## MindCipher

### May 10, 2013

For the first problem we perform a simple simulation:

```(define (flips rev)   (define (flip) (if (< (rand) 1/2) 'H 'T))   (let loop ((k 3) (fs (list (flip) (flip) (flip))))     (if (equal? fs rev) k       (loop (+ k 1) (take 3 (cons (flip) fs))))))```

Here rev is a list of coin flips in reverse order, since the coin flips are added to the front of the list. One call to `flips` computes the average number of flips to find the desired sequence. The two functions below compute the average over n trials:

```(define (monday n)   (let loop ((i 0) (k 0))     (if (= i n) (/ k n 1.0)       (loop (+ i 1) (+ k (flips '(H T H)))))))```

```(define (tuesday n)   (let loop ((i 0) (k 0))     (if (= i n) (/ k n 1.0)       (loop (+ i 1) (+ k (flips '(T T H)))))))```

Here are our results:

```> (monday 1000) 9.996 > (tuesday 1000) 8.058```

Thus Monday is likely to be higher than Tuesday. That makes sense because on Monday the new pattern starts fresh after a two-shot miss, while on Tuesday the new pattern has a head start of one flip after a two-shot miss. That is, if you flip heads-then-tails on Monday, then flip a tail to miss the pattern, you have to wait for the next heads-then-tails for the next possible hit. But on Tuesday, if you flip heads-then-tails, then flip a head to miss the pattern, you already have the first flip of the next heads-then-tails pattern.

The second problem is easily solved by splitting the number into pieces and checking each year after 1978 in order until the answer is found:

```> (let loop ((n 1979))     (let ((front (quotient n 100))           (middle (modulo (quotient n 10) 100))           (back (modulo n 100)))       (if (= (+ front back) middle) n         (loop (+ n 1))))) 2307```

If you don’t like the brute force solution, you can represent a year as 1000a + 100b + 10c + d, employ a little bit of algebra to form a diophantine equation, then solve.

The third problem is a trick: (p + q) / 2 is between p and q, and by definition there are no primes between p and q, so the answer is no, with no programming required.

We used `take` and `rand` from the Standard Prelude. You can run the programs at http://programmingpraxis.codepad.org/fpSuFV7A. Be sure to take a look at MindCipher; some of the problems there will turn your brain inside-out.

Pages: 1 2

### 20 Responses to “MindCipher”

1. […] today’s Programming Praxis exercise, our goal is to solve two exercises from the MindCipher website […]

2. My Haskell solution (see http://bonsaicode.wordpress.com/2013/05/10/programming-praxis-mindcipher/ for a version with comments):

```import Control.Monad
import Data.List
import System.Random

flipUntil :: [Bool] -> IO Int
flipUntil pattern = fmap (length . takeWhile (not . isPrefixOf pattern) .
tails . randomRs (False, True)) newStdGen

day :: [Bool] -> IO Double
day pattern = fmap (\cs -> fromIntegral (sum cs) / fromIntegral (length cs)) .
replicateM 10000 \$ flipUntil pattern

sumDay :: Maybe Integer
sumDay = find (\d -> div d 100 + mod d 100 == div (mod d 1000) 10) [1979..]

main :: IO ()
main = do print =<< liftM2 compare (day [False, True, False])
(day [False, True, True])
print sumDay
```
3. eupraxia said
```# if p and q are consecutive prime numbers then, since p<q,
# p < (p+q)/2 < q => (p+q)/2 cannot be prime.
```
4. OptimusPrime said

Ans 2) 2307
n = 1979
while n<9999:
if (n/100) + (n%100) == (n/10)%100:
print n
break

n=n+1

5. namako said

1.’s most accurate answer is probably ‘there is no way to know’ (in fact, since one can only do something a finite number of times in a day, there’s a (tiny) nonzero chance you didn’t see the sequences at all on either day). However, we can get the average time until one sees any particular 3-element code using a Markov chain with 9 states.
Source
Mean time to hit TTT = 14
Mean time to hit TTH = 8
Mean time to hit THT = 10
Mean time to hit THH = 8
Mean time to hit HTT = 8
Mean time to hit HTH = 10
Mean time to hit HHT = 8
Mean time to hit HHH = 14
E(HTH) > E(HTT), so one would probably see Monday’s average as longer than Tuesday’s.

6. wolvesatthedoor said

here’s my solution to 1978 in python:

```for a in xrange(10):
for b in xrange(10):
for c in xrange(10):
for d in xrange(10):
if ((10*a) - (9*b) + (9*c) + d)== 0:
y = int(str(a)+str(b)+str(c)+str(d))
if y > 1978:
print y
exit(1)
```
7. eupraxia said
```from random import choice
from itertools import repeat
from func_utils import repeatedly, partition

def number_tosses(pattern):
heads_tails = repeatedly(lambda: choice(['H', 'T']))
triples = partition(heads_tails, 3, 1)
indexed = enumerate(triples, 1)
return next(i for (i, (t, t1, t2)) in indexed if (t, t1, t2)==pattern)

patterns = [('T', 'T', 'T'), ('T', 'T', 'H'), ('T', 'H', 'T'),
('T', 'H', 'H'), ('H', 'T', 'T'), ('H', 'T', 'H'),
('H', 'H', 'T'), ('H', 'H', 'H')]

for p in patterns:
print p, sum(map(number_tosses, repeat(p, 1000000)))/1000000

# ('T', 'T', 'T')  12.01629   ~= 12
# ('T', 'T', 'H')  6.006479   ~=  6
# ('T', 'H', 'T')  8.003555   ~=  8
# ('T', 'H', 'H')  5.999449   ~=  6
# ('H', 'T', 'T')  5.999335   ~=  6
# ('H', 'T', 'H')  7.993025   ~=  8
# ('H', 'H', 'T')  5.99461     ~=  6
# ('H', 'H', 'H')  12.000927 ~= 12

# This gives the same ordering as Namako ie.
# TTH==THH==HTT==HHT < THT==HTH < TTT==HHH
# but different ratios.
```
8. namako said

@eupraxia
I’m not sure (not familiar with the language) but if you’re only counting triples, you’re missing the first two tosses. (It takes 3 to get a triple in the first place, which will be the first you test).

9. eupraxia said

@namako. Many thanks. You’re absolutely right, I should have started the count at three, not one.

```from random import choice
from itertools import repeat
from func_utils import repeatedly, partition

def number_tosses(pattern):
heads_tails = repeatedly(lambda: choice(['H', 'T']))
triples = partition(heads_tails, 3, 1)
# changed from: indexed = enumerate(triples, 1)
indexed = enumerate(triples, 3)
return next(i for (i, (t, t1, t2)) in indexed if (t, t1, t2)==pattern)

for p in patterns:
print p, sum(map(number_tosses, repeat(p, 1000000)))/1000000

#('T', 'T', 'T') 14.018673
#('T', 'T', 'H') 8.010297
#('T', 'H', 'T') 10.008117
#('T', 'H', 'H') 8.001807
#('H', 'T', 'T') 8.006271
#('H', 'T', 'H') 10.001372
#('H', 'H', 'T') 8.006121
#('H', 'H', 'H') 13.995327

from random import choice
from itertools import repeat

def number_tosses(pattern):
toss = lambda:choice(['H', 'T'])
a, b, c = toss(), toss(), toss()
# n = number of tosses so far
n = 3
while True:
if (a, b, c)==pattern:
return n
a, b, c = b, c, toss()
n+=1

patterns = [('T', 'T', 'T'), ('T', 'T', 'H'), ('T', 'H', 'T'),
('T', 'H', 'H'), ('H', 'T', 'T'), ('H', 'T', 'H'),
('H', 'H', 'T'), ('H', 'H', 'H')]

for p in patterns:
print p, sum(map(number_tosses, repeat(p, 1000000)))/1000000

#('T', 'T', 'T') 14.004905
#('T', 'T', 'H') 7.999236
#('T', 'H', 'T') 10.001018
#('T', 'H', 'H') 8.006784
#('H', 'T', 'T') 8.004825
#('H', 'T', 'H') 10.002537
#('H', 'H', 'T') 7.995162
#('H', 'H', 'H') 13.99838
```
10. aks said

The coin tossing problem comparing the statistics on two different
targets “HTH” versus “HTT” can be resolved with some logical
analysis:

Both targets are a specific permutation of three flips of a
coin.

Coin flipping continues until the target permutation occurs.

After each coin flip that fails to create a matching
permutation, one or none previous flips can be used as part
of the permutation that includes the next flip.

For the HTH target:

A failing sequence of “HTT” requires at least three more
tosses; none of the previous tosses can be used for a
successfully matched permutation. e.g.: HTT{HTH}

A failing sequence of “HH” requires two more tosses, because
the second toss can be part of succesful matching sequence:
H{HTH}.

For the HTT target:

A failing sequence of “HTH” can be successfully matched with
two more tosses: “HT{HTT}”.

A failing sequence of “HH” can be successfully matched with
just two more tosses: “H{HTT}”.

So, the number of tosses needed for a succesful match, including
a random number of failing tosses is smaller for the “HTT”
target than for the “HTH” target.

I wrote a Ruby program to demonstrate the problem:

https://gist.github.com/aks/5562771

```
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
#!/usr/bin/env ruby
# coins.rb
# count average flips until HTH is reached
#

def flip ; rand(2) == 1 ? 'H' : 'T' ; end

# 'HTH'.count_flips

class String
def count_flips
target = self.to_s
src = ''
count = 0
while (target != src)   # loop until we match the target
src += flip           # append new toss
count += 1            # count flips
src.slice!(0,1) if src.size > target.size
end
count                   # return flip count
end
end

# Enumerable didn't include this -- why?
class Array
def sum
self.reduce(:+)
end
def average
Float(self.sum) / self.nitems
end
end

# average_flips
#
# count coin flips until a target is reached
# do this repeated for a day, then average those counts.
#
# Simulate human-speed coin flips by estimating time for each flip, and
# assuming constant flipping for 24 hours, without a break.

\$flip_time = 4    # 4 seconds per flip
\$max_flips_per_day = 24 * 60 * 60 / \$flip_time

class String
def flip_stats
target = self

counts = []
(1..\$max_flips_per_day).each{|i| counts += [target.count_flips]}
min_count, max_count = counts.minmax
average_count = counts.average
[min_count, max_count, average_count]
end
end

['HTH', 'HTT'].each {|target|
puts "Computing #{target} stats.." if \$DEBUG
(min, max, avg) = target.flip_stats
printf "Target: %s  Coin flip stats: min: %4.1f  max: %5.1f  avg: %5.1f\n", target, min, max, avg
}
exit
```
11. aks said
```#!/usr/bin/env ruby
# year-digits-sum.rb
#
# 1978: The year 1978 is such that the sum of the first two digits
# and the latter two digits is equal to the middle two digits, i.e.
# 19 + 78 = 97. What is the next year (after 1978) for which this is
# true?
#
#    a b c d
#    1 9 7 8
#
# Formula:
#
#   [F1] a*10 + b + c*10 + d = b*10 + c
#
#   [F2] a*10 - b*9 + c*9 + d = 0   ( Simplified )
#
# Problem:
#
#  Find Min(Year) such that [F2] is true.

class Fixnum
def find_next_year
year = self.to_i
while true do
year += 1
a, b, c, d = year.to_s.split(//).map{|c| c.to_i}
if a*10 - b*9 + c*9 + d == 0
return year
end
end
end
end

puts "After 1978, the next year for which (ab) + (cd) = (bc) is.."
puts 1978.find_next_year
exit
```

Here’s the run output:

```\$ ./year-digits-sum.rb
After 1978, the next year for which (ab) + (cd) = (bc) is..
2307
```
12. aks said

For the primes problem (#3), here is a J program to show that there are no primes in the set of (p+q)/2 where p,q are consecutive primes in the first million prime numbers.

https://gist.github.com/aks/5563008

```   i. 20
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

NB. generate the first 20 primes
p: i. 20
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71

NB. box up consecutive pairs of those primes
(2 <\ ]) p: i. 20
┌───┬───┬───┬────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐
│2 3│3 5│5 7│7 11│11 13│13 17│17 19│19 23│23 29│29 31│31 37│37 41│41 43│43 47│47 53│53 59│59 61│61 67│67 71│
└───┴───┴───┴────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘

NB. sum up each pair of primes
+/ each (2 <\ ])p: i. 20
┌─┬─┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬───┬───┬───┬───┬───┐
│5│8│12│18│24│30│36│42│52│60│68│78│84│90│100│112│120│128│138│
└─┴─┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴───┴───┴───┴───┴───┘

NB. divide each sum by 2
2 %~ each +/ each (2 <\ ])p: i. 20
┌───┬─┬─┬─┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┐
│2.5│4│6│9│12│15│18│21│26│30│34│39│42│45│50│56│60│64│69│
└───┴─┴─┴─┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┘

NB. now, test each of those results for being prime.  1 p: y -- tests y for being prime

1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20
┌─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐
│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│
└─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘

NB. open the boxed results, so we can add them up
>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

NB. sum/reduce the vector of booleans.  If there's a prime, the sum will be > 0
+/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20
0

NB. ok. No primes.  Let's keep checking for larger groups

+/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 1000
0
+/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 10000
0
+/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 100000
0

NB. the previous output took a few seconds.  The next will take a few minutes

+/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 1000000
0
```
13. Mike said

For number 3:

p and q are consecutive primes.
p and q are positive integers
let m = (p + q)/2
m is the mean of p
p < m < q
therefore, m cannot be prime, because p and q are consecutive primes

14. Mike said

For number 2:

```Let the digits be labelled a,b,c,d.  So for 1978, a=1, b=9, c=7, d=8.

observation 1 : a + c = b, therefor a <= b and c <= b.
Moreover, a < b, because if a == b, then c = 0, and there is
no d such that bb + 0d == b0.
So, only need to consider centuries in which a < b.

observation 2: c = b - a if b + d = 10.  So for any given century, there
are only two possible decades.

observation 3: for any give decade there is only 1 possible year such
that (b+d)%10 == c

So:
19cd: c can be 7 or 8
197d: (9 + d)%10 == 7 => d = 8, and 19 + 78 = 97 **** 1978 is a winner
198d: (9 + d)%10 == 8 => d = 9, but 19 + 89 != 98

20cd: can't work because a >= b
21cd:            "
22cd:            "

23cd: c can be 0 or 1
230d:  (3 + d)%10 == 0  => d = 7, and 23 + 07 = 30 **** 2307 is a winner
231d:  (3 + d)%10 == 1  => d = 8, but 23 + 18 != 31

24cd: c can be 1 or 2:
241d: d = 7, and 24 + 17 == 41 **** 2417 is a winner

Based on observation 1, there are at most 8 possibilities for 1bcd, 7 for 2bcd, etc. for a total of (8*(8+1))/2 = 36 possible solutions between the years 1000 and 9999.
```
15. Mike said

The end of my previous comment should have said: Based on observation 1, there are at most 8 possibilities for 1bcd, 7 for 2bcd, etc. for a total of (8*(8+1))/2 = 36 possible solutions between the years 1000 and 9999

```aks said a,b,c,d must satisfy the formula: a*10 - b*9 + c*9 + d = 0.

if you have (a, b, c, d) that satisfy the formula, these also satisfy
the formula:

observation 4 : (   a   , b +/- k, c +/- k,    d   )

observation 5 : (a +/- k, b +/- k,    c   , d -/+ k)

observation 6 : (a +/- k,    b   , c -/+ k, d -/+ k)

where k is limited so that the new a,b,c,d are all single digits

So:
from 1978, apply observation 6, with k=1, to get a solution 2967.  Then
apply observation 4, with k=6, to get the solution 2307

by repeatedly applying observations 4 and 6, all solutions between 1000 and 9999 are easily found:

1208, 1318, 1428, 1538, 1648, 1758, 1868, 1978
2307, 2417, 2527, 2637, 2747, 2857, 2967
3406, 3516, 3626, 3736, 3846, 3956
4505, 4615, 4725, 4835, 4945
5604, 5714, 5824, 5934
6703, 6813, 6923
7802, 7912
8901

Observation 4 moves along a row, observation 6 moves up or down a column.
```
16. Vignesh M said

// solution to 1978 prob. using Java

public class Year1978Problem {

/**
* @param args
*/
public static final int REF_YEAR = 1978;
public static final int DEFAULT_VAL = 0;
private int findNextYear(){
boolean yearFound = false;
int i = REF_YEAR; // (1978)
int nextYear;
do{
i++; // starting the test with 1979
int a = i/100; // to get (19)
int b = i%100; // to get (79)
int c = a%10; // to get (9)
int d = b/10; // to get (7)
int e = Integer.parseInt((Integer.toString(c)+d)); // to get (97)
if((a+b) == e){ // checking whether 1979 is OK
yearFound = true;
return i; // return if yes,
}

}while(!yearFound); // continue till finding the year
return DEFAULT_VAL;
}
public static void main(String[] args) {
Year1978Problem obj = new Year1978Problem();
int nextYear = obj.findNextYear();
System.out.println(“Next year satisfying given condition : “+nextYear);
}

}

# Next year satisfying given condition : 2307

17. Alcriss said

bool check = false;
int _refYear = 1978;

while (check == false)
{
_refYear += 1;
check = CheckUp(_refYear);
}

Console.Write(“\n Next to 1978 is ” + _refYear + “.”);
Environment.Exit(0);

}
//METHODS

public static bool CheckUp(int entry)
{
string _year = string.Empty;
_year = Convert.ToString(entry);
int _num1 = Convert.ToInt32(_year[0].ToString() + _year[1].ToString());
int _num2 = Convert.ToInt32(_year[2].ToString() + _year[3].ToString());
int _sum = _num1 + _num2;
string _new = _year[1].ToString() + _year[2].ToString();
if (_new == Convert.ToString(_sum))
{
return true;
}
return false;
}

18. Tyler said

I have to learn Coldfusion for my job, so I’ve been using different examples from your site to help me learn. Most solutions I’ve seen to 1978 solve it by splitting it up into a,b,c,d. I chose a different route because I felt the individual digits didn’t mean as much as the two digits together. Is there anything wrong with me solving it this way?

``` Year Program Find the next year after 1978 in which the middle digits is equals to the sum of the first two digits and the last two digits #NumberFormat(a, mask)# + #NumberFormat(c, mask)# = #NumberFormat(b, mask)# ```

19. tylerchilds said

Sorry, last one I placed inside ``` instead of ```

``````, my bad.

``````
``` Year Program Find the next year after 1978 in which the middle digits is equals to the sum of the first two digits and the last two digits #NumberFormat(a, mask)# + #NumberFormat(c, mask)# = #NumberFormat(b, mask)# ```

20. For #2:

#!/usr/bin/ruby
r=[];1978.upto(9999) {|y|y=y.to_s;((“#{y[0]}#{y[1]}”.to_i+”#{y[2]}#{y[3]}”.
to_i)==(“#{y[1]}#{y[2]}”.to_i))?r.push(y):”; };puts r.to_s

That will give you all the solutions that are in a 4 digit year. You can easily quit after the first by just printing (instead of pushing to an array and exiting):

#!/usr/bin/ruby
1978.upto(9999) {|y|y=y.to_s;((“#{y[0]}#{y[1]}”.to_i+”#{y[2]}#{y[3]}”.
to_i)==(“#{y[1]}#{y[2]}”.to_i))?puts y;exit:”; }