# Calculate : 1

I"m writing some software that takes a group of users and compares each user with every other user in the group. I need khổng lồ display the amount of comparisons needed for a countdown type feature.

For example, this group <1,2,3,4,5> would be analysed lượt thích this:

1-2, 1-3, 1-4, 1-52-3, 2-4, 2-53-4, 3-54-5By creating little diagrams lượt thích this I"ve figured out the pattern which is as follows:

Users - Comparisons2 - 13 - 3 (+2)4 - 6 (+3)5 - 10 (+4)6 - 15 (+5)7 - 21 (+6)8 - 28 (+7)9 - 36 (+8)I need to be able khổng lồ take any number of users, and calculate how many comparisons it will take to lớn compare every user with every other user.

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Can someone please tell me what the formula for this is?

combinatorics summation
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edited May 2, 2014 at 15:23 MJD
asked May 2, năm trước at 15:21 OwenOwen
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## 7 Answers 7

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You want lớn know how many ways there are khổng lồ choose \$2\$ users froma phối of \$n\$ users.

Generally, the number of ways to choose \$k\$ elements from a phối oforder \$n\$ (that is, all elements in the mix are distinct) is denotedby \$\$inomnk\$\$

and is equivalent to \$\$fracn!(n-k)!k!\$\$

In the case of \$k=2\$ the latter equals to lớn \$\$fracn!(n-2)!2!=fracn(n-1)2\$\$

which is also the sum of \$1+2+...+n-1\$.

For more information see Binomial coefficient & Arithmetic progression

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answered May 2, năm trước at 15:28 BelgiBelgi
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The sum of \$0+cdots + n-1\$ is \$\$frac12(n-1)n.\$\$

Here \$n\$ is the number of users; there are 0 comparisons needed for the first user alone, 1 for the second user (to compare them to the first), 2 for the third user, và so on, up khổng lồ the \$n\$th user who must be compared with the \$n-1\$ previous users.

For example, for \$9\$ people you are adding up \$0+1+2+3+4+5+6+7+8\$, which is equal lớn \$\$frac12cdot 8cdot 9= frac722 = 36\$\$ and for \$10\$ people you may compute \$\$frac12cdot9cdot10 = frac902 = 45.\$\$

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answered May 2, năm trước at 15:25
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The following way lớn getting the solution is beautiful & said to have been found by young Gauss in school. The idea is that the order of adding \$1+2+cdots+n=S_n\$ does not change the value of the sum. Therefore:

\$\$1 + 2 + ldots + (n-1) + n=S_n\$\$\$\$n + (n-1) + ldots + 2 + 1=S_n\$\$

Adding the two equations term by term gives

\$\$(n+1)+(n+1)+ldots+(n+1)=2S_n\$\$

so \$n(n+1)=2S_n\$. For \$n\$ persons, there are \$S_n-1\$ possibilities, as others answers have shown already nicely.

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answered May 2, năm trước at 15:33 binkyhorsebinkyhorse
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The discrete sum up to lớn a finite value \$N\$ is given by,

\$\$sum_n=1^N n = frac12N(N+1)\$\$

Proof:

The proof by induction roughly boils down to:

\$\$S_N = 1+ 2 +dots+N\$\$

\$\$S_N+1= 1+ 2 + dots + N + (N+1) = underbracefrac12N(N+1)_S_N + (N+1)\$\$

assuming that the induction hypothesis is true. The right hand side:

\$\$fracN(N+1)2+(N+1)=frac(N+1)(N+2)2\$\$

which is precisely the induction hypothesis applied khổng lồ \$S_N+1\$.

Just for your own curiosity, the case \$N=infty\$ is of course divergent. However, with the use of the zeta function, it may be regularized to yield,

\$\$sum_n=1^inftyn = zeta(-1)=-frac112\$\$

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answered May 2, năm trước at 15:31 JPhyJPhy
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\$\$N=2: 1 + 2 = (1 + 2) = 1 imes3\$\$

\$\$N=4: 1 + 2 + 3 + 4 = (1 + 4) + (2 + 3) = 2 imes5\$\$

\$\$N=6: 1 + 2 + 3 + 4 + 5 + 6 = (1 + 6) + (2 + 5) + (3 + 4) = 3 imes7\$\$

\$\$N=8: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8= (1 + 8) + (2 + 7) + (3 + 6)+ (4 + 5) = 4 imes9\$\$

More generally, \$N/2 imes(N + 1)\$.

For odd \$N\$, sum the \$N-1\$ first terms (using the even formula) together with \$N\$, giving \$(N-1)/2 imes N+N=N imes(N+1)/2\$.

Xem thêm: Ioe: Hấp Dẫn - Luyện Thi Ioe Tiếng Anh Lớp 5 Vòng 1 Đến Vòng 35

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edited May 2, năm trước at 15:47
answered May 2, 2014 at 15:41
user65203user65203
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corsiKa: the formula explicitly shows the sum from 1 to lớn \$N\$ inclusive (triangular numbers) and is perfectly correct. It is NOT the formula for the number of comparisons between \$N\$ users. \$endgroup\$
–user65203
May 2, năm trước at 21:38

\$egingroup\$ The sum must be taken from \$1\$ lớn \$Users-1\$ inclusive. \$endgroup\$
–user65203
May 2, 2014 at 21:51
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Here is another way khổng lồ find the sum of the first \$n\$ squares that generalizes to lớn sums of higher powers.

\$(k+1)^2 - k^2 = 2k+1\$

\$sum_k=1^n ( (k+1)^2 - k^2 ) = sum_k=1^n (2k+1)\$

\$(n+1)^2 - 1^2 = 2 sum_k=1^n k + n\$

\$sum_k=1^n k = frac (n+1)^2 - 1 - n 2 = fracn^2+n2\$

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answered May 3, năm trước at 7:11
user21820user21820
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This is kind of a pseudo code:

Say you have n number of people, & you labeled them.

for i in (1,2,3,...,n), person i need to compare with all the people who has a number larger (strictly), so person i need to compare (n-i) times.

so adding up would be(n-1) + (n-2) + ... + 3 + 2 + 1...

which would be the sum from 1 to lớn (n-1)

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answered May 2, năm trước at 15:26
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