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I am restoring the Ascota 170 antique mechanical programmable computer. It is already working.
Now I’m looking for an algorithm to demonstrate its capabilities — like calculating trigonometric or logarithmic tables. Or something like that.
Unfortunately, from mathematical operations, a computer is only capable of adding and subtracting integers (55 registers from -1E12 to 1E12). There is not even a shift-to-digit operation — so that it can be programmatically implemented to multiply only by very small numbers.
But its logical operations are very well developed.
Could you advise me any suitable algorithm?
So what you're doing is really kinda awesome. And as it happens, I can explain quite a bit about how to implement fractional logarithms using only integer addition and subtraction! This post is going to be long, but there's lots of detail included, and a working implementation at the end, and it should be enough for you to do some fun things with your weird mechanical computer.
Implementing Comparisons
You're going to need to be able to compare numbers. While you said you can perform comparisons == 0 and > 0, that's not really quite enough for most of the interesting algorithms you'll want to implement. You need relative comparisons, which can be determined via subtraction:
isLessThan(a, b):
diff = b - a
if diff > 0 then return true
else return false
isGreaterThan(a, b):
diff = a - b
if diff > 0 then return true
else return false
isLessThanOrEqual(a, b):
diff = a - b
if diff > 0 then return false
else return true
isGreaterThanOrEqual(a, b):
diff = b - a
if diff > 0 then return false
else return true
For the rest of this post, I'm just going to write the simpler form of a > b, but if you can't do that directly, you can substitute in one of the operations above.
Implementing Shifts
Now, since you don't have digit-shifting hardware, you'll have to create "routines" to implement it. A left-shift is easy: Add a number to itself, and again, and again, and then add the original number, and then add it one more time; and that's the equivalent of shifting left by 1 digit.
So shift left by one digit, or multiply-by-ten:
shiftLeft(value):
value2 = value + value
value4 = value2 + value2
value5 = value4 + value
return value5 + value5
Shifting by many digits is just repeated invocation of shiftLeft():
shl(value, count):
repeat:
if count <= 0 then goto done
value = shiftLeft(value)
count = count - 1
done:
return value
Shifting right by one digit is a little harder: We need to do this with repeated subtraction and addition, as in the pseudocode below:
shr(value, count):
if count == 0 then return value
index = 11
shifted = 0
repeat1:
if index < 0 then goto done
adder = shl(1, index - count)
subtractor = shl(adder, count)
repeat2:
if value <= subtractor then goto next
value = value - subtractor
shifted = shifted + adder
goto repeat2
next:
index = index - 1
goto repeat1
done:
return count
Conveniently, since it's hard to shift right in the first place, the algorithm lets us directly choose how many digits to shift by.
Multiplication
It looks like your hardware might have multiplication? But if it doesn't, you can implement multiplication using repeated addition and shifting. Binary multiplication is the easiest form to implement that's actually efficient, and that requires us to first implement multiplyByTwo() and divideByTwo(), using the same basic techniques that we used to implement shiftLeft() and shr().
Once you have those implemented, multiplication involves repeatedly slicing off the last bit of one of the numbers, and if that bit is a 1, then adding a growing version of the other number to the running total:
multiply(a, b):
product = 0
repeat:
if b <= 0 then goto done
nextB = divideByTwo(b)
bit = b - multiplyByTwo(nextB)
if bit == 0 then goto skip
product = product + a
skip:
a = a + a
b = nextB
goto repeat
done:
return product
A full implementation of this is included below, if you need it.
Integer Logarithms
We can use our ability to shift right by a digit to calculate the integer part of the base-10 logarithm of a number — this is really just how many times you can shift the number right before you reach a number too small to shift.
integerLogarithm(value):
count = 0
repeat:
if value <= 9 then goto done
value = shiftRight(value)
count = count + 1
goto repeat
done:
return count
So for 0-9, this returns 0; for 10-99, this returns 1; for 100-999 this returns 2, and so on.
Integer Exponents
The opposite of the above algorithm is pretty trivial: To calculate 10 raised to an integer power, we just shift the digits left by the power.
integerExponent(count):
value = shl(1, count)
return value
So for 0, this returns 1; for 1, this return 10; for 2, this returns 100; for 3, this returns 1000; and so on.
Splitting the Integer and Fraction
Now that we can handle integer powers and logarithms, we're almost ready to handle the fractional part. But before we can really talk about how to compute the fractional part of the logarithm, we have to talk about how to divide up the problem so we can compute the fractional part separately from the integer part. Ideally, we only want to deal with computing logarithms for numbers in a fixed range — say, from 1 to 10, rather than from 1 to infinity.
We can use our integer logarithm and exponent routines to slice up the full logarithm problem so that we're always dealing with a value in the range of [1, 10), no matter what the input number was.
First, we calculate the integer logarithm, and then the integer exponent, and then we subtract that from the original number. Whatever is left over is the fractional part that we need to calculate: And then the only remaining exercise is to shift that fractional part so that it's always in a consistent range.
normalize(value):
intLog = integerLogarithm(value) // From 0 to 12 (meaningful digits)
if intLog <= 5 then goto lessThan
value = shr(value, intLog - 5)
goto done
lessThan:
value = shl(value, 5 - intLog)
done:
return value
You can convince yourself with relatively little effort that no matter what the original value was, its highest nonzero digit will be moved to column 7: So "12345" will become "000000123450" (i.e., "0000001.23450"). This allows us to pretend that there's always an invisible decimal point a little more than halfway down the number, so that now we only need to solve the problem of calculating logarithms of values in the range of [1, 10).
(Why "more than halfway"? We will need the upper half of the value to always be zero, and you'll see why in a moment.)
Fractional Logarithms
Knuth explains how to do this in The Art of Computer Programming, section 1.2.2. Our goal will be to calculate log10(x) so that for some values of b1, b2, b3 ... , where n is already 0 (because we split out the integer portion above):
log10(x) = n + b1/2 + b2/4 + b3/8 + b4/16 + ...
Knuth says that we can obtain b1, b2, b3 ... like this:
To obtain b1, b2, ..., we now set x0 = x / 10^n and, for k >= 1,
b[k] = 0, x[k] = x[k-1] ^ 2, if x[k-1] ^ 2 < 10;
b[k] = 1, x[k] = x[k-1] ^ 2 / 10, if x[k-1] ^ 2 >= 10.
That is to say, each step uses pseudocode loop something like this:
fractionalLogarithm(x):
for i = 1 to numberOfBinaryDigitsOfPrecision:
nextX = x * x
if nextX < 10 then:
b[i] = 0
else:
b[i] = 1
nextX = nextX / 10
In order for this to work using the fixed-point numbers we have above, we have to implement x * x using a shift to move the decimal point back into place, which will lose some digits. This will cause error to propagate, as Knuth says, but it will give enough accuracy that it's good enough for demonstration purposes.
So given a fractional value generated by normalize(value), we can compute its fractional binary logarithm like this:
fractionalLogarithm(value):
for i = 1 to 20:
value = shr(value * value, 6)
if value < 1000000 then:
b[i] = 0
else:
b[i] = 1
value = shr(value, 1)
But a binary fractional logarithm — individual bits! — isn't especially useful, especially since we computed an decimal version of the integer part of the logarithm in the earlier step. So we'll modify this one more time, to calculate a decimal fractional logarithm, to five places, instead of calculating an array of bits; for that, we'll need a table of 20 values that represent the conversions of each of those bits to decimal, and we'll store them as fixed-point as well:
table[1] = 1/(2^1) = 1/2 = 500000
table[2] = 1/(2^2) = 1/4 = 250000
table[3] = 1/(2^3) = 1/8 = 125000
table[4] = 1/(2^4) = 1/16 = 062500
table[5] = 1/(2^5) = 1/32 = 031250
table[6] = 1/(2^6) = 1/64 = 015625
...
table[17] = 1/(2^17) = 1/131072 = 000008
table[18] = 1/(2^18) = 1/262144 = 000004
table[19] = 1/(2^19) = 1/514288 = 000002
table[20] = 1/(2^20) = 1/1048576 = 000001
So now with that table, we can produce the whole fractional logarithm, using pure integer math:
fractionalLogarithm(value):
log = 0
for i = 1 to 20:
value = shr(value * value, 6)
if value >= 1000000 then:
log = log + table[i]
value = shr(value, 1)
return log
Putting It All Together
Finally, for a complete logarithm of any integer your machine can represent, this is the whole thing, which will compute the logarithm with six digits of precision, in the form "0000XX.XXXXXX":
log(value):
intPart = integerLogarithm(value)
value = normalize(value)
fracPart = fractionalLogarithm(value)
result = shl(intPart, 6) + fracPart
return result
Demonstration
To show that the math works — and that it works pretty well! — below is a JavaScript implementation of the above algorithm. It uses pure integer math: Only addition, subtraction, and relative comparison. Functions are used to organize the code, but they behave like subroutines: They're not recursive, and don't nest very deeply.
You can try it out live (click the 'Run' button and type 12345 in the input field). Compare the result to the standard Math.log() function, and you'll see how close the pure-integer version gets:
function shiftLeft(value) {
var value2 = value + value;
var value4 = value2 + value2;
var value5 = value4 + value;
return value5 + value5;
}
function shl(value, count) {
while (count > 0) {
value = shiftLeft(value);
count = count - 1;
}
return value;
}
function shr(value, count) {
if (count == 0) return value;
var index = 11;
var shifted = 0;
while (index >= 0) {
var adder = shl(1, index - count);
var subtractor = shl(adder, count);
while (value > subtractor) {
value = value - subtractor;
shifted = shifted + adder;
}
index = index - 1;
}
return shifted;
}
//-----------------------------------
function multiplyByTwo(value) {
return value + value;
}
function multiplyByPowerOfTwo(value, count) {
while (count > 0) {
value = value + value;
count = count - 1;
}
return value;
}
function divideByPowerOfTwo(value, count) {
if (count == 0) return value;
var index = 39; // lg(floor(pow(10, 12)))
var shifted = 0;
while (index >= 0) {
var adder = multiplyByPowerOfTwo(1, index - count);
var subtractor = multiplyByPowerOfTwo(adder, count);
while (value >= subtractor) {
value = value - subtractor;
shifted = shifted + adder;
}
index = index - 1;
}
return shifted;
}
function divideByTwo(value) {
return divideByPowerOfTwo(value, 1);
}
function multiply(a, b) {
var product = 0;
while (b > 0) {
nextB = divideByTwo(b);
bit = b - multiplyByTwo(nextB);
if (bit != 0) {
product += a;
}
a = a + a;
b = nextB;
}
return product;
}
//-----------------------------------
var logTable = {
"1": 500000,
"2": 250000,
"3": 125000,
"4": 62500,
"5": 31250,
"6": 15625,
"7": 7813,
"8": 3906,
"9": 1953,
"10": 977,
"11": 488,
"12": 244,
"13": 122,
"14": 61,
"15": 31,
"16": 15,
"17": 8,
"18": 4,
"19": 2,
"20": 1,
};
//-----------------------------------
function integerLogarithm(value) {
var count = 0;
while (value > 9) {
value = shr(value, 1);
count = count + 1;
}
return count;
}
function normalize(value) {
var intLog = integerLogarithm(value);
if (intLog > 5)
value = shr(value, intLog - 5);
else
value = shl(value, 5 - intLog);
return value;
}
function fractionalLogarithm(value) {
var log = 0;
for (i = 1; i < 20; i++) {
var squaredValue = multiply(value, value);
value = shr(squaredValue, 5);
if (value >= 1000000) {
log = log + logTable[i];
value = shr(value, 1);
}
}
return log;
}
function log(value) {
var intPart = integerLogarithm(value);
value = normalize(value);
var fracPart = fractionalLogarithm(value);
var result = shl(intPart, 6) + fracPart;
return result;
}
//-----------------------------------
// Just a little jQuery event handling to wrap a UI around the above functions.
$("#InputValue").on("keydown keyup keypress focus blur", function(e) {
var inputValue = Number(this.value.replace(/[^0-9]+/g, ''));
var outputValue = log(inputValue);
$("#OutputValue").text(outputValue / 1000000);
var trueResult = Math.floor((Math.log(inputValue) / Math.log(10)) * 1000000 + 0.5) / 1000000
$("#TrueResult").text(trueResult);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Input integer: <input type="text" id="InputValue" /><br /><br />
Result using integer algorithm: <span id="OutputValue"></span><br /><br />
True logarithm: <span id="TrueResult"></span><br />
As I mentioned in your Original question on SE/RC for pow,sqrt,n-root,log,exp see:
Power by squaring for negative exponents
and all the sub-links in there.
Once you got working *,/,<<,>> (which the other answer covers well) and may fixed point instead of floating you can also start computing goniometrics. For that the best is use Chebyshev series but as I lack the math behind them I can use only already precomputed ones ... Taylor is a common knowledge so computing that should be easy here what I code for my arithmetics template to cover math for arbitrary math data types (bignums):
// Taylor goniometric https://en.wikipedia.org/wiki/Taylor_series
friend T sin (const T &x) // = sin(x)
{
int i; T z,dz,x2,a,b;
x2=x/(pi+pi); x2-=::integer(x2); x2*=pi+pi;
for (z=x2,a=x2,b=1,x2*=x2,i=2;;)
{
a*=x2; b*=i; i++; b*=i; i++; dz=a/b; z-=dz;
a*=x2; b*=i; i++; b*=i; i++; dz=a/b; z+=dz;
if (::abs(dz)<zero) break;
}
return z;
}
friend T cos (const T &x) // = cos(x)
{
int i; T z,dz,x2,a,b;
x2=x/(pi+pi); x2-=::integer(x2); x2*=pi+pi;
for (z=1,a=1,b=1,x2*=x2,i=1;;)
{
a*=x2; b*=i; i++; b*=i; i++; dz=a/b; z-=dz;
a*=x2; b*=i; i++; b*=i; i++; dz=a/b; z+=dz;
if (::abs(dz)<zero) break;
}
return z;
}
friend T tan (const T &x) // = tan(x)
{
int i; T z0,z1,dz,x1,x2,a,b;
x1=x/pi; x1-=::integer(x1); x1*=pi; x2=x1*x1;
for (z0=1,z1=1,a=1,b=1,i=2;;)
{
a*=x2; b*=i; i++; dz=a/b; z0-=dz; // z0=cos(x)
b*=i; i++; dz=a/b; z1-=dz; // z1=sin(x)/x
a*=x2; b*=i; i++; dz=a/b; z0+=dz;
b*=i; i++; dz=a/b; z1+=dz;
if (::abs(dz)<zero) break;
}
return (x1*z1)/z0;
}
friend T ctg (const T &x) // = cotan(x)
{
int i; T z0,z1,dz,x1,x2,a,b;
x1=x/pi; x1-=::integer(x1); x1*=pi; x2=x1*x1;
for (z0=1,z1=1,a=1,b=1,i=2;;)
{
a*=x2; b*=i; i++; dz=a/b; z0-=dz; // z0=cos(x)
b*=i; i++; dz=a/b; z1-=dz; // z1=sin(x)/x
a*=x2; b*=i; i++; dz=a/b; z0+=dz;
b*=i; i++; dz=a/b; z1+=dz;
if (::abs(dz)<zero) break;
}
return z0/(x1*z1);
}
friend T asin (const T &x) // = asin(x)
{
if (x<=-1.0) return -0.5*pi;
if (x>=+1.0) return +0.5*pi;
return ::atan(x/::sqrt(1.0-(x*x)));
}
friend T acos (const T &x){ T z; z=0.5*pi-::asin(x); return z; } // = acos(x)
friend T atan (const T &x) // = atan(x)
{
bool _shift=false;
bool _invert=false;
bool _negative=false;
T z,dz,x1,x2,a,b; x1=x;
if (x1<0.0) { _negative=true; x1=-x1; }
if (x1>1.0) { _invert=true; x1=1.0/x1; }
if (x1>0.7) { _shift=true; b=::sqrt(3.0)/3.0; x1=(x1-b)/(1.0+(x1*b)); }
for (x2=x1*x1,z=x1,a=x1,b=1;;) // if x1>0.8 convergence is slow
{
a*=x2; b+=2; dz=a/b; z-=dz;
a*=x2; b+=2; dz=a/b; z+=dz;
if (::abs(dz)<zero) break;
}
if (_shift) z+=pi/6.0;
if (_invert) z=0.5*pi-z;
if (_negative) z=-z;
return z;
}
friend T actg (const T &x){ T z; z=::atan(1.0/x); return z; } // = acotan(x)
friend T atan2 (const T &y,const T &x){ return atanxy(x,y); } // = atan(y/x)
friend T atanxy (const T &x,const T &y) // = atan(y/x)
{
int sx,sy; T a;
T _zero=1.0e-30;
sx=0; if (x<-_zero) sx=-1; if (x>+_zero) sx=+1;
sy=0; if (y<-_zero) sy=-1; if (y>+_zero) sy=+1;
if ((sy==0)&&(sx==0)) return 0.0;
if ((sx==0)&&(sy> 0)) return 0.5*x.pi;
if ((sx==0)&&(sy< 0)) return 1.5*x.pi;
if ((sy==0)&&(sx> 0)) return 0.0;
if ((sy==0)&&(sx< 0)) return x.pi;
a=y/x; if (a<0) a=-a;
a=::atan(a);
if ((sx>0)&&(sy>0)) a=a;
if ((sx<0)&&(sy>0)) a=x.pi-a;
if ((sx<0)&&(sy<0)) a=x.pi+a;
if ((sx>0)&&(sy<0)) a=x.pi+x.pi-a;
return a;
}
As I mentioned you need to use floating or fixed point for this as the results are not integers !!!
But as I mentioned before CORDIC is better suited for computing on integers (if you search there where some QAs here on SE/SO with C++ code for this).
IIRC it exploit some (arc)tan angle summation identity that leads to a nicely computable on integers delta angle something like sqrt(1+x*x) which is easily computable on integers. With binary search or approximation/iteration you can compute the tan of any angle and using goniometric identities you can compute any cotan sin and cos ... But I might be wrong as I do not use CORDIC and read about it a long time ago
Anyway once you got some function its inverse can be usually computed with binary search.
I want to create a function which will give me fixed size 6 char alpha-numeric IDs with the requirement that the first and last character must be an alpha.
I want them to be generated sequentially. I think using base36 would be the way to go with an alphabet of [0-9A-Z] however I am not too sure how to ensure that they are always 6 char long with an alpha at the start and end.
E.g., if I create the IDs sequentially and start from 0, I would get 0 for output since 0 is the same in both bases.
Does anyone know of an efficient algorithm that could help here?
Thanks
You can use the standard algorithm for converting from an int to a base36 string, extracting one digit at a time by taking the modulo of the base and then dividing the remainder by the base, but add a special case for the first and last digit:
For e.g. in Java:
static String getId(int id)
{
String s = "";
for(int i = 0; i < 6; i++)
{
// compute the digit using modulo arithmetic using base 26
// for first and last character and base 36 for others
int digit;
if((i == 0) || (i == 5))
{
digit = (id % 26) + 10;
id /= 26;
}
else
{
digit = id % 36;
id /= 36;
}
// add the digit to the string:
if(digit < 10)
s = (char)('0' + digit) + s;
else
s = (char)('A' + (digit - 10)) + s;
}
return s;
}
There are 26*36*36*36*36*26 = 1135420416 possibilities, which means you only need a 32-bit integer to store them all.
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You are given as input an unsorted array of n distinct numbers, where n is a power of 2. Give an algorithm that identifies the second-largest number in the array, and that uses at most n+log₂(n)−2 comparisons.
Start with comparing elements of the n element array in odd and even positions and determining largest element of each pair. This step requires n/2 comparisons. Now you've got only n/2 elements. Continue pairwise comparisons to get n/4, n/8, ... elements. Stop when the largest element is found. This step requires a total of n/2 + n/4 + n/8 + ... + 1 = n-1 comparisons.
During previous step, the largest element was immediately compared with log₂(n) other elements. You can determine the largest of these elements in log₂(n)-1 comparisons. That would be the second-largest number in the array.
Example: array of 8 numbers [10,9,5,4,11,100,120,110].
Comparisons on level 1: [10,9] ->10 [5,4]-> 5, [11,100]->100 , [120,110]-->120.
Comparisons on level 2: [10,5] ->10 [100,120]->120.
Comparisons on level 3: [10,120]->120.
Maximum is 120. It was immediately compared with: 10 (on level 3), 100 (on level 2), 110 (on level 1).
Step 2 should find the maximum of 10, 100, and 110. Which is 110. That's the second largest element.
sly s's answer is derived from this paper, but he didn't explain the algorithm, which means someone stumbling across this question has to read the whole paper, and his code isn't very sleek as well. I'll give the crux of the algorithm from the aforementioned paper, complete with complexity analysis, and also provide a Scala implementation, just because that's the language I chose while working on these problems.
Basically, we do two passes:
Find the max, and keep track of which elements the max was compared to.
Find the max among the elements the max was compared to; the result is the second largest element.
In the picture above, 12 is the largest number in the array, and was compared to 3, 1, 11, and 10 in the first pass. In the second pass, we find the largest among {3, 1, 11, 10}, which is 11, which is the second largest number in the original array.
Time Complexity:
All elements must be looked at, therefore, n - 1 comparisons for pass 1.
Since we divide the problem into two halves each time, there are at most log₂n recursive calls, for each of which, the comparisons sequence grows by at most one; the size of the comparisons sequence is thus at most log₂n, therefore, log₂n - 1 comparisons for pass 2.
Total number of comparisons <= (n - 1) + (log₂n - 1) = n + log₂n - 2
def second_largest(nums: Sequence[int]) -> int:
def _max(lo: int, hi: int, seq: Sequence[int]) -> Tuple[int, MutableSequence[int]]:
if lo >= hi:
return seq[lo], []
mid = lo + (hi - lo) // 2
x, a = _max(lo, mid, seq)
y, b = _max(mid + 1, hi, seq)
if x > y:
a.append(y)
return x, a
b.append(x)
return y, b
comparisons = _max(0, len(nums) - 1, nums)[1]
return _max(0, len(comparisons) - 1, comparisons)[0]
The first run for the given example is as follows:
lo=0, hi=1, mid=0, x=10, a=[], y=4, b=[]
lo=0, hi=2, mid=1, x=10, a=[4], y=5, b=[]
lo=3, hi=4, mid=3, x=8, a=[], y=7, b=[]
lo=3, hi=5, mid=4, x=8, a=[7], y=2, b=[]
lo=0, hi=5, mid=2, x=10, a=[4, 5], y=8, b=[7, 2]
lo=6, hi=7, mid=6, x=12, a=[], y=3, b=[]
lo=6, hi=8, mid=7, x=12, a=[3], y=1, b=[]
lo=9, hi=10, mid=9, x=6, a=[], y=9, b=[]
lo=9, hi=11, mid=10, x=9, a=[6], y=11, b=[]
lo=6, hi=11, mid=8, x=12, a=[3, 1], y=11, b=[9]
lo=0, hi=11, mid=5, x=10, a=[4, 5, 8], y=12, b=[3, 1, 11]
Things to note:
There are exactly n - 1=11 comparisons for n=12.
From the last line, y=12 wins over x=10, and the next pass starts with the sequence [3, 1, 11, 10], which has log₂(12)=3.58 ~ 4 elements, and will require 3 comparisons to find the maximum.
I have implemented this algorithm in Java answered by #Evgeny Kluev. The total comparisons are n+log2(n)−2. There is also a good reference:
Alexander Dekhtyar: CSC 349: Design and Analyis of Algorithms. This is similar to the top voted algorithm.
public class op1 {
private static int findSecondRecursive(int n, int[] A){
int[] firstCompared = findMaxTournament(0, n-1, A); //n-1 comparisons;
int[] secondCompared = findMaxTournament(2, firstCompared[0]-1, firstCompared); //log2(n)-1 comparisons.
//Total comparisons: n+log2(n)-2;
return secondCompared[1];
}
private static int[] findMaxTournament(int low, int high, int[] A){
if(low == high){
int[] compared = new int[2];
compared[0] = 2;
compared[1] = A[low];
return compared;
}
int[] compared1 = findMaxTournament(low, (low+high)/2, A);
int[] compared2 = findMaxTournament((low+high)/2+1, high, A);
if(compared1[1] > compared2[1]){
int k = compared1[0] + 1;
int[] newcompared1 = new int[k];
System.arraycopy(compared1, 0, newcompared1, 0, compared1[0]);
newcompared1[0] = k;
newcompared1[k-1] = compared2[1];
return newcompared1;
}
int k = compared2[0] + 1;
int[] newcompared2 = new int[k];
System.arraycopy(compared2, 0, newcompared2, 0, compared2[0]);
newcompared2[0] = k;
newcompared2[k-1] = compared1[1];
return newcompared2;
}
private static void printarray(int[] a){
for(int i:a){
System.out.print(i + " ");
}
System.out.println();
}
public static void main(String[] args) {
//Demo.
System.out.println("Origial array: ");
int[] A = {10,4,5,8,7,2,12,3,1,6,9,11};
printarray(A);
int secondMax = findSecondRecursive(A.length,A);
Arrays.sort(A);
System.out.println("Sorted array(for check use): ");
printarray(A);
System.out.println("Second largest number in A: " + secondMax);
}
}
the problem is:
let's say, in comparison level 1, the algorithm need to be remember all the array element because largest is not yet known, then, second, finally, third. by keep tracking these element via assignment will invoke additional value assignment and later when the largest is known, you need also consider the tracking back. As the result, it will not be significantly faster than simple 2N-2 Comparison algorithm. Moreover, because the code is more complicated, you need also think about potential debugging time.
eg: in PHP, RUNNING time for comparison vs value assignment roughly is :Comparison: (11-19) to value assignment: 16.
I shall give some examples for better understanding. :
example 1 :
>12 56 98 12 76 34 97 23
>>(12 56) (98 12) (76 34) (97 23)
>>> 56 98 76 97
>>>> (56 98) (76 97)
>>>>> 98 97
>>>>>> 98
The largest element is 98
Now compare with lost ones of the largest element 98. 97 will be the second largest.
nlogn implementation
public class Test {
public static void main(String...args){
int arr[] = new int[]{1,2,2,3,3,4,9,5, 100 , 101, 1, 2, 1000, 102, 2,2,2};
System.out.println(getMax(arr, 0, 16));
}
public static Holder getMax(int[] arr, int start, int end){
if (start == end)
return new Holder(arr[start], Integer.MIN_VALUE);
else {
int mid = ( start + end ) / 2;
Holder l = getMax(arr, start, mid);
Holder r = getMax(arr, mid + 1, end);
if (l.compareTo(r) > 0 )
return new Holder(l.high(), r.high() > l.low() ? r.high() : l.low());
else
return new Holder(r.high(), l.high() > r.low() ? l.high(): r.low());
}
}
static class Holder implements Comparable<Holder> {
private int low, high;
public Holder(int r, int l){low = l; high = r;}
public String toString(){
return String.format("Max: %d, SecMax: %d", high, low);
}
public int compareTo(Holder data){
if (high == data.high)
return 0;
if (high > data.high)
return 1;
else
return -1;
}
public int high(){
return high;
}
public int low(){
return low;
}
}
}
Why not to use this hashing algorithm for given array[n]? It runs c*n, where c is constant time for check and hash. And it does n comparisons.
int first = 0;
int second = 0;
for(int i = 0; i < n; i++) {
if(array[i] > first) {
second = first;
first = array[i];
}
}
Or am I just do not understand the question...
In Python2.7: The following code works at O(nlog log n) for the extra sort. Any optimizations?
def secondLargest(testList):
secondList = []
# Iterate through the list
while(len(testList) > 1):
left = testList[0::2]
right = testList[1::2]
if (len(testList) % 2 == 1):
right.append(0)
myzip = zip(left,right)
mymax = [ max(list(val)) for val in myzip ]
myzip.sort()
secondMax = [x for x in myzip[-1] if x != max(mymax)][0]
if (secondMax != 0 ):
secondList.append(secondMax)
testList = mymax
return max(secondList)
public static int FindSecondLargest(int[] input)
{
Dictionary<int, List<int>> dictWinnerLoser = new Dictionary<int, List<int>>();//Keeps track of loosers with winners
List<int> lstWinners = null;
List<int> lstLoosers = null;
int winner = 0;
int looser = 0;
while (input.Count() > 1)//Runs till we get max in the array
{
lstWinners = new List<int>();//Keeps track of winners of each run, as we have to run with winners of each run till we get one winner
for (int i = 0; i < input.Count() - 1; i += 2)
{
if (input[i] > input[i + 1])
{
winner = input[i];
looser = input[i + 1];
}
else
{
winner = input[i + 1];
looser = input[i];
}
lstWinners.Add(winner);
if (!dictWinnerLoser.ContainsKey(winner))
{
lstLoosers = new List<int>();
lstLoosers.Add(looser);
dictWinnerLoser.Add(winner, lstLoosers);
}
else
{
lstLoosers = dictWinnerLoser[winner];
lstLoosers.Add(looser);
dictWinnerLoser[winner] = lstLoosers;
}
}
input = lstWinners.ToArray();//run the loop again with winners
}
List<int> loosersOfWinner = dictWinnerLoser[input[0]];//Gives all the elemetns who lost to max element of array, input array now has only one element which is actually the max of the array
winner = 0;
for (int i = 0; i < loosersOfWinner.Count(); i++)//Now max in the lossers of winner will give second largest
{
if (winner < loosersOfWinner[i])
{
winner = loosersOfWinner[i];
}
}
return winner;
}
What is the best method to find the number of digits of a positive integer?
I have found this 3 basic methods:
conversion to string
String s = new Integer(t).toString();
int len = s.length();
for loop
for(long long int temp = number; temp >= 1;)
{
temp/=10;
decimalPlaces++;
}
logaritmic calculation
digits = floor( log10( number ) ) + 1;
where you can calculate log10(x) = ln(x) / ln(10) in most languages.
First I thought the string method is the dirtiest one but the more I think about it the more I think it's the fastest way. Or is it?
There's always this method:
n = 1;
if ( i >= 100000000 ) { n += 8; i /= 100000000; }
if ( i >= 10000 ) { n += 4; i /= 10000; }
if ( i >= 100 ) { n += 2; i /= 100; }
if ( i >= 10 ) { n += 1; }
Well the correct answer would be to measure it - but you should be able to make a guess about the number of CPU steps involved in converting strings and going through them looking for an end marker
Then think how many FPU operations/s your processor can do and how easy it is to calculate a single log.
edit: wasting some more time on a monday morning :-)
String s = new Integer(t).toString();
int len = s.length();
One of the problems with high level languages is guessing how much work the system is doing behind the scenes of an apparently simple statement. Mandatory Joel link
This statement involves allocating memory for a string, and possibly a couple of temporary copies of a string. It must parse the integer and copy the digits of it into a string, possibly having to reallocate and move the existing memory if the number is large. It might have to check a bunch of locale settings to decide if your country uses "," or ".", it might have to do a bunch of unicode conversions.
Then finding the length has to scan the entire string, again considering unicode and any local specific settings such as - are you in a right->left language?.
Alternatively:
digits = floor( log10( number ) ) + 1;
Just because this would be harder for you to do on paper doesn't mean it's hard for a computer! In fact a good rule in high performance computing seems to have been - if something is hard for a human (fluid dynamics, 3d rendering) it's easy for a computer, and if it's easy for a human (face recognition, detecting a voice in a noisy room) it's hard for a computer!
You can generally assume that the builtin maths functions log/sin/cos etc - have been an important part of computer design for 50years. So even if they don't map directly into a hardware function in the FPU you can bet that the alternative implementation is pretty efficient.
I don't know, and the answer may well be different depending on how your individual language is implemented.
So, stress test it! Implement all three solutions. Run them on 1 through 1,000,000 (or some other huge set of numbers that's representative of the numbers the solution will be running against) and time how long each of them takes.
Pit your solutions against one another and let them fight it out. Like intellectual gladiators. Three algorithms enter! One algorithm leaves!
Test conditions
Decimal numeral system
Positive integers
Up to 10 digits
Language: ActionScript 3
Results
digits: [1,10],
no. of runs: 1,000,000
random sample: 8777509,40442298,477894,329950,513,91751410,313,3159,131309,2
result: 7,8,6,6,3,8,3,4,6,1
CONVERSION TO STRING: 724ms
LOGARITMIC CALCULATION: 349ms
DIV 10 ITERATION: 229ms
MANUAL CONDITIONING: 136ms
Note: Author refrains from making any conclusions for numbers with more than 10 digits.
Script
package {
import flash.display.MovieClip;
import flash.utils.getTimer;
/**
* #author Daniel
*/
public class Digits extends MovieClip {
private const NUMBERS : uint = 1000000;
private const DIGITS : uint = 10;
private var numbers : Array;
private var digits : Array;
public function Digits() {
// ************* NUMBERS *************
numbers = [];
for (var i : int = 0; i < NUMBERS; i++) {
var number : Number = Math.floor(Math.pow(10, Math.random()*DIGITS));
numbers.push(number);
}
trace('Max digits: ' + DIGITS + ', count of numbers: ' + NUMBERS);
trace('sample: ' + numbers.slice(0, 10));
// ************* CONVERSION TO STRING *************
digits = [];
var time : Number = getTimer();
for (var i : int = 0; i < numbers.length; i++) {
digits.push(String(numbers[i]).length);
}
trace('\nCONVERSION TO STRING - time: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
// ************* LOGARITMIC CALCULATION *************
digits = [];
time = getTimer();
for (var i : int = 0; i < numbers.length; i++) {
digits.push(Math.floor( Math.log( numbers[i] ) / Math.log(10) ) + 1);
}
trace('\nLOGARITMIC CALCULATION - time: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
// ************* DIV 10 ITERATION *************
digits = [];
time = getTimer();
var digit : uint = 0;
for (var i : int = 0; i < numbers.length; i++) {
digit = 0;
for(var temp : Number = numbers[i]; temp >= 1;)
{
temp/=10;
digit++;
}
digits.push(digit);
}
trace('\nDIV 10 ITERATION - time: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
// ************* MANUAL CONDITIONING *************
digits = [];
time = getTimer();
var digit : uint;
for (var i : int = 0; i < numbers.length; i++) {
var number : Number = numbers[i];
if (number < 10) digit = 1;
else if (number < 100) digit = 2;
else if (number < 1000) digit = 3;
else if (number < 10000) digit = 4;
else if (number < 100000) digit = 5;
else if (number < 1000000) digit = 6;
else if (number < 10000000) digit = 7;
else if (number < 100000000) digit = 8;
else if (number < 1000000000) digit = 9;
else if (number < 10000000000) digit = 10;
digits.push(digit);
}
trace('\nMANUAL CONDITIONING: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
}
}
}
This algorithm might be good also, assuming that:
Number is integer and binary encoded (<< operation is cheap)
We don't known number boundaries
var num = 123456789L;
var len = 0;
var tmp = 1L;
while(tmp < num)
{
len++;
tmp = (tmp << 3) + (tmp << 1);
}
This algorithm, should have speed comparable to for-loop (2) provided, but a bit faster due to (2 bit-shifts, add and subtract, instead of division).
As for Log10 algorithm, it will give you only approximate answer (that is close to real, but still), since analytic formula for computing Log function have infinite loop and can't be calculated precisely Wiki.
Use the simplest solution in whatever programming language you're using. I can't think of a case where counting digits in an integer would be the bottleneck in any (useful) program.
C, C++:
char buffer[32];
int length = sprintf(buffer, "%ld", (long)123456789);
Haskell:
len = (length . show) 123456789
JavaScript:
length = String(123456789).length;
PHP:
$length = strlen(123456789);
Visual Basic (untested):
length = Len(str(123456789)) - 1
conversion to string: This will have to iterate through each digit, find the character that maps to the current digit, add a character to a collection of characters. Then get the length of the resulting String object. Will run in O(n) for n=#digits.
for-loop: will perform 2 mathematical operation: dividing the number by 10 and incrementing a counter. Will run in O(n) for n=#digits.
logarithmic: Will call log10 and floor, and add 1. Looks like O(1) but I'm not really sure how fast the log10 or floor functions are. My knowledge of this sort of things has atrophied with lack of use so there could be hidden complexity in these functions.
So I guess it comes down to: is looking up digit mappings faster than multiple mathematical operations or whatever is happening in log10? The answer will probably vary. There could be platforms where the character mapping is faster, and others where doing the calculations is faster. Also to keep in mind is that the first method will creats a new String object that only exists for the purpose of getting the length. This will probably use more memory than the other two methods, but it may or may not matter.
You can obviously eliminate the method 1 from the competition, because the atoi/toString algorithm it uses would be similar to method 2.
Method 3's speed depends on whether the code is being compiled for a system whose instruction set includes log base 10.
For very large integers, the log method is much faster. For instance, with a 2491327 digit number (the 11920928th Fibonacci number, if you care), Python takes several minutes to execute the divide-by-10 algorithm, and milliseconds to execute 1+floor(log(n,10)).
import math
def numdigits(n):
return ( int(math.floor(math.log10(n))) + 1 )
Regarding the three methods you propose for "determining the number of digits necessary to represent a given number in a given base", I don't like any of them, actually; I prefer the method I give below instead.
Re your method #1 (strings): Anything involving converting back-and-forth between strings and numbers is usually very slow.
Re your method #2 (temp/=10): This is fatally flawed because it assumes that x/10 always means "x divided by 10". But in many programming languages (eg: C, C++), if "x" is an integer type, then "x/10" means "integer division", which isn't the same thing as floating-point division, and it introduces round-off errors at every iteration, and they accumulate in a recursive formula such as your solution #2 uses.
Re your method #3 (logs): it's buggy for large numbers (at least in C, and probably other languages as well), because floating-point data types tend not to be as precise as 64-bit integers.
Hence I dislike all 3 of those methods: #1 works but is slow, #2 is broken, and #3 is buggy for large numbers. Instead, I prefer this, which works for numbers from 0 up to about 18.44 quintillion:
unsigned NumberOfDigits (uint64_t Number, unsigned Base)
{
unsigned Digits = 1;
uint64_t Power = 1;
while ( Number / Power >= Base )
{
++Digits;
Power *= Base;
}
return Digits;
}
Keep it simple:
long long int a = 223452355415634664;
int x;
for (x = 1; a >= 10; x++)
{
a = a / 10;
}
printf("%d", x);
You can use a recursive solution instead of a loop, but somehow similar:
#tailrec
def digits (i: Long, carry: Int=1) : Int = if (i < 10) carry else digits (i/10, carry+1)
digits (8345012978643L)
With longs, the picture might change - measure small and long numbers independently against different algorithms, and pick the appropriate one, depending on your typical input. :)
Of course nothing beats a switch:
switch (x) {
case 0: case 1: case 2: case 3: case 4: case 5: case 6: case 7: case 8: case 9: return 1;
case 10: case 11: // ...
case 99: return 2;
case 100: // you get the point :)
default: return 10; // switch only over int
}
except a plain-o-array:
int [] size = {1,1,1,1,1,1,1,1,1,2,2,2,2,2,... };
int x = 234561798;
return size [x];
Some people will tell you to optimize the code-size, but yaknow, premature optimization ...
log(x,n)-mod(log(x,n),1)+1
Where x is a the base and n is the number.
Here is the measurement in Swift 4.
Algorithms code:
extension Int {
var numberOfDigits0: Int {
var currentNumber = self
var n = 1
if (currentNumber >= 100000000) {
n += 8
currentNumber /= 100000000
}
if (currentNumber >= 10000) {
n += 4
currentNumber /= 10000
}
if (currentNumber >= 100) {
n += 2
currentNumber /= 100
}
if (currentNumber >= 10) {
n += 1
}
return n
}
var numberOfDigits1: Int {
return String(self).count
}
var numberOfDigits2: Int {
var n = 1
var currentNumber = self
while currentNumber > 9 {
n += 1
currentNumber /= 10
}
return n
}
}
Measurement code:
var timeInterval0 = Date()
for i in 0...10000 {
i.numberOfDigits0
}
print("timeInterval0: \(Date().timeIntervalSince(timeInterval0))")
var timeInterval1 = Date()
for i in 0...10000 {
i.numberOfDigits1
}
print("timeInterval1: \(Date().timeIntervalSince(timeInterval1))")
var timeInterval2 = Date()
for i in 0...10000 {
i.numberOfDigits2
}
print("timeInterval2: \(Date().timeIntervalSince(timeInterval2))")
Output
timeInterval0: 1.92149806022644
timeInterval1: 0.557608008384705
timeInterval2: 2.83262193202972
On this measurement basis String conversion is the best option for the Swift language.
I was curious after seeing #daniel.sedlacek results so I did some testing using Swift for numbers having more than 10 digits. I ran the following script in the playground.
let base = [Double(100090000000), Double(100050000), Double(100050000), Double(100000200)]
var rar = [Double]()
for i in 1...10 {
for d in base {
let v = d*Double(arc4random_uniform(UInt32(1000000000)))
rar.append(v*Double(arc4random_uniform(UInt32(1000000000))))
rar.append(Double(1)*pow(1,Double(i)))
}
}
print(rar)
var timeInterval = NSDate().timeIntervalSince1970
for d in rar {
floor(log10(d))
}
var newTimeInterval = NSDate().timeIntervalSince1970
print(newTimeInterval-timeInterval)
timeInterval = NSDate().timeIntervalSince1970
for d in rar {
var c = d
while c > 10 {
c = c/10
}
}
newTimeInterval = NSDate().timeIntervalSince1970
print(newTimeInterval-timeInterval)
Results of 80 elements
0.105069875717163 for floor(log10(x))
0.867973804473877 for div 10 iterations
Adding one more approach to many of the already mentioned approaches.
The idea is to use binarySearch on an array containing the range of integers based on the digits of the int data type.
The signature of Java Arrays class binarySearch is :
binarySearch(dataType[] array, dataType key) which returns the index of the search key, if it is contained in the array; otherwise, (-(insertion point) – 1).
The insertion point is defined as the point at which the key would be inserted into the array.
Below is the implementation:
static int [] digits = {9,99,999,9999,99999,999999,9999999,99999999,999999999,Integer.MAX_VALUE};
static int digitsCounter(int N)
{
int digitCount = Arrays.binarySearch(digits , N<0 ? -N:N);
return 1 + (digitCount < 0 ? ~digitCount : digitCount);
}
Please note that the above approach only works for : Integer.MIN_VALUE <= N <= Integer.MAX_VALUE, but can be easily extended for Long data type by adding more values to the digits array.
For example,
I) for N = 555, digitCount = Arrays.binarySearch(digits , 555) returns -3 (-(2)-1) as it's not present in the array but is supposed to be inserted at point 2 between 9 & 99 like [9, 55, 99].
As the index we got is negative we need to take the bitwise compliment of the result.
At last, we need to add 1 to the result to get the actual number of digits in the number N.
In Swift 5.x, you get the number of digit in integer as below :
Convert to string and then count number of character in string
let nums = [1, 7892, 78, 92, 90]
for i in nums {
let ch = String(describing: i)
print(ch.count)
}
Calculating the number of digits in integer using loop
var digitCount = 0
for i in nums {
var tmp = i
while tmp >= 1 {
tmp /= 10
digitCount += 1
}
print(digitCount)
}
let numDigits num =
let num = abs(num)
let rec numDigitsInner num =
match num with
| num when num < 10 -> 1
| _ -> 1 + numDigitsInner (num / 10)
numDigitsInner num
F# Version, without casting to a string.
I need help solving problem N from this earlier competition:
Problem N: Digit Sums
Given 3 positive integers A, B and C,
find how many positive integers less
than or equal to A, when expressed in
base B, have digits which sum to C.
Input will consist of a series of
lines, each containing three integers,
A, B and C, 2 ≤ B ≤ 100, 1 ≤ A, C ≤
1,000,000,000. The numbers A, B and C
are given in base 10 and are separated
by one or more blanks. The input is
terminated by a line containing three
zeros.
Output will be the number of numbers,
for each input line (it must be given
in base 10).
Sample input
100 10 9
100 10 1
750000 2 2
1000000000 10 40
100000000 100 200
0 0 0
Sample output
10
3
189
45433800
666303
The relevant rules:
Read all input from the keyboard, i.e. use stdin, System.in, cin or equivalent. Input will be redirected from a file to form the input to your submission.
Write all output to the screen, i.e. use stdout, System.out, cout or equivalent. Do not write to stderr. Do NOT use, or even include, any module that allows direct manipulation of the screen, such as conio, Crt or anything similar. Output from your program is redirected to a file for later checking. Use of direct I/O means that such output is not redirected and hence cannot be checked. This could mean that a correct program is rejected!
Unless otherwise stated, all integers in the input will fit into a standard 32-bit computer word. Adjacent integers on a line will be separated by one or more spaces.
Of course, it's fair to say that I should learn more before trying to solve this, but i'd really appreciate it if someone here told me how it's done.
Thanks in advance, John.
Other people pointed out trivial solution: iterate over all numbers from 1 to A. But this problem, actually, can be solved in nearly constant time: O(length of A), which is O(log(A)).
Code provided is for base 10. Adapting it for arbitrary base is trivial.
To reach above estimate for time, you need to add memorization to recursion. Let me know if you have questions about that part.
Now, recursive function itself. Written in Java, but everything should work in C#/C++ without any changes. It's big, but mostly because of comments where I try to clarify algorithm.
// returns amount of numbers strictly less than 'num' with sum of digits 'sum'
// pay attention to word 'strictly'
int count(int num, int sum) {
// no numbers with negative sum of digits
if (sum < 0) {
return 0;
}
int result = 0;
// imagine, 'num' == 1234
// let's check numbers 1233, 1232, 1231, 1230 manually
while (num % 10 > 0) {
--num;
// check if current number is good
if (sumOfDigits(num) == sum) {
// one more result
++result;
}
}
if (num == 0) {
// zero reached, no more numbers to check
return result;
}
num /= 10;
// Using example above (1234), now we're left with numbers
// strictly less than 1230 to check (1..1229)
// It means, any number less than 123 with arbitrary digit appended to the right
// E.g., if this digit in the right (last digit) is 3,
// then sum of the other digits must be "sum - 3"
// and we need to add to result 'count(123, sum - 3)'
// let's iterate over all possible values of last digit
for (int digit = 0; digit < 10; ++digit) {
result += count(num, sum - digit);
}
return result;
}
Helper function
// returns sum of digits, plain and simple
int sumOfDigits(int x) {
int result = 0;
while (x > 0) {
result += x % 10;
x /= 10;
}
return result;
}
Now, let's write a little tester
int A = 12345;
int C = 13;
// recursive solution
System.out.println(count(A + 1, C));
// brute-force solution
int total = 0;
for (int i = 1; i <= A; ++i) {
if (sumOfDigits(i) == C) {
++total;
}
}
System.out.println(total);
You can write more comprehensive tester checking all values of A, but overall solution seems to be correct. (I tried several random A's and C's.)
Don't forget, you can't test solution for A == 1000000000 without memorization: it'll run too long. But with memorization, you can test it even for A == 10^1000.
edit
Just to prove a concept, poor man's memorization. (in Java, in other languages hashtables are declared differently) But if you want to learn something, it might be better to try to do it yourself.
// hold values here
private Map<String, Integer> mem;
int count(int num, int sum) {
// no numbers with negative sum of digits
if (sum < 0) {
return 0;
}
String key = num + " " + sum;
if (mem.containsKey(key)) {
return mem.get(key);
}
// ...
// continue as above...
// ...
mem.put(key, result);
return result;
}
Here's the same memoized recursive solution that Rybak posted, but with a simpler implementation, in my humble opinion:
HashMap<String, Integer> cache = new HashMap<String, Integer>();
int count(int bound, int base, int sum) {
// No negative digit sums.
if (sum < 0)
return 0;
// Handle one digit case.
if (bound < base)
return (sum <= bound) ? 1 : 0;
String key = bound + " " + sum;
if (cache.containsKey(key))
return cache.get(key);
int count = 0;
for (int digit = 0; digit < base; digit++)
count += count((bound - digit) / base, base, sum - digit);
cache.put(key, count);
return count;
}
This is not the complete solution (no input parsing). To get the number in base B, repeatedly take the modulo B, and then divide by B until the result is 0. This effectively computes the base-B digit from the right, and then shifts the number right.
int A,B,C; // from input
for (int x=1; x<A; x++)
{
int sumDigits = 0;
int v = x;
while (v!=0) {
sumDigits += (v % B);
v /= B;
}
if (sumDigits==C)
cout << x;
}
This is a brute force approach. It may be possible to compute this quicker by determining which sets of base B digits add up to C, arranging these in all permutations that are less than A, and then working backwards from that to create the original number.
Yum.
Try this:
int number, digitSum, resultCounter = 0;
for(int i=1; i<=A, i++)
{
number = i; //to avoid screwing up our counter
digitSum = 0;
while(number > 1)
{
//this is the next "digit" of the number as it would be in base B;
//works with any base including 10.
digitSum += (number % B);
//remove this digit from the number, square the base, rinse, repeat
number /= B;
}
digitSum += number;
//Does the sum match?
if(digitSum == C)
resultCounter++;
}
That's your basic algorithm for one line. Now you wrap this in another For loop for each input line you received, preceded by the input collection phase itself. This process can be simplified, but I don't feel like coding your entire answer to see if my algorithm works, and this looks right whereas the simpler tricks are harder to pass by inspection.
The way this works is by modulo dividing by powers of the base. Simple example, 1234 in base 10:
1234 % 10 = 4
1234 / 10 = 123 //integer division truncates any fraction
123 % 10 = 3 //sum is 7
123 / 10 = 12
12 % 10 = 2 //sum is 9
12 / 10 = 1 //end condition, add this and the sum is 10
A harder example to figure out by inspection would be the same number in base 12:
1234 % 12 = 10 //you can call it "A" like in hex, but we need a sum anyway
1234 / 12 = 102
102 % 12 = 6 // sum 16
102/12 = 8
8 % 12 = 8 //sum 24
8 / 12 = 0 //end condition, sum still 24.
So 1234 in base 12 would be written 86A. Check the math:
8*12^2 + 6*12 + 10 = 1152 + 72 + 10 = 1234
Have fun wrapping the rest of the code around this.