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Is there a performant way to generate an unbiased 64b random integer without 3 set bits in a row, assuming a fast-and-unbiased input PRNG? I don't care about 'wasting bits' of the input source.
That is, something better than the naive rejection-sampling approach:
uint64_t r;
do {
r = get_rand_64();
} while (r & (r >> 1) & (r >> 2));
...which "works", but is very slow. It looks like it's iterating ~187x on average or so.
One possibility I've explored is roughly:
bool p2 = get_rand_bit();
bool p1 = get_rand_bit();
uint64_t r = (p1 << 1) | p2;
for (int i = 2; i < 64; i++) {
bool p0 = (p1 && p2) ? false : get_rand_bit();
r |= p0 << i;
p2 = p1;
p1 = p0;
}
...however, this is still slow. Mainly because using this approach the entire calculation is bit-serial. EDIT: and it's also biased. Easiest to see with a 3-bit integer - 0b011 occurs 1/8th of the time, which is wrong (should be 1/7th).
I've tried doing various parallel fixups, but haven't been able to come up with anything unbiased. It's useful to play around with 4-bit integers first - e.g. setting all bits involved in a conflict to random values ends up biased, and drawing out the Markov chain for 4 bits makes that obvious
Is there a better way to do this?
I optimized the lexicographic decoder, resulting in a four-fold speedup relative to my previous answer. There are two new ideas:
Use the one-to-one correspondence implied by the recurrence T(n) = T(k−1) T(n−k) + T(k−2) T(n−k−1) + T(k−2) T(n−k−2) + T(k−3) T(n−k−1) to avoid working one bit at a time;
Cache the small words without 111 in addition to the recurrence values, incurring an L1 cache hit to save a number of arithmetic operations.
#include <assert.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
enum { kTribonacci14 = 5768 };
static uint64_t g_tribonacci[65];
static void InitTribonacci(void) {
for (unsigned i = 0; i < 65; i++) {
g_tribonacci[i] =
i < 3 ? 1 << i
: g_tribonacci[i - 1] + g_tribonacci[i - 2] + g_tribonacci[i - 3];
}
assert(g_tribonacci[14] == kTribonacci14);
}
static uint16_t g_words_no_111[kTribonacci14];
static void InitCachedWordsNo111(void) {
unsigned i = 0;
for (unsigned word = 0; word < ((unsigned)1 << 14); word++) {
if ((word & (word >> 1) & (word >> 2)) == 0) {
assert(i < kTribonacci14);
g_words_no_111[i++] = (uint16_t)word;
}
}
assert(i == kTribonacci14);
}
static bool CaseNo111(uint64_t *restrict result, unsigned *restrict n,
uint64_t *restrict index, unsigned left_n,
unsigned right_n) {
uint64_t left_count = g_tribonacci[left_n];
uint64_t right_count = g_tribonacci[right_n];
uint64_t product = left_count * right_count;
if (*index >= product) {
*index -= product;
return false;
}
*result = (*result << left_n) + g_words_no_111[*index / right_count];
*n = right_n;
*index %= right_count;
return true;
}
static void Append(uint64_t *result, uint64_t bit) {
*result = (*result << 1) + bit;
}
static uint64_t DecodeNo111(unsigned n, uint64_t index) {
assert(0 <= n && n <= 64);
assert(index < g_tribonacci[n]);
uint64_t result = 0;
while (n > 14) {
assert(g_tribonacci[n] == g_tribonacci[12] * g_tribonacci[n - 13] +
g_tribonacci[11] * g_tribonacci[n - 14] +
g_tribonacci[11] * g_tribonacci[n - 15] +
g_tribonacci[10] * g_tribonacci[n - 14]);
if (CaseNo111(&result, &n, &index, 12, n - 13)) {
Append(&result, 0);
} else if (CaseNo111(&result, &n, &index, 11, n - 14)) {
Append(&result, 0);
Append(&result, 1);
Append(&result, 0);
} else if (CaseNo111(&result, &n, &index, 11, n - 15)) {
Append(&result, 0);
Append(&result, 1);
Append(&result, 1);
Append(&result, 0);
} else if (CaseNo111(&result, &n, &index, 10, n - 14)) {
Append(&result, 0);
Append(&result, 1);
Append(&result, 1);
Append(&result, 0);
} else {
assert(false);
}
}
return (result << n) + g_words_no_111[index];
}
static void PrintWord(unsigned n, uint64_t word) {
assert(0 <= n && n <= 64);
while (n-- > 0) {
putchar('0' + ((word >> n) & 1));
}
putchar('\n');
}
int main(void) {
InitTribonacci();
InitCachedWordsNo111();
if ((false)) {
enum { kN = 20 };
for (uint64_t i = 0; i < g_tribonacci[kN]; i++) {
PrintWord(kN, DecodeNo111(kN, i));
}
}
uint64_t sum = 0;
uint64_t index = 0;
for (uint32_t i = 0; i < 10000000; i++) {
sum += DecodeNo111(64, index % g_tribonacci[64]);
index = (index * 2862933555777941757) + 3037000493;
}
return sum & 127;
}
From #John Coleman's comment, here's the start of an approach based on Tribonacci numbers. Basic idea:
Generate an unbiased number in the range [0..T(bits)), where T(0) = 1, T(1) = 2, T(2) = 4, T(n) = T(n-1) + T(n-2) + T(n-3).
Convert to Tribonacci representation.
You're done.
A minimal example is as follows:
// 1, 2, 4, TRIBO[n-3]+TRIBO[n-2]+TRIBO[n-1]
// possible minor perf optimization: reverse TRIBO
static const uint64_t TRIBO[65] = {1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852, 2082876103, 3831006429, 7046319384, 12960201916, 23837527729, 43844049029, 80641778674, 148323355432, 272809183135, 501774317241, 922906855808, 1697490356184, 3122171529233, 5742568741225, 10562230626642, 19426970897100, 35731770264967, 65720971788709, 120879712950776, 222332455004452, 408933139743937, 752145307699165, 1383410902447554, 2544489349890656, 4680045560037375, 8607945812375585, 15832480722303616, 29120472094716576, 53560898629395777, 98513851446415969];
// exclusive of max
extern uint64_t get_rand_64_range(uint64_t max);
uint64_t get_rand_no111(void) {
uint64_t idx = get_rand_64_range(TRIBO[64]);
uint64_t ret = 0;
for (int i = 63; i >= 0; i--) {
if (idx >= TRIBO[i]) {
ret |= ((uint64_t) 1) << i;
idx -= TRIBO[i];
}
// optional: if (idx == 0) {break;}
}
return ret;
}
(Warning: retyped from Python code. I suggest testing.)
This satisfies the 'unbiased' portion, and is indeed faster than the naive rejection-sampling approach, but unfortunately is still pretty slow, because it's looping ~64 times.
The idea behind the code below is to generate the upper 32 bits with the proper (non-uniform!) distribution, then generate the lower 32 conditional on the upper. On my laptop, it’s significantly faster than the baseline, and slightly faster than lexicographic decoding.
You can see the logic behind the non-uniform upper distribution with 4-bit outputs: 00 and 10 have four 2-bit lowers, 01 has three lowers, and 11 has two lowers.
#include <cstdint>
#include <random>
namespace {
using Generator = std::mt19937_64;
template <int bits> std::uint64_t GenerateUniform(Generator &gen) {
static_assert(0 <= bits && bits <= 63);
return gen() & ((std::uint64_t{1} << bits) - 1);
}
template <> std::uint64_t GenerateUniform<64>(Generator &gen) { return gen(); }
template <int bits> std::uint64_t GenerateNo111Baseline(Generator &gen) {
std::uint64_t r;
do {
r = GenerateUniform<bits>(gen);
} while (r & (r >> 1) & (r >> 2));
return r;
}
template <int bits> struct Tribonacci {
static constexpr std::uint64_t value = Tribonacci<bits - 1>::value +
Tribonacci<bits - 2>::value +
Tribonacci<bits - 3>::value;
};
template <> struct Tribonacci<0> { static constexpr std::uint64_t value = 1; };
template <> struct Tribonacci<-1> { static constexpr std::uint64_t value = 1; };
template <> struct Tribonacci<-2> { static constexpr std::uint64_t value = 0; };
template <int bits> std::uint64_t GenerateNo111(Generator &gen) {
constexpr int upper_bits = 16;
constexpr int lower_bits = bits - upper_bits;
const std::uint64_t upper = GenerateNo111Baseline<upper_bits>(gen);
for (;;) {
if ((upper & 1) == 0) {
return (upper << lower_bits) + GenerateNo111<lower_bits>(gen);
}
std::uint64_t outcome = std::uniform_int_distribution<std::uint64_t>{
0, Tribonacci<upper_bits>::value - 1}(gen);
if ((upper & 2) == 0) {
if (outcome < Tribonacci<upper_bits - 2>::value) {
return (upper << lower_bits) + (std::uint64_t{1} << (lower_bits - 1)) +
GenerateNo111<lower_bits - 2>(gen);
}
outcome -= Tribonacci<upper_bits - 2>::value;
}
if (outcome < Tribonacci<lower_bits - 1>::value) {
return (upper << lower_bits) + GenerateNo111<lower_bits - 1>(gen);
}
}
}
#define BASELINE(bits) \
template <> std::uint64_t GenerateNo111<bits>(Generator & gen) { \
return GenerateNo111Baseline<bits>(gen); \
}
BASELINE(0)
BASELINE(1)
BASELINE(2)
BASELINE(3)
BASELINE(4)
BASELINE(5)
BASELINE(6)
BASELINE(7)
BASELINE(8)
BASELINE(9)
BASELINE(10)
BASELINE(11)
BASELINE(12)
BASELINE(13)
BASELINE(14)
BASELINE(15)
BASELINE(16)
#undef BASELINE
static const std::uint64_t TRIBO[65] = {1,
2,
4,
7,
13,
24,
44,
81,
149,
274,
504,
927,
1705,
3136,
5768,
10609,
19513,
35890,
66012,
121415,
223317,
410744,
755476,
1389537,
2555757,
4700770,
8646064,
15902591,
29249425,
53798080,
98950096,
181997601,
334745777,
615693474,
1132436852,
2082876103,
3831006429,
7046319384,
12960201916,
23837527729,
43844049029,
80641778674,
148323355432,
272809183135,
501774317241,
922906855808,
1697490356184,
3122171529233,
5742568741225,
10562230626642,
19426970897100,
35731770264967,
65720971788709,
120879712950776,
222332455004452,
408933139743937,
752145307699165,
1383410902447554,
2544489349890656,
4680045560037375,
8607945812375585,
15832480722303616,
29120472094716576,
53560898629395777,
98513851446415969};
std::uint64_t get_rand_no111(Generator &gen) {
std::uint64_t idx =
std::uniform_int_distribution<std::uint64_t>{0, TRIBO[64] - 1}(gen);
std::uint64_t ret = 0;
for (int i = 63; i >= 0; --i) {
if (idx >= TRIBO[i]) {
ret |= std::uint64_t{1} << i;
idx -= TRIBO[i];
}
}
return ret;
}
} // namespace
int main() {
Generator gen{std::random_device{}()};
std::uint64_t sum = 0;
for (std::int32_t i = 0; i < 10000000; i++) {
if constexpr (true) {
sum += GenerateNo111<64>(gen);
} else {
sum += get_rand_no111(gen);
}
}
return sum & 127;
}
What about following simple idea:
Generate random r.
Find within this r window(s)-mask, contains 3 or more sequenced 1s.
If mask is 0 (no 3 or more sequenced bits) - return the r.
Substitute "incorrect" bits under that mask to new random ones.
Goto 2
Code sample (did not tested, compiled only):
uint64_t rand_no3() {
uint64_t r, mask;
for(r = get_rand_64() ; ; ) {
mask = r & (r >> 1) & (r >> 2);
mask |= (mask << 1) | (mask << 2);
if(mask == 0)
return r;
r ^= mask & get_rand_64();
}
}
Another variant of same code, with just single call get_rand_64():
uint64_t rand_no3() {
uint64_t r, mask = ~0ULL;
do {
r ^= mask & get_rand_64();
mask = r & (r >> 1) & (r >> 2);
mask |= (mask << 1) | (mask << 2);
} while(mask != 0);
return r;
}
I know, the last code does not init the r, but it is not matter, because of this variable will be overwritten in 1st loop iteration.
You could generate the number one bit at a time, keeping track of the number of consecutive set bits. Whenever you have two consecutive set bits, you insert an unset bit and set the count back to 0.
How do you convert a numerical number to an Excel column name in C# without using automation getting the value directly from Excel.
Excel 2007 has a possible range of 1 to 16384, which is the number of columns that it supports. The resulting values should be in the form of excel column names, e.g. A, AA, AAA etc.
Here's how I do it:
private string GetExcelColumnName(int columnNumber)
{
string columnName = "";
while (columnNumber > 0)
{
int modulo = (columnNumber - 1) % 26;
columnName = Convert.ToChar('A' + modulo) + columnName;
columnNumber = (columnNumber - modulo) / 26;
}
return columnName;
}
If anyone needs to do this in Excel without VBA, here is a way:
=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")
where colNum is the column number
And in VBA:
Function GetColumnName(colNum As Integer) As String
Dim d As Integer
Dim m As Integer
Dim name As String
d = colNum
name = ""
Do While (d > 0)
m = (d - 1) Mod 26
name = Chr(65 + m) + name
d = Int((d - m) / 26)
Loop
GetColumnName = name
End Function
You might need conversion both ways, e.g from Excel column adress like AAZ to integer and from any integer to Excel. The two methods below will do just that. Assumes 1 based indexing, first element in your "arrays" are element number 1.
No limits on size here, so you can use adresses like ERROR and that would be column number 2613824 ...
public static string ColumnAdress(int col)
{
if (col <= 26) {
return Convert.ToChar(col + 64).ToString();
}
int div = col / 26;
int mod = col % 26;
if (mod == 0) {mod = 26;div--;}
return ColumnAdress(div) + ColumnAdress(mod);
}
public static int ColumnNumber(string colAdress)
{
int[] digits = new int[colAdress.Length];
for (int i = 0; i < colAdress.Length; ++i)
{
digits[i] = Convert.ToInt32(colAdress[i]) - 64;
}
int mul=1;int res=0;
for (int pos = digits.Length - 1; pos >= 0; --pos)
{
res += digits[pos] * mul;
mul *= 26;
}
return res;
}
Sorry, this is Python instead of C#, but at least the results are correct:
def ColIdxToXlName(idx):
if idx < 1:
raise ValueError("Index is too small")
result = ""
while True:
if idx > 26:
idx, r = divmod(idx - 1, 26)
result = chr(r + ord('A')) + result
else:
return chr(idx + ord('A') - 1) + result
for i in xrange(1, 1024):
print "%4d : %s" % (i, ColIdxToXlName(i))
I discovered an error in my first post, so I decided to sit down and do the the math. What I found is that the number system used to identify Excel columns is not a base 26 system, as another person posted. Consider the following in base 10. You can also do this with the letters of the alphabet.
Space:.........................S1, S2, S3 : S1, S2, S3
....................................0, 00, 000 :.. A, AA, AAA
....................................1, 01, 001 :.. B, AB, AAB
.................................... …, …, … :.. …, …, …
....................................9, 99, 999 :.. Z, ZZ, ZZZ
Total states in space: 10, 100, 1000 : 26, 676, 17576
Total States:...............1110................18278
Excel numbers columns in the individual alphabetical spaces using base 26. You can see that in general, the state space progression is a, a^2, a^3, … for some base a, and the total number of states is a + a^2 + a^3 + … .
Suppose you want to find the total number of states A in the first N spaces. The formula for doing so is A = (a)(a^N - 1 )/(a-1). This is important because we need to find the space N that corresponds to our index K. If I want to find out where K lies in the number system I need to replace A with K and solve for N. The solution is N = log{base a} (A (a-1)/a +1). If I use the example of a = 10 and K = 192, I know that N = 2.23804… . This tells me that K lies at the beginning of the third space since it is a little greater than two.
The next step is to find exactly how far in the current space we are. To find this, subtract from K the A generated using the floor of N. In this example, the floor of N is two. So, A = (10)(10^2 – 1)/(10-1) = 110, as is expected when you combine the states of the first two spaces. This needs to be subtracted from K because these first 110 states would have already been accounted for in the first two spaces. This leaves us with 82 states. So, in this number system, the representation of 192 in base 10 is 082.
The C# code using a base index of zero is
private string ExcelColumnIndexToName(int Index)
{
string range = string.Empty;
if (Index < 0 ) return range;
int a = 26;
int x = (int)Math.Floor(Math.Log((Index) * (a - 1) / a + 1, a));
Index -= (int)(Math.Pow(a, x) - 1) * a / (a - 1);
for (int i = x+1; Index + i > 0; i--)
{
range = ((char)(65 + Index % a)).ToString() + range;
Index /= a;
}
return range;
}
//Old Post
A zero-based solution in C#.
private string ExcelColumnIndexToName(int Index)
{
string range = "";
if (Index < 0 ) return range;
for(int i=1;Index + i > 0;i=0)
{
range = ((char)(65 + Index % 26)).ToString() + range;
Index /= 26;
}
if (range.Length > 1) range = ((char)((int)range[0] - 1)).ToString() + range.Substring(1);
return range;
}
This answer is in javaScript:
function getCharFromNumber(columnNumber){
var dividend = columnNumber;
var columnName = "";
var modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnName = String.fromCharCode(65 + modulo).toString() + columnName;
dividend = parseInt((dividend - modulo) / 26);
}
return columnName;
}
Easy with recursion.
public static string GetStandardExcelColumnName(int columnNumberOneBased)
{
int baseValue = Convert.ToInt32('A');
int columnNumberZeroBased = columnNumberOneBased - 1;
string ret = "";
if (columnNumberOneBased > 26)
{
ret = GetStandardExcelColumnName(columnNumberZeroBased / 26) ;
}
return ret + Convert.ToChar(baseValue + (columnNumberZeroBased % 26) );
}
I'm surprised all of the solutions so far contain either iteration or recursion.
Here's my solution that runs in constant time (no loops). This solution works for all possible Excel columns and checks that the input can be turned into an Excel column. Possible columns are in the range [A, XFD] or [1, 16384]. (This is dependent on your version of Excel)
private static string Turn(uint col)
{
if (col < 1 || col > 16384) //Excel columns are one-based (one = 'A')
throw new ArgumentException("col must be >= 1 and <= 16384");
if (col <= 26) //one character
return ((char)(col + 'A' - 1)).ToString();
else if (col <= 702) //two characters
{
char firstChar = (char)((int)((col - 1) / 26) + 'A' - 1);
char secondChar = (char)(col % 26 + 'A' - 1);
if (secondChar == '#') //Excel is one-based, but modulo operations are zero-based
secondChar = 'Z'; //convert one-based to zero-based
return string.Format("{0}{1}", firstChar, secondChar);
}
else //three characters
{
char firstChar = (char)((int)((col - 1) / 702) + 'A' - 1);
char secondChar = (char)((col - 1) / 26 % 26 + 'A' - 1);
char thirdChar = (char)(col % 26 + 'A' - 1);
if (thirdChar == '#') //Excel is one-based, but modulo operations are zero-based
thirdChar = 'Z'; //convert one-based to zero-based
return string.Format("{0}{1}{2}", firstChar, secondChar, thirdChar);
}
}
Same implementation in Java
public String getExcelColumnName (int columnNumber)
{
int dividend = columnNumber;
int i;
String columnName = "";
int modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
i = 65 + modulo;
columnName = new Character((char)i).toString() + columnName;
dividend = (int)((dividend - modulo) / 26);
}
return columnName;
}
int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol >= 26)
{
int nChar = nCol % 26;
nCol = (nCol - nChar) / 26;
// You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;
Update: Peter's comment is right. That's what I get for writing code in the browser. :-) My solution was not compiling, it was missing the left-most letter and it was building the string in reverse order - all now fixed.
Bugs aside, the algorithm is basically converting a number from base 10 to base 26.
Update 2: Joel Coehoorn is right - the code above will return AB for 27. If it was real base 26 number, AA would be equal to A and the next number after Z would be BA.
int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol > 26)
{
int nChar = nCol % 26;
if (nChar == 0)
nChar = 26;
nCol = (nCol - nChar) / 26;
sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
sCol = sChars[nCol] + sCol;
..And converted to php:
function GetExcelColumnName($columnNumber) {
$columnName = '';
while ($columnNumber > 0) {
$modulo = ($columnNumber - 1) % 26;
$columnName = chr(65 + $modulo) . $columnName;
$columnNumber = (int)(($columnNumber - $modulo) / 26);
}
return $columnName;
}
Just throwing in a simple two-line C# implementation using recursion, because all the answers here seem far more complicated than necessary.
/// <summary>
/// Gets the column letter(s) corresponding to the given column number.
/// </summary>
/// <param name="column">The one-based column index. Must be greater than zero.</param>
/// <returns>The desired column letter, or an empty string if the column number was invalid.</returns>
public static string GetColumnLetter(int column) {
if (column < 1) return String.Empty;
return GetColumnLetter((column - 1) / 26) + (char)('A' + (column - 1) % 26);
}
Although there are already a bunch of valid answers1, none get into the theory behind it.
Excel column names are bijective base-26 representations of their number. This is quite different than an ordinary base 26 (there is no leading zero), and I really recommend reading the Wikipedia entry to grasp the differences. For example, the decimal value 702 (decomposed in 26*26 + 26) is represented in "ordinary" base 26 by 110 (i.e. 1x26^2 + 1x26^1 + 0x26^0) and in bijective base-26 by ZZ (i.e. 26x26^1 + 26x26^0).
Differences aside, bijective numeration is a positional notation, and as such we can perform conversions using an iterative (or recursive) algorithm which on each iteration finds the digit of the next position (similarly to an ordinary base conversion algorithm).
The general formula to get the digit at the last position (the one indexed 0) of the bijective base-k representation of a decimal number m is (f being the ceiling function minus 1):
m - (f(m / k) * k)
The digit at the next position (i.e. the one indexed 1) is found by applying the same formula to the result of f(m / k). We know that for the last digit (i.e. the one with the highest index) f(m / k) is 0.
This forms the basis for an iteration that finds each successive digit in bijective base-k of a decimal number. In pseudo-code it would look like this (digit() maps a decimal integer to its representation in the bijective base -- e.g. digit(1) would return A in bijective base-26):
fun conv(m)
q = f(m / k)
a = m - (q * k)
if (q == 0)
return digit(a)
else
return conv(q) + digit(a);
So we can translate this to C#2 to get a generic3 "conversion to bijective base-k" ToBijective() routine:
class BijectiveNumeration {
private int baseK;
private Func<int, char> getDigit;
public BijectiveNumeration(int baseK, Func<int, char> getDigit) {
this.baseK = baseK;
this.getDigit = getDigit;
}
public string ToBijective(double decimalValue) {
double q = f(decimalValue / baseK);
double a = decimalValue - (q * baseK);
return ((q > 0) ? ToBijective(q) : "") + getDigit((int)a);
}
private static double f(double i) {
return (Math.Ceiling(i) - 1);
}
}
Now for conversion to bijective base-26 (our "Excel column name" use case):
static void Main(string[] args)
{
BijectiveNumeration bijBase26 = new BijectiveNumeration(
26,
(value) => Convert.ToChar('A' + (value - 1))
);
Console.WriteLine(bijBase26.ToBijective(1)); // prints "A"
Console.WriteLine(bijBase26.ToBijective(26)); // prints "Z"
Console.WriteLine(bijBase26.ToBijective(27)); // prints "AA"
Console.WriteLine(bijBase26.ToBijective(702)); // prints "ZZ"
Console.WriteLine(bijBase26.ToBijective(16384)); // prints "XFD"
}
Excel's maximum column index is 16384 / XFD, but this code will convert any positive number.
As an added bonus, we can now easily convert to any bijective base. For example for bijective base-10:
static void Main(string[] args)
{
BijectiveNumeration bijBase10 = new BijectiveNumeration(
10,
(value) => value < 10 ? Convert.ToChar('0'+value) : 'A'
);
Console.WriteLine(bijBase10.ToBijective(1)); // prints "1"
Console.WriteLine(bijBase10.ToBijective(10)); // prints "A"
Console.WriteLine(bijBase10.ToBijective(123)); // prints "123"
Console.WriteLine(bijBase10.ToBijective(20)); // prints "1A"
Console.WriteLine(bijBase10.ToBijective(100)); // prints "9A"
Console.WriteLine(bijBase10.ToBijective(101)); // prints "A1"
Console.WriteLine(bijBase10.ToBijective(2010)); // prints "19AA"
}
1 This generic answer can eventually be reduced to the other, correct, specific answers, but I find it hard to fully grasp the logic of the solutions without the formal theory behind bijective numeration in general. It also proves its correctness nicely. Additionally, several similar questions link back to this one, some being language-agnostic or more generic. That's why I thought the addition of this answer was warranted, and that this question was a good place to put it.
2 C# disclaimer: I implemented an example in C# because this is what is asked here, but I have never learned nor used the language. I have verified it does compile and run, but please adapt it to fit the language best practices / general conventions, if necessary.
3 This example only aims to be correct and understandable ; it could and should be optimized would performance matter (e.g. with tail-recursion -- but that seems to require trampolining in C#), and made safer (e.g. by validating parameters).
I wanted to throw in my static class I use, for interoping between col index and col Label. I use a modified accepted answer for my ColumnLabel Method
public static class Extensions
{
public static string ColumnLabel(this int col)
{
var dividend = col;
var columnLabel = string.Empty;
int modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnLabel = Convert.ToChar(65 + modulo).ToString() + columnLabel;
dividend = (int)((dividend - modulo) / 26);
}
return columnLabel;
}
public static int ColumnIndex(this string colLabel)
{
// "AD" (1 * 26^1) + (4 * 26^0) ...
var colIndex = 0;
for(int ind = 0, pow = colLabel.Count()-1; ind < colLabel.Count(); ++ind, --pow)
{
var cVal = Convert.ToInt32(colLabel[ind]) - 64; //col A is index 1
colIndex += cVal * ((int)Math.Pow(26, pow));
}
return colIndex;
}
}
Use this like...
30.ColumnLabel(); // "AD"
"AD".ColumnIndex(); // 30
private String getColumn(int c) {
String s = "";
do {
s = (char)('A' + (c % 26)) + s;
c /= 26;
} while (c-- > 0);
return s;
}
Its not exactly base 26, there is no 0 in the system. If there was, 'Z' would be followed by 'BA' not by 'AA'.
if you just want it for a cell formula without code, here's a formula for it:
IF(COLUMN()>=26,CHAR(ROUND(COLUMN()/26,1)+64)&CHAR(MOD(COLUMN(),26)+64),CHAR(COLUMN()+64))
In Delphi (Pascal):
function GetExcelColumnName(columnNumber: integer): string;
var
dividend, modulo: integer;
begin
Result := '';
dividend := columnNumber;
while dividend > 0 do begin
modulo := (dividend - 1) mod 26;
Result := Chr(65 + modulo) + Result;
dividend := (dividend - modulo) div 26;
end;
end;
A little late to the game, but here's the code I use (in C#):
private static readonly string _Alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static int ColumnNameParse(string value)
{
// assumes value.Length is [1,3]
// assumes value is uppercase
var digits = value.PadLeft(3).Select(x => _Alphabet.IndexOf(x));
return digits.Aggregate(0, (current, index) => (current * 26) + (index + 1));
}
In perl, for an input of 1 (A), 27 (AA), etc.
sub excel_colname {
my ($idx) = #_; # one-based column number
--$idx; # zero-based column index
my $name = "";
while ($idx >= 0) {
$name .= chr(ord("A") + ($idx % 26));
$idx = int($idx / 26) - 1;
}
return scalar reverse $name;
}
Though I am late to the game, Graham's answer is far from being optimal. Particularly, you don't have to use the modulo, call ToString() and apply (int) cast. Considering that in most cases in C# world you would start numbering from 0, here is my revision:
public static string GetColumnName(int index) // zero-based
{
const byte BASE = 'Z' - 'A' + 1;
string name = String.Empty;
do
{
name = Convert.ToChar('A' + index % BASE) + name;
index = index / BASE - 1;
}
while (index >= 0);
return name;
}
More than 30 solutions already, but here's my one-line C# solution...
public string IntToExcelColumn(int i)
{
return ((i<16926? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<2730? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<26? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + ((char)((i%26)+65)));
}
After looking at all the supplied Versions here, I decided to do one myself, using recursion.
Here is my vb.net Version:
Function CL(ByVal x As Integer) As String
If x >= 1 And x <= 26 Then
CL = Chr(x + 64)
Else
CL = CL((x - x Mod 26) / 26) & Chr((x Mod 26) + 1 + 64)
End If
End Function
Refining the original solution (in C#):
public static class ExcelHelper
{
private static Dictionary<UInt16, String> l_DictionaryOfColumns;
public static ExcelHelper() {
l_DictionaryOfColumns = new Dictionary<ushort, string>(256);
}
public static String GetExcelColumnName(UInt16 l_Column)
{
UInt16 l_ColumnCopy = l_Column;
String l_Chars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String l_rVal = "";
UInt16 l_Char;
if (l_DictionaryOfColumns.ContainsKey(l_Column) == true)
{
l_rVal = l_DictionaryOfColumns[l_Column];
}
else
{
while (l_ColumnCopy > 26)
{
l_Char = l_ColumnCopy % 26;
if (l_Char == 0)
l_Char = 26;
l_ColumnCopy = (l_ColumnCopy - l_Char) / 26;
l_rVal = l_Chars[l_Char] + l_rVal;
}
if (l_ColumnCopy != 0)
l_rVal = l_Chars[l_ColumnCopy] + l_rVal;
l_DictionaryOfColumns.ContainsKey(l_Column) = l_rVal;
}
return l_rVal;
}
}
Here is an Actionscript version:
private var columnNumbers:Array = ['A', 'B', 'C', 'D', 'E', 'F' , 'G', 'H', 'I', 'J', 'K' ,'L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
private function getExcelColumnName(columnNumber:int) : String{
var dividend:int = columnNumber;
var columnName:String = "";
var modulo:int;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnName = columnNumbers[modulo] + columnName;
dividend = int((dividend - modulo) / 26);
}
return columnName;
}
JavaScript Solution
/**
* Calculate the column letter abbreviation from a 1 based index
* #param {Number} value
* #returns {string}
*/
getColumnFromIndex = function (value) {
var base = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split('');
var remainder, result = "";
do {
remainder = value % 26;
result = base[(remainder || 26) - 1] + result;
value = Math.floor(value / 26);
} while (value > 0);
return result;
};
These my codes to convert specific number (index start from 1) to Excel Column.
public static string NumberToExcelColumn(uint number)
{
uint originalNumber = number;
uint numChars = 1;
while (Math.Pow(26, numChars) < number)
{
numChars++;
if (Math.Pow(26, numChars) + 26 >= number)
{
break;
}
}
string toRet = "";
uint lastValue = 0;
do
{
number -= lastValue;
double powerVal = Math.Pow(26, numChars - 1);
byte thisCharIdx = (byte)Math.Truncate((columnNumber - 1) / powerVal);
lastValue = (int)powerVal * thisCharIdx;
if (numChars - 2 >= 0)
{
double powerVal_next = Math.Pow(26, numChars - 2);
byte thisCharIdx_next = (byte)Math.Truncate((columnNumber - lastValue - 1) / powerVal_next);
int lastValue_next = (int)Math.Pow(26, numChars - 2) * thisCharIdx_next;
if (thisCharIdx_next == 0 && lastValue_next == 0 && powerVal_next == 26)
{
thisCharIdx--;
lastValue = (int)powerVal * thisCharIdx;
}
}
toRet += (char)((byte)'A' + thisCharIdx + ((numChars > 1) ? -1 : 0));
numChars--;
} while (numChars > 0);
return toRet;
}
My Unit Test:
[TestMethod]
public void Test()
{
Assert.AreEqual("A", NumberToExcelColumn(1));
Assert.AreEqual("Z", NumberToExcelColumn(26));
Assert.AreEqual("AA", NumberToExcelColumn(27));
Assert.AreEqual("AO", NumberToExcelColumn(41));
Assert.AreEqual("AZ", NumberToExcelColumn(52));
Assert.AreEqual("BA", NumberToExcelColumn(53));
Assert.AreEqual("ZZ", NumberToExcelColumn(702));
Assert.AreEqual("AAA", NumberToExcelColumn(703));
Assert.AreEqual("ABC", NumberToExcelColumn(731));
Assert.AreEqual("ACQ", NumberToExcelColumn(771));
Assert.AreEqual("AYZ", NumberToExcelColumn(1352));
Assert.AreEqual("AZA", NumberToExcelColumn(1353));
Assert.AreEqual("AZB", NumberToExcelColumn(1354));
Assert.AreEqual("BAA", NumberToExcelColumn(1379));
Assert.AreEqual("CNU", NumberToExcelColumn(2413));
Assert.AreEqual("GCM", NumberToExcelColumn(4823));
Assert.AreEqual("MSR", NumberToExcelColumn(9300));
Assert.AreEqual("OMB", NumberToExcelColumn(10480));
Assert.AreEqual("ULV", NumberToExcelColumn(14530));
Assert.AreEqual("XFD", NumberToExcelColumn(16384));
}
Sorry, this is Python instead of C#, but at least the results are correct:
def excel_column_number_to_name(column_number):
output = ""
index = column_number-1
while index >= 0:
character = chr((index%26)+ord('A'))
output = output + character
index = index/26 - 1
return output[::-1]
for i in xrange(1, 1024):
print "%4d : %s" % (i, excel_column_number_to_name(i))
Passed these test cases:
Column Number: 494286 => ABCDZ
Column Number: 27 => AA
Column Number: 52 => AZ
For what it is worth, here is Graham's code in Powershell:
function ConvertTo-ExcelColumnID {
param (
[parameter(Position = 0,
HelpMessage = "A 1-based index to convert to an excel column ID. e.g. 2 => 'B', 29 => 'AC'",
Mandatory = $true)]
[int]$index
);
[string]$result = '';
if ($index -le 0 ) {
return $result;
}
while ($index -gt 0) {
[int]$modulo = ($index - 1) % 26;
$character = [char]($modulo + [int][char]'A');
$result = $character + $result;
[int]$index = ($index - $modulo) / 26;
}
return $result;
}
Another VBA way
Public Function GetColumnName(TargetCell As Range) As String
GetColumnName = Split(CStr(TargetCell.Cells(1, 1).Address), "$")(1)
End Function
Here's my super late implementation in PHP. This one's recursive. I wrote it just before I found this post. I wanted to see if others had solved this problem already...
public function GetColumn($intNumber, $strCol = null) {
if ($intNumber > 0) {
$intRem = ($intNumber - 1) % 26;
$strCol = $this->GetColumn(intval(($intNumber - $intRem) / 26), sprintf('%s%s', chr(65 + $intRem), $strCol));
}
return $strCol;
}
I have a sequence of n letters: e.g. A B C A D , or A B F A
What I want is to get every possible variant with a comma between the letters.
i.e. for A B F A
A B F A
A,B F A
A B,F A
A B F,A
A,B,F A
A B,F,A
A,B F,A
A,B,F,A
Can anyone recommend a good algorithm for doing this? Language not important.
Simple solution to use binary array to represent if there is a comma or not.
A B F A contains three positions where comma may be (AB, BF, FA)
That means if you create 3-element array and try every possible combination of 0 and 1 you'll get the desired result. 000, 001, 010, 011, 100, 101, 110, 111
Simple program in java prints all binary permutation for n bits:
String s = "ABFA";
int bits = s.length() - 1;
int lastNumber = (int)Math.pow(2, bits);
System.out.println(lastNumber);
for (int i = 0; i < lastNumber; i++) {
System.out.println(completeZeros(Integer.toString(i, 2), bits));
}
static String completeZeros(String s, int bits) {
String result = s;
for (int i = 0; i < bits - s.length(); i++) {
result = "0" + result;
}
return result;
}
To apply binary permutation "010" to string "ABFA" use next function:
static String applyBinary(String s, String binary) {
String result = "" + s.charAt(0);
for (int i = 0; i < binary.length(); i++) {
if (binary.charAt(i) == '1') result += ", ";
result += s.charAt(i + 1);
}
return result;
}
The output is:
ABFA
ABF, A
AB, FA
AB, F, A
A, BFA
A, BF, A
A, B, FA
A, B, F, A
Use binary system for this task.
1 means comma is present, 0 means comma is not present. Each position in number informs about presence of another comma. For example for AFA:
00 : A F A
01 : A F,A
10 : A,F A
11 : A,F,A
Numbers must be taken from range [0 .. (n-1)^2-1]
Here's a simple JavaScript demonstration.
var str = "ABFA";
function algo(str) {
var result = [];
var n = str.length;
var total = Math.pow(n-1, 2) - 1;
for(var mask = 0; mask < total; mask++) {
var bits = mask;
var newstr = "";
for(var i=0; i<str.length - 1; i++, bits>>>=1) {
var hasComma = (bits & 1) == 1;
newstr += str.charAt(i);
newstr += (hasComma ? "," : " ");
}
newstr += str.charAt(str.length - 1);
result.push(newstr);
}
return result;
}
algo(str);
You calculate the total number of combinations "total"
You count up to that number "mask"
You use the binary bits of your counter "bits" to add commas
There are two approaches to this problem.
1.Recursive ( Start = "printAllComb ( s , "" , 0 );" )
printAllComb(string s, string const, int i)
{
if ( i == s.length() )
print const
printAllComb(s,const+string.at(i)+',',i+1);
printAllComb(s,const+string.at(i),i+1);
}
2.Dynamic Programming
char binaryS[s.length]="0000";
//Basically no. of zeros = no. of Alphabets in the string
//define a function AddOne() which adds 1 to the character representation
//AddOne() modifies the character array such that it stays in the bit representation
//Characters because to save space
while ( All the bits are not one )
{
for ( int i=0; i<s.length(); i++ )
{
print s.at(i)
if ( binaryS.at(i) == '1' )
print ","
}
print "\n"
AddOne();
}
how do I perform ( A div B ) mod C where I am calculating A in one function,B in another function,and C is say ( 10 pow 9 plus 7 ) but both A and B may be greater than C,or INT_MAX
if the only problem is the size of the numbers you can use long. if the number may be larger then long.MAX_VALUE then you need some function to calculate using strings, or use two or more long types, and make your own functions. for example, the plus function will get two long types, check if the sum of them is smaller then one of them(meaning they went over the MAX_SIZE), then return an array of long, containing to numbers, one for the carry bit, and one for the sum.
here is an example in c#, but it's easy to translate it to c++
public static string sum(long a, long b)
{
string sum;
if (a + b > a && a + b > b)
{
sum = (a + b).ToString();
}
else
{
string aStr = a.ToString();
string bStr = b.ToString();
if (bStr.Length > aStr.Length)
{
string tmp = aStr;
aStr = bStr;
bStr = tmp;
}
sum = new string('0', aStr.Length + bStr.Length);
char[] arr = sum.ToCharArray();
for (int i = 0; i < bStr.Length; i++)
{
int loc = sum.Length - 1 - i;
arr[loc] += (char)(aStr[aStr.Length - 1 - i] + bStr[bStr.Length - 1 - i] - '0' * 2);
if (arr[loc] > '9')
{
arr[loc - 1] = '1';
arr[loc] = (char)(arr[loc] - '9' - 1);
}
}
for (int i = bStr.Length ; i < aStr.Length; i++)
{
int loc = sum.Length - 1 - i;
arr[loc] += (char)(aStr[aStr.Length - 1 - i] - '0');
if (arr[loc] > '9')
{
arr[loc - 1] = '1';
arr[loc] = (char)(arr[loc] - '9' - 1);
}
}
sum = new string(arr);
}
return sum;
}
}
Given a phone keypad as shown below:
1 2 3
4 5 6
7 8 9
0
How many different 10-digit numbers can be formed starting from 1? The constraint is that the movement from 1 digit to the next is similar to the movement of the Knight in a chess game.
For eg. if we are at 1 then the next digit can be either 6 or 8
if we are at 6 then the next digit can be 1, 7 or 0.
Repetition of digits are allowed - 1616161616 is a valid number.
Is there a polynomial time algorithm which solves this problem? The problem requires us to just give the count of 10-digit numbers and not necessarily list the numbers.
EDIT: I tried modeling this as a graph with each digit having 2 or 3 digits as its neighbors. Then I used DFS to navigate upto the depth of 10 nodes and then increment the count of numbers each time I reached the depth of 10. This obviously is not polynomial time. Assuming each digit had just 2 neighbors, this would have required at least 2^10 iterations.
The variable here is the number of digits. I have taken the eg. of 10 digit numbers. It could as well be n-digits.
Sure it can be done in polynomial time. It's an excellent exercise in dynamic programming or memoization.
Lets assume N (the number of digits) equals 10 for the example.
Think of it recursively like this: How many numbers can I construct using 10 digits starting from 1?
Answer is
[number of 9-digit numbers starting from 8] +
[number of 9-digit numbers starting from 6].
So how many "9-digit numbers starting from 8" are there? Well,
[number of 8-digit numbers starting from 1] +
[number of 8-digit numbers starting from 3]
and so on. Base case is reached when you get the question "How many 1-digit numbers are there starting from X" (and the answer is obviously 1).
When it comes to complexity, the key observation is that you reuse previously computed solutions. That is for instance, the answer to "how many 5-digit numbers starting from 3" there are, can be used both when answering "how many 6-digit numbers are there starting from 8" AND "how many 6-digit numbers are there starting from 4". This reuse make the complexity collapse from exponential to polynomial.
Let's take a closer look at the complexity of a dynamic programming solution:
Such implementation would fill in a matrix in the following way:
num[1][i] = 1, for all 0<=i<=9 -- there are one 1-digit number starting from X.
for digits = 2...N
for from = 0...9
num[digits][from] = num[digits-1][successor 1 of from] +
num[digits-1][successor 2 of from] +
...
num[digits-1][successor K of from]
return num[N][1] -- number of N-digit numbers starting from 1.
The algorithm simply fills the matrix one cell at a time, and the matrix is of dimension 10*N, and thus runs in linear time.
Wrote it down from the top of my head, please correct me if there are any typos.
I decided to tackle this problem and make it as extensible as I can. This solution allows you to:
Define your own board (phone pad, chess board, etc.)
Define your own chess piece (Knight, Rook, Bishop, etc.); you will have to write the concrete class and generate it from the factory.
Retrieve several pieces of information through some useful utility methods.
The classes are as follows:
PadNumber: Class defining a button on the phone pad. Could be renamed to 'Square' to represent a board square.
ChessPiece: Abstract class that defines fields for all chess pieces.
Movement: Interface that defines movement methods and allows for factory generation of pieces.
PieceFactory: Factory class to generate Chess pieces.
Knight: Concrete class that inherits from ChessPiece and implements Movement
PhoneChess: Entrance class.
Driver: Driver code.
OK, here's the code :)
package PhoneChess;
import java.awt.Point;
public class PadNumber {
private String number = "";
private Point coordinates = null;
public PadNumber(String number, Point coordinates)
{
if(number != null && number.isEmpty()==false)
this.number = number;
else
throw new IllegalArgumentException("Input cannot be null or empty.");
if(coordinates == null || coordinates.x < 0 || coordinates.y < 0)
throw new IllegalArgumentException();
else
this.coordinates = coordinates;
}
public String getNumber()
{
return this.number;
}
public Integer getNumberAsNumber()
{
return Integer.parseInt(this.number);
}
public Point getCoordinates()
{
return this.coordinates;
}
public int getX()
{
return this.coordinates.x;
}
public int getY()
{
return this.coordinates.y;
}
}
ChessPiece
package PhoneChess;
import java.util.HashMap;
import java.util.List;
public abstract class ChessPiece implements Movement {
protected String name = "";
protected HashMap<PadNumber, List<PadNumber>> moves = null;
protected Integer fullNumbers = 0;
protected int[] movesFrom = null;
protected PadNumber[][] thePad = null;
}
Movement Interface:
package PhoneChess;
import java.util.List;
public interface Movement
{
public Integer findNumbers(PadNumber start, Integer digits);
public abstract boolean canMove(PadNumber from, PadNumber to);
public List<PadNumber> allowedMoves(PadNumber from);
public Integer countAllowedMoves(PadNumber from);
}
PieceFactory
package PhoneChess;
public class PieceFactory
{
public ChessPiece getPiece(String piece, PadNumber[][] thePad)
{
if(thePad == null || thePad.length == 0 || thePad[0].length == 0)
throw new IllegalArgumentException("Invalid pad");
if(piece == null)
throw new IllegalArgumentException("Invalid chess piece");
if(piece.equalsIgnoreCase("Knight"))
return new Knight("Knight", thePad);
else
return null;
}
}
Knight class
package PhoneChess;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
public final class Knight extends ChessPiece implements Movement {
/**Knight movements
* One horizontal, followed by two vertical
* Or
* One vertical, followed by two horizontal
* #param name
*/
public Knight(String name, PadNumber[][] thePad)
{
if(name == null || name.isEmpty() == true)
throw new IllegalArgumentException("Name cannot be null or empty");
this.name = name;
this.thePad = thePad;
this.moves = new HashMap<>();
}
private Integer fullNumbers = null;
#Override
public Integer findNumbers(PadNumber start, Integer digits)
{
if(start == null || "*".equals(start.getNumber()) || "#".equals(start.getNumber()) ) { throw new IllegalArgumentException("Invalid start point"); }
if(start.getNumberAsNumber() == 5) { return 0; } //Consider adding an 'allowSpecialChars' condition
if(digits == 1) { return 1; };
//Init
this.movesFrom = new int[thePad.length * thePad[0].length];
for(int i = 0; i < this.movesFrom.length; i++)
this.movesFrom[i] = -1;
fullNumbers = 0;
findNumbers(start, digits, 1);
return fullNumbers;
}
private void findNumbers(PadNumber start, Integer digits, Integer currentDigits)
{
//Base condition
if(currentDigits == digits)
{
//Reset
currentDigits = 1;
fullNumbers++;
return;
}
if(!this.moves.containsKey(start))
allowedMoves(start);
List<PadNumber> options = this.moves.get(start);
if(options != null)
{
currentDigits++; //More digits to be got
for(PadNumber option : options)
findNumbers(option, digits, currentDigits);
}
}
#Override
public boolean canMove(PadNumber from, PadNumber to)
{
//Is the moves list available?
if(!this.moves.containsKey(from.getNumber()))
{
//No? Process.
allowedMoves(from);
}
if(this.moves.get(from) != null)
{
for(PadNumber option : this.moves.get(from))
{
if(option.getNumber().equals(to.getNumber()))
return true;
}
}
return false;
}
/***
* Overriden method that defines each Piece's movement restrictions.
*/
#Override
public List<PadNumber> allowedMoves(PadNumber from)
{
//First encounter
if(this.moves == null)
this.moves = new HashMap<>();
if(this.moves.containsKey(from))
return this.moves.get(from);
else
{
List<PadNumber> found = new ArrayList<>();
int row = from.getY();//rows
int col = from.getX();//columns
//Cases:
//1. One horizontal move each way followed by two vertical moves each way
if(col-1 >= 0 && row-2 >= 0)//valid
{
if(thePad[row-2][col-1].getNumber().equals("*") == false &&
thePad[row-2][col-1].getNumber().equals("#") == false)
{
found.add(thePad[row-2][col-1]);
this.movesFrom[from.getNumberAsNumber()] = this.movesFrom[from.getNumberAsNumber()] + 1;
}
}
if(col-1 >= 0 && row+2 < thePad.length)//valid
{
if(thePad[row+2][col-1].getNumber().equals("*") == false &&
thePad[row+2][col-1].getNumber().equals("#") == false)
{
found.add(thePad[row+2][col-1]);
this.movesFrom[from.getNumberAsNumber()] = this.movesFrom[from.getNumberAsNumber()] + 1;
}
}
if(col+1 < thePad[0].length && row+2 < thePad.length)//valid
{
if(thePad[row+2][col+1].getNumber().equals("*") == false &&
thePad[row+2][col+1].getNumber().equals("#") == false)
{
found.add(thePad[row+2][col+1]);
this.movesFrom[from.getNumberAsNumber()] = this.movesFrom[from.getNumberAsNumber()] + 1;
}
}
if(col+1 < thePad[0].length && row-2 >= 0)//valid
{
if(thePad[row-2][col+1].getNumber().equals("*") == false &&
thePad[row-2][col+1].getNumber().equals("#") == false)
found.add(thePad[row-2][col+1]);
}
//Case 2. One vertical move each way follow by two horizontal moves each way
if(col-2 >= 0 && row-1 >= 0)
{
if(thePad[row-1][col-2].getNumber().equals("*") == false &&
thePad[row-1][col-2].getNumber().equals("#") == false)
found.add(thePad[row-1][col-2]);
}
if(col-2 >= 0 && row+1 < thePad.length)
{
if(thePad[row+1][col-2].getNumber().equals("*") == false &&
thePad[row+1][col-2].getNumber().equals("#") == false)
found.add(thePad[row+1][col-2]);
}
if(col+2 < thePad[0].length && row-1 >= 0)
{
if(thePad[row-1][col+2].getNumber().equals("*") == false &&
thePad[row-1][col+2].getNumber().equals("#") == false)
found.add(thePad[row-1][col+2]);
}
if(col+2 < thePad[0].length && row+1 < thePad.length)
{
if(thePad[row+1][col+2].getNumber().equals("*") == false &&
thePad[row+1][col+2].getNumber().equals("#") == false)
found.add(thePad[row+1][col+2]);
}
if(found.size() > 0)
{
this.moves.put(from, found);
this.movesFrom[from.getNumberAsNumber()] = found.size();
}
else
{
this.moves.put(from, null); //for example the Knight cannot move from 5 to anywhere
this.movesFrom[from.getNumberAsNumber()] = 0;
}
}
return this.moves.get(from);
}
#Override
public Integer countAllowedMoves(PadNumber from)
{
int start = from.getNumberAsNumber();
if(movesFrom[start] != -1)
return movesFrom[start];
else
{
movesFrom[start] = allowedMoves(from).size();
}
return movesFrom[start];
}
#Override
public String toString()
{
return this.name;
}
}
PhoneChess entrant class
package PhoneChess;
public final class PhoneChess
{
private ChessPiece thePiece = null;
private PieceFactory factory = null;
public ChessPiece ThePiece()
{
return this.thePiece;
}
public PhoneChess(PadNumber[][] thePad, String piece)
{
if(thePad == null || thePad.length == 0 || thePad[0].length == 0)
throw new IllegalArgumentException("Invalid pad");
if(piece == null)
throw new IllegalArgumentException("Invalid chess piece");
this.factory = new PieceFactory();
this.thePiece = this.factory.getPiece(piece, thePad);
}
public Integer findPossibleDigits(PadNumber start, Integer digits)
{
if(digits <= 0)
throw new IllegalArgumentException("Digits cannot be less than or equal to zero");
return thePiece.findNumbers(start, digits);
}
public boolean isValidMove(PadNumber from, PadNumber to)
{
return this.thePiece.canMove(from, to);
}
}
Driver Code:
public static void main(String[] args) {
PadNumber[][] thePad = new PadNumber[4][3];
thePad[0][0] = new PadNumber("1", new Point(0,0));
thePad[0][1] = new PadNumber("2", new Point(1,0));
thePad[0][2] = new PadNumber("3",new Point(2,0));
thePad[1][0] = new PadNumber("4",new Point(0,1));
thePad[1][1] = new PadNumber("5",new Point(1,1));
thePad[1][2] = new PadNumber("6", new Point(2,1));
thePad[2][0] = new PadNumber("7", new Point(0,2));
thePad[2][1] = new PadNumber("8", new Point(1,2));
thePad[2][2] = new PadNumber("9", new Point(2,2));
thePad[3][0] = new PadNumber("*", new Point(0,3));
thePad[3][1] = new PadNumber("0", new Point(1,3));
thePad[3][2] = new PadNumber("#", new Point(2,3));
PhoneChess phoneChess = new PhoneChess(thePad, "Knight");
System.out.println(phoneChess.findPossibleDigits(thePad[0][1],4));
}
}
This can be done in O(log N). Consider the keypad and the possible moves on it as a graph G(V, E) where vertices are the available digits and edges say which digits can follow which. Now for each output position i we can form a vector Paths(i) containing the number of different paths each vertex can be reached in. Now it's pretty easy to see that for a given position i and digit v, the possible paths that it can be reached through is the sum of the different paths that possible preceding digits could be reached through, or Paths(i)[v] = sum(Paths(i-1)[v2] * (1 if (v,v2) in E else 0) for v2 in V ). Now, this is taking the sum of each position the preceding vector times a corresponding position in a column of the adjacency matrix. So we can simplify this as Paths(i) = Paths(i-1) · A, where A is the adjacency matrix of the graph. Getting rid of the recursion and taking advantage of associativity of matrix multiplication, this becomes Paths(i) = Paths(1) · A^(i-1). We know Paths(1): we have only one path, to the digit 1.
The total number of paths for an n digit number is the sum of the paths for each digit, so the final algorithm becomes: TotalPaths(n) = sum( [1,0,0,0,0,0,0,0,0,0] · A^(n-1) )
The exponentiation can be calculated via squaring in O(log(n)) time, given constant time multiplies, otherwise O(M(n) * log(n)) where M(n) is the complexity of your favorite arbitrary precision multiplication algorithm for n digit numbers.
A simpler answer.
#include<stdio.h>
int a[10] = {2,2,2,2,3,0,3,2,2,2};
int b[10][3] = {{4,6},{6,8},{7,9},{4,8},{0,3,9},{},{1,7,0},{2,6},{1,3},{2,4}};
int count(int curr,int n)
{
int sum = 0;
if(n==10)
return 1;
else
{
int i = 0;
int val = 0;
for(i = 0; i < a[curr]; i++)
{
val = count(b[curr][i],n+1);
sum += val;
}
return sum;
}
}
int main()
{
int n = 1;
int val = count(1,0);
printf("%d\n",val);
}
celebrate!!
Run time constant time solution:
#include <iostream>
constexpr int notValid(int x, int y) {
return !(( 1 == x && 3 == y ) || //zero on bottom.
( 0 <= x && 3 > x && //1-9
0 <= y && 3 > y ));
}
class Knight {
template<unsigned N > constexpr int move(int x, int y) {
return notValid(x,y)? 0 : jump<N-1>(x,y);
}
template<unsigned N> constexpr int jump( int x, int y ) {
return move<N>(x+1, y-2) +
move<N>(x-1, y-2) +
move<N>(x+1, y+2) +
move<N>(x-1, y+2) +
move<N>(x+2, y+1) +
move<N>(x-2, y+1) +
move<N>(x+2, y-1) +
move<N>(x-2, y-1);
}
public:
template<unsigned N> constexpr int count() {
return move<N-1>(0,1) + move<N-1>(0,2) +
move<N-1>(1,0) + move<N-1>(1,1) + move<N-1>(1,2) +
move<N-1>(2,0) + move<N-1>(2,1) + move<N-1>(2,2);
}
};
template<> constexpr int Knight::move<0>(int x, int y) { return notValid(x,y)? 0 : 1; }
template<> constexpr int Knight::count<0>() { return 0; } //terminal cases.
template<> constexpr int Knight::count<1>() { return 8; }
int main(int argc, char* argv[]) {
static_assert( ( 16 == Knight().count<2>() ), "Fail on test with 2 lenght" ); // prof of performance
static_assert( ( 35 == Knight().count<3>() ), "Fail on test with 3 lenght" );
std::cout<< "Number of valid Knight phones numbers:" << Knight().count<10>() << std::endl;
return 0;
}
Method returns list of 10 digit numbers starting with 1. Again the count is 1424.
public ArrayList<String> getList(int digit, int length, String base ){
ArrayList<String> list = new ArrayList<String>();
if(length == 1){
list.add(base);
return list;
}
ArrayList<String> temp;
for(int i : b[digit]){
String newBase = base +i;
list.addAll(getList(i, length -1, newBase ));
}
return list;
}
I'm not sure if I missed something, but reading the description of the problem I came to this solution. It has O(n) time complexity and O(1) space complexity.
I figured that number 1 is at a corner, right? In each corner you can either move to one of the sides (4 from 9 and 3, or 6 from 7 an 1) or one of the 'vertical' sides (8 from 3 and 1, or 2 from 9 and 7). So, corners add two moves: a side move and a 'vertical' move. This is true for all four corners (1,3,9,7).
From each side, you can either move to two corners (7 and 1 from 6, 9 and 3 from 4) or you can reach the bottom key (0). That's three moves. Two corners and one bottom.
On the bottom key (0), you can move to both sides (4 and 6). So, in each step, you check out all possible endings for the path of the previous length (that is, how many ended on a corner, a side, a 'vertical' or the 'bottom' zero key) and then generate new ending counts according to the generation rules stated before.
Each corner ending adds a side and a vertical.
Each side ending adds 2 corners and a bottom.
Each vertical ending adds 2 corners.
Each bottom ending adds 2 sides.
If you start from the '1' key, you start with one possible corner solution, in each step you count the number of corner, side, vertical and bottom endings of the previous step and then apply the rules to generate the next count.
In plain javascript code.
function paths(n) {
//Index to 0
var corners = 1;
var verticals = 0;
var bottom = 0;
var sides = 0;
if (n <= 0) {
//No moves possible for paths without length
return 0;
}
for (var i = 1; i < n; i++) {
var previousCorners = corners;
var previousVerticals = verticals;
var previousBottom = bottom;
var previousSides = sides;
sides = 1 * previousCorners + 2 * previousBottom;
verticals = 1 * previousCorners;
bottom = 1 * previousSides;
corners = 2 * previousSides + 2 * previousVerticals;
//console.log("Moves: %d, Length: %d, Sides: %d, Verticals: %d, Bottom: %d, Corners: %d, Total: %d", i, i + 1, sides, verticals, bottom, corners, sides+verticals+bottom+corners);
}
return sides + verticals + bottom + corners;
}
for (var i = 0; i <= 10; i++) {
console.log(paths(i));
}
This problem may be also modelled as a Constraint satisfaction problem (aka CSP for short).
I suggest to use the Minion solver (fast and scalable) that you can find here.
Modelling maybe tedious and time consumming (steep learning curve).
Instead of using Minion language input, my advice is to formulate the model with solver independent modelling language such as ESSENCE and find a converter accordingly.
//Both the iterative and recursive with memorize shows count as 1424 for 10 digit numbers starting with 1.
int[][] b = {{4,6},{6,8},{7,9},{4,8},{0,3,9},{},{1,7,0},{2,6},{1,3},{2,4}};
public int countIterative(int digit, int length) {
int[][] matrix = new int[length][10];
for(int dig =0; dig <=9; dig++){
matrix[0][dig] = 1;
}
for(int len = 1; len < length; len++){
for(int dig =0; dig <=9; dig++){
int sum = 0;
for(int i : b[dig]){
sum += matrix[len-1][i];
}
matrix[len][dig] = sum;
}
}
return matrix[length-1][digit];
}
public int count(int index, int length, int[][] matrix ){
int sum = 0;
if(matrix[length-1][index] > 0){
System.out.println("getting value from memoize:"+index + "length:"+ length);
return matrix[length-1][index];
}
if( length == 1){
return 1;
}
for(int i: b[index] ) {
sum += count(i, length-1,matrix);
}
matrix[length-1][index] = sum;
return sum;
}
Recursive memoization approach:
vector<vector<int>> lupt = { {4, 6}, {6, 8}, {9, 7}, {4, 8}, {3, 9, 0},
{}, {1,7,0}, {6, 2}, {1, 3}, {2, 4} };
int numPhoneNumbersUtil(int startdigit, int& phonenumberlength, int currCount, map< pair<int,int>,int>& memT)
{
int noOfCombs = 0;
vector<int> enddigits;
auto it = memT.find(make_pair(startdigit,currCount));
if(it != memT.end())
{
noOfCombs = it->second;
return noOfCombs;
}
if(currCount == phonenumberlength)
{
return 1;
}
enddigits = lupt[startdigit];
for(auto it : enddigits)
{
noOfCombs += numPhoneNumbersUtil(it, phonenumberlength, currCount + 1, memT);
}
memT.insert(make_pair(make_pair(startdigit,currCount), noOfCombs));
return memT[make_pair(startdigit,currCount)];
}
int numPhoneNumbers(int startdigit, int phonenumberlength)
{
map<pair<int,int>,int> memT;
int currentCount = 1; //the first digit has already been added
return numPhoneNumbersUtil(startdigit, phonenumberlength, currentCount, memT);
}
I implemented both brute force and dynamic programming models
import queue
def chess_numbers_bf(start, length):
if length <= 0:
return 0
phone = [[7, 5], [6, 8], [3, 7], [9, 2, 8], [], [6, 9, 0], [1, 5], [0, 2], [3, 1], [5, 3]]
total = 0
q = queue.Queue()
q.put((start, 1))
while not q.empty():
front = q.get()
val = front[0]
len_ = front[1]
if len_ < length:
for elm in phone[val]:
q.put((elm, len_ + 1))
else:
total += 1
return total
def chess_numbers_dp(start, length):
if length <= 0:
return 0
phone = [[7, 5], [6, 8], [3, 7], [9, 2, 8], [], [6, 9, 0], [1, 5], [0, 2], [3, 1], [5, 3]]
memory = {}
def __chess_numbers_dp(s, l):
if (s, l) in memory:
return memory[(s, l)]
elif l == length - 1:
memory[(s, l)] = 1
return 1
else:
total_n_ways = 0
for number in phone[s]:
total_n_ways += __chess_numbers_dp(number, l+1)
memory[(s, l)] = total_n_ways
return total_n_ways
return __chess_numbers_dp(start, 0)
# bf
for i in range(0, 10):
print(i, chess_numbers_bf(3, i))
print('\n')
for i in range(0, 10):
print(i, chess_numbers_bf(9, i))
print('\n')
# dp
for i in range(0, 10):
print(i, chess_numbers_dp(3, i))
print('\n')
# dp
for i in range(0, 10):
print(i, chess_numbers_dp(9, i))
print('\n')
Recursive function in Java:
public static int countPhoneNumbers (int n, int r, int c) {
if (outOfBounds(r,c)) {
return 0;
} else {
char button = buttons[r][c];
if (button == '.') {
// visited
return 0;
} else {
buttons[r][c] = '.'; // record this position so don't revisit.
// Count all possible phone numbers with one less digit starting
int result=0;
result = countPhoneNumbers(n-1,r-2,c-1)
+ countPhoneNumbers(n-1,r-2,c+1)
+ countPhoneNumbers(n-1,r+2,c-1)
+ countPhoneNumbers(n-1,r+2,c+1)
+ countPhoneNumbers(n-1,r-1,c-2)
+ countPhoneNumbers(n-1,r-1,c+2)
+ countPhoneNumbers(n-1,r+1,c-2)
+ countPhoneNumbers(n-1,r+1,c+2);
}
buttons[r][c] = button; // Remove record from position.
return result;
}
}
}