<< problem 363 - Bézier Curves Minimum of subsequences - problem 375 >>

# Problem 371: Licence plates

Oregon licence plates consist of three letters followed by a three digit number (each digit can be from [0..9]).
While driving to work Seth plays the following game:
Whenever the numbers of two licence plates seen on his trip add to 1000 that's a win.

E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too. (as long as he sees them in the same trip).

Find the expected number of plates he needs to see for a win.
Give your answer rounded to 8 decimal places behind the decimal point.

Note: We assume that each licence plate seen is equally likely to have any three digit number on it.

# My Algorithm

The problem description tries hard to confuse the reader ... the plates' letters (e.g. RYU or SET) don't matter at all.
The only relevant part of a plate is its number. And I've never seen a leading zero on a plate - though it's possible according to the text.
In the end it doesn't matter as well.

A simple Monte-Carlo simulation can find the first digits (see monteCarlo()):

• generate a random number plate
• if 1000 - plate was seen before then the game is over
• add plate to the set of seen plates
There are three groups of numbers:
(a) 500 can only be paired with a second 500
(b) 0 doesn't pair with any number
(c) all other numbers

The whole problem can be mapped to states (not the meaning "countries" but "character"):
• there are 499 pairs in group c → 499 states
• for each of those state there are two options: 500 wasn't seen or it was seen once
→ multiplying both results in 998 states
I decided to create two arrays no500 and with500, each with 499 elements (for a total of 998 values).

And their state transitions:
• the "zero" plate (group b) doesn't change the current state
The third group (1 - 499 and 501 - 999) has an interesting property:
for each pair (x, y) where x + y = 1000 is doesn't matter whether x < y or x > y.
Therefore I assume that the first number of any pair in group c is always smaller than 500.
The exact numbers on the plates don't affect the result as long as they belong to group c: observing (12,430,299) is identical to (1,2,3).
In the end I am only concerned whether a plate's number is new (→ choose the next free number 1,2,3,...), is a duplicate or pairs with an existing number.
The probabilities are (assuming have denoted the number of unpaired plates in group c):
(1) p_{new} = (998 - 2 * have) / 1000
(2) p_{finished} = have / 1000
(3) p_{duplicate} = have / 1000
(4) p_{zero} = 1 / 1000
(5) p_{500} = 1 / 1000

Stupid me needs to verify that I got everything right: 1 = p_{new} + p_{finished} + p_{duplicate} + p_{zero} + p_{500} → it's okay !

If I progress from one state to another then the expected value is
(6) E = 1 + p_{E_1} * E_1 + p_{E_2} * E_2 + p_{E_3} * E_3 + ...

I have to be careful with state transitions that lead back to the same state they started from:
(7) E = 1 + p_E * E
(8) E - p_E * E = 1
(9) E * (1 - p_E) = 1
(10) E = 1 / (1 - p_E)
→ e.g. assume I throw a fair coin p_E = 0.5 => E = 1 / (1 - 0.5) = 1 / 0.5 = 2 so I expect to need two attempts to see the other side
→ e.g. assume I throw a biased coin p_E = 0.6 that has a higher probability to show the last side => E = 1 / (1 - 0.6) = 1 / 0.4 = 2.5 attempts to see the other side

And the generalized equation is E = (1 + p_{E_1} * E_1 + p_{E_2} * E_2 + ...) / (1 - p_E)

I found it much easier to reverse the problem:
if I have seen all plates 1 to 499 (and there was no pair yet), then
1. there is a chance 1 / 1000 to see "zero"
2. there is a chance 499 / 1000 to see an old plate
3. there is a chance 499 / 1000 to finally see a matching plate
4. there is a chance 1 / 1000 to see "500"

The first two cases keep the player in its current state.
The game is won if case 3 happens-
If the current state includes that "500" was seen before, then the game is over, too.
If the current state includes that "500" was NOT seen before, then the state changes to "have500 with 499 plates".

Now the expected number for have500[499] is:
(11) E_{499,have500} = 1 / (1 - (p_{zero} + p_{duplicate})) = 2

And no500[499]:
(12) E_{499,no500} = (1 + 1/1000 * E_{499,have500}) / (1 - (p_{zero} + p_{duplicate})) = 2.4

Now E_{498,have500} can be computed by using the probabilities of staying in the current state and both state transitions (finish or advance to 499 plates).
With similar argument E_{498,no500} depends only on E_{498,have500} and E_{499,no500}.
A for-loop computes all values.

Initially, the player hasn't seen a plate at all. That means E_{0,no500} is the result.

## Alternative Approaches

A Dynamic Programming approach sounds viable but I highly doubt it can match the efficiency of the plain for-loop.

## Note

search() computes bad results if the number of plates is odd because then there is no equivalent to the unique features of the "500" plates.

It took me a few minutes to realize that "zero" is special: first, I didn't think it's a valid plate number at all and second, I'm dumb. Sad but true ...

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 10 | ./371

Output:

Note: the original problem's input 1000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <iomanip>
#include <vector>

// a simple pseudo-random number generator
// (produces the same result no matter what compiler you have - unlike rand() from math.h)
unsigned int myrand()
{
// copied from problem 227
static unsigned long long seed = 0;
seed = 6364136223846793005ULL * seed + 1;
return (unsigned int)(seed >> 30);
}

// simulate random journeys ...
double monteCarlo(unsigned int numPlates, unsigned int iterations)
{
unsigned int numPlatesSeen = 0;
for (unsigned int i = 0; i < iterations; i++)
{
// true if plate was seen
std::vector<bool> seen(numPlates, false);

while (true)
{
// generate a new plate
auto plate = myrand() % numPlates;
numPlatesSeen++;

// look for matching plate (so that sum is 1000)
if (plate != 0) // 0 isn't matched by any other plate
{
auto other = numPlates - plate;
if (seen[other])
break;
}

// mark current plate as "seen"
seen[plate] = true;
}
}

return numPlatesSeen / double(iterations);
}

// and now smarter, faster, more precise ...
double search(unsigned int numPlates)
{
// how I compute the expected number:
// E              = 1 + fail * E
// E - fail * E   = 1
// E * (1 - fail) = 1
// E              = 1 / (1 - fail)
// if there is another state transition E2 then with similar argument
// E = 1 + fail * E + good * E2
// E = (1 + good * E2) / (1 - fail)

// Seth may observe at most 499 "ordinary" plates
// => zero and 500 are "special"
const auto maxHave = numPlates / 2 - 1;

// several times I have to divide by numPlates and have to make sure the result is a double,
// so let's convert it to double once and keep using plates instead of numPlates from here on
const double plates = numPlates; // actually I thought about invPlates = 1 / double(numPlates)

// track expected number of plates you still have to see if n unpaired plates were already observed
// there is a significant difference whether the "500" plate was seen or not !
std::vector<double> have500(maxHave + 1, 0);
std::vector<double> no500  (maxHave + 1, 0);

// probabilities:
// observe an "old" plate, nothing changes (but the expected number)
auto probDuplicate = maxHave / plates;
// observe a  "000" plate, nothing changes (but the expected number)
auto probZero      =    1    / plates;
// observe a  "500" plate, stop if already saw a "500", otherwise store it
auto prob500       =    1    / plates;

// solve the base case: all 499 plates already seen
// compute probability that the next plate still doesn't finish the game
auto probUnchanged = probDuplicate + probZero;
// now find expected number in case Seth 499 plates plus the "500" plate
have500[maxHave]   =  1                               / (1 - probUnchanged);
// and find expected number in case Seth 499 plates BUT NOT the "500" plate
no500  [maxHave]   = (1 + prob500 * have500[maxHave]) / (1 - probUnchanged);

// continue the computation for 498, 497, ..., 1, 0 observed plates
for (int have = maxHave - 1; have >= 0; have--)
{
// determine number of unpaired plates (exclude zero and 500)
auto numNew  = plates - 2*have - 2;
// probability that the plate wasn't seen before but pairs with no other plate yet
auto probNew = numNew / plates;

// it becomes less probable to see an "old" plate ...
probDuplicate = have / plates;            // same as probDuplicate -= 1 / plates
probUnchanged = probDuplicate + probZero; // same as probUnchanged += 1 / plates

// same as above: E = (1 + good * E2) / (1 - fail)
have500[have] = (1                           + probNew * have500[have + 1]) / (1 - probUnchanged);
no500  [have] = (1 + prob500 * have500[have] + probNew * no500  [have + 1]) / (1 - probUnchanged);
}

// no plate seen yet, not even "500"
return no500[0];
}

int main()
{
unsigned int numPlates = 1000;
std::cin >> numPlates;

std::cout << std::fixed << std::setprecision(8);
std::cout << search(numPlates) << std::endl;

// run simulation
//while (true)
//std::cout << monteCarlo(numPlates, 1000000) << std::endl;

return 0;
}


This solution contains 20 empty lines, 40 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

September 21, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 30% (out of 100%).

# Heatmap

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I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
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My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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