Hornswoggling Swordsman: Newsletter Challenge (May 2019)

First round odd numbers kill even numbers.

Second round, 1 kills 3, 5 kills 7, etc leaving 1,5,9,...97 who kills 99

Third round, 1 kills 5, 9 kills 13, etc leaving 1,9,13,17,...,97 who kills 1

Fourth round, 9 kills 17, 21 kills 29, 37 kills 45, 53 kills 61, 69 kills 77, 85 kills 93

Fifth round 9 kills 21, 37 kills 53, 69 kills 85

Sixth round 9 kills 37, 69 kills 9

#37 survives

Correction:

1: Even numbers eliminated, odd numbers remain

2: 1, 5, 9, 13, ... 97 remain, 97 kills 99

3: 1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97 remain, 97 kills 1

4: 9, 25, 41, 57, 73, 89 remain, 89 kills 97

5: 9, 41, 73 remain, 73 kills 89

6: 9, 73 remain, 73 kills 9

7: 73 remains

I think this is right...

I did it with a graphics program - 7Revs (pages) and we're down to the last person standing... at position 73.

Nice graphical representation. That must have been a long knife to reach across the circle!

I did a spreadsheet. I agree that 73 is the right answer.

Here's a screen capture.

Yellow survives, white doesn't.

I do like Mikerho graphic. Clearly depicts the challenge.

I say #65 is last man standing...

No wait... it's #73....

log base 1.065116862 x n

step 1 : For a given value of N, find the “Power of 2” immediately greater than N. Let’s call it P
Step 2 : Subtract N from (P-1). Lets call it M, i.e, M = (P-1)- N
Step 3: N - M = answer

proof n= 100

P= 128

M= 27

100 - 27 = 73

I haven't checked -does it work for other lemmas?

From your formula, Ans = 2N-P+1;

I ran it through my little program (#13) and every value I tried checked out with your formula. It's going to drive me nuts until I figure out why it works...

It's a miracle!!

OK, here's a proof that is easy to visualize.

1. For number left, N = 2^k, a power of 2, then #1 wins. Half are eliminated on each revolution. For example, if you start with 16, 8 are left, then 4, then 2, each time with the sword returning to #1. It's proof by mathematical induction. If it's true for N=2^k, and N=2^k reduces to N=2^k-1, then it's true for N=2^k for any k.

2. Generally, N = 2^k + r, where r is the remainder after the greatest power of 2 less than N. For example, if N=100, N=2⁶+ 36. After r=36 turns, 36 have been eliminated and the sword has been passed 2 x 36 positions, or to position #1 + (2 x 36) = position #73. There are now N=2⁶ left, so position #73 is the winner.

Ans = 2N-P+1

Yours is the simplest formula, you win the Occam's Razor Award.

Let me just add the following constraint.

Let P=2^K; Where K is a positive integer and P>N.

Works even when N is an even power of 2.

1. (1+1).... (1x3x5x7x9x)(11x13x15x17x19) ->(100-1=99)
2. (3+4).... (99x3x7x11x)(15x19x23x27x31) ->(100-3=97)
3. (7+8).... (97x7x15x23x)(31x39x47x55x63)->(100-5=95)
4. (15+16) (95x15x31x47x63x79x91)-------- ->(100-9=91)
5. (31+32) (91x31x55x87)--------------------- ->(100-13=87)
6. (55+64) (87x55)
7. answer 87

For 2ⁿ participants, number 1 is always the survivor.

By inspection, 8 reduces to 4 which reduces to 2 leaving 1. Extend for any 2ⁿ.

If all 100 people are equally martially-capable, each combat would be a toss-up, and if the numericallyfirst combatant was allowed to lose the first match, then any numeral of the sole-survivor would have an equal likelihood of being statistically the 1-in-99 winner, and the entire process was repeated an infinite number of times...

...but, if the numeral one survivor always won his single match during each sub-round of each complete cycle, then the numeral one survivor would win every time...

Extra credit: How would you set up a solution when the number of participants is in the form 2ⁿ? How is this approach unique?

This is easier. First round all the even numbers get killed, then the even survivors, etc.

Example: with 32

1st round survivors: 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31

2nd round survivors: 1,5,9,13,17,21,25,29

3rd round survivors: 1,9,17,25

4th round survivors: 1,17

5th round survivor: #1

#1 always wins when number is 2ⁿ

No one has calculated how many would say "I'm outa here!" and left.

Would you stick around such an event?

Why couldn't he just tap the guy on the shoulder and ask him to leave, no need for all this bloodshed...

Step 1 : For a given value of N, find the “Power of 2” immediately greater than N. Let’s call it P
Step 2 : Subtract N from (P-1). Lets call it M, i.e, M = (P-1)- N
Step 3 : Multiply M by 2. i.e M*2
Step 4 : Subtract M*2 from P-1. Let’s call it ans, i.e, ans = (P-1) – (M*2)
So, the person with number “ans” will survive till last.

Blurp

If anyone's got access to Matlab or Octave, here's a Matlab function that calculates the winner of the game "circlekill" with the number of players passed in as a parameter.

For example, the present problem would be called as circlekill(100) and would indicate 73 as the winner. (% after the command statement is comment and can be omitted.)

Enjoy...

function circlekill(S)

A=1:S;

n=S; % n=number left

k=1; % index into A

toggle=0; % toggle determines killer and killee

while n>1

v=A(k);

if (v>0) % still alive

if toggle==1 % killee

A(k)=0; % killed, replaced by 0

printf('%d dies\n',v);

n=n-1; % decrement number left

toggle=0; % next one is killer

else

toggle=1; % next one is killee

end

end

k=1+mod(k,S); % circular index

endwhile

printf('%d is left\n',sum(A));

end

My previous answer is wrong; I have assumed that the group is in a straight line and the next action requires that the last one standing starts the next round? And I have not taken into account an even or odd number left standing at the end of each round. And also the alternative of always starting back at the beginning, a point already raised, would result with number 1 left standing at the end. So my next solution will assume that the group is standing in a circle, closing ranks at the end of each round? Also taking into account, odd and even numbers left standing.

1. 1x3x5x7x9x11x13x15x17 +2, etc, 99 being the last of an even number of 50 left standing after 99 removed the last number 100, leaving number 1 as the next in line to start the next round.
2. 1x5x9x13x17x21x25x29 +4, etc, 95 being the last of an odd number of 25 left standing after 95 removed number 99, leaving number 1 as the next in line to start the next round.
3. 1x9x17x25x33x41x47 +8, etc, 95 being the last of an odd number of 13 left standing after number 81 removed number 87, number 95 removes number 1 leaving number 9 as the next in line to start the next round.
4. 9x25x41x57x73x87, number 87 being the last of an even number 6 left standing after 73 removed number 81, number 87 removes number 9 leaving number 25 in line to start the next round.
5. 25x57x89, number 89 being the last of an odd number of 3 left standing, after number 57 removed number 73, number 89 removes number 25 leaving number 57 in line to start the next round.
6. 57x Answer number 57 the last standing.

One hundred people stand in a circle in order, numbered 1 to 100.

So this is a question about a person killing another, then giving away an instrument to kill another and another,,, ?

Like there isn't enough of that.in the world,

Now, here, death is used as an amusement. Who comes up with these questions

It's not as if it really worries anyone at all, after all, we have it rammed down our throats daily with Iraq, Syria, Sri Lanka and many more. We wake up to death each day so why worry about some hypothetical question. We are all immune and de-sensitised to it now after 18+ years of America flexing its muscles in the Middle East and rest of the world.

Did, julius robert oppenheimer ask himself the answer to a hypothetical question ?

https://en.wikipedia.org/wiki/J._Robert_Oppenheimer

Good read...

Perhaps as a model for presidential primary season?

Let's go with 73 but I prefer tonyhemet's answer

No 73 stays alive. But dies the next day. His wife wants the insurance money.

73?

Funny thing about numbers, we all understand the basics, but can get lost in the formula.

#73 -- see table below with casualties numbered by order of elimination in five rounds leaving #73 as sole survivor,

https://www.youtube.com/watch?v=uCsD3ZGzMgE

# 41 is the sole survivor. He killed #73.

I used Excel and it's 73.

Unites

1 2 3 4 5 6 7 8 9 10

1 x 3 x 5 x 7 x 9 xx

1 3 5 7 9

1 x 5 x 9

1 5 9

x 5 x

5

Tens

0 1 2 3 4 5 6 7 8 9

0 x 2 x 4 x 6 x 8 x

0 2 4 6 8

0 x 4 x 8

0 4 8

x 4 x

Answer 40+5 = number 45 left standing

see #26

99

see #26

If it is 2^^n then 1 always wins.

But what if the combatant with the sword loses a random combat ?...

0+2 = 2

1+4 = 5

3+8 = 11

5+16 = 25

7+32 = 39

9+64 = 73 answer 73

0+128 out of range.

5+16 = 21

We got 100 elves together in a circle with a marker (no swords, elves are too valuable), had them mark each other up and decrease the size of the circle, so they could reach, (that is an awful big circle to reach across if you are an elf), until we got to the last unmarked elf. #87

No one, after the first death, will stand around to be killed!

Therefore #3 will remain with the sword unless #1 doesn't hand it off.

#57 is the last swordsman standing.

When the sword ends up in the hands of the last swordsman in the circle it starts over at one and it continues. All even numbered swordsman are killed off the first time around the circle. #99 kills #100 then #1 kills #3 and #5 kills #7 and so on.

99

But if only 1 survives then isn't 1 the answer?

Because '57 was a great Vintage for wine, cars, women, the basic math principle of SWMBO must be applied....57 is the last man standing if he knows what's good for him.

I still say 73 but only when SWMBO is out of range.
SWMBO = She Who Must Be Obeyed

Left over will be the number 1 as in the each cycle the even number in the chain starting from number 1 will be killed :( and the sword will be back to number 1 always.

When do we get the correct answer?

There are many different answers here, unusual with an Engineering forum I feel.....

Answer is now posted.

Many thanks.

I am pleased to be correct!!

I was amazed at just how many wrong answers there were.....I did in my head twice, to get my answer, not that difficult, but it helps in keeping the old grey matter working.....

Have you some others of that ilk please?

Look out for the next one, posting on May 31 -- we think it's going to be a good one!

Great!

Here is a quick and easy one. Can anyone tell me exactly why this is mathematically correct? It fooled a Yale Math professor that I met many years ago too!

Its addition!

----44,11,07,0.25
----12,17,06,0.50+
--------------------
-2,13,09,01,0.75
--------------------

The "minuses" on the LH side are only to get everything lined up vertically, so just ignore them please.....they have nothing to do with the math.

The commas are just being used as separators, nothing to do with the math either....

Enjoy!!

So nobody attempted the math question I posted?

Shame on you all!!

What is a Hornswoggler? Sounds like old cowboy slang?

Just going back over old Challenge questions I've missed.

44 pounds 11 shillings 7 pence and a farthing PLUS

12 pounds 17 shillings 6 pence and a hapeny

EQUALS 57 pounds 9 shillings 1 penny and 3 farthing

I can't think of any slang for £22 which would make your sum correct.

Am I on the right lines?

Almost correct.

21 Shillings is a Guinea.....

And if you have 20 shillings left over, then you also have a Pound Sterling, but ONLY then!!

If No.1 had the sword, kills person#2 and so on, by the end person #99 kills person 100, so there is no one left to kill person #1. So person #1 would survive.

Answer is 42.