Find the shortest distance route starting from City A and visiting all the remaining 5 cities just once and then return back to original starting point City A. There are 15 input distances and 120 possible combination to get the shortest route distance.
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The reverse direction is also correct: Solution 109, AFEDCBA
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Change distance b to e from 33 to 3. What is the shortest route? Answer = 42, solution 13, ABECDFA or the reverse direction solution 105, AFDCEBA
Change distance b to d from 44 to 4. What is the shortest route? Answer = 41, solution 12, ABDCEFA or the reverse direction solution 112, AFECDBA
Change distance a to b from 5 to 59. What is the shortest route? Answer = 86, solution 59, ADCBEFA or the reverse direction solution 113, AFEBCDA
1e9 in the input box.
Tip: Use 1e9 to represent missing connections—this ensures the algorithm treats them as non-viable paths.
This is interactive so try it. if the edge or line has no distance enter 1e9 (meaning 1,000,000,000) there is no available road to connect the two nodes. For example go to Drawing #1 example node A to node D no line so you enter 1e9 as input. (line is also called as edge, path, circuit in graph of traveling salesman analysis).
In the source material, the term "line" is used informally and is said to be interchangeable with edge, path, and circuit in graph of traveling salesman analysis. However, in formal graph theory, these terms have distinct meanings:
While the source uses "line" as a general term, it's important to use precise terminology when teaching or analyzing graphs to avoid confusion.
ENTER YOUR NEW DATA HERE
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a to b |
a to e |
a to d |
a to c |
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b to e |
b to c |
c to d |
c to e |
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d to e |
b to d |
e to f |
a to f |
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b to f |
d to f |
c to f |
shortest distance |
In a travelling salesman problem with 6 cities, the goal is to find the shortest possible route that starts at City A, visits each of the remaining five cities exactly once, and returns to City A.
The number of possible directed tours is given by:
P(n) = (n − 1)!
For six cities:
P(6) = 5! = 120
This matches the 120 possible combinations shown in the calculator (each representing a distinct ordered route).
In an undirected TSP, a route and its reverse direction represent the same path. To remove these duplicates, the number of unique tours is:
U(n) = (n − 1)! / 2
For six cities:
U(6) = 5! / 2 = 60
This means there are 60 unique undirected tours when reverse paths are treated as identical.
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solution 2 A B C D F E A |
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solution 18 A B E F C D A |
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solution 63 A D E B C F A |
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solution 98 A F C D B E A |
solution 99 A F C E B D A |
solution 100 A F C E D B A |
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solution 102 A F C B D E A |
solution 103 A F D E B C A |
solution 104 A F D E C B A |
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solution 106 A F D C B E A |
solution 107 A F D B E C A |
solution 108 A F D B C E A |
| solution 109 A F E D C B A |
solution 110 A F E D B C A |
solution 111 A F E C B D A |
solution 112 A F E C D B A |
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solution 114 A F E B D C A |
solution 115 A F B C E D A |
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solution 118 A F B D C E A |
solution 119 A F B E D C A |
solution 120 A F B E C D A |
| Input A | Input B | Input C | Input D |
|---|---|---|---|
| AB =4 | AE =1e9 | AD =1e9 | AC =7 |
| BE =1e9 | BC =4 | CD =5 | CE =9 |
| DE =3 | BD =1e9 | EF =5 | AF =5 |
| BF =1e9 | DF =1e9 | CF =1e9 |
| Input A | Input B | Input C | Input D |
|---|---|---|---|
| AB =4 | AE =1e9 | AD =6 | AC =7 |
| BE =1e9 | BC =4 | CD =5 | CE =9 |
| DE =3 | BD =6 | EF =5 | AF =5 |
| BF =1e9 | DF =5 | CF =1e9 |
| Input A | Input B | Input C | Input D |
|---|---|---|---|
| AB =5 | AE =30 | AD =1e9 | AC =1e9 |
| BE =1e9 | BC =20 | CD =2 | CE =1e9 |
| DE =2 | BD =1e9 | EF =15 | AF =7 |
| BF =10 | DF =9 | CF =15 |