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some suggested a thread in which the algorithm multiplies the 2 matrices with lowest values first. Mine is different: it divides by parenthesis the 2 matrices. And continues to the next section.

The problem is: finding the most efficient way to multiply a series of matrices together, using an algorithm of some sort, and comparing these algorithms to find the most efficient one.

The dynamic algorithm approach works at a time complexity of theta of N^3.

My question is, what is the runtime of this algorithm:

A= 5x2 B= 2x7 C= 7x3

1) First, find the matrix with the lowest dimension ( a matrix which has the lower number from the rows or columns of all the matrices).

*If the lowest number is in the last dimension, it is the same like putting the entire sequence in parentheses.

*There is no guidance as to what to do if the lowest number appears twice in the sequence, so I suggest the algorithm just takes 1 randomly.

2) Then divide the sequence to 2: (A)(B•C). The parentheses will close on the matrix from the right and open on the next matrix. This way they will divide the series to 2 parts.

Then repeat the process for the 2 parts. Stop when you have 1 (or 2) matrices in the sequence.

Is this algorithm optimal? It has to be better than N^3 (the usual algorithm)

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  • $\begingroup$ cs.stackexchange.com/questions/48280/… $\endgroup$ – eru-cs Dec 15 '19 at 11:38
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    $\begingroup$ Does this answer your question? Matrix Chain Multiplication Greedy Approach $\endgroup$ – Yuval Filmus Dec 15 '19 at 11:53
  • $\begingroup$ No. It actually gives a good example to my algorithm.. $\endgroup$ – John Appleboim Dec 15 '19 at 12:28
  • $\begingroup$ Does "find the lowest number in the lines / rows column" mean finding lowest number from the numbers of rows or columns? You find 2 in your example, right? What if lowest number appears twice, e.g. A=3x2 B=2x2 C=2x5? $\endgroup$ – Ryoji Dec 15 '19 at 14:47
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    $\begingroup$ I hope I understand your approach. What about 1x2, 2x4, 4x2? $\endgroup$ – Hendrik Jan Dec 15 '19 at 16:28
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Consider the product $ABC$, where

  • $A$ is $5\times 2$
  • $B$ is $2\times 3$
  • $C$ is $3\times 100$

Your algorithm first computes $BC$ (600 products) and then $A(BC)$ (1000 products), for a total of 1600 products.

The optimal solution first computes $AB$ (30 products) and then $(AB)C$ (1500 products), for a total of only 1530 products.


Suppose that $A$ is $a\times b$, that $B$ is $b \times c$, and that $C$ is $c\times d$. There are two possibilities:

  • Compute $AB$ and then $(AB)C$: cost is $abc + acd = ac(b+d)$.
  • Compute $BC$ and then $A(BC)$: cost is $bcd + abd = bd(a+c)$.

The first option is preferable if $$ \frac{b+d}{bd} < \frac{a+c}{ac} $$ or equivalently $$ \frac{1}{b} + \frac{1}{d} < \frac{1}{a} + \frac{1}{c}. $$ For example, above we have $a = 5$, $b = 2$, $c = 3$, $d = 100$. Since $$ \frac{1}{2} + \frac{1}{100} < \frac{1}{5} + \frac{1}{3}, $$ the order $(AB)C$ is preferable.

Your decision procedure would prefer this order if $c < b$, that is, if $$ \frac{1}{b} < \frac{1}{c}. $$ Your condition misses the contribution of $a,d$.

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  • $\begingroup$ Thank you! Is there a more general rule that applies to longer chains of matrices? Very insightful! $\endgroup$ – John Appleboim Dec 17 '19 at 14:03
  • $\begingroup$ You can do the calculation yourself and see whether a simple condition arises. $\endgroup$ – Yuval Filmus Dec 17 '19 at 15:06
  • $\begingroup$ Nice formulas! And I was right about 1x2, 2x4, 4x2. $\endgroup$ – Hendrik Jan Dec 18 '19 at 14:15

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