Take se[a,e].

Step 1:-
Take se[a,e],sv{a}.
pc(se[a,e],sv{a}) = (a,e) = PC1(HC7)
pc(se[a,e],sv{a}) = (d,a,e) = PC2(HC7)
pc(se[a,e],sv{a}) = (c,d,a,e) = PC3(HC7)
pc(se[a,e],sv{a}) = (b,c,d,a,e) = PC4(HC7)
PC4(HC7) was formed by skipping the edge [c,e]
hc(se[a,e],sv{a}) = (b,c,d,a,e) = HC7
w(HC7) = 10+7+6+9+5 = 37

Step 2:-
Take se[a,e],sv{e}.
pc(se[a,e],sv{e}) = (a,e) = PC1(HC8)
pc(se[a,e],sv{e}) = (b,c,d) = PC2(HC8)
pc(se[a,e],sv{e}) = (b,c,d,a) = PC3(HC8)
pc(se[a,e],sv{e}) = (b,c,d,a,e) = PC4(HC8)
hc(se[a,e],sv{e}) = (b,c,d,a,e) = HC8
w(HC8)=10+5+9+6+7=37
No edges were skipped while forming HC8.

Take se[a,c].

Step 1:-
Take se[a,c],sv{a}.
pc(se[a,c],sv{a}) = (a,c) = PC1(HC9)
pc(se[a,c],sv{a}) = (d,a,c) = PC2(HC9)
pc(se[a,c],sv{a}) = (e,d,a,c) = PC3(HC9)
PC3(HC9) was formed by skipping the edge [c,d]
pc(se[a,c],sv{a}) = (b,e,d,a,c) = PC4(HC9)
hc(se[a,c],sv{a}) = (b,e,d,a,c) = HC9
w(HC9) = 12+7+11+5+9 = 44

Step 2:-
Take se[a,c],sv{c}.
pc(se[a,c],sv{c}) = (a,c) = PC1(HC10)
pc(se[a,c],sv{c}) = (a,c,d) = PC2(HC10)
pc(se[a,c],sv{c}) = (a,c,d,e) = PC3(HC10)
PC3(HC10) was formed by skipping the edge [a,d]
pc(se[a,c],sv{c}) = (a,c,d,e,b) = PC4(HC10)
hc(se[a,c],sv{c}) = (a,c,d,e,b) = HC10
HC10 was formed by skipping the edges [b,c] and [b,d]
w(HC10)=10+6+11+5+14=46

Stage 2 :-

The edges which were skipped in Stage 1 were [b,c] , [b,d] , [c,e] , [c,d] and [a,d] . Now sorting the skipped edges according to the weights , we get the sorted order as [c,d] , [a,d] , [c,e] , [b,c] and [b,d]. Edge [c,e] was processed in the beginning as the first starter edge. [c,d] and [b,c] was processed in Stage 1. So now the edges left for processing in Stage 2 are [a,d] and [b,d].

Take se[a,d].

Step 1:-
Take se[a,d],sv{a}.
pc(se[a,d],sv{a}) = (a,d) = PC1(HC11)
pc(se[a,d],sv{a}) = (e,a,d) = PC2(HC11)
pc(se[a,d],sv{a}) = (b,e,a,d) = PC3(HC11)
pc(se[a,d],sv{a}) = (c,b,e,a,d) = PC4(HC11)
hc(se[a,d],sv{a}) = (c,b,e,a,d) = HC11
w(HC11) = 7+10+5+9+6=37
No edges were skipped while forming HC11.

Step 2:-
Take se[a,d],sv{d}.
pc(se[a,d],sv{d}) = (a,d) = PC1(HC12)
pc(se[a,d],sv{d}) = (a,d,c) = PC2(HC12)
pc(se[a,d],sv{d}) = (a,d,c,e) = PC3(HC12)
pc(se[a,d],sv{d}) = (a,d,c,e,b) = PC4(HC12)
hc(se[a,d],sv{d}) = (a,d,c,e,b) = HC12
HC12 was formed by skipping the edges [b,c] and [b,d]
w(HC12)=7+6+8+5+14=40

Take se[b,d].

Step 1:-
Take se[b,d],sv{b}.
pc(se[b,d],sv{b}) = (b,d) = PC1(HC14)
pc(se[b,d],sv{b}) = (e,b,d) = PC2(HC14)
pc(se[b,d],sv{b}) = (c,e,b,d) = PC3(HC14)
pc(se[b,d],sv{b}) = (a,c,e,b,d) = PC4(HC14)
PC4(HC14) was formed by skipping the edges [c,d] and [b,c]
hc(se[b,d],sv{b}) = (a,c,e,b,d) = HC14
w(HC14) = 13+5+8+12+7=45

Step 2:-
Take se[b,d],sv{d}.
pc(se[b,d],sv{d}) = (b,d) = PC1(HC15)
pc(se[b,d],sv{d}) = (b,d,c) = PC2(HC15)
pc(se[b,d],sv{d}) = (b,d,c,e) = PC3(HC15)
pc(se[b,d],sv{d}) = (b,d,c,e,a) = PC4(HC15)
PC4(HC15) was formed by skipping the edge [b,e]
hc(se[b,d],sv{d}) = (b,d,c,e,a) = HC15
HC15 was formed by skipping the edge [a,d] and [a,c]
w(HC15)=13+6+8+10+14=51

Stage 3 :-

The edges which were skipped in Stage 2 were [b,c] , [b,d] , [c,d] , [b,e] , [a,d] and [a,c] . Now sorting the skipped edges according to the weights , we get the sorted order as [b,e], [c,d] , [a,d] , [b,c] , [a,c] and [b,d]. Edges [c,d] , [b,c] and [a,c] was processed in Stage 1. Edges [a,d] and [b,d] was processed in Stage 2 . So now the edge left for processing in Stage 3 is [b,e].

Take se[b,e].

Step 1:-
Take se[b,e],sv{b}.
pc(se[b,e],sv{b}) = (c,b,e) = PC1(HC16)
pc(se[b,e],sv{b}) = (c,b,e) = PC2(HC16)
pc(se[b,e],sv{b}) = (d,c,b,e) = PC3(HC16)
pc(se[b,e],sv{b}) = (a,d,c,b,e) = PC4(HC16)
hc(se[b,e],sv{b}) = (b,e) = HC16
w(HC16) = 5+9+6+7+10=37
No edges were skipped while forming HC16.

Continued in "Travelling Salesman Problem - Part 5"