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32 people were invited at a party and started exchanging handshakes. Because of the confusion, each of them shook hands with each other multiple times: at least twice and up to X times. However, every invitee exchanged different number of handshakes from every other invitee. What is the minimum possible number X, so that the above condition is met?
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You mean, the number of handshakes each of them had was different?
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Correct.
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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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How?
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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Only as an image I think.
Can you explain method and calculations?
Happy 1st of May, by the way
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[hide=chart] I made a chart on excel sheet but can not be produced completely here. Only a part of it is shown.
I used names A,B,C,D......X,Y,Z,a,b,c,d,e,f for 32 people. The chart looks warped because of single and double digit numbers.Only columns upto T are shown For the remaining it is symmetrical about the long diagonal.
A B C D E F G H I J K L M N O P Q R S T
A x 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
B 2 x 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
C 3 4 x 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
D 4 5 6 x 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
E 5 6 7 8 x 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
F 6 7 8 9 10 x 12 13 14 15 16 17 18 19 20 21 22 23 24 25
G 7 8 9 10 11 12 x 14 15 16 17 18 19 20 21 22 23 24 25 26
H 8 9 10 11 12 13 14 x 16 17 18 19 20 21 22 23 24 25 26 27
I 9 10 11 12 13 14 15 16 x 18 19 20 21 22 23 24 25 26 27 28
J 10 11 12 13 14 15 16 17 18 x 20 21 22 23 24 25 26 27 28 29
K 11 12 13 14 15 16 17 18 19 20 x 22 23 24 25 26 27 28 29 30
L 12 13 14 15 16 17 18 19 20 21 22 x 24 25 26 27 28 29 30 31
M 13 14 15 16 17 18 19 20 21 22 23 24 x 26 27 28 29 30 31 32
N 14 15 16 17 18 19 20 21 22 23 24 25 26 x 28 29 30 31 32 33
O 15 16 17 18 19 20 21 22 23 24 25 26 27 28 x 30 31 32 33 2
P 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 x 32 33 2 3
Q 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 x 2 3 4
R 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 2 x 4 5
S 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 2 3 4 x 6
T 20 21 22 23 24 25 26 27 28 29 30 31 32 33 2 3 4 5 6 x
U 21 22 23 24 25 26 27 28 29 30 31 32 33 2 3 4 5 6 7 8
V 22 23 24 25 26 27 28 29 30 31 32 33 2 3 4 5 6 7 8 9
W 23 24 25 26 27 28 29 30 31 32 33 2 3 4 5 6 7 8 9 10
X 24 25 26 27 28 29 30 31 32 33 2 3 4 5 6 7 8 9 10 11
Y 25 26 27 28 29 30 31 32 33 2 3 4 5 6 7 8 9 10 11 12
Z 26 27 28 29 30 31 32 33 2 3 4 5 6 7 8 9 10 11 12 13
a 27 28 29 30 31 32 33 2 3 4 5 6 7 8 9 10 11 12 13 14
b 28 29 30 31 32 33 2 3 4 5 6 7 8 9 10 11 12 13 14 15
c 29 30 31 32 33 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
d 30 31 32 33 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
e 31 32 33 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
f 32 33 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Starting with A he has to go upto 32 handshakes when he reaches "f". for B he has to go upto 33. C however gets a chance to use 2 handshakes with "f". D makes use of 2 & 3 for e & f and so on.So the minimum number stands at 33.
Last edited by thickhead (2016-04-30 18:41:02)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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All the possible pairs of 32 people are 32!/2!30! = 496. If each of them shook hands (i.e. "participated" in handshaking) at least twice and up to X times, then the total number T of handshakes must be from 2*496 to X*496. On the other hand, the total number of "single" handshakes must be a series of different numbers.
Their sum must be 2*T, correct?
Maybe you can combine the above to give a solution through calculations (and also verify your above outcome)?
Much appreciated!!
BTW I don't know the correct answer!!
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The chart looks warped because of single and double digit numbers.
Here's the same chart in 'code' tags.
If you want to align the single-digit numbers with the right digit of the double-digit numbers, you'd have to include spaces where needed...but this is the simple solution.
Last edited by phrontister (2016-04-30 21:40:01)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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But how do you use code tags ?
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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There is a list of BBCodes here...or, when you are logged in, you can get there by clicking on the BBCode button near the bottom of the thread page, on the left-hand side. The code tags are near the bottom of the list.
Also, again when you are logged in, you can click on the 'Quote' button in the bottom right-hand corner of my previous post to see how I used it.
Last edited by phrontister (2017-02-23 11:20:40)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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The total number of handshakes is 8672.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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The first of them (say A) exchanged 2*31=62 handshakes. Then if we consider that all others exchanged a different number of handshakes, we have a sum of
62+63+...+93 = 2480.
Are we dividing this by 496 (i.e. the total possible number of pairs)?
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A had 2 handshakes with B,3 handshakes with C, 4 handshakes with D .... 32 handshakes with f (32nd person).So total handshakes by A=2+3+3+4+.......+31+32=527 handshakes. Add handshakes of all people and divide by 2.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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We are not told that each people has different number of handshakes with each of the others; only that each one "participates" in a different number of handshakes in total. Therefore, if A participates in 2*31=62 (because we are told that each pair exchanges at least two handshakes), then all others from B to f participate in a different number of handshakes (but I don't know how to find them). I guess their sum must be 496*X*2.
A had 2 handshakes with B,3 handshakes with C, 4 handshakes with D .... 32 handshakes with f (32nd person).So total handshakes by A=2+3+3+4+.......+31+32=527 handshakes. Add handshakes of all people and divide by 2.
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Correct Anna.
A for example, may have 3 handshakes with B, 3 with C and 5 with D and so on.
We are not told that each people has different number of handshakes with each of the others; only that each one "participates" in a different number of handshakes in total. Therefore, if A participates in 2*31=62 (because we are told that each pair exchanges at least two handshakes), then all others from B to f participate in a different number of handshakes (but I don't know how to find them). I guess their sum must be 496*X*2.
thickhead wrote:A had 2 handshakes with B,3 handshakes with C, 4 handshakes with D .... 32 handshakes with f (32nd person).So total handshakes by A=2+3+3+4+.......+31+32=527 handshakes. Add handshakes of all people and divide by 2.
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So finally what is the correct number for X?
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what I assumed is that a must have different number of handshakes from others but these numbers could be used by others also. e.g. if A had 10 handshakes with P ,R can have 10 handshakes with C ;there is no bar for that but obviously P can not have 10 handshakes with anybody other than A.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Why not?
He could possibly have 10 handshakes with someone else, as long as his total number of handshakes with everyone is different from everybody else's.
what I assumed is that a must have different number of handshakes from others but these numbers could be used by others also. e.g. if A had 10 handshakes with P ,R can have 10 handshakes with C ;there is no bar for that but obviously P can not have 10 handshakes with anybody other than A.
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32 people were invited at a party and started exchanging handshakes. Because of the confusion, each of them shook hands with each other multiple times: at least twice and up to X times. However, every invitee exchanged different number of handshakes from every other invitee. What is the minimum possible number X, so that the above condition is met?
As per the terms of problem setter, If A has 10 handshakes with P, he can not have 10 handshakes with anybody else. At the same time P also had 10 handshakes with A ;so P also can not have 10 handshakes with anybody else. here 10 handshakes means 10 and 10 only.Neither more nor less.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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As per the terms, we only have the following condition: A must have a different number of TOTAL handshakes from B or C etc. It is possible, however, that he has 10 handshakes with P and 10 with someone else, provided that his handshakes in TOTAL are not the same with anybody else's.
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I see. I misunderstood the conditions.Now I have to start afresh.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hi;
I think I agree with thickhead's answer in post #6.
Last edited by phrontister (2016-05-02 01:21:40)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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I change my answer. minimum X=3
In phrontister's excel table in the first line (corresponding to B) under f 3 is o.k.
In the 2nd line we will keep 3 under "e" and 3 under "f".
In the 3rd line 3 under "d","e' and "f".
"f" will not have heavy burden. the totals under A to e will remain same. Only total of "f" will decrease to 2+30*3=92
I am wondering why it is not 93.i have to check the table.
Last edited by thickhead (2016-05-02 02:09:02)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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