edited by brianfry713 on Tue Jan 27, 2015 9:26 pm, edited 1 time in total.
Reason: Added code blocks
Reason: Added code blocks

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 Posts: 5947
 Joined: Thu Sep 01, 2011 9:09 am
 Location: San Jose, CA, USA
Re: 10079  Pizza Cutting
Don't double post.
Initialize n.
Initialize n.
Check input and AC output for thousands of problems on uDebug!