Shikamaru's Maths Challange

Click here to see the question and some correct answers .
Click here to see current challenge .



Your Answers ----------------------------------------------------------------


(20/06) Blazafan : There will always be 2 ninjas fighting the same amount of people. This is because if it is a 1 on 1 fight their will be 2 ninjas fighting one person if it is a mismatch of 10 on 32 there will be at least 2 ninjas on the group of 32 fighting one ninja. No matter what there will always be 2 ninjas fighting one person whether they are opponents or allies.

Shika: I don't see how your argument makes my assertion inevitable? There could be a large no. of conflicting groups. The situation is more complicated than you think. 


(22/06) Josh Klindienst : Mentally tag one of the genins and follow his/her actions.  Watch who he/she goes for consistantly and they are his enemies (accidents can happen, make it 3 or so attacks).  Those he/she avoids or aids can be considered allies.

Shika : ??????????????


(15/06) Warbringer : G = "At least 2 ninjas are fighting the same number of ninjas"

Case with x ninjas:

n[1] |{n[2], n[3], n[4] ... n[x-2], n[x-1], n[x] } (x-1)
n[2] |{n[1], n[3], n[4] ... n[x-2], n[x-1], n[x] } (x-1)
n[3] |{n[1], n[2], n[4] ... n[x-2], n[x-1], n[x] } (x-1)
... | ... (x-1)
... | ... (x-1)
... | ... (x-1)
n[x-2]|{n[1], n[2], n[3] ... n[x-3], n[x-1], n[x] } (x-1)
n[x-1]|{n[1], n[2], n[3] ... n[x-3], n[x-2], n[x] } (x-1)
n[x] |{n[1], n[2], n[3] ... n[x-3], n[x-2], n[x-1]} (x-1)

This chart applies to a situation where everyone fights against everyone (commonly called deathmatch). We have x ninjas (n[1], n[2], etc..), and each one of them is fighting against the other x-1 ninjas.

I've taken this as a model because it is the simplest and it can lead to all other situations.

It is obvious that if that's the case, then all ninjas are fighting the same number of ninjas, which ends the problem.

Now let's treat the other cases. What differs is the opponents that each ninja has. We can change that by taking out one ninja from the opponent list of someone. When doing so, 2 lists are modified, because we can't have a ninja that accepts being hit by another and not retaliating. Either they are allies (or neutral) or they are enemies. The length of these 2 lists will decrease by 1, thus changing the number of enemies for these to ninjas to x-2. This leaves us x-2 ninjas fighting x-1 ninjas, and 2 ninjas fighting x-2 ninjas. Now we can modify the lists again by taking out another element
from a list. We can either change ones we haven't touched yet, or change one of the lists we modified earlier along with an untouched one. There are 2 outcomes, that are the following :

a. x-4 ninjas fighting x-1 ninjas, 4 ninjas fighting x-2 ninjas (modify 2 unchanged lists).
b. x-3 ninjas fighting x-1 ninjas, 2 ninjas fighting x-2 ninjas, 1 ninja fighting x-3 ninjas (modify 1 unchanged list and 1 changed one).

&&&Note: We can't change 2 simultaneously modified lists again (That's obvious... See example).&&&

In the list of number of opponents of each ninja, the following happened :

a. We lost 2*(x-1) for 2*(x-2)
b. We lost 1*(x-1) for 1*(x-3) (actually it is 1*(x-1) and 1*(x-2) for
1*(x-2) and 1*(x-3) but we can simplify)

In both possibilities, we are sure that G is verified:

In a., we have 2 additional ninjas that are fighting x-2 ninjas.
In b., we still have 2 ninjas that are fighting x-2 ninjas (we initially had 2, we lost one and gained one).

Obviously, we can create whatever situation possible from the initial chart by simply removing elements from the lists again and again. However, we will always have only 2 possible types of modifications (a, b). Needless to say, there has to be at least one element in each list (nobody is just watching the others apparently).

This concludes the problem.

Maybe I haven't explained very well, so here's my point in short:

We build the initial chart, then for whatever situation we need to create, we begin by taking out one element from a list (and apply either change a. or b.), then another one (another a. or b. modification), and another one again, etc... until we reach the corresponding situation in which each ninja is fighting against his given opponents. G will be verified because it is true at the end of each a. or b. modification process.

That's theory for the general case, so here's an example for everyone's convenience:

Suppose there are 5 ninjas fighting, here's the base chart :

A|{B, C, D, E} (4)
B|{A, C, D, E} (4)
C|{A, B, D, E} (4)
D|{A, B, C, E} (4)
E|{A, B, C, D} (4)

That's a situation where G is true. Let's try another one :

A|{B, C, D, } (3)
B|{A, C, D, E} (4)
C|{A, B, D, E} (4)
D|{A, B, C, E} (4)
E|{ , B, C, D} (3)

G is true here too. Let's change again :

&&&Note: We can't change the list for A and E again.&&&

A|{B, C, , } (2)
B|{A, C, D, E} (4)
C|{A, B, D, E} (4)
D|{ , B, C, E} (3)
E|{ , B, C, D} (3)

G is true. Let's try another situation :

A|{ , C, , } (1)
B|{ , C, D, E} (3)
C|{A, B, D, E} (4)
D|{ , B, C, E} (3)
E|{ , B, C, D} (3)

G is true again. Here's another one :

A|{ , C, , } (1)
B|{ , C, D, E} (3)
C|{A, B, , E} (3)
D|{ , B, , E} (2)
E|{ , B, C, D} (3)

G is still true.

That's all, hope I made myself clear.

Shika: Nice try but the correct answer is shorter. Much shorter. Effort points.



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