Can the Last Pigeon Group Enter Selected Holes if M of Them Must Be Empty?

In summary, the question is about determining if all pigeons can fit into K pigeonholes given that N groups of pigeons, each with M pigeons, select Z holes each where K >= Z >= M. The capacity of each hole is one pigeon and it is unknown which holes are selected by each group. This may require a recursive routine to solve, as the selection of holes by one group may affect the ability of other groups to fit into the remaining holes.
  • #1
mostaphaamini
4
0
Hi all,
I have a question and hope to solve.
problem terms :
1- there are K holes.
2- there are N group of Pigeon.
3- there are M Pigeon in any Pigeon group.
4- each group of Pigeon can chose Z hole so that K >= Z >= M.
5- we don`t know which hole selected by a Pigeon only we know M Pigeon in group select Z hole.
6- the capacity of each hole is exact one Pigeon.
How we can determine last Pigeon group can not enter to the selected holes (because the selected hole are full)?
 
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  • #2
The explanation is a bit unclear, but this is how I interpreted it:

There are K pigeonholes, N groups and M pigeons in each group. Z is a number between M and N, and is the number of holes each group occupies. The question is to determine if all pigeons can fit in the holes or not.

If that is so, then obviously, since each group occupies Z holes, and they all fit in, M*Z must be less than or equal to N. If M*Z is larger than N, the last group will not be able to choose Z holes. I can hardly see how this is a more complex situation than the classical pigeonhole principle, so I assume my interpretation is incorrect. Besides, what is the relevance of 6- if every group need Z holes each.
 
  • #3
Jarle said:
The explanation is a bit unclear, but this is how I interpreted it:

There are K pigeonholes, N groups and M pigeons in each group. Z is a number between M and N, and is the number of holes each group occupies. The question is to determine if all pigeons can fit in the holes or not.

If that is so, then obviously, since each group occupies Z holes, and they all fit in, M*Z must be less than or equal to N. If M*Z is larger than N, the last group will not be able to choose Z holes. I can hardly see how this is a more complex situation than the classical pigeonhole principle, so I assume my interpretation is incorrect. Besides, what is the relevance of 6- if every group need Z holes each.

Jarle, read more carefully. Z is number between K and M or can be equal (term 4) no between N and M
this question is a bit hard and need more think
 
  • #4
mostaphaamini said:
Jarle, read more carefully. Z is number between K and M or can be equal (term 4) no between N and M
this question is a bit hard and need more think

Yes, I meant between or equal to K or M. I confused a few variables. It should be (N/M)*Z, not M*Z. And this must be compared to K, not N. Explain how it matters how man pigeons that fit into a hole when all you know is that a group of M pigeons occupies Z holes?
 
  • #5
Jarle said:
Yes, I meant between or equal to K or M. I confused a few variables. It should be (N/M)*Z, not M*Z. And this must be compared to K, not N. Explain how it matters how man pigeons that fit into a hole when all you know is that a group of M pigeons occupies Z holes?
I am sorry for my English
Let me for an example :
there is 4 holes, first pigeons group arrive and this group have 2 pigeons, pigeons select first and second holes now another group is coming and in the new group there is only one pigeon that select first hole, there are totally 3 pigeons and 4 holes and with your formula all must be true but in fact pigeons can not enter to holes because first hole selected by 2 pigeons now let's change the situation, first pigeons*group select first, second and third holes and now we have the answer because 2 pigeons in the first group can enter to holes 2 and 3 and pigeon in the second group can enter to hole 2. this problem will be solved by a recursive routine but I need a formula
this forum disappoint me I think it have many mathematician and I expect more reply
 
  • #6
You can't expect a lot of replies when it's hard to figure out what you're asking. So the pigeons select Z holes that they are allowed to go into, and the requirement is that M of them are empty?

This is going to depend heavily on which Z holes the pigeons select... you can't answer it based only on M,N,Z and K.

As an example with two pigeons and two holes: M=1,Z=1, K=2, N=2:

First pigeon selects first hole. Second pigeon selects first hole. They fail to fit

First pigeon selects first hole. Second pigeon selects second hole. They fit.

What kind of criterion do you have for which holes the pigeons are allowed to select?
 
  • #7
Office_Shredder said:
You can't expect a lot of replies when it's hard to figure out what you're asking. So the pigeons select Z holes that they are allowed to go into, and the requirement is that M of them are empty?

This is going to depend heavily on which Z holes the pigeons select... you can't answer it based only on M,N,Z and K.

As an example with two pigeons and two holes: M=1,Z=1, K=2, N=2:

First pigeon selects first hole. Second pigeon selects first hole. They fail to fit

First pigeon selects first hole. Second pigeon selects second hole. They fit.

What kind of criterion do you have for which holes the pigeons are allowed to select?

please think more higher, it is completely clear that holes capacity is exact one pigeon so when the first pigeon select a hole other pigeons of group must select another holes.
in the older post there is an example but more explain : we know any group of pigeons select Z holes from K holes and K >= Z >= M but we don`t know which hole are selected by pigeons so we can suppose if next group select one hole of previous group holes, pigeons of the previous group did not enter to this specific hole (if K > M and no K = M). this condition may be check every time that new group arrive for all older groups
 
  • #8
It's not easy to understand the situation after the first reading..the pigeons select Z holes that they are allowed to go into, and the requirement is that M of them are empty..how can we predict? they are unpredictable))
 

1. What is the Advanced Pigeonhole principle?

The Advanced Pigeonhole principle is a mathematical concept that states that if there are n pigeons and m pigeonholes, with n > m, then at least one pigeonhole must contain more than one pigeon. This principle is often used in combinatorics and other areas of mathematics to prove the existence of certain patterns or solutions.

2. How is the Advanced Pigeonhole principle different from the basic Pigeonhole principle?

The basic Pigeonhole principle states that if there are n pigeons and m pigeonholes, with n > m, then at least one pigeonhole must contain at least one pigeon. The Advanced Pigeonhole principle is a stronger version of this principle, as it guarantees the presence of multiple pigeons in at least one pigeonhole. This makes it more useful for solving problems that require the existence of a certain number of objects.

3. Can the Advanced Pigeonhole principle be applied to real-life situations?

Yes, the Advanced Pigeonhole principle can be applied to real-life situations. For example, if there are 10 people in a room and only 9 chairs, then at least one chair must be occupied by more than one person. This also applies to scenarios such as scheduling, where there are more tasks than available time slots.

4. How is the Advanced Pigeonhole principle used in mathematics?

The Advanced Pigeonhole principle is often used in combinatorics, which is the branch of mathematics that deals with counting and arranging objects. It can also be used in other areas such as number theory, graph theory, and computer science to prove the existence of certain patterns or solutions.

5. Are there any limitations to the Advanced Pigeonhole principle?

Yes, there are limitations to the Advanced Pigeonhole principle. It can only be applied when there are more objects than containers, and it does not provide any information about the distribution of the objects. Additionally, it may not be applicable in certain scenarios where the objects are not discrete and cannot be easily divided into groups.

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