see Answer
- SANE
- gPhoto2
- CUPS
- wvdial
see Answer
see Answer
see Answer
Soumen Nandy has accumulated 640 points.
Points on this battle were won as follows:
Quercus macrocarpa Quercus virginiana Magnolia soulangiana Magnolia grandiflora Gingko biloba Pseudotsuga menziesii Larix laricina Pinus ponderosa Ilex verticillatta Ilex opaca
Points on this battle were won as follows:
Points on this battle were won as follows:
Points on this battle were won as follows:
short explanation:
in a square with respective distances of a, b, c, and d from a
point P to each sequential vertice, it can be shown that the
following relation holds:
(a * a) + (c * c) = (b * b) + (d * d)
in this particular case, a = 7, b = 35, and c = 49. thus:
d = sqrt((7 * 7) + (49 * 49) - (35 * 35)) = 35
longer explanation:
let a, b, c, and d be the respective distances from a point P
(located somewhere within a square of side s) to the four
sequential vertices. let us further define two more lengths, x and
y, according to the picture below:
|--------s---------|
- --------------------
| |\ __/|
| | \ c __/ |
| | b\ __/ |
| | \ __/ |
| | \/ |
s | P - - - - - -| -
| | /\__ | |
| | / \__d | |
| | a/ | \__ | y
| | / \__ | |
| |/ | \| |
- -------------------- -
|-----x------|
from the pythagorean theorem, the following four equations hold:
1) (s-x)*(s-x) + y*y = a*a
2) (s-x)*(s-x) + (s-y)*(s-y) = b*b
3) x*x + (s-y)*(s-y) = c*c
4) x*x + y*y = d*d
we want to solve these for d in terms of a, b, and c. to do so,
simply add 1) to 3) and subtract 2):
[(s-x)*(s-x) + y*y] + [x*x + (s-y)*(s-y)] -
[(s-x)*(s-x) + (s-y)*(s-y)] = a*a + c*c - b*b
simplifying this gives:
x*x + y*y = a*a + c*c - b*b
but the left side of this equation is simply equal to d*d (via 4)).
thus:
a*a + c*c = b*b + d*d
now apply the "short explanation"...
below is a listing of sixty sequential spaces where "x"'s mark the
bad apples and "p"'s mark the perfect apples:
apple 1 apple 10
| |
| |
p p x x p x p x x x
p x p p x x p x p x -- apple 20
x p p x p p x x p x -- apple 30
p x x p p x p p x x
p x p x x p p x p x
x x p x p x x p p x -- apple 60
|
|
apple 51
Soumen Nandy disagrees with the above answer. Here are his comments:
I disagree on the answer to #5. I get 2700, not 2800. IMHO, the method described needs a sample of 60 or 120, rather than 100. 100 is not divisible by 3, guaranteeing a round-off error. 60 and 120 are divisible by all the given fractions My more formal, but simpler, solution (based on the Poisson method) is given below. This kind of problem is often wrongly approached by subtracting the number "spoiled" due to each cause (2000 too small, 1500 too green, and 600 bruised). However, there is overlap in these subsets. Some of the apples that are too small are also too green or bruised. Perhaps the easiest method (one I use quite often) is the Poisson method of "inverting the probability": 2/3 are NOT too small, 3/4 are NOT too green and 9/10 are NOT bruised. The fraction that's not too small AND not too green AND not bruised is: 2/3 * 3/4 * 9/10 = 54/120 (or 2700 perfect apples out of 6000)Response from Duane:
I think you are wrong since the answer posted is simple in its logic and seems right and since your answer is different I would conclude that yours must be wrong. That means that you must be throwing out some perfect apples. Actually one perfect apple out of every 60. I guess the next step, if I get time, will be to go through the 60 apples as diagrammed on the web page and figure out which perfect one you are rejecting. Conversely, one could go through those 60 and figure out which one was listed as perfect in error thus proving your answer.Soumen Nandy replied:
I stand by my answer of 2700, based on the "complement of probability" (not "inverse", as I wrote earlier) - the method you'll find in any probability text for this kind of problem. I can't tell you "which extra apple I would discard", because the other poster didn't discard *any* apples. He discarded integers. We often use numbers an analogs for object, but it can and does get us in trouble This is a very subtle point which I find intriguing, and which I hope you will enjoy, as well. I don't mean to be quarrelsome or repetitive. "The set of integers from 1-60" (or any set of integers) is NOT a representative random sample of a batch of apples. In this case, integers have properties that relate them to one another, which apples don't share. One example: Integer multiples of four CAN'T be consecutive, while it's downright LIKELY that many consecutive real-world apples will be ripe (spoiled, bruised, etc.): every housewife knows that each apple influences the ripeness and softness of its neighbors. Conversely, apples have relations between their properties that integers don't: a green apple may be slightly smaller than it would be if it had finished growing to full ripeness; a green apple is also firmer and harder to bruise. Also, saying "one in four" apples, isn't the same thing as "X modulo 4" in integers, even though "one integer in four" will be X modulo 4. "Every fourth integer" implies a specific structure because integers have an implicit order, Apples don't have implicit order, so "one apple in four" simply means "any one-fourth of the total". In a probability problem, the hypothetical apples must be generic independent widgets, and the three criteria must be independent, or we can't answer at all without the correlation figures relating the properties/criteria [an analysis we often do in medicine]. The flaw in the solution provided is that 2 is a common factor between the NUMBERS 4 and 10, so among INTEGERS, 4 and 10 will share twice as many multiples as independent factors would. (One out of every FIVE multiples of 4 is a multiple of 10, but one in TEN of the green apples would also be bruised) The integer-mapping solution given over-reports the fraction of the fixed number of "bruised" apples that are already "green". ****Of the 15 "green" integers in the sample of 60, three (20, 40, 60) are also "bruised" (multiples of 10). Instead of predicting 150 bruised apples among the 1500 green ones, the integer-map predicts 300 green apples "would've been discarded anyway for bruising". Since there is a fixed number of unsuitable fruit, overestimating the overlap will cause you to overestimate the amount of good fruit remaining Clearly, there is another 50 apples worth of error in the mapping as well, but I haven't had time to think about it. [Besides, the problem wasn't to account for all of someone else's error, it was only to provide the right answer] Since my approach used bulk ratios (and says nothing about any individual item) and calculated the number of good fruit directly, it gives the more accurate bulk result. Again, I'm not trying to give anyone a headache. (If we can't trust integers, what can we trust?) It's just one of those times when 'counting on your fingers' doesn't work. One could construct a problem where integer mapping would give a correct answer by forcing apples to have the same rules of sequence and relation that consecutive integers have. E.g. "I have a bin conveyor with one apple per bin, and a geared device that knocks over every third bin; then later every fourth bin (whether or not that bin still has an apple in it); then knocks over every tenth bin" That's not how we usually sort apples, though, is it?Response from Duane:
I do find this discussion intriguing. I have a few comments. First you stated that 100 is not evenly divisible by 3 guaranteeing a round-off error. The sample was not 100 but 6000 which is evenly divisible by 3. As for the difference between apples and integers, this is basically a math problem on a web site not a pile of actual apples and in general such problems assume that the reader will make certain assumptions implicit in the problem such as the assumption that every 3rd apple may be too small although we know that in real life the apples that were too small would, most likely, be randomly distributed throughout the sample. (In the same way we can postulate a World Series between the Red Sox and the Cubs although we know that in real life, hell probably won't freeze over.) So we essentially have two questions here: Is the integer solution given correct for the mathematical question asked? To what extent can we use abstract mathematics to solve real world problems and what inconsistencies or errors do we introduce by doing so? Of course, you will probably say that math works if we just use your method of "complementing the probability". I say that since apples are distinct objects they can be represented by integers and if two mathematical methods of solving the problem give different answers then either one of the methods is incorrectly applied or a simple math mistake has been made. I suspect that your method does not properly account for overlap.Response from Aaron:
i’m not quite sure that i exactly follow soumen’s logic. i cannot see how the
concept of a “complement of probability” (at least as i learned it in the
realm of binomial or poisson probability) applies in this case. as i learned
it, a “complement of probability” applies only to mutually exclusive events.
but, in this case, the set of “bad” apples is made up of many groups on non-
mutually exclusive “bad” apples. for example, the set of all “too small”
apples (which are divisible by 3) has some overlap with the ones that are “too
green” (which are disivible by 4). apple #12 (divisible by both 3 and 4), for
example, proves that the sets are not mutually exclusive.
but i am a practical guy; perhaps soumen is using some more advanced math than
i am privy to. the best way to settle this problem is not to rely on fancy
arguments but simply to count out 6000 sequential apples and see how many are
“perfect”. i have written a short code (a C++ program) which will do just
that. here is the code:
#include<stdio.h>
int main(){
int goodAppleCount=0, badAppleCount=0, isAppleBad;
for(int apple=1;apple<=6000;apple++){
isAppleBad=0; //each apple is initially considered “good”
if((apple% 3)==0) isAppleBad=1;
if((apple% 4)==0) isAppleBad=1;
if((apple%10)==0) isAppleBad=1;
if(isAppleBad==1) badAppleCount++;
else goodAppleCount++;
if((apple%60)==0)
printf("Total apples: %6d , good apples: %6d , bad apples: %6d\n",
apple,goodAppleCount,badAppleCount);
}
}
this code counts the good and bad apples and prints out the following results along the way:
Total apples: 60 , good apples: 28 , bad apples: 32
Total apples: 120 , good apples: 56 , bad apples: 64
Total apples: 180 , good apples: 84 , bad apples: 96
Total apples: 240 , good apples: 112 , bad apples: 128
Total apples: 300 , good apples: 140 , bad apples: 160
Total apples: 360 , good apples: 168 , bad apples: 192
Total apples: 420 , good apples: 196 , bad apples: 224
Total apples: 480 , good apples: 224 , bad apples: 256
Total apples: 540 , good apples: 252 , bad apples: 288
.
.
.
Total apples: 5760 , good apples: 2688 , bad apples: 3072
Total apples: 5820 , good apples: 2716 , bad apples: 3104
Total apples: 5880 , good apples: 2744 , bad apples: 3136
Total apples: 5940 , good apples: 2772 , bad apples: 3168
Total apples: 6000 , good apples: 2800 , bad apples: 3200
you can clearly see that, in every group of sixty apples, 28 are “perfect” and 32 are “bad”. this pattern continues all the way up to 6000 total apples. thus, i am forced to conclude that 2800 “perfect” apples is the correct answer.
Response from Daniel:
To rebut some of the comments made by Soumen Nandy 1. The sample size was 60. The figure '100' mentioned referred to "100 groups of sixty". Yes, 60 or 120 are appropriate smallest group sizes. So are 180, 240, or any multiple of 60 because 60 is the smallest number that 3, 4, and 10 will divide into evenly. Thus after 60 apples have cycled through, you are back to the start of the integer map pattern. 2. Probability. Why is probability and the concept of identifying a sample incorrect for this answer? Because the way the question was written suggests that we are not taking a sample. Certainly, if 10% of the apples were bruised, and 25% of the apples were green etc then this would be a candidate for sampling and the use of statistics. However, as every third, fourth and tenth apple is involved and all the apples are checked, (assumptions, i suppose) my concept of visualising all the apples coming past you on a conveyor is also appropriate. As Soumen notes, each item must be a 'generic independant widget' Perhaps this question would have been better put as one where a mechanical device was produced with three components. If every fourth component A was faulty, every third component B was faulty and every tenth component C was faulty, then you could automatically assume that a certain number of certain devices would contain faulty components and also identify which number they were on the conveyor. The analoge nature of problems with apples (greenish or affected by the rest of the barrel) and the fact that 'real' apples won't appear on a conveyor in model order require certain assumptions to be made. Manufactured goods require less assumptions because you can build the faults in. Small only - 1400 Green only - 800 Bruised only - 200 S&G only - 400 (lower number occur more often so are more likely to co-incide) S&B only - 100 G&B only - 200 (both have 2 as a factor so co-incidence occurs more frequently) S&G&B only - 100 S(total) = 1400 + 400 + 100 + 100 = 2000 (2000/6000 = 1/3) G(total) = 800 + 400 + 200 + 100 = 1500 (1500/6000 = 1/4) B(total) = 200 + 100 + 200 + 100 = 600 (600/6000 = 1/10) where (total = all apples that are defective in this way, even if they are also defective in other ways) Summary I think that this argument comes down to models and assumptions. I made the assumption that the model was relatively simplistic and that each 'apple' was in fact an 'independant generic widget'. Thus integer mapping is appropriate and a useful tool. Soumen made the assumption we were modelling 'real' apples and used stastics and probability - better for 'real' apples, less so for widgets. I won't be as dismissive as old (lies, damned lies and statistics) Winston Churchill but I would be interested to see why or if there is a difference between a statistical result and the integer mapping (or counting :D ) techniques.Soumen Nandy replied:
If "taking [literally] every fourth apple" works, it must work equally well for all equivalent cases of "every fourth apple", but if I take every fourth apple, beginning with the first, second, third or fourth apple, I get four different answers.
Here are the possible sequential integer maps for this method (In case you didn't have the "New Math" in school "N mod 4" means the remainder after integer division of N by 4) I've expanded each entry to indicate each reason for discarding an apple, so that the overlaps can be plainly seen.
N MOD 4 = 0 [multiples of 4) 28 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: PPP PPP 3-- -4- PPP 3-- PPP -4- 3-- --X
10: PPP 34- PPP PPP 3-- -4- PPP 3-- PPP --X
20: 3-- PPP PPP 34- PPP PPP 3-- -4- PPP -4X
30: PPP -4- 3-- PPP PPP 34- PPP PPP 3-- 3-x
40: PPP 3-- PPP -4- 3-- PPP PPP 34- PPP -4X
50: 3-- -4- PPP 3-- PPP -4- 3-- PPP PPP 34X
N MOD 4 = 1 [Case A] 26 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: -A- PPP 3-- PPP -A- 3-- PPP PPP 3A- --X
01: PPP 3-- -A- PPP 3-- PPP -A- 3-- PPP --X
02: 3A- PPP PPP 3-- -A- PPP 3-- PPP -A- 3-X
03: PPP PPP 3A- PPP PPP 3-- -A- PPP 3-- --X
04: -A- 3-- PPP PPP 3A- PPP PPP 3-- -A- --X
05: 3-- PPP -A- 3-- PPP PPP 3A- PPP PPP 3-X
N MOD 4 = 2 [Case B] 29 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: PPP -B- 3-- PPP PPP 3B- PPP PPP 3-- -BX
01: PPP 3-- PPP -B- 3-- PPP PPP 3B- PPP --X
02: 3-- -B- PPP 3-- PPP -B- 3-- PPP PPP 3BX
03: PPP PPP 3-- -B- PPP 3-- PPP -B- 3-- --X
04: PPP 3B- PPP PPP 3-- -B- PPP 3-- PPP -BX
05: 3-- PPP PPP 3B- PPP PPP 3-- -B- PPP 3-X
N MOD 4 = 3 [Case C] 25 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: --- PPP 3C- PPP PPP 3-- -C- PPP 3-- --X
01: -C- 3-- PPP PPP 3C- PPP ppp 3-- -C- --X
02: 3-- PPP -C- 3-- PPP PPP 3C- PPP PPP 3-X
03: -C- PPP 3-- PPP -C- 3-- PPP PPP 3C- --X
04: PPP 3-- -C- PPP 3-- PPP -C- 3-- PPP --X
05: 3C- PPP PPP 3-- -C- PPP 3-- PPP -C- 3-X
None of the four cases agrees, yet each must be equally valid if the logic
holds. The average value [(28+26+29+25)/4 = 27] happens to agree which the
value that I calculated using the standard complement probability method. (probability that a random apple isn't too small) x (probability that a random apple isn't too green) x (probability that a random apple isn't bruised) --------------------------------------------------- (probability that a random apple isn't too small, green or bruised)If there is any remaining doubt about the flaw in fixed interval integer mapping, let me suggest the simplest possible problem as a test case.
Points on this battle were won as follows:
Then the Lord sent venomous snakes among them; they bit the people and many Israelites died. The people came to Moses and said, "We sinned when we spoke against the Lord and against you. Pray that the Lord will take the snakes away from us." So Moses prayed for the people. The Lord said to Moses, "Make a snake and put it up on a pole; anyone who is bitten can look at it and live." So Moses made a bronze snake and put it up on a pole. Then whenany one was bitten by a snake and looked at the bronze snake, he lived.Representation in Art
Officer, officer, there they are-- Dolores Haze and her lover! Whip out your gun and follow that car. Now tumble out and take cover.
Points on this battle were won as follows:
A$="Saddam Hussein"
T=0
FOR Z=1 TO LEN(A$)
T=T+ASC(MID$(A$,Z,1))
NEXT Z
PRINT T
Points on this battle were won as follows:
Well, that mockingbird's gonna sail away, We're gonna forget it. That big, fat moon is gonna shine like a spoon, But we're gonna let it, You won't regret it.
Points on this battle were won as follows:
Points on this battle were won as follows:
Points on this battle were won as follows:
short explanation:
in a square with respective distances of a, b, c, and d from a
point P to each sequential vertice, it can be shown that the
following relation holds:
(a * a) + (c * c) = (b * b) + (d * d)
in this particular case, a = 7, b = 35, and c = 49. thus:
d = sqrt((7 * 7) + (49 * 49) - (35 * 35)) = 35
longer explanation:
let a, b, c, and d be the respective distances from a point P
(located somewhere within a square of side s) to the four
sequential vertices. let us further define two more lengths, x and
y, according to the picture below:
|--------s---------|
- --------------------
| |\ __/|
| | \ c __/ |
| | b\ __/ |
| | \ __/ |
| | \/ |
s | P - - - - - -| -
| | /\__ | |
| | / \__d | |
| | a/ | \__ | y
| | / \__ | |
| |/ | \| |
- -------------------- -
|-----x------|
from the pythagorean theorem, the following four equations hold:
1) (s-x)*(s-x) + y*y = a*a
2) (s-x)*(s-x) + (s-y)*(s-y) = b*b
3) x*x + (s-y)*(s-y) = c*c
4) x*x + y*y = d*d
we want to solve these for d in terms of a, b, and c. to do so,
simply add 1) to 3) and subtract 2):
[(s-x)*(s-x) + y*y] + [x*x + (s-y)*(s-y)] -
[(s-x)*(s-x) + (s-y)*(s-y)] = a*a + c*c - b*b
simplifying this gives:
x*x + y*y = a*a + c*c - b*b
but the left side of this equation is simply equal to d*d (via 4)).
thus:
a*a + c*c = b*b + d*d
now apply the "short explanation"...
Points on this battle were won as follows:
below is a listing of sixty sequential spaces where "x"'s mark the
bad apples and "p"'s mark the perfect apples:
apple 1 apple 10
| |
| |
p p x x p x p x x x
p x p p x x p x p x -- apple 20
x p p x p p x x p x -- apple 30
p x x p p x p p x x
p x p x x p p x p x
x x p x p x x p p x -- apple 60
|
|
apple 51
Soumen Nandy disagrees with the above answer. Here are his comments:
I disagree on the answer to #5. I get 2700, not 2800. IMHO, the method described needs a sample of 60 or 120, rather than 100. 100 is not divisible by 3, guaranteeing a round-off error. 60 and 120 are divisible by all the given fractions My more formal, but simpler, solution (based on the Poisson method) is given below. This kind of problem is often wrongly approached by subtracting the number "spoiled" due to each cause (2000 too small, 1500 too green, and 600 bruised). However, there is overlap in these subsets. Some of the apples that are too small are also too green or bruised. Perhaps the easiest method (one I use quite often) is the Poisson method of "inverting the probability": 2/3 are NOT too small, 3/4 are NOT too green and 9/10 are NOT bruised. The fraction that's not too small AND not too green AND not bruised is: 2/3 * 3/4 * 9/10 = 54/120 (or 2700 perfect apples out of 6000)Response from Duane:
I think you are wrong since the answer posted is simple in its logic and seems right and since your answer is different I would conclude that yours must be wrong. That means that you must be throwing out some perfect apples. Actually one perfect apple out of every 60. I guess the next step, if I get time, will be to go through the 60 apples as diagrammed on the web page and figure out which perfect one you are rejecting. Conversely, one could go through those 60 and figure out which one was listed as perfect in error thus proving your answer.Soumen Nandy replied:
I stand by my answer of 2700, based on the "complement of probability" (not "inverse", as I wrote earlier) - the method you'll find in any probability text for this kind of problem. I can't tell you "which extra apple I would discard", because the other poster didn't discard *any* apples. He discarded integers. We often use numbers an analogs for object, but it can and does get us in trouble This is a very subtle point which I find intriguing, and which I hope you will enjoy, as well. I don't mean to be quarrelsome or repetitive. "The set of integers from 1-60" (or any set of integers) is NOT a representative random sample of a batch of apples. In this case, integers have properties that relate them to one another, which apples don't share. One example: Integer multiples of four CAN'T be consecutive, while it's downright LIKELY that many consecutive real-world apples will be ripe (spoiled, bruised, etc.): every housewife knows that each apple influences the ripeness and softness of its neighbors. Conversely, apples have relations between their properties that integers don't: a green apple may be slightly smaller than it would be if it had finished growing to full ripeness; a green apple is also firmer and harder to bruise. Also, saying "one in four" apples, isn't the same thing as "X modulo 4" in integers, even though "one integer in four" will be X modulo 4. "Every fourth integer" implies a specific structure because integers have an implicit order, Apples don't have implicit order, so "one apple in four" simply means "any one-fourth of the total". In a probability problem, the hypothetical apples must be generic independent widgets, and the three criteria must be independent, or we can't answer at all without the correlation figures relating the properties/criteria [an analysis we often do in medicine]. The flaw in the solution provided is that 2 is a common factor between the NUMBERS 4 and 10, so among INTEGERS, 4 and 10 will share twice as many multiples as independent factors would. (One out of every FIVE multiples of 4 is a multiple of 10, but one in TEN of the green apples would also be bruised) The integer-mapping solution given over-reports the fraction of the fixed number of "bruised" apples that are already "green". ****Of the 15 "green" integers in the sample of 60, three (20, 40, 60) are also "bruised" (multiples of 10). Instead of predicting 150 bruised apples among the 1500 green ones, the integer-map predicts 300 green apples "would've been discarded anyway for bruising". Since there is a fixed number of unsuitable fruit, overestimating the overlap will cause you to overestimate the amount of good fruit remaining Clearly, there is another 50 apples worth of error in the mapping as well, but I haven't had time to think about it. [Besides, the problem wasn't to account for all of someone else's error, it was only to provide the right answer] Since my approach used bulk ratios (and says nothing about any individual item) and calculated the number of good fruit directly, it gives the more accurate bulk result. Again, I'm not trying to give anyone a headache. (If we can't trust integers, what can we trust?) It's just one of those times when 'counting on your fingers' doesn't work. One could construct a problem where integer mapping would give a correct answer by forcing apples to have the same rules of sequence and relation that consecutive integers have. E.g. "I have a bin conveyor with one apple per bin, and a geared device that knocks over every third bin; then later every fourth bin (whether or not that bin still has an apple in it); then knocks over every tenth bin" That's not how we usually sort apples, though, is it?Response from Duane:
I do find this discussion intriguing. I have a few comments. First you stated that 100 is not evenly divisible by 3 guaranteeing a round-off error. The sample was not 100 but 6000 which is evenly divisible by 3. As for the difference between apples and integers, this is basically a math problem on a web site not a pile of actual apples and in general such problems assume that the reader will make certain assumptions implicit in the problem such as the assumption that every 3rd apple may be too small although we know that in real life the apples that were too small would, most likely, be randomly distributed throughout the sample. (In the same way we can postulate a World Series between the Red Sox and the Cubs although we know that in real life, hell probably won't freeze over.) So we essentially have two questions here: Is the integer solution given correct for the mathematical question asked? To what extent can we use abstract mathematics to solve real world problems and what inconsistencies or errors do we introduce by doing so? Of course, you will probably say that math works if we just use your method of "complementing the probability". I say that since apples are distinct objects they can be represented by integers and if two mathematical methods of solving the problem give different answers then either one of the methods is incorrectly applied or a simple math mistake has been made. I suspect that your method does not properly account for overlap.Response from Aaron:
i’m not quite sure that i exactly follow soumen’s logic. i cannot see how the
concept of a “complement of probability” (at least as i learned it in the
realm of binomial or poisson probability) applies in this case. as i learned
it, a “complement of probability” applies only to mutually exclusive events.
but, in this case, the set of “bad” apples is made up of many groups on non-
mutually exclusive “bad” apples. for example, the set of all “too small”
apples (which are divisible by 3) has some overlap with the ones that are “too
green” (which are disivible by 4). apple #12 (divisible by both 3 and 4), for
example, proves that the sets are not mutually exclusive.
but i am a practical guy; perhaps soumen is using some more advanced math than
i am privy to. the best way to settle this problem is not to rely on fancy
arguments but simply to count out 6000 sequential apples and see how many are
“perfect”. i have written a short code (a C++ program) which will do just
that. here is the code:
#include<stdio.h>
int main(){
int goodAppleCount=0, badAppleCount=0, isAppleBad;
for(int apple=1;apple<=6000;apple++){
isAppleBad=0; //each apple is initially considered “good”
if((apple% 3)==0) isAppleBad=1;
if((apple% 4)==0) isAppleBad=1;
if((apple%10)==0) isAppleBad=1;
if(isAppleBad==1) badAppleCount++;
else goodAppleCount++;
if((apple%60)==0)
printf("Total apples: %6d , good apples: %6d , bad apples: %6d\n",
apple,goodAppleCount,badAppleCount);
}
}
this code counts the good and bad apples and prints out the following results along the way:
Total apples: 60 , good apples: 28 , bad apples: 32
Total apples: 120 , good apples: 56 , bad apples: 64
Total apples: 180 , good apples: 84 , bad apples: 96
Total apples: 240 , good apples: 112 , bad apples: 128
Total apples: 300 , good apples: 140 , bad apples: 160
Total apples: 360 , good apples: 168 , bad apples: 192
Total apples: 420 , good apples: 196 , bad apples: 224
Total apples: 480 , good apples: 224 , bad apples: 256
Total apples: 540 , good apples: 252 , bad apples: 288
.
.
.
Total apples: 5760 , good apples: 2688 , bad apples: 3072
Total apples: 5820 , good apples: 2716 , bad apples: 3104
Total apples: 5880 , good apples: 2744 , bad apples: 3136
Total apples: 5940 , good apples: 2772 , bad apples: 3168
Total apples: 6000 , good apples: 2800 , bad apples: 3200
you can clearly see that, in every group of sixty apples, 28 are “perfect” and 32 are “bad”. this pattern continues all the way up to 6000 total apples. thus, i am forced to conclude that 2800 “perfect” apples is the correct answer.
Response from Daniel:
To rebut some of the comments made by Soumen Nandy 1. The sample size was 60. The figure '100' mentioned referred to "100 groups of sixty". Yes, 60 or 120 are appropriate smallest group sizes. So are 180, 240, or any multiple of 60 because 60 is the smallest number that 3, 4, and 10 will divide into evenly. Thus after 60 apples have cycled through, you are back to the start of the integer map pattern. 2. Probability. Why is probability and the concept of identifying a sample incorrect for this answer? Because the way the question was written suggests that we are not taking a sample. Certainly, if 10% of the apples were bruised, and 25% of the apples were green etc then this would be a candidate for sampling and the use of statistics. However, as every third, fourth and tenth apple is involved and all the apples are checked, (assumptions, i suppose) my concept of visualising all the apples coming past you on a conveyor is also appropriate. As Soumen notes, each item must be a 'generic independant widget' Perhaps this question would have been better put as one where a mechanical device was produced with three components. If every fourth component A was faulty, every third component B was faulty and every tenth component C was faulty, then you could automatically assume that a certain number of certain devices would contain faulty components and also identify which number they were on the conveyor. The analoge nature of problems with apples (greenish or affected by the rest of the barrel) and the fact that 'real' apples won't appear on a conveyor in model order require certain assumptions to be made. Manufactured goods require less assumptions because you can build the faults in. Small only - 1400 Green only - 800 Bruised only - 200 S&G only - 400 (lower number occur more often so are more likely to co-incide) S&B only - 100 G&B only - 200 (both have 2 as a factor so co-incidence occurs more frequently) S&G&B only - 100 S(total) = 1400 + 400 + 100 + 100 = 2000 (2000/6000 = 1/3) G(total) = 800 + 400 + 200 + 100 = 1500 (1500/6000 = 1/4) B(total) = 200 + 100 + 200 + 100 = 600 (600/6000 = 1/10) where (total = all apples that are defective in this way, even if they are also defective in other ways) Summary I think that this argument comes down to models and assumptions. I made the assumption that the model was relatively simplistic and that each 'apple' was in fact an 'independant generic widget'. Thus integer mapping is appropriate and a useful tool. Soumen made the assumption we were modelling 'real' apples and used stastics and probability - better for 'real' apples, less so for widgets. I won't be as dismissive as old (lies, damned lies and statistics) Winston Churchill but I would be interested to see why or if there is a difference between a statistical result and the integer mapping (or counting :D ) techniques.Soumen Nandy replied:
If "taking [literally] every fourth apple" works, it must work equally well for all equivalent cases of "every fourth apple", but if I take every fourth apple, beginning with the first, second, third or fourth apple, I get four different answers.
Here are the possible sequential integer maps for this method (In case you didn't have the "New Math" in school "N mod 4" means the remainder after integer division of N by 4) I've expanded each entry to indicate each reason for discarding an apple, so that the overlaps can be plainly seen.
N MOD 4 = 0 [multiples of 4) 28 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: PPP PPP 3-- -4- PPP 3-- PPP -4- 3-- --X
10: PPP 34- PPP PPP 3-- -4- PPP 3-- PPP --X
20: 3-- PPP PPP 34- PPP PPP 3-- -4- PPP -4X
30: PPP -4- 3-- PPP PPP 34- PPP PPP 3-- 3-x
40: PPP 3-- PPP -4- 3-- PPP PPP 34- PPP -4X
50: 3-- -4- PPP 3-- PPP -4- 3-- PPP PPP 34X
N MOD 4 = 1 [Case A] 26 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: -A- PPP 3-- PPP -A- 3-- PPP PPP 3A- --X
01: PPP 3-- -A- PPP 3-- PPP -A- 3-- PPP --X
02: 3A- PPP PPP 3-- -A- PPP 3-- PPP -A- 3-X
03: PPP PPP 3A- PPP PPP 3-- -A- PPP 3-- --X
04: -A- 3-- PPP PPP 3A- PPP PPP 3-- -A- --X
05: 3-- PPP -A- 3-- PPP PPP 3A- PPP PPP 3-X
N MOD 4 = 2 [Case B] 29 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: PPP -B- 3-- PPP PPP 3B- PPP PPP 3-- -BX
01: PPP 3-- PPP -B- 3-- PPP PPP 3B- PPP --X
02: 3-- -B- PPP 3-- PPP -B- 3-- PPP PPP 3BX
03: PPP PPP 3-- -B- PPP 3-- PPP -B- 3-- --X
04: PPP 3B- PPP PPP 3-- -B- PPP 3-- PPP -BX
05: 3-- PPP PPP 3B- PPP PPP 3-- -B- PPP 3-X
N MOD 4 = 3 [Case C] 25 PERFECTS
----------------------------------------------------
1 2 3 4 5 6 7 8 9 10
00: --- PPP 3C- PPP PPP 3-- -C- PPP 3-- --X
01: -C- 3-- PPP PPP 3C- PPP ppp 3-- -C- --X
02: 3-- PPP -C- 3-- PPP PPP 3C- PPP PPP 3-X
03: -C- PPP 3-- PPP -C- 3-- PPP PPP 3C- --X
04: PPP 3-- -C- PPP 3-- PPP -C- 3-- PPP --X
05: 3C- PPP PPP 3-- -C- PPP 3-- PPP -C- 3-X
None of the four cases agrees, yet each must be equally valid if the logic
holds. The average value [(28+26+29+25)/4 = 27] happens to agree which the
value that I calculated using the standard complement probability method. (probability that a random apple isn't too small) x (probability that a random apple isn't too green) x (probability that a random apple isn't bruised) --------------------------------------------------- (probability that a random apple isn't too small, green or bruised)If there is any remaining doubt about the flaw in fixed interval integer mapping, let me suggest the simplest possible problem as a test case.
Points on this battle were won as follows:
Last revised November 19, 2003.
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