Lemma 7.6 - Revision history

Tomtom2357 at 20:11, 1 March 2021

2021-03-01T20:11:58Z

← Older revision		Revision as of 13:11, 1 March 2021
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	If <math>\mathcal{A}</math> contains a size 5 set, then <math>\mathcal{A}</math> is Frankl's.		If <math>\mathcal{A}</math> contains a size 5 set, then <math>\mathcal{A}</math> is Frankl's.

	WLOG let that set be 12345. Let w(x)=6 is x=1,2,3,4, or 5 and w(x)=1 otherwise. The target weight is 35.		WLOG let that set be 12345. Let w(x)=6 if x=1,2,3,4, or 5 and w(x)=1 otherwise. The target weight is 35.

	==\|K\|=0:==		==\|K\|=0:==

Tomtom2357 at 12:14, 2 December 2016

2016-12-02T12:14:35Z

← Older revision		Revision as of 05:14, 2 December 2016
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	This page proves a lemma for the [[m=13 case of FUNC]]. It is a sublemma needed for the proof of [[Lemma 7]]		This page proves a lemma for the [[m=13 case of FUNC]]. It is a sublemma needed for the proof of [[Lemma 7]].

	==Lemma 7.6:==		==Lemma 7.6:==

Tomtom2357: /* |K|=5: */

2016-12-02T11:49:43Z

|K|=5:

← Older revision		Revision as of 04:49, 2 December 2016
Line 67:		Line 67:

	3: 123		3: 123

	4: 124		4: 124

	5: 125		5: 125

Tomtom2357: /* |K|=4: */

2016-12-02T11:49:31Z

|K|=4:

← Older revision		Revision as of 04:49, 2 December 2016
Line 42:		Line 42:

	23: 123		23: 123

	24: 124		24: 124

	25: 125		25: 125

	34: 1234		34: 1234

	35: 1235		35: 1235

	45: 1245		45: 1245

	The other sets 13, 14, 15 can only have one set contributing to the deficit, due to:		The other sets 13, 14, 15 can only have one set contributing to the deficit, due to:
	13 and 14: 134		13 and 14: 134

	13 and 15: 135		13 and 15: 135

	14 and 15: 145		14 and 15: 145

Tomtom2357: /* |K|=3: */

2016-12-02T11:49:07Z

|K|=3:

← Older revision		Revision as of 04:49, 2 December 2016
Line 24:		Line 24:

	134, 234: 1234		134, 234: 1234

	135, 235: 1235		135, 235: 1235

	Any 2 of 124, 125, 145: 1245 (this cancels out the deficit of all but at most one of them)		Any 2 of 124, 125, 145: 1245 (this cancels out the deficit of all but at most one of them)

	245 and 345: 2345		245 and 345: 2345

Tomtom2357 at 10:58, 2 December 2016

2016-12-02T10:58:31Z

← Older revision		Revision as of 03:58, 2 December 2016
Line 1:		Line 1:
	This page proves a lemma for the [[m=13 case of FUNC]].		This page proves a lemma for the [[m=13 case of FUNC]]. It is a sublemma needed for the proof of [[Lemma 7]]

	==Lemma 7.6:==		==Lemma 7.6:==

Tomtom2357: Finished the lemma

2016-12-02T10:55:24Z

Finished the lemma

@@ Line 13: / Line 13: @@
 ==|K|=1:==
-The only possible set in this case is the <math>C_5</math> sets 12345 (as any smaller sets would violate the assumption).
+The only possible set in this case is the <math>C_5</math> set 12345 (as any smaller sets would violate the assumption), and it has weight exactly 35, so there is no deficit in this case
-==
+==|K|=2:==

Tomtom2357: /* Lemma 7.6: */

2016-12-02T08:13:15Z

Lemma 7.6:

← Older revision		Revision as of 01:13, 2 December 2016
Line 5:		Line 5:
	If <math>\mathcal{A}</math> contains a size 5 set, then <math>\mathcal{A}</math> is Frankl's.		If <math>\mathcal{A}</math> contains a size 5 set, then <math>\mathcal{A}</math> is Frankl's.

	WLOG let that set be 12345. Let w(x)=6 is x=1, 2, 3, 4, 5 and w(x)=1 otherwise. The target weight is 35.		WLOG let that set be 12345. Let w(x)=6 is x=1,2,3,4, or 5 and w(x)=1 otherwise. The target weight is 35.

	==\|K\|=0:==		==\|K\|=0:==

Tomtom2357: Created page with "This page proves a lemma for the m=13 case of FUNC. ==Lemma 7.6:== If $\mathcal{A}$ contains a size 5 set, then $\mathcal{A}$ is Frankl's. WLOG le..."

2016-12-02T07:07:31Z

Created page with "This page proves a lemma for the m=13 case of FUNC. ==Lemma 7.6:== If <math>\mathcal{A}</math> contains a size 5 set, then <math>\mathcal{A}</math> is Frankl's. WLOG le..."

New page

This page proves a lemma for the [[m=13 case of FUNC]].

==Lemma 7.6:==

If <math>\mathcal{A}</math> contains a size 5 set, then <math>\mathcal{A}</math> is Frankl's.

WLOG let that set be 12345. Let w(x)=6 is x=1, 2, 3, 4, 5 and w(x)=1 otherwise. The target weight is 35.

==|K|=0:==

In this case there are only two sets, the empty set and the full set 12345. The deficit is therefore 35+5=40.

==|K|=1:==

The only possible set in this case is the <math>C_5</math> sets 12345 (as any smaller sets would violate the assumption).

==