ADVANCED Pidgeonhole Principle Formula

thereddiamanthe

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How to calculate --

how many random positive integers have to form - for at least 2 of them to sum up to - the highest integer in the pool plus one.

Eg.

The pool is 27 [1,2,.....,26,27].
Target Sum is 28.

How many integers are required for at least two of them to sum to 28.
 

thereddiamanthe

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& to cherry the pie ..

is there a formula for 'at least three integers' to form the sum of the 'Total Poll Max +1'
 

mmm4444bot

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Hello. Is this homework? Have you made any progress that you can share? Thanks.

🤖

 

thereddiamanthe

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Yeah well, I ain't a student in a sense of being in school.. I need some math advice, occasionally.
For my own purposes.


[Rant removed] I found the solution long way around by manual research -- which is exactly what I wanted to supersede by getting the answer, & moving on utilizing it as a piece in much bigger framework ..

So the solution is simply the

x=(y/2-1)+(y/2+1); y=total pool of numbers.

I don't know how to exactly write in math terms when the y pool is an odd number,;
as the pair sum is only Total Pool (& not Total Pool + 1, as initially outlined)
effectively the y/2 natural positive integer being a standalone pair, meaning it has to repeat itself to sum up to the y itself.
 
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topsquark

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I'd appreciate getting straight answers from now on to speed up my (larger) work.
I sympathize. I've had this problem occasionally on sites, too. I have been teaching myself QFT and am about to embark (again) on String Theory. The problem here, specifically, is that we are geared to use a Socratic method to instruct. I am not saying we can't make exceptions and it is quite possible now that you have told us what you are doing we may be able to just answer your questions. But only the staff can make that decision.

If they decide not to change things for you, please understand that the decision is not personal. We can still help and it probably won't take us more than a post or two longer. And hey, the more information you can give us, the better we can help you.

-Dan
 

Cubist

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So the solution is simply the

x=(y/2-1)+(y/2+1); y=total pool of numbers.

Note that "x=(y/2-1)+(y/2+1)" simplifies to "x=y"

But, personally I don't fully understand your requirement.

If y=4 then do you think the possible sums are { 1+4, 2+3 } which would imply that x=2 because the set contains 2 items? Or do you require the possible sums to be { 1+4, 4+1, 2+3, 3+2 }?

Likewise if y=8, do we have { 1+2+6, 1+3+5, 2+3+4, 1+8, 2+7, 3+6, 4+5 } which implies x ought to be 7? Or perhaps you'd count all of the following permutations { 1+2+6, 1+6+2, 2+1+6, 2+6+1, 6+1+2, 6+2+1, 1+3+5, 1+5+3, 3+1+5, 3+5+1, 5+1+3, 5+3+1, 2+3+4, 2+4+3, 3+2+4, 3+4+2, 4+2+3, 4+3+2, 1+8, 8+1, 2+7, 7+2, 3+6, 6+3, 4+5, 5+4 } ?
 

mmm4444bot

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I ain't a student…in school … I'd appreciate getting straight answers from now on
You likely would have received more direct help, had you told us up front that you're not doing homework. Informing members about forum policies is the reason why we had provided you with a link to the guidelines when you'd joined the forum.

Guidelines Summary said:
Unless you say otherwise, we will treat you like a student working on a school assignment who is stuck at one of the steps. We don't do homework…
 

thereddiamanthe

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All fine.

@Cubist
I wrote the formula wrongly, just shows that I need math advice ..
x=((y/2)-1)+y((y/2)+1)

Or in long ..
If the pool is 0 to 36 positive integers y=37 then x=19
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
17+19=36

Or in max 17 random positive integers there will be at least one pair summing up to 36.

Could be written as well as y=37, x=(y/2)+1 rounded up.
 

thereddiamanthe

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Those with the brightest math heads ..

I had an intuition of running a layered PhP, where y differs in two layered groups.
Let's say that one group y1=6 y2=12, or y1=12 y2=18, or y1=18 y=2=37.

Can two-layered PhP groups run effectively simultaneously narrow & improve the precision,
as the same outcomes pertaining to both of the two groups are the same but labeled differently;
obviously the pigeons & holes are defined differently in each group but run as a whole.
 

Jomo

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& to answer your question, by the time checked here again & got a non-answer/ignore due to who knows which reason,

I'd appreciate getting straight answers from now on to speed up my (larger) work.
Excuse me, but i do not appreciate what you said to a helper on this forum.

You chose not to read the posting guideline which states that we are a help forum and do not solve problems for any posters.

Then you have the nerve to wrote I'd appreciate getting straight answers from now on to speed up my (larger) work.

You are absolutely the most entitled person who ever came to this forum. Now do me a favor and go away--far away.
 

lookagain

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User thereddiamanthe first posted in the forum on Oct. 16, 2019,
where he/she posted information and three questions with no
work shown.

Super Moderator Subhotosh Khan was the immediate next (and only)
reply to this new user to state:

"Please show your work/thoughts about the assignment.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING [It's normally seen as a link in blue letters.]"

-----------------------

This new user never responded in that 2019 thread. Just recently,
since just September 24, 2021, this same user posted posts numbers 2
through number 10 in only a couple of threads, completely blowing off
anything that was stated to him/her in the first reply back in 2019 by
Super Moderator Subhotosh Khan.

I am requesting at least a temporary ban on thereddiamanthe, given
what I pointed out and the behavior/attitude shown in the recent posts.
 
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