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# Siamese method proof

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## Hongbo Fang

on 28 February 2013

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#### Transcript of Siamese method proof

Siamese method proof
1.induction proof: each row add up to correct sum
2.induction proof: each column add up to correct sum
3.ignore diagonal (# -_-)\(^-^ )
4.it is a magic square proof process Required sum: 1 2 3
4 5 6
7 8 9 = (1 + n^2) * n^2 2 From 1 to n^2 1 n * = (1 + n^2) * n 2 sum of all the numbers sum of each row/column Induction proof: each row add up to correct sum:
1. first row equals the that sum equation
2. sum of each row equals sum of first row 3 9 15 16 22
21 2 8 14 20
19 25 1 7 13
12 18 24 5 6
10 11 17 23 4 3 14 30 31 42
41 2 13 24 40
34 100 1 12 23
22 33 44 10 11
20 21 32 43 4 0*n+3 1*n+4 2*n+n 3*n+1 4*n+2
4*n+1 0*n+2 1*n+3 2*n+4 3*n+n
3*n+4 4*n+n 0*n+1 1*n+2 2*n+3
2*n+2 3*n+3 4*n+4 0*n+n 1*n+1
1*n+n 2*n+1 3*n+2 4*n+3 0*n+4 Decimal notation base-5 notation additional notation 18(base-10) = 3*5+3 = 3*n+3 0*n+1 1*n+2 2*n+3 3*n+4 4*n+n
0*n+2 1*n+3 2*n+4 3*n+n 4*n+1
0*n+3 1*n+4 2*n+n 3*n+1 4*n+2
0*n+4 1*n+n 2*n+1 3*n+2 4*n+3
0*n+n 1*n+1 2*n+2 3*n+3 4*n+4 0*n+3 1*n+4 2*n+n 3*n+1 4*n+2 First row: =(1 + .. + n-1)*n + (1 + 2 + .. + n-1 + n)
=(1+n-1)*(n-1)/2 * n + (1+n)*n/2

=(1+n^2)*n/2 Looks familiar (^. ^) n=5 j= 1 2 3 4 n i=1
2
3
4
n E(i,j)=(j-1)*n + ((i+j-2)mod n + 1). E(i+1,j)=(j-1)*n + ((i+j-1)mod n + 1) two cases: E(4,2)=1*n+n
E(5,2)=1*n+1 E(i,j)>E(i+1,j) 1: E(i,j) = (j-1)*n + ((i+j-2)mod n + 1) = (j-1)*n + n (i+1,j) therefore satisfy (i+j-1)mod n = 0 E(i+1,j)= (j-1)*n + ((i+j-1)mod n + 1) = (j-1)*n + 1 (i, j) satisfy (i+j-2) mod n = n-1 E(i+1,j)=E(i,j)-(n-1) 2: Not satisfy the condition above. E(i,j) = (j-1)*n + ((i+j-2)mod n + 1) E(i+1,j) = (j-1)*n + ((i+j-1)mod n + 1) Example: E(i+1,j)=E(i,j)+1 0*n+1 1*n+2 2*n+3 3*n+4 4*n+n
0*n+2 1*n+3 2*n+4 3*n+n 4*n+1
0*n+3 1*n+4 2*n+n 3*n+1 4*n+2
0*n+4 1*n+n 2*n+1 3*n+2 4*n+3
0*n+n 1*n+1 2*n+2 3*n+3 4*n+4 only 1 case n-1 cases case 1: E(I+1,j)=E(I,j)-(n-1) once
case 2: E(I+1,j)=E(I,j)+1 (n-1) times Induction proof: each row add up to correct sum:
1. first row equals the that sum equation
2. sum of each row equals sum of first row
set w(i) = sum of row i in case 1
set m(i) = sum of row i in case 2

w(i+1)=w(i)-(n-1)
m(i+1)=m(i)+1*(n-1) set s(i) = sum of row i

s(i)=w(i)+m(i)
s(i+1)=w(i+1)+m(i+1)
=w(i)-(n-1)+m(i)+1*(n-1)
=w(i)+m(i) So, s(i)=s(i+1) 1 2 3 4 n
n+3 n+4 n+n n+1 n+2
2n+n 2n+1 2n+2 2n+3 2n+4
3n+2 3n+3 3n+4 3n+n 3n+1
4n+4 4n+n 4n+1 4n+2 4n+3 j=1 2 3 4 n i= 1
2
3
4
n E(i,j) = (i-1)n + ((j+2(i-1)-1)mod n +1)
E(i,j)+1 = (i-1)n + ((j+2(i-1))mod n +1) In case of j+2(i-1)-1 = n - 1 ==> j+2(i-1) = n

E(i,j) = (i-1)n + n, E(i,j)+1 = 1(i-1)n + 1
E(i,j+1)+1 = E(i,j)-(n-1)

In other n-1 cases, E(i,j+1) = E(i,j)+1 set s(j) = sum of column j
w=value of first case
m=sum of other case

s(j)=w+m
s(j+1)=w-(n-1)+m+1*(n-1)
=w+m

so, s(j)=s(j+1) sum of first column
(1+n-1)*(n-1)/2*n+(1+n)*n/2
= (n^2+1)*n/2
equals to the required
sum of a magic square Proof process:
1.induction proof: each row add up to correct sum
2.induction proof: each column add up to correct sum
3.ignore diagonal (# -_-)\(^-^ )
4.it is a magic square E(i,j)=(j-i)*n+(3(i-1)+j-1)mod n + 1) one equation to rule them all! i= j=
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