On Erd˝os-Gallai and Havel-Hakimi algorithms

On Erdős-Gallai and Havel-Hakimi algorithms 9
elements of the pairs formed from the tail and from the beginning of the head,
that is
Xi,1 ≤ min(Hhi , Tn − Ti, hi(n − i)). (6)
A similar train of thought results
Xi,2 ≤ min(Hi − Hhi , Tn − Ti, (i − hi)(n − i)). (7)
Xi,3 is at most hi(i − hi) and at most Hi, implying
Xi,3 ≤ min(hi(i − hi), Hi). (8)
Xi,4 is at most hi
2 and at most Hhi , implying
hi
Xi,4 ≤ min( , Hhi ), (9)
2
while Xi,5 is at most i−hi
2 and at most Hi − Hhi , implying
i − hi
Xi,5 ≤ . (10)
2
A requirement is also, that the tail can overrun its capacity, that is
Summing of (6), (7), (8), (9), and (10) results
Xi,1 + Xi,2 ≤ Ti. (11)
Hi ≤ Xi,1 + Xi,2 + Xi,3 + 2Xi,4 + 2Xi,5. (12)
Substituting of (6), (7), (8), (9), and (10) into (12) results (4), while (11) is
equivalent with (5).
The following algorithm executes the test based on Lemma 4.
Input. n: the number of vertices (n ≥ 1);
b = (b1, . . . , bn): an n-even sequence, accepted by Binomial-Test;
H = (H1, . . . , Hn): Hi the sum of the first i elements of b;
T = (T1, . . . , Tn): Ti the sum of the last n − i elements of b.
Output. L: logical variable (L = False signals,that b is not graphical, while
L = True shows, that the test could nor decide, whether b is graphical).
Working variables. i: cycle variable;
h: the actual value of hi X = (X1, X2, X3, X4, X5): Xj is the value of the actual
Xi,j of the end of the head.