I'm a total rookie and I'm trying to write a simple code that compares 3 random numbers for a slot machine type game.
In this case only between 1 and 3 since I'm encountering a problem.
It seems that only the combinations of 1,1,1 and 2,2,1 and 3,3,1 win.
Does anybody understand why 2,2,1 and 3,3,1 are winning and 2,2,2 and 3,3,3 do not?
let num1 = Math.floor(Math.random() * 3 + 1);
let num2 = Math.floor(Math.random() * 3 + 1);
let num3 = Math.floor(Math.random() * 3 + 1);
function guess(){
console.log(num1, num2, num3)
if(num1==num2==num3){
console.log('JACKPOT!');
}
else{
console.log('try again');
}
}
guess()
Thanks in advance!
I'm not JS enthusiast, but I think it would be better if you compare the 3 numbers in a different way like that for example:
num1==num2 && num2==num3
I think that in your code works even if just num1 and num2 compare goes well
It is because when using a==b==c you are using booblean algebra, so 1==1 is true and true==1 will be true, when evaluating you will end up with only 1 1 1 working or 1 1 X working or 1 2 2 working. If going to use booblean algebra you have to evaluate true = true & true = true, the AND will evaluate that if in the left side is true AND the right side is true then JACKPOT.
function guess(){
let num1 = Math.floor(Math.random() * 3 + 1);
let num2 = Math.floor(Math.random() * 3 + 1);
let num3 = Math.floor(Math.random() * 3 + 1);
console.log(num1, num2, num3)
if(num1==num2&&num2==num3){
console.log('JACKPOT!');
}
else{
console.log('try again');
}
}
guess()
Related
So, for my group work, we were tasked to write pseudocode. We had just been learning it for, three weeks (almost a month) now.
The question is:
First two numbers are 1 and 2. After that, each succeeding number is the multiple of the two preceding.
1,2,2,4,8,32....
Write pseudocode that will display the set of numbers up to 32.
My groupmates and I basically get the concept of it. So it's like num1 x num2 = num3. Then num3 x num2 = num4, and it loops for 32 times, ect...I was wondering if there is an easier way to 'overwrite' the numbers so that instead of putting 'num4' 'num5' and more, we can just use num 1,2,3 repeatedly since the process is the same.
Set 1 = 1 x 2 = 2
Set 2 = 2 x 2 = 4
Set 3 = 2 x 4 = 8
Set 4 = 8 x 4 = 32
This shall do the trick:
int num1 = 1;
int num2 = 2;
while( num2 < 32 )
{
int temp = num1 * num2;
print(temp);
num1 = num2;
num2 = temp;
}
I took two ways to round numbers to decimals. First function just rounds the number:
function round(num)
local under = math.floor(num)
local over = math.floor(num) + 1
local underV = -(under - num)
local overV = over - num
if overV > underV then
return under
else
return over
end
end
The next two functions use this function to round a number to decimals:
function roundf(num, dec)
return round(num * (1 * dec)) / (1 * dec)
end
function roundf_alt(num, dec)
local r = math.exp(1 * math.log(dec));
return round(r * num) / r;
end
Why not simply
function round(num)
return num >= 0 and math.floor(num+0.5) or math.ceil(num-0.5)
end
Instead of math.floor(num) + 1 you can simply use math.ceil(num) btw.
Why do you multiply with 1 multiple times?
There are many things to consider when rounding numbers. Please do some research on how to handle special cases.
With d3.js we can achieve eased time out of normalized time t, typically in the range [0,1]
For example:
d3.easeCubic(0.25) = 0.0625
How can we reverse that, how can we find x given known y ?
d3.easeCubic(X) = 0.0625,
X ???
The answer here is cubic root, but still.
The problem is in reusability, ease function can change to d3.easeExpIn, or `d3.easeCircleOut, or any other, do you need to invent reverse functions on your own, or are they hidden anywhere ?
Firstly, your math is wrong. d3.easeCubic(0.25) will give you 0.0625:
var easy = d3.easeCubic(0.25);
console.log(easy);
<script src="https://d3js.org/d3.v4.min.js"></script>
Now, back to your question:
How can we reverse that, how can we find x given known y?
There is no native solution, but we can create our own function to find X given a known Y. The problem, of course, is that we have to invert the math for each specific easing... But, since you asked about d3.easeCubic, which is the same of d3.easeCubicInOut, let's try to create an inverted function for that particular easing.
First step, let's have a look at the source code:
export function cubicInOut(t) {
return ((t *= 2) <= 1 ? t * t * t : (t -= 2) * t * t + 2) / 2;
}
You can easily see that this is the correct function, giving us the same value as the first snippet:
function cubicInOut(t) {
return ((t *= 2) <= 1 ? t * t * t : (t -= 2) * t * t + 2) / 2;
}
console.log(cubicInOut(0.25))
Now, let's try to invert it.
The math here is somehow complicated, but for values less than 1, here is the function:
function inverseEaseCubic(t){
return Math.cbrt(t * 2) / 2;
}
And here is the demo. We pass 0.0625 to the function, and it returns 0.25:
function inverseEaseCubic(t){
return Math.cbrt(t * 2) / 2;
}
console.log(inverseEaseCubic(0.0625))
If you want to deal with numbers bigger than 1, this is the complete function:
function InverseEaseCubic(t){
return t <= 1 ? Math.cbrt(t * 2) / 2 : (Math.cbrt(2 * t - 2) + 2) / 2;
}
PS: In his comment, #altocumulus just reminded us that, sometimes, it's even impossible to find the value. Here is a very simple example. Suppose this function:
function exponentiation(a){
return a*a;
}
Now imagine that, when called with an unknown argument, the function returned 4. What's the argument? Can we find out? Impossible to determine, because second degree equations, like this one, have 2 roots:
console.log(exponentiation(2))//returns 4
console.log(exponentiation(-2))//also returns 4
I used the #Gerardo Furtado answer but the inverse function didn't work well so I wrote another
function cubicInOut(t) {
return ((t *= 2) <= 1 ? t * t * t : (t -= 2) * t * t + 2) / 2;
}
function inverseEaseCubic(x) {
return x < .5 ? Math.cbrt(x / 4) : (2 - Math.cbrt(2 - 2 * x)) / 2;
}
console.log(inverseEaseCubic(cubicInOut(1)) === 1);
console.log(inverseEaseCubic(cubicInOut(0.6)) === 0.6);
console.log(inverseEaseCubic(cubicInOut(0.4)) === 0.4);
console.log(inverseEaseCubic(cubicInOut(0.1)) === 0.1);
console.log(inverseEaseCubic(cubicInOut(0)) === 0);
I am trying to do a fast exponentiation. But the result does not seem to produce the correct result. Any help would be appreciated.
EDIT: Manage to solve it thanks for all the help.
if (content[i] == '1')
s1 = (int)(po1 * (Math.pow(po1, 2)));
else
s1 = po1 * po1;
final_result *= temp;
Check out this Exponation by squaring
You probably want to bit-shift right and square your base each time you encounter a 1 bit in the exponent
int pow(int base, int e)
{
int retVal = 1;
while (e)
{
if (e % 2 == 1)//i.e. last bit of exponent is 1
retVal *= base;
e >>= 1; //bitshift exponent to the right.
base *= base; // square base since we shifted 1 bit in our exponent
}
return retVal ;
}
A good way of thinking about it is that your exponent is being broken down: say, 6^7 (exponent in bits is 1, 1, 1) = 6^1 * 6^2 * 6^4 = 6 * 36 * 36^2 = 6 * 36 * 1296. Your base is always squaring itself.
temp = (int)(g1 * (Math.pow(g1, 2)));
This basically just boils down to g13. I'm not familiar with this algorithm but this can't be right.
Also, as a side note, don't ever call Math.pow(<var>, 2), just write <var> * <var>.
There are several problems with your code, starting with the fact that you are reading the exp string in the wrong direction, adding extra multiplications by the base, and not considering the rank of the 1 when raising the powers of 2.
Here is a python quick sketch of what you are trying to achieve:
a = int(raw_input("base"))
b = "{0:b}".format(int(raw_input("exp")))
res = 1
for index, i in enumerate(b[::-1]):
if i == '1':
res *= a**(2**index)
print res
Alternatively, you could square a at every iteration instead:
for index, i in enumerate(b[::-1]):
if i == '1':
res *= a
a *= a
I need to calculate 5-star ratings like the one on Amazon website. I have done enough search to find what is the best algorithm, but I am not able to get a proper answer. For example, if these are the ratings
5 star - 252
4 star - 124
3 star - 40
2 star - 29
1 star - 33
totally 478 reviews
Amazon has calculated this to be "4.1 out of 5 stars". Can anyone tell me how this figure is arrived at? I am not able to get this just by doing average.
That's a weighted average, where you weigh each rating with the number of votes it got:
(5*252 + 4*124 + 3*40 + 2*29 + 1*33) / (252+124+40+29+33) = 4.11 and change
If you are start calculation of overall rating from beginning then this formula will help you.
Formula
((Overall Rating * Total Rating) + new Rating) / (Total Rating + 1)
Example
suppose you have no ratings till now then formula is like,
overall rating is "0" till now.
total rating "0"
and given rating is "4"
((0*0)+4)/1 = 4
If overall rating is "4.11" Total rating is "478" And new rating
giving by one user is "2"
then formula is like
((4.11 * 478)+ 2)/479 // 479 is increment of new rating from 478
a better way to do this,
rating = (sum_of_rating * 5)/sum_of_max_rating_of_user_count
example:
total users rated: 6
sum_of_max_rating_of_user_count: 6 x 5 = 30
sum_of_rating: 25
rating = (25 * 5) / 30
Done!
Yes, you can average them out:
(5 * 252 + 4 * 124 + 3 * 40 + 2 * 29 + 1 * 33) / 478 = 4.11
Super helpful reply by Blindy, here's the PHP code that's based on it. Some may find useful. The results will be 4.11 as per OP's example:
$ratings = array(
5 => 252,
4 => 124,
3 => 40,
2 => 29,
1 => 33
);
function calcAverageRating($ratings) {
$totalWeight = 0;
$totalReviews = 0;
foreach ($ratings as $weight => $numberofReviews) {
$WeightMultipliedByNumber = $weight * $numberofReviews;
$totalWeight += $WeightMultipliedByNumber;
$totalReviews += $numberofReviews;
}
//divide the total weight by total number of reviews
$averageRating = $totalWeight / $totalReviews;
return $averageRating;
}
How to build the above $ratings array
Example pseudo code, but which should work that explains how to build the $ratings array when info is stored in DB assuming you have a table called "ratings" and a column called "rating". In this case it's 1 join, you would need to do 4 joins to get all ratings, but this should get you started:
SELECT count(c1.rating) as one_star, count(c2.rating) as two_star
FROM ratings c1
LEFT OUTER JOIN
ratings c2
ON
c1.id = c2.id
WHERE
c1.rating = 1
AND
c2.rating = 2
another approach suggested in comments
SELECT SUM(rating = 1) AS one_s ,SUM(rating = 2) AS two_s ,SUM(rating = 3) as three_s FROM reviews where product_id = 9
This rating system is based on a weighted average or weighted mean. That is, they used the weight in terms of stars to compute a decimal value which rounds to 4.1. For example:
Sum of (weight * number of reviews at that weight) / total number of reviews
(5*252 + 4*124 + 3*40 + 2*29 + 1*33) / 478 = 4.1
Weighted average, sum the number of stars times its weight, and then divide it through by the total number of reviews.
According to your question your solution will be like this.
Sum of (Rate*TotalRatingOfThatRate)/ TotalNumberOfReviews
((5*252)+(4*124)+(3*40)+(2*29)+(1*33)) / (252+124+40+29+33)
output will be 4.1
in Javascript
function calcAverageRating(ratings) {
let totalWeight = 0;
let totalReviews = 0;
ratings.forEach((rating) => {
const weightMultipliedByNumber = rating.weight * rating.count;
totalWeight += weightMultipliedByNumber;
totalReviews += rating.count;
});
const averageRating = totalWeight / totalReviews;
return averageRating.toFixed(2);
}
const ratings = [
{
weight: 5,
count: 252
},
{
weight: 4,
count: 124
},
{
weight: 3,
count: 40
},
{
weight: 2,
count: 29
},
{
weight: 1,
count: 33
}
];
console.log(calcAverageRating(ratings));
In addition, I am just trying to make practical and full code for all.
My Json Object Array
var yourRatingData =[{
review_id:1,
customer_id:5,
customer_name:"Faysal",
rating:5,
review_content:"I like this product it's cool and best in quality"
},
{
review_id:2,
customer_id:6,
customer_name:"Adams",
rating:4,
review_content:"It's quality product though price a bit high"
},
{
review_id:3,
customer_id:8,
customer_name:"Jane",
rating:3,
review_content:"I like but should improve quality"
},
{
review_id:4,
customer_id:9,
customer_name:"Julia",
rating:1,
review_content:"It's not good"
}];
Rating Calculation
let _5star = yourRatingData.filter(r=>r.rating==5).length;
let _4star = yourRatingData.filter(r=>r.rating==4).length;
let _3star = yourRatingData.filter(r=>r.rating==3).length;
let _2star = yourRatingData.filter(r=>r.rating==2).length;
let _1star = yourRatingData.filter(r=>r.rating==1).length;
//Sum of individual star.
let sumOfRating = parseInt( _5star + _4star + _3star + _2star + _1star );
//Total number of rating
let overallRating = parseInt( 5*_5star + 4*_4star + 3*_3star + 2*_2star +1*_1star );
//Average of all rating
let averageRating = parseFloat(overallRating/sumOfRating);
//Percentage of each star rating
let _5starPercentage = parseInt((_5star/totalRating)*100);
let _4starPercentage = parseInt((_4star/totalRating)*100);
let _3starPercentage = parseInt((_3star/totalRating)*100);
let _2starPercentage = parseInt((_2star/totalRating)*100);
let _1starPercentage = parseInt((_1star/totalRating)*100);
I think it's helpful.
This is an example method using flutter,
double starRating = 5.0;
getRating() {
// ! Check Already Accepted by others or Not -----------------------------------
//
int totalof5s = 0;
int totalof4s = 0;
int totalof3s = 0;
int totalof2s = 0;
int totalof1s = 0;
//
FirebaseFirestore.instance
.collection('usersRating')
.where("passengerUid", isEqualTo: "GNblJJJsjicaA2vkXJNJ6XCAiwa2")
.snapshots()
.forEach((querySnapshot) {
if (querySnapshot.size > 0) {
querySnapshot.docs.forEach((element) async {
if (element["rating"] == 5) {
totalof5s++;
} else if (element["rating"] == 4) {
totalof4s++;
} else if (element["rating"] == 3) {
totalof3s++;
} else if (element["rating"] == 2) {
totalof2s++;
} else if (element["rating"] == 1) {
totalof1s++;
}
//
if (this.mounted) {
setState(() {
starRating = (5 * totalof5s +
4 * totalof4s +
3 * totalof3s +
2 * totalof2s +
1 * totalof1s) /
(totalof5s + totalof4s + totalof3s + totalof2s + totalof1s);
});
}
});
} else {
// This is default one in any case these user doesn't have any rating document exists
if (this.mounted) {
setState(() {
starRating = 5.0;
});
}
}
});
}
(Total nunber of star / total number of persons who review * 5 ) * 5
= Answer
Fixed decimals in js to 1.
answer.toFixed(1);
Example the total reviews of 5 person is 20 star.
(20/5*5)*5 = 4.0
I used this way to calculate ratings and it is working perfectly
let one_star = 0;
let two_star = 0;
let three_star = 0;
let four_star = 0;
let five_star = 0;
Object.values(reviews).forEach(({ rating }) => {
switch (rating) {
case 1: one_star++; break;
case 2: two_star++; break;
case 3: three_star++; break;
case 4: four_star++; break;
case 5: five_star++; break;
}
});
let sum = one_star + two_star + three_star + four_star + five_star;
let average = ((5 * five_star) + (4 * four_star) + (3 * three_star) + (2 * two_star) + (1 * one_star)) / (sum);
console.log(average);