Post 874 has been clarified and moved here (Hoping for criticism):
This post is the promised proof / explanation as to why my Room Assignment Algorithm, RAA, using only ONE integer squared and at worst, only ONE finite list of passengers, only 2N-1 long, where N is the number of the just arriving bus, fills all rooms in the infinite hotel, even when an infinite number of buses, each with an infinite number of passenger brings new guests to hotel with all even number rooms empty. B*P is notation (not Cartesian coordinates) naming the passenger number P on bus number B. The buses are given their number in the order they park in the infinite lot and open their exit door. Passenger always gets off in order. I. e. P before passenger (P+1) does. At worst, the determination of the room assigned to any passengers uses a finite list of ordered passengers, such as ONE of the red colored blocks below, and probably only one simple calculation, directly assigns each a room, but that has not yet been proven.
An infinite set of squares are named by the positive integers, N = 1, 2, 3, 4, … and have areas: S1 = 1, S2 = 4, S3 = 9, S4 = 16, etc. This is the same set of squares as I defined in my quickly written post 811 describing my RAA. The "S-numbers" tell the number of passengers inside or on the boundary of square N. All these “in or on boundary” passengers get the first N^2 even room numbers before any passenger leaves bus N+1 to (or a higher number bus) gets his. The first out of bus N+1 gets room # = 2{(N+1)^2 +2} not 2{(N+1)^2 +1} as that room goes to a still unassigned passenger from bus 1, who is 1*N. The order of room assignments my RAA produces is illustrated below for the first 6^2 +1 = 37 even numbers:
1*1 . 1*2 . 2*1 . 2*2 . 1*3 . 3*1 . 2*3 . 3*2 . 3*3 . 1*4 . 4*1 . 2*4 . 4*2 . 3*4 . 4*3 . 4*4 . 1*5 . 5*1 . 2*5 . 5*2 . 3*5 . 5*3 . 4*5 . 5*4 . 5*5 . 1*6 . 6*1 … etc in B+P pairs until 6*6 . 1*7 ….
Where color changes (between none and red) when a new group of "square boundary" passengers begins to fill and for more convenient reading, dots separate the passengers.
Note that with this "lowest B+P sum pairs first" (in any boundary group) ordering plain:
(1) The boundary of square N always has an odd integer of passengers on it. I. e. each of the colored (or not) blocks above has 2n -1 passengers where n is an integer. E.g. 1, 3, 5, 7, 9, 11 etc. and are "boundary passengers" of different "boundary passengers blocks or groups."
(2) That the last passenger filling the Nth square is always N*N and that the first passenger joining a new group of "square boundaries" passengers added to make square N+1, always comes from bus 1. He is 1*(N+1). The second second passenger helping fill the new square's boundary is always the first off the new bus N+1. I. e. he is (N +1)*1.
(3) That the RAA rules assign two consecutive room #s to passengers with the same B+P sum except for the last of each square boundary group block, N*N who has no "partner." E. g. a*b and b*a are consecutively assigned rooms with the one from the lower bus number getting the room # less by 2 than his "partner."
(4) That the RAA rule that when P + B from two different buses are equal, the lower B is given a room before the passenger on the later to arrive bus is applied ONLY WITH IN the same boundary group of passengers; not when the passenger "belong to" different boundary groups. For example: 4*4, with B+P= 8 gets a lower room # than 1*5 did even thought 1*5's B+P is only 6.
In general, note that when N*1, the first off of bus N, gets his Room#, (N-1)^2 +1 Room #s have already been assigned to the first (N-1)^2 +1 passengers. So his Room # is 2x[(N-1)^2 + 2], as is illustrated in the order assignment list above. This ordering of the 2N -1 passengers in the boundary group of the Nth square could be different, as I noted in post 811. For example, the N passengers, all with bus # + N could be assigned in order of their increasing P, starting with P =1, then N*N is assigned his room, and then N remaining passengers of that boundary group, could be assigned starting with the highest bus number first, and then decreasing to B = 1. This would be working thru the 2N+1 boundary passengers going "counter clockwise" around the boundary. I need / want to know what assignment order plan corresponds to the formulae of the RAA given.
The "lowest B+P sum pairs first" ordering plain illustrated with alternating red / black boundary groups above had the N = 5 boundary group order as:
1*5 . 5*1 . 2*5 . 5*2 . 3*5 . 5*3 . 4*5 . 5*4 . 5*5. With the "counter clockwise" plain, the ordering would be:
5*1 . 5*2 . 5*3 . 5*4 . 5*5 . 4*5 . 3*5 . 2*5 . 1*5. I. e. passenger 1*5 is the Last, not the First, of the group so would have a longer wait, even though his bus was the first to arrive! However the "best" ordering plain is the one which makes the RAA most simple. AFAIK, that is the same as the first N+1 passengers as in the "counter clockwise" plain and last N with order reversed or "clockwise" so that ordering plain will be used and again illustrated with boundary group 5, which is numerically the same as the "just arriving" bus B = 5. I. e. the following room assignment order: 5*1 . 5*2 . 5*3 . 5*4 . 5*5 . 1*5 . 2*5 . 3*5 . 4*5. Then the RAA becomes only one short sentence with very easy to calculate terms, which is:
When P is < or = B, passenger B*P's room is 2[(B-1)^2 +P], otherwise it is 2[(B-1)^2 + B + P].
Note that P is not larger than B when this RAA is applied and passenger with P = B, B*P, (or B*B or P*P or P*B as all four point to same guy) gets his room by the "pre-otherwise" part.
The "otherwise part" assigns room to Ps = 1 ,2,3, ... up to P = B-1 after the "first part" gave rooms to B passengers (P numbers = 1 thru B) from bus B. The "B" in last part of the RAA notes / accounts for fact / that the first part has just given out B more rooms. In earlier, poorly worded, post 811, I tried to do this accounting with reference to a B by P rectangle.
As stated earlier this post was to show the room numbers assigned do fill EVERY even number room, only with one passenger. That objective has been achieved. However, it concluded the simple RAA of the bold, larger type, sentence above.
SUMMARY: It is the fact than the total number of passengers in or on boundary of smaller squares is simply an integer squared that makes it so easy to immediately tell the next group of boundary passengers their room numbers via a simple formula sentence since there are always just 2B+1 of them in their Bth boundary group to assign to the next not yet assigned rooms. I probably will make a post numerically illustrating how easy a couple of B*P passengers with large value of B & P, like a billion or more, with this RAA, and no "snake thru" all even numbers procedure as someguy1 first suggested (or Trippy first implemented in an infinite look-up table), can easily do that, I think.
This post is the promised proof / explanation as to why my Room Assignment Algorithm, RAA, using only ONE integer squared and at worst, only ONE finite list of passengers, only 2N-1 long, where N is the number of the just arriving bus, fills all rooms in the infinite hotel, even when an infinite number of buses, each with an infinite number of passenger brings new guests to hotel with all even number rooms empty. B*P is notation (not Cartesian coordinates) naming the passenger number P on bus number B. The buses are given their number in the order they park in the infinite lot and open their exit door. Passenger always gets off in order. I. e. P before passenger (P+1) does. At worst, the determination of the room assigned to any passengers uses a finite list of ordered passengers, such as ONE of the red colored blocks below, and probably only one simple calculation, directly assigns each a room, but that has not yet been proven.
An infinite set of squares are named by the positive integers, N = 1, 2, 3, 4, … and have areas: S1 = 1, S2 = 4, S3 = 9, S4 = 16, etc. This is the same set of squares as I defined in my quickly written post 811 describing my RAA. The "S-numbers" tell the number of passengers inside or on the boundary of square N. All these “in or on boundary” passengers get the first N^2 even room numbers before any passenger leaves bus N+1 to (or a higher number bus) gets his. The first out of bus N+1 gets room # = 2{(N+1)^2 +2} not 2{(N+1)^2 +1} as that room goes to a still unassigned passenger from bus 1, who is 1*N. The order of room assignments my RAA produces is illustrated below for the first 6^2 +1 = 37 even numbers:
1*1 . 1*2 . 2*1 . 2*2 . 1*3 . 3*1 . 2*3 . 3*2 . 3*3 . 1*4 . 4*1 . 2*4 . 4*2 . 3*4 . 4*3 . 4*4 . 1*5 . 5*1 . 2*5 . 5*2 . 3*5 . 5*3 . 4*5 . 5*4 . 5*5 . 1*6 . 6*1 … etc in B+P pairs until 6*6 . 1*7 ….
Where color changes (between none and red) when a new group of "square boundary" passengers begins to fill and for more convenient reading, dots separate the passengers.
Note that with this "lowest B+P sum pairs first" (in any boundary group) ordering plain:
(1) The boundary of square N always has an odd integer of passengers on it. I. e. each of the colored (or not) blocks above has 2n -1 passengers where n is an integer. E.g. 1, 3, 5, 7, 9, 11 etc. and are "boundary passengers" of different "boundary passengers blocks or groups."
(2) That the last passenger filling the Nth square is always N*N and that the first passenger joining a new group of "square boundaries" passengers added to make square N+1, always comes from bus 1. He is 1*(N+1). The second second passenger helping fill the new square's boundary is always the first off the new bus N+1. I. e. he is (N +1)*1.
(3) That the RAA rules assign two consecutive room #s to passengers with the same B+P sum except for the last of each square boundary group block, N*N who has no "partner." E. g. a*b and b*a are consecutively assigned rooms with the one from the lower bus number getting the room # less by 2 than his "partner."
(4) That the RAA rule that when P + B from two different buses are equal, the lower B is given a room before the passenger on the later to arrive bus is applied ONLY WITH IN the same boundary group of passengers; not when the passenger "belong to" different boundary groups. For example: 4*4, with B+P= 8 gets a lower room # than 1*5 did even thought 1*5's B+P is only 6.
In general, note that when N*1, the first off of bus N, gets his Room#, (N-1)^2 +1 Room #s have already been assigned to the first (N-1)^2 +1 passengers. So his Room # is 2x[(N-1)^2 + 2], as is illustrated in the order assignment list above. This ordering of the 2N -1 passengers in the boundary group of the Nth square could be different, as I noted in post 811. For example, the N passengers, all with bus # + N could be assigned in order of their increasing P, starting with P =1, then N*N is assigned his room, and then N remaining passengers of that boundary group, could be assigned starting with the highest bus number first, and then decreasing to B = 1. This would be working thru the 2N+1 boundary passengers going "counter clockwise" around the boundary. I need / want to know what assignment order plan corresponds to the formulae of the RAA given.
The "lowest B+P sum pairs first" ordering plain illustrated with alternating red / black boundary groups above had the N = 5 boundary group order as:
1*5 . 5*1 . 2*5 . 5*2 . 3*5 . 5*3 . 4*5 . 5*4 . 5*5. With the "counter clockwise" plain, the ordering would be:
5*1 . 5*2 . 5*3 . 5*4 . 5*5 . 4*5 . 3*5 . 2*5 . 1*5. I. e. passenger 1*5 is the Last, not the First, of the group so would have a longer wait, even though his bus was the first to arrive! However the "best" ordering plain is the one which makes the RAA most simple. AFAIK, that is the same as the first N+1 passengers as in the "counter clockwise" plain and last N with order reversed or "clockwise" so that ordering plain will be used and again illustrated with boundary group 5, which is numerically the same as the "just arriving" bus B = 5. I. e. the following room assignment order: 5*1 . 5*2 . 5*3 . 5*4 . 5*5 . 1*5 . 2*5 . 3*5 . 4*5. Then the RAA becomes only one short sentence with very easy to calculate terms, which is:
When P is < or = B, passenger B*P's room is 2[(B-1)^2 +P], otherwise it is 2[(B-1)^2 + B + P].
Note that P is not larger than B when this RAA is applied and passenger with P = B, B*P, (or B*B or P*P or P*B as all four point to same guy) gets his room by the "pre-otherwise" part.
The "otherwise part" assigns room to Ps = 1 ,2,3, ... up to P = B-1 after the "first part" gave rooms to B passengers (P numbers = 1 thru B) from bus B. The "B" in last part of the RAA notes / accounts for fact / that the first part has just given out B more rooms. In earlier, poorly worded, post 811, I tried to do this accounting with reference to a B by P rectangle.
As stated earlier this post was to show the room numbers assigned do fill EVERY even number room, only with one passenger. That objective has been achieved. However, it concluded the simple RAA of the bold, larger type, sentence above.
SUMMARY: It is the fact than the total number of passengers in or on boundary of smaller squares is simply an integer squared that makes it so easy to immediately tell the next group of boundary passengers their room numbers via a simple formula sentence since there are always just 2B+1 of them in their Bth boundary group to assign to the next not yet assigned rooms. I probably will make a post numerically illustrating how easy a couple of B*P passengers with large value of B & P, like a billion or more, with this RAA, and no "snake thru" all even numbers procedure as someguy1 first suggested (or Trippy first implemented in an infinite look-up table), can easily do that, I think.
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