Destination T reached (820). Stop.
Unvisited min = D(510). Current = D. Neighbors: A(no), B(no), C(510+120=630 vs 530 no), F(510+300=810), T(510+500=1010). Update T tentative = 1010. Visited S,A,B,D.
Unvisited min = F(730). Current = F. Neighbors: D(no), E(no), T(730+90=820 vs 1010 → update T=820). Visited add F.
I used Google Maps for weights and could extend to Excel/Python. graph theory math ia
1. Introduction Aim: To determine the most efficient (shortest) route for a delivery driver in a local suburban network using graph theory, and to compare the effectiveness of Dijkstra’s algorithm against simple visual inspection.
(sort edges by weight, add if no cycle):
Unvisited min = C(530). Current = C. Neighbors: A(no), B(no), D(no), E(530+250=780 vs 630 no). Visited S,A,B,D,C. Destination T reached (820)
Unvisited min = B(350). Current = B. Neighbors: S(no), A(350+150=500 vs 200 no), C(350+180=530 vs 600 → update C=530), D(350+220=570 vs 510 no), E(350+280=630). Visited S,A,B.
Unvisited min = E(630). Current = E. Neighbors: B(no), C(no), F(630+100=730 vs 810 → update F=730). Visited add E.
I defined terms clearly, used consistent notation (( G=(V,E) )), and showed step-by-step tables. Current = D
Unvisited min = A(200). Current = A. Neighbors: S(200+200 no better), B(200+150=350 vs current 350 tie), C(200+400=600), D(200+310=510). Update: C=600, D=510. Visited S,A.
Current = S (distance 0). Neighbors: A(200), B(350). Update: A=200, B=350. Visited = S.
| Edge | Weight | Edge | Weight | |------|--------|------|--------| | S–A | 200 | B–C | 180 | | S–B | 350 | C–D | 120 | | A–B | 150 | C–E | 250 | | A–C | 400 | D–F | 300 | | B–D | 220 | E–F | 100 | | B–E | 280 | F–T | 90 | | A–D | 310 | D–T | 500 (direct but long) |