344
CHAPTER 9. RECOMMENDATION SYSTEMS


2.6
1
1
1
1
1
1
1
1
1


×
 1.617
1
1
1
1
1
1
1
1
1

=


5.204
3.6
3.6
3.6
3.6
2.617
2
2
2
2
2.617
2
2
2
2
2.617
2
2
2
2
2.617
2
2
2
2


Figure 9.14: Replace y by 1.617
The solution for y is y = 17.4/10.76 = 1.617. The improved estimates of U and
V are shown in Fig. 9.14.
We shall do one more change, to illustrate what happens when entries of M
are blank. We shall vary u31, calling it z temporarily. The new U and V are
shown in Fig. 9.15. The value of z aﬀects only the entries in the third row.


2.6
1
1
1
z
1
1
1
1
1


×
 1.617
1
1
1
1
1
1
1
1
1

=


5.204
3.6
3.6
3.6
3.6
2.617
2
2
2
2
1.617z + 1
z + 1
z + 1
z + 1
z + 1
2.617
2
2
2
2
2.617
2
2
2
2


Figure 9.15: u31 becomes a variable z
We can express the sum of the squares of the errors as
2 −(1.617z + 1)
2 +
3 −(z + 1)
2 +
1 −(z + 1)
2 +
4 −(z + 1)
2
Note that there is no contribution from the element in the second column of
the third row, since this element is blank in M. The expression simpliﬁes to
(1 −1.617z)2 + (2 −z)2 + (−z)2 + (3 −z)2
The usual process of setting the derivative to 0 gives us
−2 ×
1.617(1 −1.617z) + (2 −z) + (−z) + (3 −z)

= 0
whose solution is z = 6.617/5.615 = 1.178. The next estimate of the decompo-
sition UV is shown in Fig. 9.16.
✷
9.4.4
Optimizing an Arbitrary Element
Having seen some examples of picking the optimum value for a single element in
the matrix U or V , let us now develop the general formula. As before, assume
