Puzzles with numbers and things

[davis_greatest]

Resize it in photoshop, then copy it to paint and save it there (as a jpeg)

OK - I should be able to do that tomorrow at some point. But why is that different from resizing and saving it in Photoshop? Does Paint reduce the file size more than Photoshop does?

[Mitsuko]

Resize it in photoshop, then copy it to paint and save it there (as a jpeg)

[davis_greatest]

I'm not used to Paint. How do I resize? I'm trying it now, to make the image 60x80 pixels, but it seems just to be taking the top-left 60x80 pixels instead of resizing the image. Where is the Resize option?

[snookersfun]

I keep telling him...

[Mitsuko]

Put it on paint!!!

[April madness]

Maybe make it a few pixels smaller than allowed?

[davis_greatest]

d_g, I guess you can set quality for jpeg, that will reduce the size in KBs too

I did! I put it on the lowest quality (that's what I mean by "maximum compression") - and still it was too big

[snookersfun]

why go complicated, use simple MS Paint, resize and save as jpeg. Good enough for Avatar quality...

[April madness]

d_g, I guess you can set quality for jpeg, that will reduce the size in KBs too

[davis_greatest]

now that makes sense! Good explanation. Thanks for clearing that up.

but DG, just noticed, still no Avatar???

Didn't have much time last night. Even with maximum compression, and resizing the picture to only 80x60 pixels, my Photoshop was making the jpeg file about 30KB or so, and I think the maximum allowed for the Avatar is something like 18KB*. I'll have another go next time I get the chance.

* Edit: just looked - maximum allowed is 19.5KB - still smaller than I could get the file size.

[snookersfun]

The remaining group of 136 - 72 = 64 matches will be all those matches in which left-handers play left-handers or right-handers play right-handers. Then take any 3 left-handers OR any 3 right-handers - the three of them will all play each other.

now that makes sense! Good explanation. Thanks for clearing that up.

but DG, just noticed, still no Avatar???

[davis_greatest]

I got to those 72 as well, but small concern here, are we now on the way to proove, that you could NOT have found a group of three matches, such that only three players are in it (because somehow, I got the feeling, that the remaining group, also doesn't have any)?

The remaining group of 136 - 72 = 64 matches will be all those matches in which left-handers play left-handers or right-handers play right-handers. Then take any 3 left-handers OR any 3 right-handers - the three of them will all play each other.

[davis_greatest]

Congratulations, Vidas!

Indeed, it is possible to choose as many as 72 matches and still not be able to find among them any group of 3 matches involving only 3 players. Here is my orang-utan's proof, along similar lines, in two sentences:

I make 8 players play left-handed and the other 9 play right-handed, and then choose the 8x9 = 72 matches in which a left-hander plays a right-hander. If we choose any one of these matches, it must include one left-hander and one right-hander, so it will be impossible to find a 3rd player who can play both of them.


SO HERE IS THE SCOREBOARD AFTER ROUND 17

snookersfun……………………….…..7
robert602…………………………………4
abextra……………………………..…...3½
Vidas……………………………………….3½

(scoreboard adds up to number of rounds+1 since 2 members each got a point for round 14)

[snookersfun]

Vidas, that looks good again (at least it looks like my list - used diagonal lines in my triangle),
and actually, ignore my previous post, hadn't really looked at the actual numbers in there, just assumed by symmetry), as there are of course a huge amount of possible trios in the remaining group (which is quite interesting in itself)...

[Vidas]

It depends on how to pick 72 matches. Lets try...

A B.. Q
1 2.. 17

Players with odd numbers meet players with even, and vice versa.

1 plays 2, 4, ... 16
2 plays 3, 5, ..15, 17
...
15 - 16
16 - 17.
Total: 8+8+7+7+6+6+5+5+4+4+3+3+2+2+1+1 = 72
It can be nicely represented by tournament table - it looks like chessboard.
If we pick any three players, those who are both odd or both even,
won't meet in those 72 matches.