Computationally Feasible

Given problem x, can you compute the answer? Is it feasible to brute-force the answer?

This is easy to answer experimentally. If I define computable as something that I can run on my laptop in under 1 minutes with a slow language (Python):

-------------------- Loop 2^24 in 1.30s Loop 2^25 in 2.60s Loop 2^26 in 5.21s Loop 2^27 in 10.54s Loop 2^28 in 20.83s Loop 2^29 in 41.72s -------------------- Loop 10^1 in 0.00s Loop 10^2 in 0.00s Loop 10^3 in 0.00s Loop 10^4 in 0.00s Loop 10^5 in 0.01s Loop 10^6 in 0.08s Loop 10^7 in 0.78s Loop 10^8 in 7.93s Loop 10^9 in 78.62s --------------------

So there it is, above 2^30 or 10^9 operations your computer runs in serious trouble.

We can scale the result above in order to get the results in terms of years:

1 year: ~ 10^14 ~= 2^46 10 years: ~ 10^15 ~= 2^49 100 years: ~ 10^16 ~= 2^52

Conclusion: if napkin-calculations reveal that your problem needs less than 10^9 operations, you can easily brute-force your way through it.

Code Used to Generate Data

[Inserted file: computationally_feasible.py.]

import math import timeit def timeMe(base): limit = int(math.log(1e10) / math.log(base)) for exponent in range(1, limit + 1): n = 0 cmd = 'for i in range(%d**%d): n = i+i' % (base, exponent) nSecs = timeit.timeit(cmd, number=1) print('Loop %d^%d in %.2fs' % (base, exponent, nSecs)) print('-' * 10) timeMe(2) print('-' * 10) timeMe(10) print('-' * 10)

Real-Life Implication

I was solving the Equal Sums problem of Google Code Jam. Looking at the numbers limit, it seemed like it was feasible to brute-force through all 1, 2 and 3-tuples. This is a O(n^3) algorithm with n == 500. Since it is ~10^8 operations, we know that we should be able to reasonably brute-force this. Now, I suggest that you read the math section of the problem in order to understand why this is likely to work (but not guaranteed -- if they had a set where only >= 6-tuples could be solution then I would never have been able to prove that no solution can be found). But our questions is different: how long is it going to take to go through all the possible 3-tuples and how does it match with the result we have above.

All 2-tuples (500^2 ~= 10^5): 00.66 secs All 3-tuples (500^3 ~= 10^8): 20.57 secs

Now, we should not be surprised that it takes more time than our estimate in the first section: in this problem we do MUCH more every iteration. The fudge factor is from ~60 to ~2.5. Also, as we slow down as we go which is expected because we have an ever-expanding dictionary that keeps track of all the answers in memory. But the ~10^8 limit as what is feasible to brute-force in a few seconds remains!

Given the constraints we did not have to solve the problem; brute force was enough!

Equal Sums All 3-Tuples Brute Force Solution

[Inserted file: equal_sums_bruteforce.py.]

import copy import collections import timeit def startSmall(): N = "279723182299 376876333524 998067966121 403485644442 153635442114 399189277664 230203543364 317963821285 351345530822 148151584645 684971615669 135054212116 636307404388 623840919365 742186585581 261949667274 262985372168 456696918121 936200521159 262459964519 259367083572 405184756642 104095058407 52695575854 40109000701 786793234470 734315378491 76992992268 343786892924 697121153437 725714221703 895499931981 805467280153 196880330941 912165217021 70727548109 716021033204 53114833551 922877781847 773905963417 522513695105 812014766958 957151120167 534064397354 911573744097 998468086343 89252943844 453000312834 130161459074 569342457998 741809185948 46156906233 653624819175 944102929433 440009999356 484138801997 545648063331 555647844367 132201400220 235093862869 48840881731 949992883063 565627740876 212811725246 379870929286 206687302384 176371061529 117940879646 206204221514 25847501412 475235734321 728265674918 579394496800 864821221996 434116969604 996414146286 461599230089 69222744440 454690170897 9332645752 517144199870 323451331247 442177489557 968905696256 955683650226 110223094609 707994487197 968530741843 852218104020 720002104839 725351165376 9027153496 335247652855 954562251546 808073535166 218393270859 599421082705 153343365071 216088953944 55834519300 714266890546 973433485465 431747834536 331538991874 16887455023 919086107825 579623133938 853255424721 224134281919 974004216022 269441870173 531878276261 209771521469 409569176727 287282320121 649495488479 962543296720 899988343728 685706115716 256427996423 947841194523 221988840996 591466177571 676266006532 805088660087 814389168694 370749563588 916801120284 139145883894 996783134695 711357521523 342786470645 488923106867 375357995366 993887191145 383560524196 276316552464 973090377739 564282848803 499952346473 587462551177 314927306722 925760832003 290454868288 995135389077 548596045607 18136150640 284044933438 462874905432 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768420573336 75894745397 427137517822 501095542606 15666855955 573879228968 155904593898 227765828793 724786686470 428729034263 631949427550 303444387613 397472911452 56807290224 192578593237 461136029508 629644711830 965162215333 964805593622 866489274136 727566795152 923578697230 813260995968 807164500327 843775845627 682230751102 266037288904 302425231558 807228641715 672007702571 54316762137 328925386368 491239950009 281419905721 657332715130 745066995960 693927837925 753243333720 483109392031 899842291847 654658217180" N = [int(n) for n in N.split(' ')] sumMap = {} stack = collections.deque([([], 0, 0)]) maxDepth = 3 while len(stack) > 0: #pds = stack.popleft() pds = stack.pop() for i in range(len(N)): P, sm, depth = pds if N[i] in P: continue PP = copy.copy(P) PP.append(N[i]) PP.sort() depth += 1 sm += N[i] if sumMap.get(sm) is None: sumMap[sm] = PP else: if PP != sumMap[sm]: return (sumMap[sm], PP) if depth < maxDepth: stack.append((PP, sm, depth)) return None nSecs = timeit.timeit( "startSmall()", 'from __main__ import startSmall', number=1) print('Took %.2f secs.' % nSecs)

Page Metadata

date: ['2014-02-12']

category: ['CS']

tags: ['CS', 'Computer Science Interview Question', 'csiq_elementary', 'viewA']



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