13
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I have a specific number in mind, but it's part of a challenge I'm doing, and I don't want people to do (all) the work for me.

Here is a number which has the same digits, but shuffled:

5713167915926167134578399473447223554460066674314639815391281352328315313091488448321843
8892917486601064146636679920143691047671721184150386045081532202458651561779976236919751
5521854951599379666116678853267398393892536121049731949764192014193648608210652358947001
6332620900065461061195026191178967128001712341637591690941978871368243245270800684616029
6679555942849366434586090627998161441134473428845367022486230724219981658438108844675033
4461550796750244527407413996606134735852639191026103378962082622204359677030054592798927
4145951979523473408718011778751084514127053772614511042703365596651912104541233491744530
87457854312602843967491787086250478422477028164189

The number has 666 digits (decimal). Because I'm using Python, integers (or technically longs) are automatically bignums.

I have 255 chars to use, and I need to describe the same number. The description is meant to be run through eval() to produce the original number.

What strategies should I be looking into?

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  • \$\begingroup\$ base64 (or higher) encoding \$\endgroup\$ – Luis Mendo Jan 25 '16 at 22:00
  • 2
    \$\begingroup\$ Are you sure that the actual number from your challenge doesn't have some property that makes it easier to compress that may be lost due to the reshuffling? I don't think Luis's suggestions is gonna cut it. Even in base 256, this still has 277 digits. Of course you said you've got "255 characters", so I guess in principle you could use a much bigger base like 2^16, and go into Unicode. \$\endgroup\$ – Martin Ender Jan 25 '16 at 22:08
  • 4
    \$\begingroup\$ This is asking for the shortest code to produce a number, which is absolutely asking for golfing advice. My concern is the source is uncredited -- the challenge should be linked if possible so we have attribution and can check it's OK to give outside help. \$\endgroup\$ – xnor Jan 25 '16 at 22:15
  • \$\begingroup\$ I have 255 chars to use, and I need to describe the same number. The description is meant to be run through eval() to produce the original number: is it acceptable for you to read the number from an external resource, such as a web page? \$\endgroup\$ – Luis Mendo Jan 26 '16 at 17:07
  • \$\begingroup\$ @LuisMendo No, it has to be self contained. Additionally, it can only use chars that are legal in a filename. \$\endgroup\$ – Christian Sonne Jan 26 '16 at 19:01
12
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Base Encoding

A standard technique for compressing numbers is to express them in a big base and encode the digits as characters. E.g. if you encode the number in base 256, it would have only 277 digits:

[12 24 156 48 101 149 235 32 96 92 20 203 202 164 144 71 193 127 112 77 141 79 210 183 98 155 16 151 65 198 26 236 83 221 220 129 169 254 43 124 245 25 176 182 167 124 95 191 77 25 233 139 190 7 135 2 149 90 163 163 106 193 220 253 109 129 57 219 91 157 218 18 223 11 171 113 209 173 207 123 110 220 79 139 176 143 171 7 30 35 231 151 172 83 120 114 119 47 217 227 50 105 236 91 161 226 112 16 170 57 162 147 36 89 26 9 122 164 15 15 243 108 30 14 233 139 103 137 82 169 2 57 54 71 154 136 23 203 137 10 219 153 24 168 42 218 165 125 185 183 241 91 193 85 195 71 186 18 98 34 196 78 6 193 252 8 177 94 5 24 137 183 127 129 9 77 149 73 148 193 62 220 146 33 130 21 209 153 229 105 100 188 87 235 203 104 207 161 20 17 102 150 252 120 242 222 233 248 114 217 142 31 196 42 161 173 0 244 9 213 178 152 122 170 136 230 135 132 245 69 9 196 231 147 8 175 48 98 101 23 162 144 190 200 62 226 61 27 200 15 232 12 105 187 184 4 121 252 171 240 230 94 161 151 131 209 205 130 193 9 4 155 92 48 59 130 93]

Or expressed as a string

"0eë `\ËʤGÁpMOÒ·bAÆìSÝÜ©þ+|õ°¶§|_¿Mé¾Z££jÁÜým9Û[Úß«qÑ­Ï{nÜO°«#ç¬Sxrw/Ùã2iì[¡âpª9¢$Y  z¤ólégR©96GË
Û¨*Ú¥}¹·ñ[ÁUÃGºb\"ÄNÁü±^· MIÁ>Ü!Ñåid¼WëËhÏ¡füxòÞéørÙÄ*¡­ô  Õ²zªæõE Äç¯0be¢¾È>â=Èèi»¸yü«ðæ^¡ÑÍÁ  \0;]"

(Plus some unprintable characters which are stripped by SE.)

Of course, that's still too long for your 255 character allowance. If you're actually talking about characters though (as opposed to bytes), you can go into Unicode and use a much bigger base. How about 216? That's only 139 digits:

[12 6300 12389 38379 8288 23572 52170 42128 18369 32624 19853 20434 46946 39696 38721 50714 60499 56796 33193 65067 31989 6576 46759 31839 48973 6633 35774 1927 661 23203 41834 49628 64877 33081 56155 40410 4831 2987 29137 44495 31598 56399 35760 36779 1822 9191 38828 21368 29303 12249 58162 27116 23457 57968 4266 14754 37668 22810 2426 41999 4083 27678 3817 35687 35154 43266 14646 18330 34839 52105 2779 39192 43050 55973 32185 47089 23489 21955 18362 4706 8900 19974 49660 2225 24069 6281 46975 33033 19861 18836 49470 56466 8578 5585 39397 26980 48215 60363 26831 41236 4454 38652 30962 57065 63602 55694 8132 10913 44288 62473 54706 39034 43656 59015 34037 17673 50407 37640 44848 25189 6050 37054 51262 57917 7112 4072 3177 48056 1145 64683 61670 24225 38787 53709 33473 2308 39772 12347 33373]

(I can't include the actual string here, because it contains some CJK characters that are banned by SE.)

Now that seems more doable. You just need to be able to decode it in 116 characters. If you can't, Unicode has a lot more than 216 characters, so you could try using an even larger base.

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4
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Prime Factorization

If the number has no interesting features, base encoding is the best way to do it. The next thing to do is look for interesting features of the number. The first one that comes to mind is that it may have factors of small primes (2,3,5,7, etc) raised to quite large powers. IF you have nothing else to go on, keep trying to divide by small primes and see what happens. If its factors include 2**4, 3**4, and 7**4, you can write big number *42**4 which is a few bytes shorter than big number * 3111696

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  • 4
    \$\begingroup\$ I'd also try factoring the number plus or minus small integers to see if one of them has a nicer factorization. Also, if your language has a short way to get the nth prime, you can save a digit or so per prime by storing its index instead of the prime itself. \$\endgroup\$ – 2012rcampion Jan 26 '16 at 2:15
4
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Recursive removal of largest square

This approach removes the largest square number from N, repeatedly, until there's no value in continuing.

while(n>999*999):
    s = sqrt(n,2)
    print s,"** 2 +"
    n = n - s**2
print n

If you ignore the "**2+" characters, this one comes out to approximately the same number of digits as the original number, on average. Making up for those 4 extra characters per iteration requires a bit of luck. In the case of your number, the result has 670 digits of square numbers, plus 7x"**2+", another failure:

755855006990505232214298076833020140623897728341856142793250050184099570268569900389346192358073922001480310798643405893673501405667458785677166605919485512157948819102093414848159820683798554799982163455753292781944741934237780592730586508786425528910736750640071037094033497266578109597923654387813828207885510302579581252831537751**2+
33300095205899066129442737321270515378501483166974896029394675779096351509514355500527819871697116193238261137790928953798777695127752032484956608505929119246433389165**2+
187763197402063683206154659623192450644818397963460986292088297442441704645626089130**2+
278760215056365252005927060531480627653626**2+
639191600506542558482**2+
25777519523**2+
106673**2+
103405

By almost breaking even on average, this algorithm lends itself well to being used in conjunction with other algorithms (or even itself) to further reduce the numbers in the expression (at the cost of some parentheses). Those other algorithms can be more expensive, as they will be operating on significantly smaller numbers than the original. In the given example, a net gain could be had if a more expensive and effective algorithm could cut out 25% of the characters of 33300095205899066129442737321270515378501483166974896029394675779096351509514355500527819871697116193238261137790928953798777695127752032484956608505929119246433389165 (the second large value in the result)

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  • \$\begingroup\$ This approach can be slightly improved by checking for cubes, and very rarely fourth powers, as well. \$\endgroup\$ – Sparr Feb 19 '16 at 20:50
0
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Nearby large powers

This approach looks for [relatively] small numbers raised to some power that come close to the target number. In most cases restating N as A**B+C will not be an improvement, but in some cases it will be.

def nearest_power(n):
    mindiff = 1
    best = (n,1)
    for a in xrange(2,10000):
        b = math.log(n,a)
        if math.ceil(b)-b<mindiff:
            mindiff = math.ceil(b)-b
            print a,"**",b
            best = (a,b)
        if b-math.floor(b)<mindiff:
            mindiff = b-math.floor(b)
            print a,"**",b
            best = (a,b)
    return best

10000 is an arbitrary constant. The bail-out condition could also be based on some target mindiff.

In the case of your sample number N with 666 digits, this function (with the 10k cap increased a bit) finds that N ~= 165661162**81.0000000025, so N-165661162**81 is a 659 digit number, cutting 7 digits off the number to be handled at a cost of 14 characters of expression, a failure.

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