20050528, 12:42  #1 
Jun 2003
The Texas Hill Country
10001000001_{2} Posts 
Generating 2005
The folks at mathzone.net.tc posted a (set of) challenge(s) to generate the numeric value 2005 using a single digit and various operators.
As an example of what they had in mind, we could generate the number 12 as (44+4)/4 Their metric for the quality of the solution was the number of times the required digit occurred. Paul (Xilman on this forum) presented a solution that used a long chain of square root functions to generate any arbitrary integer. Since each sqrt function did not require any digits, his answer has a very low count. I wish to propose a similar challenge that does not rule out Paul's approach but makes it less desirable because I am specifying a different metric. Your goal for each of the ten challenges is to write the shortest C++ code fragment to be inserted in the program specified below. The code must meet the following criteria: The submitted code fragment will be compressed as follows: All comments will be removed All literal strings will be replaced with the null string "" All whitespace, including newlines, will be removed The resulting entry must include at least one instance of the digit for which it is entered. When inserted in place of the line "2005" in the code below, the program must compile and execute properly to produce (just) the answer "ACCEPTABLE". Entries will be ranked on two levels First, the use of Distinct Digits. We count the number of distinct elements of the set { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' } that are used. In case of a tie, we count Total Characters. Remember that only nonblank characters will be counted. Feel free to use multiple lines, spaces, and comments to make the method more readable. These are all stripped before evaluation and therefore do not penalize you. In both cases, an entry using fewer will be ranked higher. My example entry qualifies in the "0", "2", and "5" categories. In each case, it is ranked on the basis of (3,4) Let the Challenge Begin! Code:
#include <iostream> int main (int argc, char * const argv[]) { int answer = // BEGIN inserted code on the next line 2005 // END inserted code on the above line ; if (answer != 2005) { std::cout << "REJECTED"; } else { std::cout << "ACCEPTABLE"; } std::cout << std::endl; return 0; } Last fiddled with by Wacky on 20050528 at 13:47 Reason: Typo 
20050528, 13:37  #2 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
Scoreboard
Code:
Digit Rank Date By Fragment 0 1,10 Jun 05 Ken_g6 0xFAA>>~0 1 1,16 Jun 01 Ken_g6 1~1111111<<11 // min is probably > 13 chars 2 1,13 Jun 01 Ken_g6 22222222~2 // minimal 3 1,14 Jun 01 Ken_g6 3333+~3*333~3 // By computer, probably minimal 4 1,12 Jun 01 Ken_g6 ~4*(44+~444) // By computer, minimal 5 1,12 Jun 01 Ken_g6 5*5*5<<~55 // By computer, minimal  sorry, Richard. :( 6 1,12 Jun 01 Ken_g6 ~6*666~6666 // By computer, minimal 7 1,13 Jun 01 Ken_g6 77~77~7<<7 // By computer, minimal 8 1,15 Jun 02 Ken_g6 8888/8^~8~888 // By computer, minimal 9 1,11 May 31 Ken_g6 999+(9^999) // By computer, minimal Also note: He also has provided (1,9) solutions for each of the digits using 2character literals, Since I haven't found any way to get the compiler to reject them, they are the actual "winners". But I'm leaving the other solutions on the scoreboard because they are, at least to me, more interesting. Congratulations, Ken, on a good job. (But the contest remains open as long as there is a moderator who will accept shorter answers) Wacky Last fiddled with by Wacky on 20050605 at 22:03 
20050528, 22:27  #3 
Jan 2005
Caught in a sieve
5·79 Posts 
I get it. The idea is the shortest expression (in terms of numerals) for 2005 containing digit x.
Well, for x=4, "5*401" (or 401*5) seems minimal in total size. For x=1,6 try "20061". For x=7 try "20072". For x=3,8 try "20083". For x=9 try "20094". 
20050528, 23:07  #4  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
As another example, you offered "5*401" for the digit 4. It has 4 distinct digits, '0', '1', '4' and '5' with a total of 5 characters and is scored as (4,5) I will beat that with "5*400+5" which scores (3,7) because I used only three distinct digits even though I used 2 additional characters. 

20050529, 04:31  #5 
Jan 2005
Caught in a sieve
5×79 Posts 
Okay, I get it now. First off, I got a solution for x=1 using only addition (and subtraction):
x=1: 1111+1111111111+1+1+1+1+1 That means for any 0<x<10 you can take this, multiply it by x (which changes all the 1's to x's) and divide by x again to get a solution for x. But then I found better solutions for all of them! :D I think x=0 requires at least one other digit. Can someone prove me wrong? x=0: 2002+2+2/2 "<<" is the leftshift operator. Too bad it needs parentheses. :( But it's still sometimes better than anything else I could find. x=1: (1<<11)11111111+1 x=2: 2222222+2+2+2/2 x=3: 3*333+3*333+3+3+3/3 x=4: 44*44+4*4*4+4+4/4 x=5: 555*555555*5+55+5 x=7: (7<<7)+777+7*7*777/7 x=6: 666+666+666+6+6/6 x=8: 888+888+88+88+8*888/8 x=9: 999+999+9(9+9)/9 
20050529, 05:22  #6 
Jan 2005
Caught in a sieve
5×79 Posts 
Well, I proved myself wrong! There is a way to do x=0 with no digit but 0. But it's not easy or short:
Code:
0; // What other number could I use here? // One of the few functions that get something out of 0 is exp. int a = exp(exp(0)); // Now a=2 (rounded down from e). // This line is 2^11  2^5  8  2  1 answer = (a<<a*a*a+a)(a<<a*a)  a*a*a  a  exp(0) 
20050529, 09:16  #7  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
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20050529, 09:31  #8  
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
Quote:
BTW, thanks for the comments to explain your work. They were not what the compiler disliked. 

20050529, 10:32  #9 
Mar 2005
2·5·17 Posts 
I think there is a better solution for 5:
(55+5*5)*5*5+5 Richard Cameron 
20050529, 11:53  #10  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:


20050529, 12:17  #11 
May 2004
2^{4}×5 Posts 
How about this for a digitfree solution?
Code:
'.'*'''A' Code:
x=0: '.'*'''A'+0 x=1: '.'*'''B'+1 ... x=9: '.'*'''J'+9 Dave Last fiddled with by dave_dm on 20050529 at 12:18 
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