Puzzles with numbers and things

[davis_greatest]

ill go for 208

Wow! That seems high. Is that really possible? If I ever get the time and the patience, I might work out what the maximum possible is. But then, I might never get round to it!

[Semih_Sayginer]

ill go for 208

[snookersfun]

ok, here it comes....
1st bid: 192

[davis_greatest]

Hint / clarification - normally, of course, in snooker,the maximum break would be 147 (with no free ball).

In Big Ape Break, however, you cannot pot the same colour after two reds in succession - i.e. cannot go Red Black Red Black.... (or even Red Black Red Pink Red Black)

Therefore, the maximum might be less than 147.

On the other hand, potting colours into the right pockets makes the colours score double points. So the maximum might be more than 147!

[davis_greatest]

Final rule: davis_greatest is also eligible to compete. Whoever correctly bids the highest (excluding davis_greatest) will get one point. If davis_greatest bids higher than that person, davis_greatest will get one point too.

[davis_greatest]

Round 58 - Big Ape Break

This round is a bit different from my other questions, for two reasons:

1) To get the point, you don't need to come up with THE (single) solution - you only need to come up with a better solution than anyone else, so anyone who wants to can have a go.

2) Usually when setting a question on this thread, I have an idea of what I expect to be the best way to solve it before / at the same time as setting the question. This time, I haven't thought of the (best) answer yet!


Round 58 - Gordon the gorilla is hosting a new Saturday night snooker game show, Big Ape Break. Oliver the orang utan does trick shots and tells the contestants to "Pot as many balls as you can."

On the show, contestants play a frame of snooker, just like any normal frame of snooker except that:

a) each of the 6 pockets is coloured. The colours of the pockets are:
yellow for the left-centre pocket and then, moving clockwise, blue, brown, green, pink, black - a bit like this:

0------0
!.........!
!.........!
!.........!
0------0
!.........!
!.........!
!.........!
0------0

b) Once a colour has been potted, the same colour cannot be potted following the next red, nor following the red after that. (Once the 15th red and colour have been potted, this rule no longer applies - the final colours may and must be potted in the usual order of yellow, green, brown, blue, pink, black, regardless of the colours potted with the final reds.)

Example 1: Red Brown Red Yellow Red Blue Red Brown IS allowed
BUT
Example 2: Red Brown Red Yellow Red Brown IS NOT

Example 3: For the 14th and 15th red,
Red Black Red Yellow Yellow Green Brown Blue Pink Black IS allowed


c) Whenever a colour has been potted, the following colour cannot be potted into any pocket that lies along the same edge of the table. That means that it cannot be potted into a pocket on the same side of the table (left or right) and, if it is a corner pocket, cannot be potted into a pocket at the same end of the table either.

This applies even when down to the final 6 colours.

For example, after potting a colour into the pink pocket, it would not be permissible to pot the next colour into the pink, green or brown pockets (same side), nor into the black pocket (same end)

d) None of these rules apply to reds. It makes no difference into which pockets reds are potted.

e) And this is the important bit: potting a colour into a pocket of the same colour as the ball (e.g. pink into pink pocket) scores double points (in this example 2 x 6 = 12).


Your question is, what is the highest break (ignoring free balls) you can make?

You don't need to say the highest theoretically possible - you just need to give a bid of the highest that YOU can find. Whoever has bid the highest by 23:00 GMT on Friday 24 November will be invited to explain how it is possible.

I.e. if you have the highest bid, you would then need to explain how. For example, you would then say:

Red
Green into Yellow pocket (or whatever)
Red
Pink into Brown pocket (or whatever)
....
...

and after all 15 reds and colours...

Yellow into Yellow pocket (or whatever)
Green into Pink pocket (or whatever)
Brown into Blue pocket (or whatever)
....

[davis_greatest]

Oh dear, of course I understand your argument. I wish I could state my case half as logical.
Just a few more thoughts (and then I will shut up, before my points get docked): Ultimately doing all your switches you'll end up in the reverse way you started, so you basically counted over how many neighbors you’ll have to jump (in steps of one in your case). Now, seems to me, I did basically the same thing, except, that I grouped moving over neighbors into steps of 7. Therefore we reached the same combined sum of 105.

Yes, more or less! Just be careful with using the argument that IF we use jumps of 7, then it is impossible - because the game is not played in jumps of 7. This argument is not the same logically as saying it can't be done with jumps of 2. If Charlie slaps Dott, then Dott will utter a high-pitched squeal; but if Dott utters a high-pitched squeal, that doesn't mean that Charlie has slapped him. Yet

Now I need to think what to ask next!

[snookersfun]

Oh right, I think I understand what you meant now - I hadn't realised that you were replacing each jump of 2 by a single jump of 7 - I thought you meant multiple jumps of 2. Of course, showing that 15 jumps of 7 are needed does not in itself prove that it can't be done in jumps of 2.




The important thing about the argument I gave is that I was not switching their absolute positions - only their relative (i.e. left / right) positions. This means that we are NOT saying that Ape A should stand where Ape B was and Ape B should stand where Ape A was. We are only saying that whoever was on the left of the other should finish on the right with each "switch" - not necessarily the same distance apart. Does that make sense?

Oh dear, of course I understand your argument. I wish I could state my case half as logical.
Just a few more thoughts (and then I will shut up, before my points get docked): Ultimately doing all your switches you'll end up in the reverse way you started, so you basically counted over how many neighbors you’ll have to jump (in steps of one in your case). Now, seems to me, I did basically the same thing, except, that I grouped moving over neighbors into steps of 7. Therefore we reached the same combined sum of 105.

[davis_greatest]

now, this I don't understand

the 8 is definitely in the right place, because this is the part, where I tried to show, how many jumps of a certain length are needed (or odd/even sum). So, it can be successfully done, but using jumps of 7 instead of jumps of 2 (similar to you showing how each couple can be switched with its neighbor in so an so many steps). But, one needs 15 jumps of 7, showing that an odd number in total 'jumpsum' is needed.

Oh right, I think I understand what you meant now - I hadn't realised that you were replacing each jump of 2 by a single jump of 7 - I thought you meant multiple jumps of 2. Of course, showing that 15 jumps of 7 are needed does not in itself prove that it can't be done in jumps of 2.

(similar to you showing how each couple can be switched with its neighbor in so an so many steps).

The important thing about the argument I gave is that I was not switching their absolute positions - only their relative (i.e. left / right) positions. This means that we are NOT saying that Ape A should stand where Ape B was and Ape B should stand where Ape A was. We are only saying that whoever was on the left of the other should finish on the right with each "switch" - not necessarily the same distance apart. Does that make sense?

[snookersfun]

Is your last 8 in the wrong place? Otherwise it looks like it is possible!

PS Of course, even if you show that you end with
15-14-13-12-11-10-8-9-7-6-5-4-3-2-1
that would not be a proof of impossibility

now, this I don't understand

the 8 is definitely in the right place, because this is the part, where I tried to show, how many jumps of a certain length are needed (or odd/even sum). So, it can be successfully done, but using jumps of 7 instead of jumps of 2 (similar to you showing how each couple can be switched with its neighbor in so an so many steps). But, one needs 15 jumps of 7, showing that an odd number in total 'jumpsum' is needed.

[davis_greatest]

good that the numbers were all there
Explanation of some sentences- for native English speakers:
I just switched the couples in jumps of 7 instead:
line up: 1-2-3-4-5-6-7-8-9-10-11-12-13-14-15
after first jump: 2-3-4-5-6-7-8-1-9-10-11-12-13-14-15
after 2nd jump: 2-3-4-5-6-7-8-15-1-9-10-11-12-13-14
after 4th: 3-4-5-6-7-8-15-14-2-1-9-10-11-12-13
after 6th: 4-5-6-7-8-15-14-13-3-2-1-9-10-11-12
after 8th: 5-6-7-8-15-14-13-12-4-3-2-1-9-10-11
after 10th: 6-7-8-15-14-13-12-11-5-4-3-2-1-9-10
after 12th: 7-8-15-14-13-12-11-10-6-5-4-3-2-1-9
after 14th: 8-15-14-13-12-11-10-9-7-6-5-4-3-2-1
after 15th: 15-14-13-12-11-10-9-8-7-6-5-4-3-2-1

Is your last 8 in the wrong place? Otherwise it looks like it is possible!

PS Of course, even if you show that you end with
15-14-13-12-11-10-8-9-7-6-5-4-3-2-1
that would not be a proof of impossibility

[snookersfun]

good that the numbers were all there
Explanation of some sentences- for native English speakers:
I just switched the couples in jumps of 7 instead:
line up: 1-2-3-4-5-6-7-8-9-10-11-12-13-14-15
after first jump: 2-3-4-5-6-7-8-1-9-10-11-12-13-14-15
after 2nd jump: 2-3-4-5-6-7-8-15-1-9-10-11-12-13-14
after 4th: 3-4-5-6-7-8-15-14-2-1-9-10-11-12-13
after 6th: 4-5-6-7-8-15-14-13-3-2-1-9-10-11-12
after 8th: 5-6-7-8-15-14-13-12-4-3-2-1-9-10-11
after 10th: 6-7-8-15-14-13-12-11-5-4-3-2-1-9-10
after 12th: 7-8-15-14-13-12-11-10-6-5-4-3-2-1-9
after 14th: 8-15-14-13-12-11-10-9-7-6-5-4-3-2-1
after 15th: 15-14-13-12-11-10-9-8-7-6-5-4-3-2-1

[davis_greatest]

Can snookersfun be caught?

HERE IS THE SCOREBOARD AFTER ROUND 57

snookersfun……………………….…..22½
abextra……………………………..…...11
Vidas……………………………………….10½
davis_greatest…………………..……8½
robert602…………………………………6
elvaago...............................4½
The Statman……………………..……3
Semih_Sayginer.....................2½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[davis_greatest]

Hmmmm...

Reversing the order of the line-up is not possible for 15 people by moving over two places at a time.
One can show though, that it is possible to transpose pairs succesfully by leapfrogging in steps of 7. One can basically start at one end (1 or 15) and leapfrog this number into the 'new' middle, followed by the next number at the other side and so on alternating sides (1,15,2,14....8). We need 7x15 such steps = odd number. Therefore the inversion can't be done using jumps by 2.

OK ... congratulations, snookersfun, I'll give you that. I must admit that I don't understand what you are saying in the 2nd and 3rd sentences but:

- you are right that it is impossible; and
- you have mentioned that 7 x 15 is odd, and that the jumps are 2, which is the important bit.

Here is my explanation….

I'll call everyone an "ape" for simplicity. (Looking around, that looks right anyway.) So there are 15 apes in a line.

The important thing to realise is that, if it were possible to reverse the order of all 15 apes, then for every pair, their relative positions would need to be "switched". What I mean by that is that if Ape A starts on the left of Ape B, then Ape A must finish on the right of Ape B. This applies for every pair of apes A and B.

So, how many pairs are there? Each of the 15 apes can be paired with 14 others, so there are 15 x 14 / 2 = 15 x 7 = 105 pairs. (We divide by 2 because each pair consists of 2 apes.) Note that 105 is an ODD number.

Every time someone moves in the line, he / she switches his / her relative position with 2 other apes. So the number of switches is always an even number, and can never be 105.



Scoreboard to follow...

[snookersfun]

Round 57: Stand in line!

We're still playing games at Oliver's birthday party. This is the party of all parties! This time, we all stand in a straight line, holding hands, with Oliver on the far left and Gordon on the far right, and abextra with her hands gripped by two excited apes.

In fact, the order is:

Oliver----austrian_girl----davis_greatest----April Madness----Robert602----snookersfun----Semih_Sayginer----The Statman----Vidas----elvaago----Obligation----Snooker Rocks!----Charlie----abextra----Gordon

At any time, anyone who has two or more people/apes on his or her side, can move two places to that side.
We can all move at any time, as often as we wish. Our aim in this game is for us all to be standing in the exact reverse order of how we started. So, we need to finish with Gordon on the far left, Oliver on the far right, and everyone else in between exactly reversed.

Your question is:
Is this possible? If yes, explain how we achieve it. If no, explain why not.

Reversing the order of the line-up is not possible for 15 people by moving over two places at a time.
One can show though, that it is possible to transpose pairs succesfully by leapfrogging in steps of 7. One can basically start at one end (1 or 15) and leapfrog this number into the 'new' middle, followed by the next number at the other side and so on alternating sides (1,15,2,14....8). We need 7x15 such steps = odd number. Therefore the inversion can't be done using jumps by 2.