The system of equations (64 equations -12 unknown) solution

ZeusTheMunja · January 17th, 2013, 08:35 AM

Regards,
when programming an application I faced with the following system of equations

Code:

n1=i1*i5*i9
n2=i1*i5*i10
n3=i1*i5*i11
n4=i1*i5*i12
n5=i1*i6*i9
n6=i1*i6*i10
n7=i1*i6*i11
n8=i1*i6*i12
n9=i1*i7*i9
n10=i1*i7*i10
n11=i1*i7*i11
n12=i1*i7*i12
n13=i1*i8*i9
n14=i1*i8*i10
n15=i1*i8*i11
n16=i1*i8*i12
n17=i2*i5*i9
n18=i2*i5*i10
n19=i2*i5*i11
n20=i2*i5*i12
n21=i2*i6*i9
n22=i2*i6*i10
n23=i2*i6*i11
n24=i2*i6*i12
n25=i2*i7*i9
n26=i2*i7*i10
n27=i2*i7*i11
n28=i2*i7*i12
n29=i2*i8*i9
n30=i2*i8*i10
n31=i2*i8*i11
n32=i2*i8*i12
n33=i3*i5*i9
n34=i3*i5*i10
n35=i3*i5*i11
n36=i3*i5*i12
n37=i3*i6*i9
n38=i3*i6*i10
n39=i3*i6*i11
n40=i3*i6*i12
n41=i3*i7*i9
n42=i3*i7*i10
n43=i3*i7*i11
n44=i3*i7*i12
n45=i3*i8*i9
n46=i3*i8*i10
n47=i3*i8*i11
n48=i3*i8*i12
n49=i4*i5*i9
n50=i4*i5*i10
n51=i4*i5*i11
n52=i4*i5*i12
n53=i4*i6*i9
n54=i4*i6*i10
n55=i4*i6*i11
n56=i4*i6*i12
n57=i4*i7*i9
n58=i4*i7*i10
n59=i4*i7*i11
n60=i4*i7*i12
n61=i4*i8*i9
n62=i4*i8*i10
n63=i4*i8*i11
n64=i4*i8*i12

The system of equations resulting from these matrix, with permutation of all elements

Code:

i1   i5    i9
i2   i6    i10
i3   i7    i11
i4   i8    i12

In the above equations, n (n1, n2 ..., n64) are familiar values, while the i (i1, i2, ..., i12) are unknown values

The whole day i am trying to solve this, but no success.
Is it possible to solve this system of equations?
if so. How?
Thanks in advance

mathbalarka · January 17th, 2013, 11:27 PM

If the "*" sign in your system of equations is multiplication, then there is no general method for these since the system is non-linear.

Matrices are being used to solve a system of linear equations.

JJacquelin · January 18th, 2013, 02:50 AM

Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.

mathbalarka · January 18th, 2013, 04:15 AM

Quote:

Originally Posted by JJacquelin

Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.

Yes, that could work. Nice, JJacquelin !

ZeusTheMunja · January 18th, 2013, 04:50 AM

Quote:

Originally Posted by JJacquelin

Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.

thanks very much, i was forgot to use logarithms

now i have next system

Code:

n1=log(i1)+log(i5)+log(i9)
n2=log(i1)+log(i5)+log(i10)
n3=log(i1)+log(i5)+log(i11)
n4=log(i1)+log(i5)+log(i12)
n5=log(i1)+log(i6)+log(i9)
n6=log(i1)+log(i6)+log(i10)
n7=log(i1)+log(i6)+log(i11)
n8=log(i1)+log(i6)+log(i12)
n9=log(i1)+log(i7)+log(i9)
n10=log(i1)+log(i7)+log(i10)
n11=log(i1)+log(i7)+log(i11)
n12=log(i1)+log(i7)+log(i12)
n13=log(i1)+log(i8)+log(i9)
n14=log(i1)+log(i8)+log(i10)
n15=log(i1)+log(i8)+log(i11)
n16=log(i1)+log(i8)+log(i12)
n17=log(i2)+log(i5)+log(i9)
n18=log(i2)+log(i5)+log(i10)
n19=log(i2)+log(i5)+log(i11)
n20=log(i2)+log(i5)+log(i12)
n21=log(i2)+log(i6)+log(i9)
n22=log(i2)+log(i6)+log(i10)
n23=log(i2)+log(i6)+log(i11)
n24=log(i2)+log(i6)+log(i12)
n25=log(i2)+log(i7)+log(i9)
n26=log(i2)+log(i7)+log(i10)
n27=log(i2)+log(i7)+log(i11)
n28=log(i2)+log(i7)+log(i12)
n29=log(i2)+log(i8)+log(i9)
n30=log(i2)+log(i8)+log(i10)
n31=log(i2)+log(i8)+log(i11)
n32=log(i2)+log(i8)+log(i12)
n33=log(i3)+log(i5)+log(i9)
n34=log(i3)+log(i5)+log(i10)
n35=log(i3)+log(i5)+log(i11)
n36=log(i3)+log(i5)+log(i12)
n37=log(i3)+log(i6)+log(i9)
n38=log(i3)+log(i6)+log(i10)
n39=log(i3)+log(i6)+log(i11)
n40=log(i3)+log(i6)+log(i12)
n41=log(i3)+log(i7)+log(i9)
n42=log(i3)+log(i7)+log(i10)
n43=log(i3)+log(i7)+log(i11)
n44=log(i3)+log(i7)+log(i12)
n45=log(i3)+log(i8)+log(i9)
n46=log(i3)+log(i8)+log(i10)
n47=log(i3)+log(i8)+log(i11)
n48=log(i3)+log(i8)+log(i12)
n49=log(i4)+log(i5)+log(i9)
n50=log(i4)+log(i5)+log(i10)
n51=log(i4)+log(i5)+log(i11)
n52=log(i4)+log(i5)+log(i12)
n53=log(i4)+log(i6)+log(i9)
n54=log(i4)+log(i6)+log(i10)
n55=log(i4)+log(i6)+log(i11)
n56=log(i4)+log(i6)+log(i12)
n57=log(i4)+log(i7)+log(i9)
n58=log(i4)+log(i7)+log(i10)
n59=log(i4)+log(i7)+log(i11)
n60=log(i4)+log(i7)+log(i12)
n61=log(i4)+log(i8)+log(i9)
n62=log(i4)+log(i8)+log(i10)
n63=log(i4)+log(i8)+log(i11)
n64=log(i4)+log(i8)+log(i12)

but I have still not solve this system

Denis · January 18th, 2013, 09:20 PM

Perhaps this will answer your question:

a+c+e = 14
a+c+f = 11
a+d+e = 13
a+d+f = 10
b+c+e = 13
b+c+f = 10
b+d+e = 12
b+d+f = 9

8 equations, 6 unknowns...can you solve?

JJacquelin · January 18th, 2013, 10:56 PM

Hi ZeusTheMunja !
You have not yet made the change of unknown :
x1 = log(i1)
x2 = log(i2)
etc.
Then, you have to check if it is possible that the system of 64 linear equations could have a solution (x1, x2, ..., x12), since they are too many equations compared to the number of unknowns.

Denis · January 19th, 2013, 08:19 AM

Quote:

Originally Posted by JJacquelin

You have not yet made the change of unknown :
x1 = log(i1)
x2 = log(i2)

Can't see how that will help...this equation system that I gave as shorter example (but similar):
a+c+e = 14
a+c+f = 11
a+d+e = 13
a+d+f = 10
b+c+e = 13
b+c+f = 10
b+d+e = 12
b+d+f = 9

...has 27 solutions keeping the variables > 0 ; (a,b,c,d,e,f):
1: 2,1,3,2,9,6
2: 2,1,4,3,8,5
...
26: 7,6,3,2,4,1
27: 8,7,2,1,4,1

ZeusTheMunja · January 20th, 2013, 01:48 AM

Denis, that is exactly what i want. All possible solutions.
Can you tell me how you solve it. In my case all variables are > 0

Denis · January 20th, 2013, 09:11 AM

What you are asking is not clear...are you simply writing a program to
result in obtaining 64 multiplications each of 3 integers, given 12 integers?

The 12 integers would be in an array, and you need the results of multiplying
all combinations of: [element 1 to 4] * [element 5 to 8] * [element 9 to 12]?
This would be 64 solutions: 4 * 4 * 4 = 64.

If that's all you're after, then this simple program will handle:
.................................................. .................................................. .....................
Open array N(64), I(12) : array N receives the solutions, array I contains the 12 integers

k = 0 : this variable will increase by 1 at each calculation, in order to place result in array N

loop a from 1 to 4
loop b from 5 to 8
loop c from 9 10 12

k = k + 1
N(k) = I(a) * I(b) + I(c)

print N(k) , I(a) , I(b) , I(c) : if you want results printed as you go along

endloops
.................................................. .................................................. ....................................
As example, if I(2) = 7, I(7) = 2 and I(10) = 4, and if a = 2, b = 7 and c = 10
then this will be result: N(k) = 7 * 2 * 4 = 56

If that's not what you're after, then post problem again CLEARLY,
and give an actual example.

January 17th, 2013, 11:27 PM	#2
mathbalarka Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory	Re: The system of equations (64 equations -12 unknown) solut If the "*" sign in your system of equations is multiplication, then there is no general method for these since the system is non-linear. Matrices are being used to solve a system of linear equations.
$mathbalarka is offline$

January 18th, 2013, 02:50 AM	#3
JJacquelin Senior Member Joined: Aug 2011 Posts: 333 Thanks: 8	Re: The system of equations (64 equations -12 unknown) solut Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.
$JJacquelin is offline$

January 18th, 2013, 09:20 PM	#6
Denis Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,406 Thanks: 638	Re: The system of equations (64 equations -12 unknown) solut Perhaps this will answer your question: a+c+e = 14 a+c+f = 11 a+d+e = 13 a+d+f = 10 b+c+e = 13 b+c+f = 10 b+d+e = 12 b+d+f = 9 8 equations, 6 unknowns...can you solve?
$Denis is online now$

January 18th, 2013, 10:56 PM	#7
JJacquelin Senior Member Joined: Aug 2011 Posts: 333 Thanks: 8	Re: The system of equations (64 equations -12 unknown) solut Hi ZeusTheMunja ! You have not yet made the change of unknown : x1 = log(i1) x2 = log(i2) etc. Then, you have to check if it is possible that the system of 64 linear equations could have a solution (x1, x2, ..., x12), since they are too many equations compared to the number of unknowns.
$JJacquelin is offline$

January 20th, 2013, 01:48 AM	#9
ZeusTheMunja Newbie Joined: Jan 2013 Posts: 8 Thanks: 0	Re: The system of equations (64 equations -12 unknown) solut Denis, that is exactly what i want. All possible solutions. Can you tell me how you solve it. In my case all variables are > 0
$ZeusTheMunja is offline$

log(a3/bc) log(b3/bc) log(c3/ab) formula

log a3 log a2

January 20th, 2013, 09:11 AM	#10
Denis Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,406 Thanks: 638	Re: The system of equations (64 equations -12 unknown) solut What you are asking is not clear...are you simply writing a program to result in obtaining 64 multiplications each of 3 integers, given 12 integers? The 12 integers would be in an array, and you need the results of multiplying all combinations of: [element 1 to 4] * [element 5 to 8] * [element 9 to 12]? This would be 64 solutions: 4 * 4 * 4 = 64. If that's all you're after, then this simple program will handle: .................................................. .................................................. ..................... Open array N(64), I(12) : array N receives the solutions, array I contains the 12 integers k = 0 : this variable will increase by 1 at each calculation, in order to place result in array N loop a from 1 to 4 loop b from 5 to 8 loop c from 9 10 12 k = k + 1 N(k) = I(a) * I(b) + I(c) print N(k) , I(a) , I(b) , I(c) : if you want results printed as you go along endloops .................................................. .................................................. .................................... As example, if I(2) = 7, I(7) = 2 and I(10) = 4, and if a = 2, b = 7 and c = 10 then this will be result: N(k) = 7 * 2 * 4 = 56 If that's not what you're after, then post problem again CLEARLY, and give an actual example.
$Denis is online now$

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