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M

B

L

AX

48

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10 19

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710

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Floyds Algorithm can be used to help solve Travelling Salesman

and other shortest routes problems.

Floyds uses a matrix form.

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D( 0 ) R( 0 )

This is the distance matrix.

It initially shows the direct

distance between one vertex

and the others. The infinity sign

denotes no direct route.

This is the route matrix.

It will eventually show the

next vertex that needs to be

taken on route to finding theshortest route to another vertex.

To view the process step by step click here

To view the completed matrices click here

A M L B X

A 23 10 18

M 23 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

g

g

g

g

g g

g

To look back at the network click here:

g

g

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D( 0 ) R( 0 )

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

Click here to see step:

A M L B X

A 23 10 18

M 23 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

g g

g g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

23 + 23 = 46, less than so

change.

A M L B X

A 23 10 18

M 23 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

g

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

23 + 23 = 46, less than so

change.

A M L B X

A 23 10 18

M 23 46 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

g

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

10 + 23 = 33, so we leave this

item

A M L B X

A 23 10 18

M 23 46 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

When is involved we leave

the item.g

A M L B X

A 23 10 18

M 23 46 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

18 + 23 = 41, so we replace the

item.

A M L B X

A 23 10 18

M 23 46 10 48

L 10 10 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

18 + 23 = 41, so we replace the

item.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 19 7

B 48 19 20

X 18 7 20

C A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

23 + 10 = 33, so we leave thisitem.

48 g41

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 19 7

B 48 19 20

X 18 7 20

C A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

10 + 10 = 20, so we replace thisitem.

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

Infinity is involved so we leavethis item.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

18 + 10 = 28 which is more than7 so we leave this item.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

Infinity is involved so we leavethis item.

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

Infinity is involved so we leavethis item.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

Infinity is involved so we leavethis item.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

Infinity is involved so we leavethis item.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

We replace this item with23 + 18 = 41

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

We replace this item with23 + 18 = 41

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g

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D( 0 ) R( 0 )

Click here to see step:

10 + 18 = 28, 7 is less than thisso leave it.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g

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D( 0 ) R( 0 )

Click here to see step:

Infinity is involved so leave thisitem.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

g

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D( 0 ) R( 0 )

Click here to see step:

18 + 18 = 36, so replace thisitem.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5g

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

18 + 18 = 36, so replace thisitem.

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

We highlight the first column and row of the Distance matrix and compare

all other items with the sum of the items highlighted in the same row and

column.

If the sum is less than the item then it should be replaced with the sum.

g g

g g

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D( 0 ) R( 0 )

Click here to see step:

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

A M L B X

A 1 2 3 4 5

M 1 2 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

g g

g g

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R( 0 )

Click here to see step:

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

A M L B X

A 1 2 3 4 5

M 1 1 3 4 5

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

D( 0 )

g g

g g

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R( 0 )

Click here to see step:

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

D( 0 )

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 3 4 5

B 1 2 3 4 5

X 1 2 3 4 5

g g

g g

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R( 0 )

Click here to see step:

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

D( 0 )

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 1 4 5

B 1 2 3 4 5

X 1 2 3 4 5

g g

g g

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R( 0 )

Click here to see step:

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

D( 0 )

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 1 4 5

B 1 2 3 4 5

X 1 1 3 4 5

g g

g g

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R( 0 )

Click here to see step:

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

D( 0 )

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 1 4 5

B 1 2 3 4 5

X 1 1 3 4 1

g g

g g

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We have now completed one iteration. We rename the new matrices:

Click here to see step:

Subsequent iterations are now shown completed:

Click here to see final matrices:

R( 1 )D( 1 )

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 1 4 5

B 1 2 3 4 5

X 1 1 3 4 1

g g

g g

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Click here to see step:

Subsequent iterations are now shown completed:

Click here to see final matrices:

R( 1 )D( 1 )

A M L B X

A 23 10 18

M 23 46 10 48 41

L 10 10 20 19 7

B 48 19 20

X 18 41 7 20 36

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 1 4 5

B 1 2 3 4 5

X 1 1 3 4 1

g

g g

g

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D( 1 ) R( 1 )

The items have been altered accordingly:

Click here to see step:

Subsequent iterations are now shown completed:

Click here to see final matrices:

A M L B X

A 46 23 10 71 18

M 23 46 10 48 41

L 10 10 20 19 7

B 71 48 19 96 20

X 18 41 7 20 36

A M L B X

A 1 2 3 4 5

M 1 1 3 4 1

L 1 2 1 4 5

B 1 2 3 4 5

X 1 1 3 4 1

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D( 2 ) R( 2 )

We can now rename the matrices:

Click here to see step:

Click here to see final matrices:

A M L B X

A 46 23 10 71 18

M 23 46 10 48 41

L 10 10 20 19 7

B 71 48 19 96 20

X 18 41 7 20 36

A M L B X

A 2 2 3 2 5

M 1 1 3 4 1

L 1 2 1 4 5

B 2 2 3 2 5

X 1 1 3 4 1

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D( 2 ) R( 2 )

Next iteration:

Click here to see step:

Click here to see final matrices:

A M L B X

A 46 23 10 71 18

M 23 46 10 48 41

L 10 10 20 19 7

B 71 48 19 96 20

X 18 41 7 20 36

A M L B X

A 2 2 3 2 5

M 1 1 3 4 1

L 1 2 1 4 5

B 2 2 3 2 5

X 1 1 3 4 1

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D( 2 ) R( 2 )

Next iteration, the items are altered appropriately :

Click here to see step:

Click here to see final matrices:

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 20 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 1 4 5

B 3 3 3 3 5

X 3 3 3 4 3

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D( 3 ) R( 3 )

The matrices are renamed :

Click here to see step:

Click here to see final matrices:

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 20 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 1 4 5

B 3 3 3 3 5

X 3 3 3 4 3

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The next iteration :

Click here to see step:

Click here to see final matrices:

D( 3 ) R( 3 )

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 20 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 1 4 5

B 3 3 3 3 5

X 3 3 3 4 3

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In this iteration, you do not need to change any of the items, so we go

onto the next iteration :

Click here to see step:

Click here to see final matrices:

D( 3 ) R( 3 )

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 20 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 1 4 5

B 3 3 3 3 5

X 3 3 3 4 3

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Click here to see step:

Click here to see final matrices:

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 20 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 1 4 5

B 3 3 3 3 5

X 3 3 3 4 3

D( 4 ) R( 4 )

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Click here to see step:

Click here to see final matrices:

There is only one item that we need to change in this

case:

D( 4 ) R( 4 )

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 14 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 1 4 5

B 3 3 3 3 5

X 3 3 3 4 3

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These are the final matrices, remember that you can

use them to redraw the original network. You can then

use this to help us solve travelling salesman

problems:

Click here to see new network:

D( 5 ) R( 5 )

A M L B X

A 20 20 10 29 17

M 20 20 10 29 17

L 10 10 14 19 7

B 29 29 19 38 20

X 17 17 7 20 14

A M L B X

A 3 3 3 3 3

M 3 3 3 3 3

L 1 2 5 4 5

B 3 3 3 3 5

X 3 3 3 4 3

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page 1

M

B

L

AX

29

20

10 19

17

710

20

29

17

This network now gives you a better idea of the quickest routes.

Click below to try a question:

The route matrix gives us an idea about the next vertex to visit on route -

1 represents A, 2 - M, etc.

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1

5

4

3

2

75 35

32

15

40

30

70

Try this one! Click below when you have completed it to

check the answers:

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1 2 3 4 5

1 60 30 40 45 77

2 30 30 50 15 47

3 40 50 70 35 67

4 45 15 35 30 32

5 77 47 67 32 64

1 2 3 4 5

1 2 2 3 2 2

2 1 4 4 4 4

3 1 4 4 4 4

4 2 2 3 2 5

5 4 4 4 4 4

D( 5 ) R( 5 )

These are the completed matrices. Are yours correct?

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1

5

4

3

2

20 15 12

35

50

50

10

Qu2.

Answers.

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D( 5 ) R( 5 )

These are the completed matrices. Are yours correct?

1 2 3 4 5

1 100 50 50 65 60

2 50 40 20 35 30

3 50 20 20 15 10

4 65 35 15 24 12

5 60 30 10 12 20

1 2 3 4 5

1 2 2 3 3 3

2 1 3 3 4 3

3 1 2 5 4 5

4 3 2 3 5 5

5 3 3 3 4 3

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1

4

3

2

8

3 5

4

3

2

Qu2.

Answers.

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D( 4 ) R( 4 )

These are the completed matrices. Are yours correct?

1 2 3 4

1 6 3 6 4

2 3 6 5 3

3 6 5 4 2

4 4 3 2 4

1 2 3 4

1 2 2 4 4

2 1 1 3 4

3 4 2 4 4

4 1 2 3 3