, Now it straightforward to verify that u does not receive share according to any other rule. Furthermore, as |I(u)| = 3 and |I(u + 3)| ? 3
3 Let C be an identifying code in C n (1, 3) and u ? C be a codeword such that u does not receive share according to any of the previous rules ,
Observe first that if u + 2 is a codeword, then we are immediately done since at least two of the I-sets I(u ? 1), I(u + 1) and I(u + 3) consists of at least three codewords implying s s (u) ? s(u) ? 1 + 2 · 1/2 + 2 · 1/3 = 8/3 = 11/4 ? 2/24. The same argument also applies for u ? 2 ? C. Hence, we may assume that u ? 2 and u + 2 do not belong to C. Now the proof divides into the following cases depending on the number of codewords in I(u): ? Suppose first that |I(u)| = 1, i.e., I(u) = {u}. The previous observation taken into account ,
, Consider first the case with I(u) = {u ? 3, u}. If now u + 4 ? C, then |I(u ? 3)| ? 3 and |I(u + 3)| ? 3 since I(u ? 3) = I(u) and I(u + 1) = I(u + 3) and we are done as s s (u) ? s(u) ? 1 + 2 · 1/2 + 2 · 1/3 = 11/4 ? 2/24. Hence, we may assume that u +4 / ? C. Therefore, as I(u ? 1) = I(u +1) = {u}, we have u ? 4 ? C. Furthermore, since I(u + 2) = ?, I(u + 1patternP
Now we have s(u) to any rule ,
,
Observe first that u + 7 ? C since I(u + 4) = according to any rule. Moreover, at least one of u + 6, u + 8 and u + 11, say v, is a codeword since I(u + 6) = I(u + 8). , we are done as s s (u) + s s (u + 1) ,
2 · 11/4 ? 25/24 < 2 · 11/4 ? 3/4. Hence, we may assume that u + 9 / ? C and u + 10 ? C. 14/24, and u + 7 can receive share only according to the ruleR6 (3/24 units). Therefore, we are done since s s (u) + s s (u + 7) ? (11/4 ? 11/24) + (11/4 ? 14/24 + 3/24) = 2 · 11/4 ? 22/24 < 2 · 11/4 ? 3/4. Hence, we may assume that u + 11 / ? C ,
, For n ? 0, 1, 4 (mod 6) the result for locating-dominating codes is given in, vol.17
, Note that these patterns S3 can overlap each other. We denote by P 6 the pattern xxoooo. Therefore, we can divide overlapping S3-patterns into non-overlapping patterns P 6. Without loss of generality, we can assume that the vertices 0, vol.10, 2009.
, Alice considers an arc of P ? ( e). By Observation C.1, this happens at most 2a times. This gives at most 2a marked sons by this rule, ? By rule, vol.2
, The arc e is marked right after or already marked when this happens 2k times. This gives at most 2k marked sons by this rule, ? By rule 2(b)
, This can happen only if c > b, otherwise, by rule 2(b), e is already marked, ? By rule, vol.2
, As e is unmarked, this can happen only if d > b and at most d times. This gives at most ? d>b d marked sons by this rule, ? By rule 2(d)
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