# Go deh!

Mainly Tech projects on Python and Electronic Design Automation.

## Saturday, October 05, 2013

### Project Euler #8 solution

Someone asked some innocuous question related to a Project Euler question where you had to find the largest number that is produced by multiplying five consecutive digits of a thousand digit number.

The digits were presented as twenty rows of fifty characters and the question was about how to grab the web page and extract the number.

I decided to do a straight-forward copy-paste of the digits into a string and going from there.

The algorithm I use recognises that finding a zero in the digits zeroes out any product and just traverses the digits in order performing appropriate actions based on accumulated state:

```# http://projecteuler.net/problem=8#!

xstring = '''73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450'''
x = int(xstring.replace('\n', ''))
xl = [int(ch) for ch in xstring if ch != '\n']

# current product; current maximum and index of where seen; number of digits contributing to prod
prod, mx, digits = 0, (0, -1), 0
for i, digit in enumerate(xl):
if digit == 0:
# Need to start again at next non-zero
prod = digits = 0
else:
digits += 1
if digits == 1:
prod = digit
else:
prod *= digit
if digits > 5:
# Remove influence of the first of six digits currently in product
prod //= xl[i - 5]
digits -= 1
if digits == 5 and prod > mx[0]:
# New max at position i
mx = (prod, i)
if mx[0]:
print ("The greatest product of five consecutive digits in the 1000-digit number")
print ("is %i for ...%s... occurring at digits in position up to %i" %
(mx[0], ''.join(str(i) for i in xl[mx[1] - 5: mx[1]]), mx[1]))
else:
print ("The greatest product of five consecutive digits in the 1000-digit number does not occur")
```

The output is:

The greatest product of five consecutive digits in the 1000-digit number
is 40824 for ...39987... occurring at digits in position up to 368

- END.

1. There is an off-by-one bug in the final output - it shows five characters, but shifted left by one.
Try it with xl=[9,8,7,6,5,4,3,2,1] to see what I mean.
You need this:
print ("is %i for ...%s... occurring at digits in position up to %i" % (mx[0], ''.join(str(i) for i in xl[mx[1] - 4: mx[1] + 1]), mx[1]))

Then you get this more sensible output:
is 40824 for ...99879... occurring at digits in position up to 368

2. Or just `x = ''' ... whatever ... '''.replace("\n", ""); print(max(eval("*".join(x[i:i+5])) for i in range(len(x)-4)))`