harey
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Posts posted by harey
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If our hardware uses fewer clock cycles to perform three additions than a single multiplication, we may well gain overall processing speed by using Eq. (2-4) and Eq. (2-5) instead of Eq. (1-2) for complex multiplication
If it worked, it would be great for fractal pictures. The bad news is that multiplications are in the processor heavily optimized, a fair guess 4-5 additions. Googling:
The latency is 1 cycle for an integer addition and 3 cycles for an integer multiplication. You can find the latencies and thoughput in Appendix C of the "Intel 64 and IA-32 Architectures Optimization Reference Manual", which is located on http://www.intel.com/products/processor/manuals/.
The big question is how the 3 temporary variables were treated by the compiler. If we have to access the RAM...
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@bonanova - sorry, but I believe you have a sign error with the factor bd:
I = (a+b)(c+d) - R =ac+bc+ad+bd - (ac - bd) ... = ... +2*bd ....
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@bmad
As nobody found the solution, can you post it?
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Yes, I should have started at 0. (a,b)=(0,6) -> (a+b)=6
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1) sum of ages=boys+girls+teacher=b*g+g*b+42=2*b*g+42
2) sum of ages=average_age*number_of_persons=(b+g)*(b+g+1)
1=2: 2*b*g+42=(b+g)*(b+g+1)
I fed it into Wolfram Alpha and got no positive integer (number of persons) solution.
A small program to be sure:
for b in range(1,99):
for g in range(1,99):
sum1=2*b*g+42
sum2=(b+g)*(b+g+1)
if(sum1==sum2):
print(b,g,sum1)
(b,g)=(3,5) => b+g=8
I would be interested if someone finds the solution with Wolfram Alpha.
Sorry, unable to hide. -
... B.
If we play the diagonal, I win. The losing places are A1, C3, D5, E7.
If A plays right (up), I play right (up), too. Sooner or later these moves will be compensated by my opponents up (right) and mine up (right). -
Bonus question for those who are at ease with formulae: Using this proceeding, in how many parts can you cut the cake at most?
- a pentagon has a center: cut a concentric cylinder representing 2 parts
- the remaing object can be easily cut in 5 parts -
What does mean "randomly"?
If I remember well, there is a controversy how to choose randomly points on a circle.
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as 1st on the list, 5$ (per email)
as 2nd on the list, 5*5$ ()
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as 7th on the list, 5**7$ ()
Total 5+5**2+5**3+5**4+5**5+5**6+5**7=97,655$ ()
As 74 people read this message up to now, 74*97,655=7,226,470
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Is there a more succinct solution?
There is: just the formula giving the number of boys. But it is quite unreadable and to understand it, you have more or less to go thru what I have written.
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If you take n=8 families:
4 families have b => 4b
2 families have gb => 2g 2b
1 family has ggb => 2g 1b
1 family has gggb => 3g 1b (or gggg, but does not matter much)
Total: 7g 8b
For a large n, g -> b.
Limits: A woman cannot have an infinity of children and twins may not obey to the rule.
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a) Cut in 3 pieces and burn from both sides.
b) When one piece is consumed, cut that one that seems to burn most slowly in two pieces and put on fire the ends that do not burn.
c) Repeat b) until everything burned.
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There are 60 marks, so each represents a minute.
The hour hand is on a mark when the minute hand is on {0, 12, 24, 36, 48}.
When the minute hand is on a mark, the second hand is on noon.
Blue cannot be the second hand. (It would be on noon and it would be 18 minutes past/before a full hour.)
1) Yellow is on 12. Mauve indicates 24 minutes. Blue indicates 42 [minutes] = 8 + 2/5 [hours] = 8:24
2) Yellow is on 12. Mauve indicates 36 minutes. Blue indicates 18 [minutes] = 3 + 3/5 [hours] = 3:363) Mauve is on 12 ... -> 3:36
4) Mauve is on 12 ... -> 8:24.
The watch could be rotated upside down, too.
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Place the pawns from the column H to C3, G3 and G7.
Jumps: E3 - C1 - A3 - C5 - A7 - F2 - H4 - E7 - B4 - D2 - H6 - F8
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There can be a solution in one pass disregarding the answers of C. The numbers are a little bit high for practical purposes, so I will illustrate it on a sum of 12 or 13, everyone having 4.
Notation:
"!n" means "I do not see the number n"
"<-(xy)" means "because otherwise I would know the combination is (a=x b=y)"
Both A and B see c=4, they both know a+b=8 or a+b=9.
They also know that the other one knows that.
By saying "I do not know", they in fact tell:
1 A: !8 <- (18)
B: !8 <- (81)
!1 <- (17); (18) excluded in 1A
2 A: !1 <- (71); (81) excluded in 1B
!7 <- (27); (17) excluded in 1B
B: !7 <- (72); (71) excluded in 2A
!2 <- (26); (27) excluded in 2A
3 A: !2 <- (62); (72) excluded in 2B
!6 <- (36); (26) excluded in 2B
B: !6 <- (63); (62) excluded in 3A
!3 <- (35); (36) excluded in 3A
4 A: !3 <- (53); (63) excluded in 3B
!5 <- (45); (35) excluded in 3B
--> B knows A sees 4.
--> A realises he never will get more information
A already knows that b=4, he can imagine this dialogue (the answers of B will not change whether a=4 or a=5), so his very first answer is:
1 A: I will never be able to tell.
B thinks:
if b=4, this makes sense
if b=5, A would have announced he would be able to tell.From the answers of A and B, C deduces c=4 (if necessary, I will explain).
The remaining problem is whether the answers of C are redundant or whether they lead to another distribution.
Comments?
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There might be a trap. Back to the secondary school for the definition of the measurement error.
I have rulers which are divided in cm and subdivided in 1/10 of cm.
If I tell you I mesured 2.4 cm, you assume the the value is between 2.35 and 2.45. Error in percentace=0.1/2.4*100
If I tell you I mesured 7.4 cm, you assume the the value is between 7.35 and 7.45. Error in percentace=0.1/7.4*100
I happen to have a curiosity of a ruler in inches. It's subdivisions are 6 (for 6 lines per inch or 12 cpi), 8 (for 8 lines per inch or 8 cpi) and 10 (for 10 cpi).
(Supposing this ruler shrinked by 1/24, I get a headache.)
If antelOpe has such a ruler, the error is 1/subdivison/4*100.
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A priori, the p of any card being red is the same, so it does not matter on which card you bet. (Therefore the best strategy is to bet on the first card because you are not losing time.)
Roulette gamblers will not agree with this, they will argue that if there remain 3 red cards and 1 black card, p=3/4.
Is there an easy way to calculate the p that R red cards and B black cards remain?
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Well, I have never been to LA, but I know what happens in Germany. Bus driving 80 km/h, behind a car driving 90 km/h that begins to overtake without caring much about the cars arriving at 160-260 km/h.
When it comes to a stop, I often observed that the first pack of 5 cars needs 13 seconds to move, the second pack 12, the third 11 and the remaining 10. As if the drivers on the begin would need more time to take the decision.
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(at a slower initial speed)
There is no mathematical reason for that, yet you are right. Any idea why?
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9=3**2 - I count ** (exponentiation) as one symbol
6=3*2 -
I started with the left side. IMHO, the only symbols that can be reversed are operators (I do not believe i.e. that -8 is reversed 8).
Just that there cannot be an operator alone.
Any other idea?
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Wording made some of the rules sort of shaky.
Assuming the ant must travel on hands, I understand the ant has to:
a) wait for the Minute hand
b) switch to the Minute hand
c) walk to the end of the Minute handd) wait till the Minute hand reaches 12
which should happen at 7 o'clock.
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Now, I have a problem.
witzar was the first to answer and his answer is correct.
ThunderCloud answered 1 minute later and his explication is clearer.Can I attribute two Best Answers?
Anyway, congratulations to both that they did not fell into the trap.
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Let's toss a fair coin and note the events.
When H H H appears, I pay three bucks
When T H H appears, you pay one buck.And then we start over.
Who is willing to play?
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Majority Rules
in New Logic/Math Puzzles
Posted · Edited by bonanova
spolier
We already had this problem, something like drawing two different fruits from a handbag. I just cannot locate it.